Freshman Organic Chemistry

Professor McBride outlines the course with its goals and requirements, including the required laboratory course. To the course's prime question "How do you know" he proposes two unacceptable answers (divine and human authority), and two acceptable answers (experiment and logic). He illustrates the fruitfulness of experiment and logic using the rise of science in the seventeenth century. London's Royal Society and the "crucial" experiment on light by Isaac Newton provide examples. In his correspondence with Newton Samuel Pepys, diarist and naval purchasing officer, illustrates the attitudes and habits which are most vital for budding scientists - especially those who would like to succeed in this course. The lecture closes by introducing the underlying goal for the first half of the semester: understanding the Force Law that describes chemical bonds.

Transcript



September 3, 2008



Professor Michael McBride: Okay. So for 150 years organic chemistry courses have tended to acquire a daunting reputation. So you need help. I know you're very able, but trust me, you need help. So where do you [get] the help? The PowerPoints are available on the Web. How many of you've already seen the PowerPoint for today, just so I have some idea? So about a quarter of you maybe. Okay, but anyhow, so your lecture notes are important, but you don't have to worry about getting everything down because you can download it from the Web. And I do it on a Mac, but I try my best to make it compatible with PCs and even with the free PC Viewer for PowerPoint. So you should be able to see it. But I don't see it on a PC. So if anything doesn't come through, let me know so that I can fix it. Okay then, in-class discussion is very important, and if you're really, really shy and can't participate in discussion in class, then email me a question. Okay?



There's the course website, which is our substitute for a text. It also includes the PowerPoint, and there's the link for it, and when go there you'll see it develop. The current website is mostly last year's course, so it'll change a little bit as we go along, but fundamentally it's the same. If you want to look ahead you'll see pretty much what's coming up. There'll be assigned problems and questions, and also there are previous exams and answers keys. All these things are on the course website so you'll get help there. But one thing that's really special is the course Wiki. This is the third year we've done it and the second year in a really systematic way. So you get assigned to do, to cover a couple frames of the PowerPoint. So those are the ones you really need to take careful notes on, and write them up, and help other people too; that's the nature of a Wiki, as you know. How many of you have participated in a Wiki? Well, by next week it will be all of you. Okay, but in order to get credit for it you have to get it by the night after the lecture. So for the lecture today you have to get it by late tomorrow night, 36 hours after the lecture. This is so other students can use it. Okay, in the spring there'll probably be a textbook. I haven't really decided yet. These things cost an arm and a leg. Maybe we can find one that's used, an older edition. It doesn't make any difference except to the publishers.



Okay, also there's personal help, like from me, and there's--and you can find my phone number, email and so on, on the website. Also the two graduate student TAs who are assigned to the course, who are Filip Kolundzic--Filip, back there--and Nathan Schley, not Schlay, Schley. So these are graduate students in chemistry and they run these discussion sections. Typically you have a 50-minute discussion section. But the way we run it in this course is that on two different nights a week there are two-hour sessions. You can come to any part of it you want to. You can go to both of them, you can go to four hours a week if you want to, or you can go to none at all if you want to. So really, for the bookkeeping purposes of the department you have to sign up for a section. Sign up for any section you want to and then come to what's useful for you. But also, the reason you pay the big bucks to come here, is not to hear me, it's to interact with the other students. That's a really big help. So, form study groups. And in fact you can get advice from previous people who've taken the course. That's on the Web. Also there's some of them, there's a list of them on the Web who would be happy to talk to you if you need it.



And we're blessed with three alumni, seniors who took this course as freshmen, who act as what are called peer tutors, and they'll run a session Sunday evening, from eight to ten p.m. is the current plan. We'll announce the rooms for these things on the website and probably by email to you as well. So let me introduce Tina Ho and Drew Klein and Justin Kim. So they'll be a big help to you too. So there's plenty of personal help, so use it.



These are the dates we're going to have exams. There are ten lectures and then an exam; nine lectures, exam; nine lectures, exam. Actually, if you check, you'll find that--and also you get 50 points for participation in the Wiki, and the total is 650 points, that's what your exam is based on. Actually this doesn't cover--it's nine lectures that are covered on the exam, but the previous Wednesday part of the lecture is going to be a guest lecturer that's going to be here just that day. So we're putting the exam off and it'll only cover the previous material; not that that's a big deal. Okay, and the semester grade is biased; that is, it's based on this, your total score here, out of 650 points. But if you're near a cutoff and you were very good about turning in your problem sets and so on, then we boost you up. We don't grade problem sets but it's worthwhile to do them, and they might make a difference.



So where are we going with this? What are the goals of our Freshman Organic Chemistry? In fact, if you click that in your PowerPoint you'll get taken to that site, but it's right on the website, you'll see it anyhow. First is to learn the crucial facts and vocabulary of Organic Chemistry--after all that's what we think we're here for--and to develop a theoretical intuition about how bonding works. This is the goal for, the primary goal of the first half of the fall semester is to learn how bonding works really, and that relates then to molecular structure; and also how bonding changes, and that of course is reactivity. But under the line there are a lot of other things that we do in Freshman Organic Chemistry that are arguably just as important, like make the scientific transition from school to university. In school they try to teach you what people know. In the University you try to develop new knowledge. So you need a different mindset for that, and we hope this course helps you develop that. So learn from Organic Chemistry, which is really in my view a model science, how to be a creative scientist.



So here's a creative scientist by anybody's measure, Louis Pasteur. And in the 1880s he said this in French, but in English it says, "Knowing to be astonished by something is the mind's first step toward discovery." Another way of putting that is that the characteristic comment on making a real discovery is not "Eureka", it's "Huh, that's funny." So that's what you really have to learn; learn enough about how chemistry works and form this picture in your mind that when something happens that doesn't fit, you know to be astonished so that you can discover something. That's exactly what Pasteur did, and we'll talk about that in the course. And even perhaps more important, to develop good taste, so that you can distinguish sense from nonsense; there's certainly more nonsense floating around than sense, and being able to tell the difference is important. The way you do it is to develop good taste by looking at a lot of good examples and then you're aware of how crummy the bad examples are. So we're going to try to emphasize good examples, and have fun. So and as we go along, if you have questions, break in. You'll do this much more as we go along, I know.



So the class really is mostly about theory, although we describe the basis for the theory and spend a lot of time trying to make it real. But we require Chemistry 126L, the lab. This is the only chemistry course that requires you to take a lab simultaneously. So I hope you're all enrolling in that because there'll be a certain day that you want to be able to take it. It's just one afternoon a week, three hours or whatever it is, but you want to get your first choice, so line up soon. You'll be accommodated but it's just more convenient if you get it arranged earlier.



But why? Because lab answers the really big question. And the big question was brought home to me by my son, John McBride, in his third year. This was the beginning of the third year, and his mother and I didn't know what was coming. For the next year, maybe 15, maybe 20 times a day, he said, "How do you know?" So here's John this last summer. He's now 38 and he has his own three-year-olds to say that to him. And he doesn't say, "How do you know?" anymore. He now says, "How do you know?" Okay? But that is the main question, how do you know what these things that they told you in school?



Well, there are four ways we can talk about of knowing, and two of them are shown on this manuscript from the Carolingian book painter. If we zoom in on the top frame, here's Moses on Mount Sinai. So the first way of knowing is divine authority. Here he's going to be the--here's the graduate student here getting the word. Here's the teaching assistant over on the left perhaps. [Laughter] Aaron, right? And then he comes down from Sinai, to see the class, the Children of Israel. So here's another kind of authority, which is human authority interpreting the scriptures. And here you can see the class; the guys are going like "huh." And the teaching assistant is off on the side still. But this doesn't make it. Science is not faith based. There may be other things you know that way but not science. Science ignores divine authority and it ignores human authority; not that they might not exist but they don't relate to science.



Now as you walked in today, did you notice these things over here? There's an Honor Roll of Chemists, and in fact we'll use that a lot this semester, and in particular one of the people on there is Michael Faraday, who started in a very humble way. He was a book binder's apprentice and he bound this book--not this particular copy but this book--which is called Conversations on Chemistry. He bound the first edition. This one is a later edition. So you see it's Conversations on Chemistry in which Elements of that Science are Familiarly Explained and Illustrated by Experiments. And who's the author? J. L. Comstock; he's actually not the author, he's the guy who stole it. He stole it from a woman, Mrs. Marcet, in England, who wrote this book, which was the most popular textbook--it was written for girls--but it was the most popular textbook in all chemistry, for the first half of the nineteenth century. It went through like 20-some editions. And here you see at the beginning it's a dialogue, a conversation between Mrs. B and Caroline and Emily, and it's fun to see this here, what Emily says at the beginning. "To confess the truth Mrs. B, I'm not disposed to form a very favorable idea of chemistry, nor do I expect to derive much entertainment from it." But in the long run, as you can imagine, they have a lot of fun with chemistry. It was a wonderful book, and still is. But he was binding it, and read it. And look what he says about this, as his introduction to be the leading experimental scientist of the nineteenth century:



"Do not suppose I was a very deep thinker or was marked as a precocious person. I was a very lively, imaginative person and could believe in the Arabian Knights as easily as the encyclopedia, but facts were important to me and saved me. I could trust a fact and always cross-examined an assertion. So when I questioned Mrs. Marcet's book by such little experiments as I could find means to perform, and found it true to the facts, as I could understand them, I felt I had got hold of an anchor in chemical knowledge and clung fast to it."



So the experiments were what did it. So the third way of knowing is by experimental observation. And here's Richard Feynman. How many of you've heard of Richard Feynman? He was a really great physicist, wrote a wonderful textbook as well as getting all sorts of prizes. He spoke to the National Science Teachers Association in 1966 saying,



"Learn from science that you must doubt the experts. Science is the belief in the ignorance of experts. When someone says, 'Science teaches such and such,' he's using the word incorrectly. Science doesn't teach it; experience teaches it. If they say to you, 'Science has shown such and such,' you might ask, 'How does science show it? How did the scientists find out? How, what, where?' Not science has shown, but this experiment or this effect has shown."



Now, why do we quote Feynman? Because he's an expert. [Laughter] Wrong. Though literally, expert, the etymology of expert, is it means someone who has done experiments. We quote him because what he says makes sense. So logic is the fourth way of knowing things. So the two ways that we know things in chemistry, or in science, are experiment and logic. And the lecture is a little bit more focused on logic and the lab is more focused on experiment, and you get an unbalanced view if you do one without the other.



Okay, so modern science got underway in the seventeenth century. There's the seventeenth century, 1600 to 1700. And 1638 was when New Haven Colony was founded, and 1701 was when Yale was founded. So that's when everything got underway, just when this enterprise was beginning here. Here we are. If you go back 100 years you get to quantum, quantization by Planck; and we'll talk about that. And if you go back another 100 years you get to Lavoisier and oxidation; and we'll talk about that. And if you get another 100 years you get to Newton and gravitation; and we'll talk a little bit about that. And if you go back a little more, another 100 years, you get to Copernicus and the revolution of the heavenly bodies, and Columbus and navigation, and Luther and the Reformation. And these things all have something in common. As Robert Hooke wrote, "The seventeenth century" (his age) "was an age, of all others, the most inquisitive."



All these things have to do with people inquiring into how people know things and finding out new things. And in particular an important figure was Francis Bacon and his Instauration. Now you may not know The Instauration so well, let's look at that. Here's Francis Bacon, there are his years. He was Elizabethan and Jacobean. He was almost exactly contemporary with Shakespeare, and with Galileo. He went to school, to university, at Cambridge. And here's a cartoon that shows him--it's a modern cartoon--imagining him in a class at Cambridge. Because he wrote of his tutors at Cambridge:



"They were men of sharp wits, shut up in their cells of a few authors, chiefly Aristotle, their dictator. All the philosophy of nature" (philosophy meant science in those days) "all the philosophy of nature, which is now received, is either the philosophy of the Grecians or that of the alchemists. The one is gathered out of a few vulgar" (that means ‘common' of course) "observations, and the other out of a few experiments of a furnace. The one never faileth to multiply words, and the other ever faileth to multiply gold."



So here's the book he wrote, The Instauration. That's the frontispiece for it. This picture's from the Beinecke Library; I went down and got a picture of the book. Notice it was published in 1620. What else happened then? That's when the Pilgrims came over, right? So the title of the book, rather small under his name and title as Lord Chancellor of England, is Instauratio Magna, which means the Great Restoration. Restoration of what? Of the way of knowing. A bigger, a part of it, it's called the Novum Organum, which is--and it develops the inductive scientific method, based on experiment, to replace Aristotelian deduction, which is you maybe did one experiment sometime, and then you reason everything from that. But he says no, you have to do more experiments. Now there's an interesting thing here. One of the devices on the title, on this frontispiece, is two pillars. What in the world are they doing there? Well they're the same pillars that you see on this. That's a piece of eight; you know, Treasure Island, pieces of eight? See it's eight reales, and it came from the silver of Mexico; it was minted in Mexico City. So and there you see the same pillars and on them it says plus ultra; more beyond. Beyond what? What are the pillars? Pardon me?



Student: Spain and Africa.



Professor Michael McBride: Yes, it's Africa and Spain, but it's the Pillars of Hercules, which are the mouth of the Mediterranean, the old Classical World. So there's the Mountain of Moses in Morocco and the Mountain of Tarik, which is the name of Gibraltar. So here's the Mediterranean, the Classical World of Aristotle, and you can sail out into the New World and bring back silver, for example. There's danger of course. But look at what it says at the bottom. What will be brought back? Not just silver. Multi pertransibunt & augebitur scientia --"Many will pass through and knowledge will be increased." So we go beyond Aristotle into experimentally based science and knowledge will be increased.



So here's some quotes from The Instauratio Magna. "That wisdom which we have derived principally from the Greeks" (no offense, okay?) "is but like the boyhood of knowledge, and has the characteristic property of boys: it can talk but it cannot generate;" "…it is but a device for exempting ignorance from ignominy." That means it's a way of hiding your ignorance, and we'll see examples of that. We'll talk about, in Lecture 11, about correlation energy, and we'll talk in Lecture 32 about strain energy, and you'll see that both of these are just words that are used to hide our ignorance. "…the end which this science of mine proposes is the invention, not of arguments, but of arts" (ways of doing things). "…not so much by instruments" (although new instruments are important, like microscopes and so on) "as by experiments, skillfully and artificially devised for the express purpose of determining the point in question." (So artificial experiments designed to decide a question; experiments.) "And this will lead to the restoration of learning and knowledge."



So followers of Bacon established The Royal Society in 1662, just after Charles was restored to England, after the period of Cromwell. And there was a history written of The Royal Society, a book about this thick, published in 1667, only five years after it was founded. Why did they publish a history so soon? Well let's look at this, the frontis itpiece of this book. Here's the late Francis Bacon, who was said to be Artium Instaurator, the Restorer of the Arts. And here's the President of The Royal Society, the mathematician, Viscount Brouncker, and here in the middle, being crowned with laurel, is Charles II. Why do they have him up on a pedestal? Because they're hoping, as scientists have before and ever since, to extract some money out of the government to do their research. They actually never got it from Charles but it wasn't for lack of trying. And this is why they wrote the history, to try to make the case for being supported.



Okay, now let's look at all the good things that will come from science, from The Royal Society. In the background can you see what that is? We'll blow it up. Here there's a hint to it on the bookshelf. If you look really fine on the bookshelf you can see that some of them have writing on the spine. Do you see what that one is? Can anybody read it? What? What science book do you think they might have had?



Student: Copernicus.



Professor Michael McBride: Copernicus, right? So astronomy; that's a telescope in the back. Okay, or over here on the wall, what's that thing? It's a clock. Why is it shaped like a piece of pie?



Student: Because it has a pendulum inside.



Professor Michael McBride: Ah, because it has a pendulum inside. So horology, making good clocks. Okay, or here, what's that thing? It's hard for you to know.



Student: Solar.



Professor Michael McBride: It's a wind gauge. It has a vane inside that it blows on--and you can tell from how far it goes on the scale how strong the wind is. So meteorology. And back here on the pillar, what are those things for?



Students: A compass.



Professor Michael McBride: For cartography. Now what do all these things have in common that they're going to do for Charles II? Astronomy--



Student: For the records.



Professor Michael McBride: Good clocks, meteorology, cartography, what--



[Students speak over one another]



Professor Michael McBride: They all have to do with navigation, with making England strong at sea. Now back here is another science, Chemistry. That doesn't seem to have anything to do with navigation. Why would Chemistry be important to Charles?



Student: War.



Professor Michael McBride: Look at it.



Student: War.



Professor Michael McBride: To make gun powder. There was a chapter about gun powder in The History of the Royal Society. Okay, and up here at the top is the motto, which is Nullius in Verba, which comes from this quote from Horace, and what it says is, "Lest you ask who leads me, in what household I lodge" (that is, what philosophy I advocate) "there is no master in whose words I am bound to take an oath. Wherever the storm forces me, there I put in as a guest." So it's the experiment, not the philosopher, that leads you to the conclusion. So Nullius in Verba is 'in the words of none.' And, in fact, the original name of The Royal Society was The Royal Society for the Improving of Natural Knowledge by Experiments.



Okay, so we're going to see, as the course goes along, important experiments that really decided questions, and in fact Bacon's most important kind of experiment was one that "finally decides between two rival hypotheses, proving the one and disproving the other." So you can do all sorts of experiments and just be collecting butterflies--no, I don't mean to insult people who collect butterflies, it's a fine thing to do. But there's something special about experiments that really are designed to answer a question. Now Bacon devised a name for such experiments, and they're based on this model, that you have a road that diverges and you need to know which way to go between these two hypotheses. What do you need, to know which way to go? You need a sign, or this was a cross that you mounted at a crossroads. So the Latin name for cross is crux. Do you see what they call the experiment? Crucial. That's the origin of the word crucial, right? It's the one that tells you which way to go. Okay, so here is Isaac Newton and he's holding something. Can you see what he's holding in his hand?



Student: A candle.



Professor Michael McBride: I'll give you a hint. It's a prism. Why is he holding a prism? Because that was his crucial experiment; and here's his diagram in that experiment with a prism in it. He called it the Experimentum Crucis, taking the word from Bacon. Okay, so light came in through a hole in the window, through a lens, and then got bent and dispersed into the different colors. So you get a spectrum here on this thing, the different colors. Now there's the question, how does the prism make color? Hooke and Descartes thought that light was a train of pulses and as it goes through something like a prism, or as it reflects from a thin layer of oil on water or something like that, that it changes the timing of the pulses and therefore changes the color. But Newton thought that the colors were pre-existing, and the prism just separates them. And this was his crucial experiment to decide between those two theories. You see what he did? He drilled a hole through the board and let through only the red light, and put a second prism there. And he wrote here, three times--he wrote it here, and also here, and also here--nec variat lux fracta colorem; which means "the broken light does not change its color." So this proved to Newton, at least at that time, that light is a substance, not a train of pulses. What do you think of that proof now? You think light is a substance or a train of pulses?



Students: Both.



Student: Neither.



Professor Michael McBride: But at least you can see that in terms then, that was a crucial experiment, a really important experiment, and that's the kind of experiments we'll try to talk about in the course. Experiments are indispensable in Organic Chemistry. It's an empirical science based on observation, and that's why you have to take the lab. But so is logic--that was number three and number four of the ways of knowing--logic is important too. So believe what I tell you here only when it makes sense to you. Don't just cram it in, make sure it makes sense. But what if it doesn't? Now here's how to succeed in Chem. 125, and we'll take as our model science student Samuel Pepys. How many people have heard of Samuel Pepys? Have you heard of him as a scientist?



Student: I don't remember.



Professor Michael McBride: What did you hear of him as? What do you associate with Samuel Pepys?



Student: Newton-Pepys Problems.



Professor Michael McBride: Right in this period, in the heart of the science growth. But what you know him for was his diary, which tells all about life, everyday life in Restoration London. He was actually, as a sixteen-year-old, present when Charles I was beheaded in 1649. Now what's the connection? Do you know where this is? Anybody been there? Dixwell, Goffe and Whalley Avenues, those are named for three of the 50 judges that condemned Charles I to be beheaded, and they were the only ones that lasted very long, after the Restoration, and they lasted because they fled to New England and were hidden on West Rock. All these roads are heading west toward West Rock. Okay, so that's a tie-in to the same period. But anyway, he got his B.A. in Cambridge in 1654 and a Master's in 1656. And he got a good job, he became Clerk of the Acts for the Navy Board, which meant he was the guy that purchased everything for the Royal Navy, all the rope, all the tar, all the lumber and so on. And on July Fourth, 1662--it's the fourth of July but it's more than 100 years before that became relevant--he writes in his diary, "By and by comes Mr. Cooper, mate of the Royall Charles, of whom I intend to learn mathematiques, and do begin with him to-day, he being a very able man… After an hour's being with him at arithmetique (my first attempt being to learn the multiplication-table); we then parted till tomorrow."



So here was the guy doing all the purchasing for the Royal Navy and he didn't know multiplication, let alone division. But he worked hard at it. July ninth, five days later: He's "Up by four o'clock, and at my multiplicacion-table hard, which is all the trouble I meet withal in arithmetique." He can do the other things pretty well. July eleven: "Up by four o'clock and hard at my multiplicacion-table, which I am now almost master of." Christmas--so six months later: "…so to my office, practicing arithmetique alone and making an end of last night's book with great content till eleven at night and so home to supper and to bed." Or a year later--so he was motivated and he was diligent; that's good. A year later, on a Sunday: "…I below by myself looking over my arithmetique books and timber rule. So my wife arose anon and she and I all the afternoon at arithmetique," [laughter] "and she has come to do Addition, Subtraction and Multiplicacion very well, and so I purpose not to trouble her yet with Division…" Right? [Laughter] So he worked with a study partner, and that's crucial. And Isaac Newton--does anybody recognize this book?



Student: Yes, sure.



Professor Michael McBride: Right? The Mathematical Principles of Science, of Natural Philosophy. But there's an interesting thing on the title page. Samuel Pepys is the one who gave permission to publish that book, because he was the president of The Royal Society. Now six years later Pepys encountered a problem with dice. The reason was he went to coffee shops every night for dinner and they'd gamble, and people proposed various kinds of bets, and he couldn't figure out this one, so he wrote Newton for help. So this was the problem that he wrote to Newton, twenty-second of November. So A has six dice in a box, and he has to throw a Six by throwing it; B has 12 dice and he has to fling two Sixes; and C has 18 Dice but he has to get 3 Sixes to win. The question, "Whether B and C have not as easy a Taske at A, at even luck?" That is, if the dice aren't loaded or anything, who has the better bet? [Laughter]



How many people think that it's the same? Don't be shy. How many think it's A? How many think B? How many think C? How many don't really have any opinion at all? Good, that wins. Okay, so he wrote this letter to try to get help on his bets from Newton, and Newton replied, four days later, "What is the expectation or hope of A to throw every time one six, at least, with six dyes?" So you get two sixes you still win; that wasn't clear in the original statement. So he says, "If we formulate the question that way, it appears by an easy computation that the expectation of A is greater than B or C; that is, the task of A is the easiest." There's the answer. So Pepys replied on the sixth of December: "You give it in favour of the Expectations of A, & this (as you say) by an easy Computation. But yet I must not pretend to soe much Conversation with Numbers, as presently to comprehend as I ought to doe, all the force of that wch you are pleas'd to assigne for the Reason of it, relating to their having or not having the Benefit of all their Chances." So he wasn't ashamed to admit that he didn't really understand--that's crucial. "And therefore, were it not for the trouble it must have cost you, I could have wished for a sight of the very Computation. "



Can you show me how to figure it out? And he wanted that because somebody might change the terms of the bet and then he wouldn't know. He wanted really to understand. He insisted on proof. So this is two of the pages from Newton's correspondence of the letter that he wrote in response, and you can see that Pepys certainly got more than he had bargained for. "So to compute this I set down the following progressions of numbers." So you can go through all this and you get complicated quotients here, and it turns out that A has 31,031 chances out of 46,656, or 0.6651 chance of winning, and B has this, which is 0.6187; A wins. So is Pepys satisfied with this? Pepys writes back and he says, "Why?" Right? "I cannot bear the Thought of being made Master of a Jewell I know not how to wear." So he's willing to swallow his pride to search for really solid understanding. Now compare this with a comment we got at an end-of-semester evaluation in January a year ago. "I never went to his office hours for help because I felt like he would make me feel stupid, because he's superior to me in chemistry." I hope I'm superior to you in chemistry. [Laughter] I'm not superior as a person, but I hope I'm superior in chemistry. You're paying me the big bucks because of that. So swallow your pride and ask someone for help. Follow Pepys.



So read "Pepys and Newton" and get together to do problems for Monday, and contribute to the Wiki when you're asked to do so. So here are problems. For Friday the problems are optional but very helpful. One is find out which two class members have the rooms nearest you so that you can maybe use them as study partners. You don't have to use them, use anybody, but use someone, don't try to go it alone. Two. What are the three most common items of advice from course veterans? So if you click there, or go to the webpage, you can get anonymous advice; I was careful to make sure it's anonymous. So you can get all--much of it is contradictory, so your job is to look through it and try to get some idea of what they're telling you that's worth knowing. For Monday there are problems from that webpage, "Pepys & Newton." And let me tell you that you better get together with other people to do it, because there's a lot of stuff there. So get together in a group, parcel out who works on what, discuss what went on and so on. And then for a week from Friday there's stuff about drawing Lewis Structures, from another webpage about functional groups. So we'll get to that later, I just wanted you to know what's coming up. Incidentally, this thing about the problems set that has to do with the mathematics that Newton and Pepys were working on there, has to do with isotope ratios. You know, chlorine has a funny atomic weight. Why? What is the atomic weight, does anybody know?



Student: 35.5.



Professor Michael McBride: Thirty-five and a half. Why a half? Most of the elements are pretty near integers. It's because it's a mixture of isotopes. It's a quarter of one isotope and three-quarters of the other, and the average is thirty-five and a half; thirty-five and thirty-seven. But in fact it depends on where you get chlorine from, what the ratio is. There's thirty-five and a half for a standard atomic weight, but it ranges quite a bit, depending on where you get it. So by measuring these ratios, which is a lot like these odds in the betting, you can tell where things came from. You can tell where hydrocarbons came from sometimes that way. And those are the problems that you have to deal with. So it's not really very, very relevant to the course as a whole, but it's relevant to the Pepys and Newton thing and it's a fun problem. So that's what you've got to do for Monday.



Okay, and here are the assignments. I emailed all these people. I hope you're here, and that you picked up on what we talked about today so that you can improve the Wikis from last year. Last year people developed their Wikis from scratch. This time--which is a very valuable thing to do. So I was torn this year as to whether to have you do them from scratch or whether to modify last year. I figured modifying last year is better because then you have something to read earlier on, for those who are not working on developing it. So anyhow, but the hope is--so if you modify it--it has to be something significant, not put a comma in or something like that, but add something useful to understanding. If you do it within thirty-six hours, you get two points, and if what you do is really good, you get three points. So you go to that page and you click on those things and then you can read what--or you can either modify them or read it; it's a Wiki.



Now, we come to the question that we'll address more. We have another five minutes here. The question we're really dealing with now, having seen how to work in Chem. 125, is are there atoms and molecules; how do you know? And what force holds them together? Because if we knew that they're real, and we knew what force held them together, then in principle we might be able to calculate everything. So it would all be theory. So is it springs that hold atoms together? So Robert Boyle--notice he's right at that time, 1627 to 1691, and he's the oldest person that's on the Honor Roll outside the building here, he's over that way--so Robert Boyle wrote this first important book in chemistry, New Experiments Physico-Mechanical: Touching the Spring of Air. So you know Boyle's Law, how pressure and volume change, that air in a piston acts like a spring. So PV is a constant; that's Boyle's Law. So he developed this science on the basis of a new instrument, the pneumatical engine, which was built by Robert Hooke, the guy we quoted. And there's a picture of the air pump here with that crank on it, and so on to pump things in or out of that bulb. Now a couple of years later Hooke wrote this book, Lectures, De Potentia Restitutiva, or of Spring, explaining the power of springing bodies. And this is the beginning of that book. "The Theory of Springs, though attempted by divers eminent Mathematicians of this Age has hitherto not been Published by any. It is now about eighteen years since I first found it out, but designing to apply it to some particular use, I omitted the publishing thereof." So he kept it secret so no one would steal the idea. What did he hope to do with a spring, a really important technology that he could do with a spring? There was a coil in the spring that he depended on.



Student: Flying.



Student: Pendulum clock.



Professor Michael McBride: A new kind of clock, one that could work better for navigation. It was his spring that actually allowed Harrison, 100 years later, to win the contest about making an accurate clock; some of you must know that story. So anyhow he had been hiding this so it wouldn't be stolen. In fact, it was stolen by Huygens. So "about three years since His Majesty was pleased to see the experiment that made out this theory, tried at White Hall, and also my spring watch. About two years since I printed the theory in an anagram at the end of my book on the description of helioscopes, viz." this. So this is what he published at the end of this book on helioscopes, and that is an anagram of how springs work, so that later he could prove that he knew it, if needs be, but no one could steal it in the meantime. And if you unscramble it, it's "Ut tensio sic vis; that is, the power of any spring is in the same proportion with the tension thereof. That is, if one power stretch or bend at one space, two will stress at two, three will bend at three, and so forward." So here's his figure that shows that. That's Hooke's Law, the force law. The force is proportional to the distortion. So here's his figure, and he's got plots of it here. Ut tensio (as the extension) sic vis (so the force). So it's linear, the force is how much you stretch it. And he has lots of different kinds of springs that do that. Here's this clock kind of spring, a coil spring, just stretching a wire. So the force is proportional to the extension. Or another way of saying it is, "The potential energy is proportional to the square of the extension." So it's a parabola. So that's Hooke's Law, and that's where we'll take up next time.



[end of transcript]

Professor McBride begins by following Newton's admonition to search for the force law that describes chemical bonding. Neither direct (Hooke's Law) nor inverse (Coulomb, Gravity) dependence on distance will do - a composite like the Morse potential is needed. G. N. Lewis devised a "cubic-octet" theory based on the newly discovered electron, and developed it into a shared pair model to explain bonding. After discussing Lewis-dot notation and formal charge, Professor McBride shows that in some "single-minimum" cases the Lewis formalism is inadequate and salvaging it required introducing the confusing concept of "resonance."

Transcript

September 5, 2008



Professor Michael McBride: Okay, let's start up. So last time we got to the point of asking the question, are there atoms and molecules--and well, I anticipate that the answer is yes--and what force holds the atoms together? Because if we really understand the force that holds them, we've got chemistry all solved. So when you come to the idea of forces, in the first place, one goes back to Newton and Principia Mathematica. And he had the idea, or developed the idea, the mathematics of gravity. And the interesting thing about it was there was no mechanism for it. There were no springs or strings that connected the things. It was action at a distance.



So there were other ideas. Like, Descartes had the idea that gravity resulted from blocked repulsion, that the universe was full of particles zipping around and bombarding things. So like the moon would be bombarded by particles from all directions, but there'd be no net force on it because they'd all balance out. However, if you put the earth in the way, there'd be a shadow, and now you'd have more things hitting the moon from the right than from the left. So it would seem to be attracted to the earth. So that's a great mechanism. Now, it was important for Newton that the force be proportional to r2, to inverse r2, 1/r2. Does this work for that? If you pulled the earth back, how would you change things? What do you say?



Student: I don't know. It doesn't work for this.



Professor Michael McBride: It doesn't work? Would it block more or fewer things if you pulled it back?



Students: Fewer.



Professor Michael McBride: How many fewer? Suppose you made it twice as far back, how many fewer would be blocked from hitting the moon?



Student: Wouldn't it be r2?



Professor Michael McBride: It would be 1/r2. You can think about that. So that aspect of it would work. How about--you know gravity is proportional to the mass of the things. How does that figure? You can think about that yourselves, okay? But anyhow, there were other ideas. But Newton didn't think about the mechanism, he thought about the law, that it was 1/r2 and proportional to the masses.



Now in 1717, he published the Second Edition of Opticks. He published the First Edition in 1704, and the reason he published it in 1704 was that Robert Hooke died in 1703. He'd actually--Newton had done this work in the 1660s and '70s, but he didn't publish it until Hooke died because he didn't want Hooke to get on his case, criticizing it. But in the Second Edition, in 1717, he put some additional material; you see, "with Additions. " And the additional material, even though the book is about optics, the additional material is about chemistry. And it's at the end, and posed not as hypotheses but as questions, because Baconian people, following Francis Bacon, weren't supposed to make hypotheses, but they could ask questions. So here are some of the questions. Question thirty-one, the one that has most to do with chemistry:



"Have not the small Particles of Bodies certain Powers, Virtues, or Forces by which they act at a distance, not only on the Rays of Light to reflect, refract and inflect them, but also upon one another for producing a great part of the Phaenomena of Nature?" (And this attraction between the particles or repulsion) "For it's well known that Bodies act upon one another by the Attractions of Gravity, Magnetism and Electricity; and these Instances shew the Tenor and Course of Nature, and make it not improbable but that there may be more attractive Powers than these. For Nature is very consonant and conformable to her self. How these Attractions may be perform'd, I do not here consider. What I call Attraction may be perform'd by impulse…"



Where did he get that idea? Where did he get the idea that there are impulses that can cause attraction?



Student: [Inaudible]



Professor Michael McBride: That was Descartes' idea, that we just explained. "…or by some other means unknown to me. I use that Word here to signify only in general any Force by which Bodies tend toward one another, whatsoever be the Cause. For we must learn from the Phaenomena of Nature what Bodies attract one another, and what are the Laws and Properties of attraction, before we enquire the Cause by which the Attraction is perform'd." (So get the law first so you can do mathematics on it, and then worry about how it works.)



"The Attractions of Gravity, Magnetism and Electricity, react to very sensible distances…" (He doesn't mean reasonable, he means ones that you can sense.) "…and so have been observed by vulgar Eyes." (You can see the distances over which electrostatic attraction works, obviously gravity, magnetism.) "And there may be others which reach to so small distances as hitherto to escape observation; and perhaps electrical Attraction may reach to such small distances, even without being excited by Friction." (So you know that friction can generate static electricity that'll make things attract, but maybe even without friction you can get things that'll attract a very short distance by electricity.) "The Parts of all homogeneal hard Bodies which fully touch one another, stick together very strongly." Did you ever have that experience of having two really flat things, put them together and they're very hard to get apart; what?



Student: Slides.



Student: Small things.



Professor Michael McBride: Microscope slides. Ever take out new microscope slides? They can be very hard to get apart. "And for explaining how this may be, some have invented hooked Atoms, which is begging the Question; and others tell us that Bodies are glued together by rest, that is, by an occult Quality, or rather by nothing, and others that they stick together by conspiring motions, that is by relative rest among themselves." (These are other theories that are sort of complicated; we don't need to go into it.) "I had rather infer from their Cohesion, that their Particles attract one another by some Force, which in immediate Contact is exceeding strong, at small distances performs the chymical Operations above mention'd, and reaches not far from the Particles with any sensible Effect." (So a very short-range, a very strong attraction.) "…reaches not far from the Particles with any sensible Effect."



So maybe it's even more dramatic than 1/r2. 1/r2 changes very rapidly when r gets very small, but maybe it's even a law that's more distance dependant than that, but only works at really, really small distances. "The Attraction" (and he's talking about between glass plates separated by a thin film of the Oil of Oranges; so you know, microscope slides, they put a little drop of something and stick them together and they're really hard to get apart) "may be proportionally greater" (the attraction) "and continue to increase until the thickness do not exceed that of a single Particle of Oil." (Imagine Newton, 350 years ago or 300 years ago here, thinking about a single particle of oil. Could he possibly measure that? We'll think about that later. Okay.) "There are therefore Agents in Nature able to make the Particles of Bodies stick together by very strong Attractions. And it is the business of experimental Philosophy to find them out." That's what we need to find, what is it that holds these particles together? "Experimental philosophy" means science. And we're going to engage in this business, and it'll take us about five weeks to figure out what it is that holds these things together.



Now, let's just look, from physics, at the amount of binding energy you get from different kinds of laws. First there's Coulomb's law; so it charges over the distance. I think you all know that one. Then there's magnetic interaction, two magnetic dipoles, and the energy is 1/r3, falls off very rapidly. Then there's the "strong" binding that holds particles together in the nuclei of atoms. And gravitation, of course. And the chemical bond. So here's a scale of energies, and we use, in this course, the old fashioned kilocalories/mol, which American organic chemists like to use. Some day it'll change to kilojoules/mol, but for me it's still kilocalories/mol. Okay, this is a logarithmic scale, so it covers a very wide range.



Now Coulomb, if you have a proton and an electron, as far apart as the distance of the carbon-carbon bond, that's 216 kilocalories/mol, right in the middle of this, or near the middle. Magnetic interaction is much weaker. If you have two electron spins, at the same distance from one another, their energy is what? One, two, three, four, five orders of magnitude smaller, 105th smaller. The "strong" binding that holds a proton and a neutron together is in the deuterium nucleus, that is within the nucleus, not at 1.54 Ångstroms, is 105th stronger than Coulombic interaction. And gravity is ever so much weaker. If you had two carbon atoms, that were at that distance, the energy would be 10-32 kilocalories/mol. So forget gravity as a way of holding things together. But then we have Coulombic up or down by 5 orders of magnitude.



Now what is a chemical bond; how strong is it? A chemical bond is about--the carbon-carbon bond is about 90 kilocalories/mol. So not so far from the Coulomb energy, from electrostatic attraction. So maybe this hints that electrostatic attraction has something to do with it. But maybe not, maybe it's something entirely new. And we haven't talked about kinetic energy. Does that have anything to do with the strength of the bond? So we'll get back to that later on. Okay, so is there a chemical force law, the thing that Newton was looking for? Now here's an interesting way to phrase that. Suppose you had a chain of atoms, a thread that was single atoms bonded to one another, and you started stretching it. It would stretch, and at some point it would pop.



Now let's just take three of the atoms and we'll put stiff rods on the outer ones and we'll start stretching it. So it'll stretch, stretch, stretch, stretch, pop. Now the question is how far do you stretch it before it pops? What determines how far you stretch it before it pops? Okay, this must have to do with the force law between atoms, which then will have to do with molecular structure. So it could be like springs, Hooke's law, Ut tensio, sic vis. Okay, so the force is proportional to the displacement, and the energy to the square of the displacement and the energy then, the potential energy, is a parabola. We define zero when they're at the standard, unstretched distance, and then you stretch it or compress it and the energy goes up, like a parabola. Now the slope of that parabola tells you what?



Student: The force.



Professor Michael McBride: The force. That's what's linear. As we stretch it further and further, the slope gets bigger. The force is proportional to the displacement. Okay, now there could be other kinds of force laws, like electrical charges, Coulomb, or gravity. So that's a different force law, inverse square force law, and the energy then is proportional to the displacement, in this way. So very different. Notice the zero of this one's at the top when--zero defined when they're very far apart. Here it's zero when they're at the standard distance. And so as you bring oppositely charged particles together, the energy goes down infinitely, as they get really close. Okay, and now if we look at the force here, as we go out the force gets smaller, the slope gets smaller.



Now suppose you had a particle that was connected to a spring on each side, as in this chain that we're stretching. And just to be general I made the second string--oh I say stronger, it's in fact weaker; it doesn't go up as rapidly--I'll have to change that. Okay now, but at some point the slopes of the red and the blue, at some distance, some position, will be equal and opposite. There. Okay, what's the force on the central particle at that point?



Students: Zero.



Professor Michael McBride: Zero. They balance. So if we added those two energies together to get the sum of the two, on this central particle, it looks like that. So there's a balanced minimum. There's a well defined position where the particle in the center of this chain will just sit there and it'll be hard to displace. So it can vibrate. Okay? Now how about in the other case? How about if it were electrical charges or gravity or something like that, an inverse square law that's holding things together? Then the second one is going to look like that, and that one is indeed stronger.



So they're two flanking bodies and we're interested in the position of the one in the middle. And again, there'll be some point where they balance, right there, where they're equal and opposite slopes. How will this differ from the first case? Can you see what's going to happen when we add these two together? Is it going to look more or less like the one on the left?



Student: [Inaudible]



Professor Michael McBride: It's going to look like that. And now there is no balanced minimum in the middle. There's not a position where the thing will just sit there, because it's always more strongly affected, and attracted, by the one it's closer to. There's some place where there's zero force but it's not a stable position. Because if it displaces ever so slightly, it'll keep on going. So that one is a--for the two of them--is a single minimum, and this one is a double minimum. And for the people who came early, there was a contest that if you won it, you could get an A in the course, and the contest is this. Here's a magnet, hanging here. And I'll stop it. Okay, and here are two magnets, the ones that will flank that one. And if I get it just right, it'll just sit there, the attraction to the two will balance it. So I'll put it on here, and try to get my A. So I'll get the string where it would balance right there. Ah. Okay? There is no stable position. Nobody got their A that way. Right? Because it's an inverse law, so it's always more strongly attracted to the closer thing. So you can't win with that, you can't get a balanced position. Now I'm going to get these off here. And back to the show.



So with springs you might be able to make a stable polyatomic molecule from point atoms--we saw a spring model before in the last lecture--but you can't do it with ions and you can't do it with magnets. However, Hooke's law can't do it--it can't be springs, because Hooke's law never breaks. So we need a different kind of force law. Does anybody know a force law--what it looks like--the form of the force between atoms or the energy for stretching a bond? You ever seen a picture of such a thing?



Student: [Inaudible]



Professor Michael McBride: It's called the Morse potential, or one form of it at least is called the Morse potential. And it's not something fundamental, it's not a law of nature. It was thought up by a physicist at Princeton in 1929, because it's mathematically convenient; so you could solve quantum mechanical problems if you used Morse potentials. And the idea is this. You have two neighbors and you look at the position of the one on the right and it could be there and the energy would be minimum; that's the bond distance, right? And if you move it further to the right, the energy goes up; move it to the left, the energy goes up. Okay, that's the Morse potential, and we're just holding the neighbor on the left fixed. Now we could put another neighbor on the other side and have another curve of the same sort; stretching the second bond. Or we could have it be a chain and have both neighbors there, and the sum of those two now has a minimum in the middle.



It's a single minimum, like we got with Hooke's law. So that means that atom in the middle would just stay there. And now we're going to grab those sticks on either side and pull the neighbors apart. So if we pull them apart, it's still a single minimum. And if we pull them apart further, it's still a single minimum, although it's very flat. And if we pull it further, it pops and it becomes a double minimum. And if we keep going the double minimum gets more pronounced. Now here's an interesting question. At what point--how far do you have to stretch it before it pops? That's the question we were asking, right? How can you look at those curves and tell how far apart you have to stretch it before the chain pops? Shai?



Student: The bottom is flat. So when the curve and the plot was flat, then that's as far as it could go because…



Professor Michael McBride: What do you mean flat? There's a name for a thing like that in graphs.



Student: Inflection.



Professor Michael McBride: It's the inflection point. It's when the curvature changes from being this way, like Hooke's law, to being this way, right? So when the inflection points cross, then the chain pops. Okay, so force laws are going to make a lot of difference in how atoms behave, and if really knew the force law we'd be in a great position for understanding chemistry. It snaps at the inflection point where it changes from a direct to an inverse dependence on distance for force.



Okay, but what are bonds? Newton said, "We'll look for the law." And it's sort of, the law is sort of like Morse, but we don't know where it came from, because we don't know how it works, we don't know what it really is. So can we find out what it really is? Well in the nineteenth century they discovered bonds. And this is a picture from 1861, we'll talk about this later. It's one of the first depictions that's recognizable by people today of bonds between atoms. So there are different numbers of these lines, valences, for different atoms. Hydrogen has one; carbon has four; oxygen two; nitrogen three.



Why do the elements differ? Why don't they all have the same valence? And even more complicated, sometimes nitrogen is three and sometimes it's five. For example, you have NH3 but you also have NH4Cl, where there are five things associated with the nitrogen. So how do you understand the valence? That was the challenge for people. They had figured out that there was valence, but why? How could you predict the valence of a new element? Well Gertrude and Robert Robinson published a paper in 1917 which showed this picture of ammonia, NH3. What's the loop? Pardon me?



[Students speak over one another]



Professor Michael McBride: Someone said--hold your hand up so that people can hear you. Okay yes.



Student: A lone pair on nitrogen.



Professor Michael McBride: A lone pair on nitrogen--wrong. It looks like the lone pair on nitrogen but the lone pair didn't come along that soon. What the loop means, if you look at it in the context, is what makes NH3 reactive? In this term, this is what they wrote: "It may reasonably be assumed that the partial dissociation is a stage in the complete process." So the bonds begin to break. This loop begins to break to become a weaker loop and two other partial valences, in this theory. And those valences begin to associate with partial valences from HCl. Right? The loop is a "latent" valence. You know what 'latent' means? Latent is the opposite of patent. Latent means hidden; patent is revealed. When you get a patent on something it means that you tell everybody how to do it but the government protects your right to do it for a certain period of time. Latent means it's hidden.



So there's a hidden valence, these loops, but they can become available, right? It must be assumed in some cases, as for example the combination of ammonia with hydrochloric acid. Now, so you can get a reaction and get the product, which has all five valences of nitrogen now. But how do you know there's not another loop on nitrogen, that you could break open and make it seven valent, or nine, or eleven? Or maybe you could break a loop into three. So might latent valence loop explain the trivalence and pentavalence of nitrogen, or the amine-HCl reactivity? The trouble is it's too slippery a concept. It explains everything. You could explain anything you wanted to. You could explain eighty-four valence of nitrogen or something. Anything that comes along, you'd say, "Ah ha, there are that many latent valences." But how do you know there are not more? Why do you have latent valences? When do you have them? When and why do you have partial dissociation? This thing didn't explain anything, or explained everything. Now, at this point I want you to smile so I can learn who you are.



[Professor McBride takes pictures of the class members]



Professor Michael McBride: So the electron was discovered in 1897. So maybe that has something to do with it. And the guy who had associated electrons with valence was G. N. Lewis. And this is a picture of him as a Harvard undergraduate in 1894. And this is eight years later, when he was an instructor at Harvard, and he then went to establish the Chemistry Department at the University of California at Berkley.



Student: Yay!



[Laughter]



Professor Michael McBride: But when he was at Harvard in 1902 he used these lecture notes, and what he was trying to explain was the periodic table, why you go across in eight and then another eight and another eight, and so on, and why you have electron transfers from some atoms to other atoms. And he said if the electrons in an atom are arranged at the corners of a cube, then you--eight is this very special number, because that occupies all the corners of the cube. So if there's a desire, for some reason, to complete octets, then you can explain periodicity and see why some atoms give up an electron to lose the outer shell and others gain one to complete the outer shell.



Okay, so but it also could explain how you get bonding, if you like to have octets; not just electron transfer but the formation of covalent bonds. So, for example, here's two chlorines. Each has only seven; the octet isn't complete. But if you bring them together and they share an edge, then both octets are complete. He doesn't say why octets should be so great, but if they're octets then you could explain bonds. How about double bonds? Suppose you had two oxygen atoms. What do you do now? Anybody see? What?



Student: Put the faces together.



Professor Michael McBride: Put faces together, share two edges, and then both have an octet. Now, suppose you want to put nitrogens together and form a triple bond. Now how do you do it?



Student: Squish it.



Professor Michael McBride: Push it hard, force it like a picture puzzle?



Student: Yes.



Professor Michael McBride: Put it in there.



Student: You need to push it.



Student: You did it.



Professor Michael McBride: It won't do it. But about ten years later he figured out how to do it. What you do is take the electrons--nitrogen has 5--but instead of using an octet, you contract it along opposing edges this way. So what was an octet, what was a cube, becomes what?



Student: A tetrahedron.



Professor Michael McBride: A tetrahedron. And now if you take two tetrahedra, you can put a face together and complete both octets. But this wasn't such a great development. He did this, published this, in 1916. But the tetrahedral distribution of the bonds in carbon had been known by organic chemists for 40 years by that time, that the tetrahedron was a very important structure. So a good theory should be realistic and it should be simple. But there's a tension between these things. It should at least be as simple as possible, but that may not be very simple; that's why it takes us five weeks. In regard to the facts it should allow a number of properties. It should allow prediction. You should be able to say how will these atoms react; how will this molecule react; what valence should nitrogen have; how can it be both pentavalent and trivalent? A little below this on the scale of desirability is suggestion. A theory may--even if it doesn't give you the proper thing, may at least suggest some experiment that would be really good to do.



Below that is explanation--according to my theory all these known facts fit together, but who knows about any new thing. And lowest is classification and remembering; like Roy G. Biv allows you to remember the colors of the spectrum, but there's nothing fundamental about that. So even if you have a crummy thing it might allow you to at least remember some facts and organize them maybe. Okay, but those last two are not prediction, they're post-diction; you only use it to explain things that are already known as a way of remembering them. So from the number of valence electrons we'd like to predict how many atoms of different types come together to form molecules. That's constitution--the nature and the sequence of bonds--and the valence.



We'd like to know the structure of molecules, the distance between atoms, angles. We'd like to know how charge is distributed within molecules. Are some regions positive and others negative? We'd like to know what the energy of the molecule is. Is this arrangement of the atoms better or worse than an alternative arrangement? And we'd like to know about reactivity. So Lewis explains constitution, the nature and the sequence of bonds. "Nature" means like single, double, triple, and "sequence" means which atoms are connected to which. So he has electrons, valence, and also he added unshared pairs. So you count up how many, from the atomic weight or the atomic number or the position in the periodic table. You get how many valence electrons, how many of these outer octets are there. And then you can figure that the valence of hydrogen should be one, boron three, carbon four, but then down again, three, two, one.



So here's ammonia, NH3. It works fine and it has an unshared pair. But why do you have octets? Why not sextets? And why do you have a pair for hydrogen? Why not an octet for hydrogen also? Okay, so let's just apply it to HCN. So if carbon has four, that makes possible a single bond to hydrogen, a triple bond to carbon. Everybody has their octet except hydrogen that only wants two. So everybody's happy. And we can abbreviate it this way, and leave the dot. Lewis actually thought of drawing a colon, a pair of dots, to denote an unshared pair. So that's his notation. So NH3. Bingo! How about BH3? Just fine. But here's something new. The organic chemists would already have known this, they knew the valence numbers. The electrons just added something to it but didn't say anything really new. But here's something new, that BH3 reacts with NH3, to give a new bond. Now here's a puzzle. It's also true that BH3 reacts with BH3. That doesn't look like Lewis. And the bond is reasonably strong, it's half as strong as a carbon-carbon bond. So what's the glue that holds B2H6 together? And we'll come back to that fourteen lectures from now.



But there's another thing. There's bookkeeping that you do with Lewis structures to assign what's called formal charges; not real charges, just what charges you write in the formula, and you hope they mean something. And the bookkeeping scheme you use is that each atom is assigned half interest in the bonding pair. So if there's a bonding pair, for bookkeeping purposes you assign one electron to each. It may be they aren't evenly shared, but the bookkeepers don't care.



Incidentally, speaking of bookkeepers not caring, there's--people have had a lot of nonsense trying to get signed up for lab, because the people who run the system don't allow teachers to have access to the thing to see the problems you face. So I have no way of telling--I can't get on to your system. Has everybody figured out how to sign up for lab by now?



Student: Yes.



Professor Michael McBride: If anybody still has problems, after class Dr. DiMeglio can take you to the computer center here and help you. So next year presumably it'll be easier. This is a new system this year. But anyhow, back to the bookkeepers here.



So these formal charges are not necessarily real charges, they're just what you write when you draw the formulas. Okay, so we split the BN electrons between the B and the N, and we split the hydrogen bonds as well. So nitrogen now has those four electrons, where it brought five to the table originally. So it has a positive charge. Tetravalent nitrogen is positive in the Lewis structure. Why? Because in forming a fourth bond, it gave up half-interest in one of its electron [pairs], the ones in the unshared pair. So it lost a charge, a negative charge, and becomes positive. By the same token, what else can you write in this structure?



[Students speak over one another]



Student: It's negative.



Professor Michael McBride: A negative charge on boron, because it got half-interest in a pair. But that's just bookkeeping. So when you write a Lewis structure you'd write plus on nitrogen and minus on boron. Now is it real? Is there a positive charge on nitrogen and negative charge on boron? Well we'll get later on to know about what a surface potential is. Surface potential is the energy a proton would have if it was on different points on the molecular surface. For this purpose you have to define what the surface is of BNH6. There's a way of doing that; we won't talk about it now, that's jumping the gun. But that's the surface, and it's color coded to show the energy a proton would have if it were on the surface at that point. So what you see is that the energy of the proton would be high, if it's here, where it's very blue. The energy of the proton would be low if it's in this region here. Is that consistent with what we were talking about?



Student: Yes.



Professor Michael McBride: Where would a proton be low in energy? Where there's negative charge, and it would be high in energy when there is positive charge. So this molecule, according to quantum mechanics, is positive on the left and negative on the right. And that's exactly what we have, it's consistent with what we have. And now Lewis also explains where pentavalent nitrogen comes from, because there's an ionic bond there too. Tetravalent nitrogen is plus. So there's just a Coulombic attraction between ammonium and chloride. So that's a great contribution. Now there are other things you can understand in terms of Lewis structures. Like you have an amine and it can react with an oxygen to give an amine oxide, which is positive on nitrogen, negative on oxygen. Or you can have a sulfide, which can react with an oxygen, to give a sulfoxide. Or what else can happen, with a sulfur, what additionally can happen?



Student: Another oxygen.



Professor Michael McBride: It could do another oxygen, and give a sulfone, as it's called. So Lewis explains this. So there are problems to drill on Lewis structures. They're on the Web, go look at them. One of them is draw Lewis dot structures for HNC, in that order, H then N--we did HCN; but do HNC. And then try HCNO with CNO in all six linear orders; that is, nitrogen in the middle, carbon in the middle, oxygen in the middle, and then hydrogen on either end. And it's also possible to make a ring of CNO and put the hydrogen on any of the positions. So just practice drawing Lewis structures of those.



And start memorizing functional groups. We do precious little memorizing in this course, compared to most organic chemistry courses. We try to figure out how things work so that you can predict without memorizing it. But there are some things you have to know, like the names of groups. Otherwise it takes forever to talk about them. So here are the ones you have to learn, and on the fist exam, coming up in three weeks or whatever, I'll give you a question where you have to draw one of these structures from the name, or give the name for a structure.



These are all on the webpage; you don't have to copy them down. But those are simple functional groups. [Laughter] And here are--whoops, oh there you go, sorry. [Laughter] There. They're also conjugated functional groups. So you could run it yourself. Okay, now the final question here, for today, is a thing that comes up when you talk about Lewis structures, or structures in general, Lewis structures in particular, is equilibrium and resonance. How many people know what resonance is? What is it?



Student: Like the same molecule but different structure.



Professor Michael McBride: Speak up a little bit.



Student: It's the same molecule but a different structure.



Professor Michael McBride: The same molecule but a different structure. But isn't structure a molecule?



Student: But with that I guess--



Professor Michael McBride: You know like, let me just interject a minute here. Like, one of the problems is HCNO in any order. So there would be different molecules with C in the middle, N in the middle, O in the middle. Is that what you mean?



Student: No. I mean like it's--if you have a double or a triple bond, it's in a different location, it's around the central atom.



Professor Michael McBride: Ah, now let's--actually resonance is one of those things where people try to hide ignorance from ignominy. And let me show you what I mean by that. So here's the structure for HCNO. And here it has all octets. If you go through you'll figure that out. It has charge separation though. You don't like to have that in your structures. You like not to have charges if you can avoid it. Obviously in BNH6 you can't avoid it. Okay, now but you can get rid of the charge separation by shifting that electron pair that's on oxygen to be shared between oxygen and nitrogen; then you don't have the charge on oxygen anymore. Everybody see that?



Student: Yeah sure.



Professor Michael McBride: What's the problem when you do that?



Nitrogen now has too many electrons. So you could shift a pair from--that's in the CN bond, onto the carbon, in your drawing you make. Okay, so you get this structure shown below which has all octets. But there's still charge separation, and in fact it's worse because the negative charge, instead of being on oxygen, is now on carbon. So there are these two different Lewis structures you can draw and each has its own properties, or good things and bad things.



Now what are the geometric implications of there being two structures? It could mean that in the middle one, in the lower one, nitrogen is exactly halfway between carbon and oxygen, because they are both double bonds. But in the top one, nitrogen is much closer to carbon than to oxygen. That could be the meaning of this. So that if you draw a picture of the energy, as a function of the position of the nitrogen, you could get two different structures, one where it's further to the left, where the nitrogen is further to the left you have this structure, and when the nitrogen's in the middle you have that structure. And you have to go up hill in energy, it'll click; as you start pushing the nitrogen it starts, it'll click, and it's the other one. Okay? That could be the way the atoms really behave. That's called a double minimum; we already saw a double minimum today.



But maybe in truth it's not this way. Maybe it's a single minimum. Maybe the best position for the nitrogen is neither to the left nor in the middle, but in between; maybe that's the lowest energy. And that is what resonance is. Resonance is when you--the true structure is in between the structures you draw. So really what it is, is a failure of the notation. Does everybody see how that's so? It's that the notation you use doesn't show the right structure, and when that happens there's said to be "resonance"; although actually all it is is that the true structure is a single minimum, not the double minimum you'd expect from your drawing. It's a failure of the way you draw.



And notice that for resonance you use a double headed arrow whereas if I back up a second here--equilibrium, when you have a double minimum, there are two arrows, two different structures. So the question is which is it? Is the real molecule a double minimum; or the real molecules a double minimum, are there two structures? Or is there in truth only one structure? So you need to choose. The choice between whether there's resonance or equilibrium must be based on experimental facts or on a theory that's better than Lewis's theory, to be able to know which is true; something that can distinguish between a single and a double minimum.



So equilibrium is two arrows, and there are two structures that go back and forth, and resonance is a single structure, that our notation isn't very good at drawing. So two real species, or one real species but two reasonable formulas we could draw for it, or that Lewis would draw for it. So really what it is, resonance is a failure of simplistic notation, and it's associated with unusual stability. Now when I say 'unusual,' that implies that you know what usual stability is. So compared to what is it unusual? And we'll address that question later on.



So let me just give you finally an example of equilibrium versus resonance. So this, as you'll learn when you learn functional groups, is a carboxylic acid, and you could have the hydrogen attached to this oxygen or it could be attached to the other oxygen. So these are not just the same thing upside down. It's the hydrogen being attached to one or being attached to the other of the two oxygens. And, in fact, it is that way. There are two structures and you go back and forth between them; two species.



But how about if you take the hydrogen atom away and have this thing that's called a free radical, where you don't have a complete octet on this oxygen? There are only two, four, six, seven electrons there. It has the ability to form another bond by sharing an electron. Is that two different species? Or is it one species--watch--with an intermediate geometry where you don't really have double and single bonds but something in between? And how do you know which it is? The only way is to do some experiment that tells you. And there is an experiment that tells you it's only one nuclear geometry. And we don't have time in the last thirty seconds to tell you how that proof works, but I'll tell you that the technique is electron paramagnetic resonance spectroscopy [Laughter], shows that in truth, although this is two species, this is one species. So there's equilibrium at the top and resonance at the bottom. And if you add an extra electron to make a carboxylate anion, infrared spectroscopy tells you that it's symmetrical too. There's resonance in this case. It's an intermediate structure. But that is "lore." You didn't predict that from Lewis structure ahead of time. It could've been either way, and in fact there were many people who thought it was one way or the other. It's only experiment that does it. So, next time we'll talk more about lore.



[end of transcript]

Continuing the discussion of Lewis structures and chemical forces from the previous lecture, Professor McBride introduces the double-well potential of the ozone molecule and its structural equilibrium. The inability for inverse-square force laws to account for stable arrangements of charged particles is prescribed by Earnshaw's Theorem, which may be visualized by means of lines of force. J.J. Thomson circumvented Earnshaw's prohibition on structure by postulating a "plum-pudding" atom. When Rutherford showed that the nucleus was a point, Thomson had to conclude that Coulomb's law was invalid at small distances.



Transcript



September 8, 2008



Professor Michael McBride: This is the slide from the end of last time, we didn't quite finish it. Equilibrium versus Resonance. Remember, equilibrium is a case where you have two different species, different structures, and you go back and forth between them, maybe fast, maybe slow. Like the hydrogen could be attached to the top or it could be attached to the bottom oxygen, and those are different, two different, what's called isomers; we'll get to that later on. Or you could imagine the same thing if you take the hydrogen off, you could have a short double bond to oxygen and a long single bond to oxygen, or they could exchange as to which oxygen was long and which was short. So you could have an equilibrium there. But it turns out you don't, that in fact it's not two different species, it's one species. It's a single minimum not a double minimum.



Now how do you know this? How do you know it's just one nuclear geometry with an intermediate bond distance? The only way you know is by experiment or by some really fancy calculation that you have to believe. A lot of people would believe experiments before calculations; some are the other way around. But there's evidence from a technique called electron paramagnetic resonance, or EPR, that shows that indeed this is one species, a single minimum. If you have an extra electron on it and you have a carboxylate anion, then again it's just one species, a single minimum. And there's evidence of that from infrared spectroscopy that we'll talk about next semester. But don't be disappointed that you're not able to predict this. A lot of really smart, experienced people couldn't predict it. This is lore. If you look it up in the Oxford English Dictionary it says that lore is, "That which is learned; learning, scholarship, erudition. Also, in recent use, applied to the body of traditional facts, anecdotes, or beliefs relating to some particular subject." So a lot of things are lore. You just have to learn them. You can't predict them ahead of time, they're way too subtle.



So don't be disappointed, because you haven't had enough time yet to get the lore; you're not supposed to know it yet. It would be nice if Lewis theory was so accurate and straightforward that if you draw two structures, there are two structures and it's a double minimum. But that's not true, and you have to know from something, and you don't have time to know yet. As time goes on maybe you'll figure it out. This is from a good textbook; it might be the one that we'll adopt next semester. It's a quote that says, "Empirical rules" (it's going to give) "Empirical rules for assessing the relative importance of the resonance structures of molecules and ions." That is, if you have two different pictures that you can draw for the thing, does it look halfway in between them, almost all like one, almost all like the other, or some fraction of the way between them. How far is it, one way or the other? These are two different pictures you draw to try to show different aspects of the same thing. But there is one real thing. The molecule doesn't know about resonance, it just knows what it is. The problem is with our notation. Okay, but anyhow here's what they say. So rules that will allow you to use this concept more productively by deciding which ones are better, which look more like the real thing. Okay, so it gives rules and it numbers them.



"(1) Resonance structures involve no change in the positions of nuclei; only electron distribution is involved." That is, when you draw these two structures you don't move the atoms, you just change where you draw a single bond or a double bond or a dotted bond or something like that. And in fact that's not even true because it's not--the electrons know where they want to be, it's the way we draw them that's uncertain. So when you draw two different resonance structures, you're not changing where the electrons are, you're just changing the lines you draw. Is that clear? It's our notation that's at fault.



"(2) Structures in which all first-row atoms have filled octets are generally important; however, resulting formal charges" (we talked last time about how you get formal charges) "and electronegativity differences" (and of course we have to know what electronegativity is, but you've heard about it at least) "can make appropriate nonoctet structures comparably important." So if you have a bad charge distribution, even though you have octets, it still might not be a very good structure.



"(3) The more important structures are those involving a minimum of charge separation" (so you don't want to have formal charge separation) "particularly among atoms of comparable electronegativity. Structures with negative charges assigned to electronegative atoms may also be important." So if you're going to get a charge, put it where it wants to be.



Now, look at this more carefully. Number two, it says 'generally important.' It doesn't say "always important," but "generally." That's not such a great rule because you have to know when there's going to be an exception. Or 'however;' this doesn't sound like the Ten Commandments graven in stone. Or "can make" a thing; not that "it will," but "it can." Or '"particularly" or "may also be important." These are all weasel words because the rules are not rules, and this is all lore again. So people write these rules but really the people who are writing the rules know the answers ahead of time and think "Ah, it generally works that way and mostly we can get away with it." But these are not rules like the rules you want to learn in physics. You want to learn these because it's sort of handy, but don't believe them. And notice at the top that it's "empirical" rules. It's not a fundamental theory, these are just ways of sort of correlating a bunch of the lore that's come in. And what's important is the experiments, not theories like this. And we'll get better theories later.



Okay, so the goal of this Lewis stuff was from the number of valence electrons it would be nice to be able to predict the constitution, that is, the valence numbers for the different atoms, how many atoms of one kind or another get together to make a molecule. Reactivity, we've seen a little bit of that at least, that unshared pairs can get together with vacant orbitals and make a new bond. That wasn't something that was part of the original rules of valence. And maybe something about charge distribution as well. Now let's look at the case of O2 and O3, and apply some of this stuff. So O2, you can have--complete the octets by bringing two oxygen atoms together and forming two bonds between them. So that looks pretty good, and it's a double bond, and we can draw it that way with just lines and forget the unshared pairs if we're not particularly concerned with them at a given time. Okay, now suppose you make O3. You could do the same kind of trick there and make a three-membered ring, or you could make it linear, or bent instead of a symmetrical three-membered ring. So you could have an equilateral triangle like that, or you could have it like this, have O2 and bring in another oxygen on one of the lone pairs. Now that's not an equilateral triangle; it's a triangle, or it could conceivably be a straight line, we're not quite sure what the geometric implications are. Okay, but we could call it an open structure as compared to the ring structure. And we can draw it like that, and the formal charges would be as shown. Why? Because that pair that originally belonged only to the original top oxygen of O2 is now shared with the other oxygen. So one of them loses half-interest in a pair, the other gains half-interest in the pair and it's plus, minus. So the trivalent oxygen is positive, as you would expect.



Now, what is the true structure of the molecule ozone, O3? Well you have to do some experiment to find that out or some high-faluting calculation that you believe; a quantum mechanical calculation. So it could be a ring or it could be an open structure. And notice, in the open structure you should have two resonance structures, because you could draw it that way, or you could draw it that way, and it could be two minima--it could be a double minima--double minimum situation, or it could be distorted one way or the other and click back and forth. Or it could be that the true structure is in between with equal bond distances. So we have to find out something about this.



Now one way of finding it out is to do calculations and draw a graph that shows what the answer is, and that's what we're going to do here. It's based on some fairly recent high-level calculations. But the problem is drawing the graph, because if you want to be able to show the structure, you have to--how many variables do you have to specify to say what the structure of O3 is? How many numbers would you have to have? It's three particles, right? So if you gave this distance and this distance and this distance, that would fix the triangle. Or you could give this distance and this distance and the angle here. That would also specify the triangle. But any way you slice it, you have to have three numbers to tell the structure, and then you have to have another number in your graph to say what the energy is, when it has that structure. So you have to plot four variables. And that's not trivial on a piece of paper, to plot a graph with four variables in it. So three distances plus energy, or two distances and an angle and energy you need to plot. So if you're good at plotting things and have had a lot of experience, great, but if you need a little warm-up exercise you could look at this webpage that you get by clicking up at the top there, and it uses these graphs to give you a little exercise in plotting things that are in many dimensions. But we're not going to do that in class, that's just to do on your own or with a discussion section. And we'll show the specific example of how we're going to do it in the case of O3.



So to specify O3 we need four dimensions. Now here we have a plot that shows one distance, a second distance and the angle. So that'll specify where the three oxygens are. But Rutenberg and his coworkers here, in 1997, when they did the fancy calculation, did it a little differently. They constrained it. They didn't allow everything to vary. They said we're going to require the two distances to be the same. So now instead of three variables, you have only two, and you can plot those two on a two-dimensional paper; you can plot two things. So here we're going to plot the position. If the two distances are equal, then if you know the position of the top right you know the position of the top left, because the original oxygen is at the origin 0,0. Everybody with me on that? So we're going to look only at where the top right oxygen is, and the other one will be someplace symmetrically related to it. So we're going to blow that up, and here's the plot they give then. So now any point on this will specify a precise geometry of the three oxygens. Everybody with me? We choose a point that tells where the top right one is. The bottom one is at 0,0, and the top left is just on the other side, in the corresponding mirror-image position. So any point on that thing will specify the structure. But now we still--but that's used up our two dimensions of the paper. How do we show the energy? Can you think of a three-dimensional graph you've ever used?



Student: Color codes.



Professor Michael McBride: Pardon me?



Student: Color codes.



Professor Michael McBride: You could color code it. You could make red really high and blue really low. That would be one way to do it. There are earlier ways, before printing made it--or computers made it easy to make colors.



Student: Contour lines.



Professor Michael McBride: Did you ever go hiking? Yes what?



Student: Contour lines.



Professor Michael McBride: Contour lines, like a geological map, right? Okay, so we can draw contour lines that show what the energy is for every one of these geometries. And so if that--notice that at that position, that X, it turns out to be what the red structure would be here, which is a ring; equilateral, okay? And this structure is open. And I chose those particular ones because when you do the fancy calculation, those turn out to be the ones that are at the bottom of valleys, low energy. And if you distort away from those, the energy goes up. Now you can go up--any direction you go from one of the bottoms of the valleys you go up in energy, but there's some particularly interesting positions. A particularly interesting position is that one, because that's the pass, that's the lowest you can go, the lowest energy that's required to go from one valley to the other. Everybody, I think, has probably done enough hiking to realize that that's the way you'd like to hike, or that if you spilled a cup of water up at the pass it would run down both ways.



Now in fact we can use that concept to make it even a simpler graph, where we don't need to draw contour lines, because we could take the steepest descent path--if you pour the water out at the pass and follow how it'll trickle down, it'll take the steepest way to go down. It turns out it crosses every contour perpendicularly. So you follow that and it would go down to the bottom, and then if you kept going, like if you dropped a marble or something, and it rolled down to the bottom it would keep going. So there is a particularly interesting path. Not that the true molecule would necessarily have to follow that path, but it's a well-defined path that gets from one valley to the other. So now what you could do is take a knife and slice this thing along that green line and unfold it so it's flat. Does everybody see how that would be?



Like I remember once when I was young we took a family vacation and the AAA sent you a map that showed where you would go, but it also showed--along a particular highway--and it also showed another map that showed how high you were all the time as you went along the roads. Got the idea? So this is exactly that kind of thing, where as you go along that green road there are different altitudes. So we could just--we could draw it this way. Does everyone see how that works? So it's not quite as specific about geometry anymore, the way the previous one was, but it shows how much energy you need as you go along. And a particularly important one--well there are two things that are really important. One is how high one valley is compared to the other, how much more stable one is, and the other is how much higher is the pass, how hard is it to get from one to the other?



Okay, so there now is, in a two-dimensional graph that you can draw on a piece of paper, is something that gives you this information. Okay, but this required that we choose R12 equals R23, in order to make this simplification. The guys who did the calculations said if we don't make R12 equal to R23, then it turns out that all the structures are still higher in energy. So these are the lowest energy structures, although they can't plot them on a piece of paper. So the lowest energy structure is this open form and it does have R12 equal to R23. It's a resonance structure, halfway between; not one double bond and one single bond but a symmetrical one. Okay, so it's a symmetrical single minimum, found by calculation. And there's experimental evidence that supports that too, but it's more complicated.



Now how about the charge distribution? What would the Lewis structures that we've drawn here predict about charge separation? Would there be charge anyplace, do you think, on the basis of this, in the real molecule, which is symmetrical? What do you think? Anybody got a suggestion? Yes?



Student: Well, it would be positive on the central oxygen because even converted, regardless of which of the two structures are on that side, that…



Professor Michael McBride: Both structures have it positive in the middle, so that looks like a good prediction.



Student: And the ends would be negative.



Professor Michael McBride: Right, and then partially negative but equally negative on the two ends. Now is that true? Well here's a picture that, based again on--so the prediction is positive in the middle, negative on the ends. Is it true? Well there's the structure, the minimum energy according to calculation, quantum-mechanical calculation. And we can draw what's called the surface potential of that structure, which again comes from calculation. So you put a proton--you define the molecular surface--and that is a little bit of a problem but we'll talk about that later, where the molecular surface is--but then you put a proton at a point on it; and we talked about this before, for ammonium chloride. You put a proton on and you find--or no, it was the BNH6 we talked about the surface potential. Here's the same thing for O3. And you'll notice the surface potential is high in the middle, a bad place to put a proton because there's positive charge there, and it's low on the two ends, at least some place on the two ends. So more or less the Lewis structure predicted this right. So that's great. We're not confident that it will always work but maybe the lore will build up that way; and it does. Okay, so charge distribution also.



So reactivity we saw was a special attribute of this nice Lewis theory, and charge distribution, at least qualitatively if not in detail, for O3 and also for the BNH6. Now how about specific distances and specific angles? We'll get into that later on as to whether you can do that from something like a Lewis structure. And how about the energy content? Well that's not so good actually, with just Lewis structures, but we'll test it later. Okay, so the Lewis-dot structures. It attempts to provide a physical basis for the valence rules, based on completing octets and sharing electrons for bonds. What it gives that's new is reactivity due to unshared pairs where both of the "hooks" in the bond come from the same atom, you might say. And it's convenient for electron bookkeeping. It certainly tells you what the molecular charge is. The formal atomic charges are qualitatively realistic, as we've just seen, at least in the case of O3. And there's this question of stability and resonance. But resonance actually is not something nature knows anything about. It's just a correction we try to apply to having made drawings based on Lewis theory. It's not something serious.



Now, but this leaves us with some serious questions. Why does Lewis theory work? What's so great about octets; why not have a different number? Or if you have a sestet, instead of an octet, how bad is it? Or how bad are structures that have charge separation? It said in those rules that you don't want to have charge separation. Well how bad is it? Suppose you had a choice, you either had to have a sextet instead of an octet, or you had to have charge separation. Which one would win? How bad is "bad" charge separation? Remember it said that if you put the negative charge on an electronegative atom, that's not so very bad, but how bad?



Now last year in the Wiki there came an interesting comment. Somebody said, "I have a question when drawing these structures. Is it more important to try to fill the octet or to have the lowest formal charge on as many atoms, especially carbon, as possible? And why?" That's a good question, because Lewis doesn't tell you that at all. And there's further the question, "Is this at all true?" Are there electrons--this is a really important question, we're going to spend some time on it--are there electrons in between the nuclei that hold them together, pairs of electrons? It'll take us several lectures to get to answer that question. Are there electron pairs between nuclei and are there unshared ones on some atoms? What's the nature of the force laws? That's what we're after, remember, here altogether.



Now, as to whether Lewis theory is right, there's a very fundamental theorem in physics that was developed in 1839 by Samuel Earnshaw, who was a tutor at Cambridge. And the statement of the theorem is that in systems governed by inverse-square force laws, things like gravity, magnetism, electrostatic interaction, there can be no local minimum or maximum of potential energy. He proved it mathematically. It's not our business to repeat his proof, but that's the statement of it. But we want to understand what that means. One thing that means is that if you have Coulombic interaction, positive/negative, you can't have a minimum energy structure that has a nucleus here and eight electrons at the corner of a cube, because there are inverse-square force laws and that can't be a minimum energy; if it distorted, it would keep going.



Now we can visualize Earnshaw's theorem here in terms of for electrostatics by the analogy that you've all seen of magnetic lines of force; everybody's seen something like this I think, right? And the idea, if it's electrostatic, rather than magnetic, the idea is that lines of force emerge from a positive charge and converge on a negative charge, and then you'd make them continuous by drawing these lines of force; which iron filings of course did in that case. And this was the idea of Michael Faraday, whom we met a little while back, and he thought these lines of force were real physical things. Most people don't think that now, they think they're just graphs that involve inverse-square force laws, but he thought they were real. And the neat thing about them is they not only show the direction of the force that a charged body would feel, because of the other charged body, the one that created the lines of force, they not only show the force, the direction of the force, they also show how strong the force is. And the strength of the force is by the density of lines. The denser the lines, the stronger the force.



Everybody's familiar with this idea? Speak up if things are not familiar because I'm assuming they are, if I say so. Okay? So you've seen that. Now let's just think about that a little bit. Suppose you look at the line density here. Through that little line, there are three lines of force that pass. So let's say that means there are three, the force is three at that--and it's obviously pointing to the right. Okay? Now suppose you checked it out at that distance. Stronger force or weaker force?



Students: Weaker force.



Professor Michael McBride: Weaker. In fact only one or one and a half, or something like that, lines of force are going through that. Now, how does it depend on distance, the number of lines going through this standard linked line, the blue one? Well you see that in flatland--we're just doing this in two dimensions, we'll get to three in just a second--in flatland, in two dimensions, the circumference through which all these lines of force pass, the length of the circumference is proportional to the radius. Right? So if you go out twice as far there's twice the circumference. So the density of lines passing through it is half as big. Everybody got that? Okay, so that means the force, which is proportional to the line density, must be proportional to 1/r. Okay? So then as you move out twice, there'll be half as many lines; move out three times, a third as many lines. Okay? But this is just in two dimensions. Now let's think about, how's it going to be different in three dimensions? Katelyn? Speak up a little bit, my hearing isn't so great. I said I went to my 50th high school reunion, right?



Student: Instead of a line it would be more like a square, some sort of an area.



Professor Michael McBride: Yes, it would be a two-dimensional area that this stuff's going to be. Now if you have a certain number of lines passing through at a certain distance, and you go out twice as far, what's the density of the lines going to be? Can you think? Anybody help? Yes?



Student: It'll be inverse-square.



Professor Michael McBride: Inverse-square, because now we're not talking about the circumference of a circle, we're talking about the area of a sphere as we go out. Right? So if we go to three dimensions here and take your area and move it out, the surface is going to be proportional to r2, as you go out. So the line density is going to be 1/r2. So if you have an inverse-square force law, then you can draw lines of force. The lines of force won't work if you don't have an inverse-square force law, because they have to drop off this way in order to give lines of force.



So in 3D such diagrams work only for inverse-square forces. You can't draw a thing like that for Hooke's law. Right? Now, so here's a whole bunch of charges, positive and negative, and the lines of force between them. And notice something interesting. The lines of force start on positive charges and end on negative charges. There's no other place in space where all the lines go away from it, or go toward it. Everybody see that? It's only at the charges that they all go away or all go toward. Right? And that has a very important meaning, that you can't have someplace off in free space where all the lines of force would converge. Notice if you were to have something that was positively charged, and it's at a minimum of energy, then any place you displace it, it'll get pushed back, if it's at a lowest point in energy. You push it any direction, it'll come back. That is, all the lines of force must converge on that point. Okay, everybody with me on that? But it can't be. The only place that all the lines of force converge is on a negative charge. It can't be any place--if you have inverse-square laws, you can't have someplace in space which is the lowest energy place for the charge to be, except on another charge. The same is true, you can't have a maximum either. That's a visualization of Earnshaw's theorem.



So if you have inverse-square force laws, or any combination of inverse-square force laws, like a combination of gravity and electrostatics and magnetics, you can't get a minimum energy structure, unless everything just falls together or blows apart. Okay? So Earnshaw's Theorem: "In systems governed by inverse-square force laws there can be no local maximum (or minimum) of potential energy in free space." Okay? And that's why I don't just float here, I have to be on the floor. Right? Did you ever see anything truly levitating, something that just sits in space and doesn't move, not touching anything else? That's been levitating for ten or fifteen years. It's not plugged into anything, it's just this thing right here. What do you conclude from seeing that, about the force laws that are involved?



[Students speak over one another]



Professor Michael McBride: There must be some force involved in that that's not an inverse-square force law; otherwise it wouldn't work. You can read about it on the Web, if you want to, it's not the business of this class. There is a non-inverse-square force law involved in that little magnet sitting there. The only stationary points allowed by Earnshaw's theorem are saddle points where it's flat in energy, for all directions, but you go one direction and then you go down, go the other direction and you go up. That's like a potato chip or a saddle. So you can have saddle points, but you can't have absolute minima or maxima of energy. There's the picture of that thing. Let me do the--this reminds me to get back to seeing who's here.



[Professor McBride takes attendance ]



Professor Michael McBride: Now we'll continue, got a few more minutes here. And yet it stands still. So there must be something that's not an inverse-square force law there. Okay, so J.J. Thomson, in 1897, discovered the electron. So the idea is maybe electrons have something to do with bonds; that's what brought Lewis into the game. But Thomson himself came up with what came to be called--and I don't know by who first, I've tried to find out and can't--the "plum-pudding" atom. Has anyone ever heard of that? Yes, okay. And I suspect that when people told you about the plum-pudding atom they sort of snickered. Am I right? This was sort of a naïve idea. It's not that naïve at all. Let me show you why.



Here's the book by J.J. Thomson called The Corpuscular Theory of Matter, and if you look in there you'll find out what he's talking about. He says--he has a model of electron configuration--he says: "Consider the problem of how to arrange 1, 2, 3, up to n corpuscles" (that's what he called electrons, he called them corpuscles); "Consider the problem as to how they would arrange themselves if placed in a sphere filled with positive electricity of uniform density." So that's the idea of the plum-pudding. You have this sphere of uniform positive density--why it should be like that nobody knows--but suppose you have a sphere, and you put the electrons in it, like plums in a plum-pudding, which is like a fruitcake in England. Okay, now notice that he said, "placed in a sphere filled with positive electricity." Why did he do that? Why didn't he just have--like Rutherford ultimately did it--a nucleus with positive charge and electrons around and about? Why did he put the electrons inside the positive charge? Zach?



Student: Was it lowest potential energy?



Professor Michael McBride: I can't hear very well.



Student: Lowest potential energy?



Professor Michael McBride: No. Yes?



Student: Well they were thinking [inaudible] maybe that they were going in.



Professor Michael McBride: No, no, they weren't thinking about things moving, they wanted them just sitting there.



Student: Maybe because on a macroscopic scale opposite charges attract. So maybe he might not…



Professor Michael McBride: Yes, but that's well known, Coulomb's law. Yes Keith--or Kevin, right?



Student: If you have them all, you have a positive sphere and if you have the negative corpuscle inside, then they'll cancel each other out and be a neutral body.



Professor Michael McBride: Yes, but that could be--it wouldn't have to be a big sphere of uniform density, you could have a little particle of positive charge, of the same charge, right? It'd be the same deal.



Student: They didn't know that the inverse existed, so they thought that…



Professor Michael McBride: No, none of this is it. It's what we've just been talking about. Yes?



Student: Isn't it about accounting for the bonding between the electrons and the protons and containing the electrons?



Professor Michael McBride: No, bonding hasn't come up yet. This is something very fundamental. Yes?



Student: He was a big fan of plum-pudding.



[Laughter]



Professor Michael McBride: He probably liked plum-pudding at Christmas.



Student: Did he have experimental evidence?



Professor Michael McBride: I can't hear.



Student: Did he have experimental evidence?



Professor Michael McBride: No. Yes?



Student: Because Earnshaw said that there should be no minimum or maximum…



Professor Michael McBride: Ah ha! Earnshaw said you can't have an energy minimum for separate particles, but if you put the negative ones inside the positive ones, then you could get a stable structure. It's Earnshaw's theorem that required it to be a plum-pudding. Okay? Now why a sphere, why not a doughnut or some other shape, a barbell or something, right? And he said that the positive charge is distributed in a way most amenable to mathematical calculation. He chose a sphere so it would be simple. You've heard about spherical cows and so on like that, that physicists like to calculate. Let's suppose there's a cow, let it be a sphere, right? So that's what he did. That's why it's a sphere.



Okay, we can solve the special case where the corpuscles are confined to a plane, if you do it in two dimensions--it's difficult mathematically in three dimensions--but you can do it in two dimensions. And he gives a picture in this book, like this, which is a solenoid magnet that attracts little needles that are magnetized, and those needles are stuck into corks, which float in the water, right? So they have to, the corks have to stay at the level of the surface of the water. So it's a two-dimensional problem. The needles are parallel, the magnets, so they repel one another. But the big magnet attracts them to the center. Okay? So what he does is toss a certain number of needles and corks in there and see what pattern they form. Okay? So here are some patterns. You can get these on the Web, at Greg Blonder's website here. So if you have just one, it goes to the center; no big deal. If you have two, you get a line; there's no big deal about that. Three make an equilateral triangle. Put in four, they make a square. Put in five, you make a pentagon. No one's surprised so far I suspect. Except that sometimes when you put in five and shake it up, you get a square with one in the middle. Can you see where this might be going?



Let's keep going. Okay, if you put in six, you get one inside a pentagon; seven, one inside a hexagon; eight, one inside a heptagon; nine, two inside; ten, two inside. And sometimes it's like that, and sometimes it's like that. These are experimental, right? And if you put in eleven, it's three inside, and then three inside nine; although sometimes you get two--pardon me, I'm screwing up here--sometimes you get--there are two different patterns--sometimes you get three inside eight, sometimes you get two inside nine, for that number. Okay, and then you can get four inside nine, four inside ten, five inside ten, and then after five, if you put in more, if you put in sixteen, you get one inside five inside ten. What is this reminding you of? Yes?



Student: Of orbitals.



Professor Michael McBride: The shell structure of atoms. Right? As you go down the periodic table you complete a shell. So it's a model of shells, and then you can get more and more and more. So this is what Thomson was thinking about. But that's just two dimensions. Right? Three dimensions is a bigger problem. But he could say, he was able to say something mathematically about eight. "The equilibrium of eight corpuscles at the corners of a cube is unstable." Even if you have the spherical charge--so it's not exactly Earnshaw--still you can't get eight at the corners of a cube. Now Lewis comes along and in 1923, as I've told you before, he writes: "I have ever since regarded the cubic octet as representing essentially the arrangement of electrons in the atom." This is long after Thomson had written this about not being able to do that. So was Lewis ignorant of Earnshaw's theorem? Because by 1923 they know that the nucleus is not a plum-pudding--that is the positive sphere--that it's a point. So was Lewis just naïve?



No, look what he wrote, in 1916. "The electric forces between particles which are very close together do not obey the simple laws of inverse-squares which holds at greater distances." So Coulomb's law breaks down. You don't have inverse-square. So then you don't have Earnshaw's theorem, and you don't have to worry, and you can get a structure if it's not an inverse-square force law. Okay? No trouble. But what is the force law? Well Thomson thought the same thing in 1923 in his book The Electron in Chemistry. He wrote: "If electron nuclear attraction were to vary strictly as the inverse-square, we know by Earnshaw's theorem that no stable configuration is possible with the electrons at rest or oscillating about positions of equilibrium… I shall assume that the law of force between a positive charge and an electron is expressed by this equation… Then a number of electrons can be in equilibrium about a positive charge without necessarily describing orbits around it." And look at the--we're going to end right now by looking at this equation. What's that bit of it, the first bit, the "one" part?



Students: Coulomb's law.



Professor Michael McBride: That's Coulomb's law. But he's got a correction to Coulomb's law, here at the end, c/r, where c is a distance--you divide it by r and you get a number--and c is a distance that's on the scale of atomic lengths. Right? And that means that as long as c is very small--pardon me, as long as--okay so have I got this right? Okay, so when distance r gets smaller than c, then the force changes sign. Okay? So what was attractive becomes repulsive, and then you can have the electrons sitting around the nucleus. Okay? So we'll see what happened three years later next time.



[end of transcript]

This lecture asks whether it is possible to confirm the reality of bonds by seeing or feeling them. It first describes the work of "clairvoyant" charlatans from the beginning of the twentieth century, who claimed to "see" details of atomic and molecular structure, in order to discuss proper bases for scientific belief. It then shows that the molecular scale is not inconceivably small, and that Newton and Franklin performed simple experiments that measure such small distances. In the last 25 years various realizations of Scanning Probe Microscopy have enabled chemists to "feel" individual molecules and atoms, but not bonds.



Transcript



September 10, 2008



Professor Michael McBride: Okay, so we saw last time that Lewis wasn't so dumb. He knew about Earnshaw's theorem, but he still thought there was an octet of electrons around the nucleus. How could he think such a thing? Because if you have inverse-square force laws you can't have static positions of charged things, unless they're right on top of one another or blown apart. So what did he think?



Student: It's not an inverse-square.



Professor Michael McBride: That it's not an inverse-square force law, that Coulomb's law breaks down when you get to a very small scale. And J.J. Thomson thought the same thing, and wrote here these two terms here. The first one is Coulomb's law indeed, but there's this other thing, c/r. So as long as r is very big you don't compare -- you don't care about that term, as long as it's very big compared to c. And c is a distance that's something like the size of atoms. So once r -- once the distance of the things that are interacting with one another, the electron and the nucleus say, once they get near the distance c, then this c/r thing becomes significant, if r becomes very small, and it even changes the sign. So what was attractive becomes repulsive. So that would do okay then for a structure.



That was in 1923. In 1926, just three years later, quantum mechanics came along and showed that Coulomb's law was just fine, nothing wrong with Coulomb's law. It goes to much, much, much, 1020th, smaller distances than the size of atoms. Coulomb's law still works. What was wrong was the way they treated kinetic energy, because kinetic energy quantum mechanics reformulates. So it gives you electron clouds. And so what's a cloud is not the positive charge -- remember, J.J. Thomson had a positive sphere of electricity in which he embedded the electrons -- what's a cloud is the electrons, and nuclei are put in it, to make molecules. So it really is -- he was right about plum-puddings, he just had the charges backwards. Not many people give him credit for that. So cubic octets of Lewis and ad hoc electrostatic force laws, like this c/r term in there, soon disappeared from conventional chemistry and physics; immediately in fact, I mean within a month say. But the idea of shared-pairs, and lone-pairs, that Lewis came up with, remained useful tools for discussing structure and bonding. So we still do it, and that's why you did problems today that had to do with that. Yes Niko?



Student: Yes. Can you explain the kinetic energy theorem again?



Professor Michael McBride: Yes, but I won't do it for a week still, because it takes a little while. Okay, so Earnshaw said there's no structure of minimum energy for point charges -- that is, no classical one -- but if you come to quantum mechanics and fiddle around with kinetic energy, then you can get it. But despite Earnshaw, might Lewis have been right? Might there still be shared-pair bonds and lone-pairs? What are bonds? And that's what we're going to try to find out. And how do you know, or how do you know? How are we going to know whether there are electron pairs that are shared between atoms and lone-pairs on atoms? How can we find out?



Student: Experiment.



Professor Michael McBride: Right. You've got to do an experiment. So what kind of experiment? Now these experiments are tough. Why are the experiments tough?



Student: It's so small.



Student: It's really small.



Professor Michael McBride: Because things are -- well we could see, or we could feel whether these electron pairs are there. But there's a problem, as you say, that it's inconceivably small, the thing we're looking for, to see or to feel. Okay? But is it really inconceivable how small they are? When I was in your place everybody thought yes, they're inconceivably small, we'll forget it. Right? But there are more recent techniques that suggest that you can really see things that small; at least you can think about them clearly. Now, the first idea of seeing them came from this book here. Actually this book was published -- this is a Third Edition of the book. The First Edition was 1909 and the Second was 1919, and this is the Third Edition, which is still in print. You can go on the web and buy this baby, Occult Chemistry, A Series of Clairvoyant Observations on the Chemical Elements, by Annie Besant and Charles Leadbeater. So these are the occult chemists who started practicing their trade in 1895. "Bishop" Charles Webster Leadbeater; he was a bishop of a church he founded that had -- he was the head bishop; there were like 100 bishops and 500 members of the religion. This is Annie Besant, who is one of the most interesting people in all of history. They closed the stock exchange in India the day she died, for example. This is in her -- the outfit she designed to lead the Boy Scouts in India. And she also proposed marriage to George Bernard Shaw. There are all sorts of interesting things about Annie Besant. But one of the things we're interested in is that she could see atoms. And then their helper, who became leader of the operation later on, was Jinarajadasa.



Okay, so this is a page from this book, from the paper that became the book in 1895, which shows what they could see for hydrogen, oxygen and nitrogen. Now there are a number of levels here. The very lowest phase is solid, H O N, then liquid, then gas. So this is what, when they went into sort of a trance, they could see. This is the gas. But then there's Ethereal 4, Ethereal 3, Ethereal 2 and Ethereal 1 phases as well, with higher and higher resolution. You can see these. So these, for example, these little dots that you see here are what get blown up here, and the little things inside this are what get blown up here, and the things inside this get blown up here, and the things inside this are this; and that's the same for all atoms, that's the fundamental unit. Okay, so here's hydrogen at the gas resolution level. So there are wheels in wheels in wheels here. And here's what oxygen looks like; it's these little helices. And now that's the thing you finally get to, the thing called anu. And that they actually ripped off from this book, The Principles of Color, which was published in America in 1878 by Edwin D. Babbitt. And here's his picture of this thing. And lo and behold, that's what they saw as the fundamental constituent of matter. It's fun to look at in detail. You can look at it at the Web, if you want to. It also has pictures of the Angel of Innocence here, and the psycho-magnetic curves surrounding your head.



Okay, now lo and behold the real confirmation of this thing was that if you counted up the anu in the different elements, there were eighteen anu in hydrogen, 290 in oxygen and 261 in nitrogen, and if you divide that by eighteen, you get the atomic weights, Q.E.D. Okay, so they have pictures in this book of a bunch of the atoms. This is helium, which has seventy-two anu. This is lithium with 127 anu. Not the bottom part; this is a model and this is the wooden base it sits on. Here's iron, 1008 anu. Here's neon, 360 anu. Now look at neon, and that is a 4f orbital electron density, calculated by quantum mechanics. Now the question is, why would you believe me, or believe a textbook, or believe a quantum mechanician, and not believe them? Because they said they saw it too. Okay? Well it takes time. We've talked about this before, the basis for scientific belief: evidence, you always cross-examine an assertion; logic; and taste, but taste matures with experience and at the beginning you don't know which people to believe and which not to believe. It's a problem in politics too. Okay, there's sodium, 418. They not only said they saw atoms, they also saw molecules. For example, here is sodium carbonate. What do you see in there? Two sodiums, right? Okay? In the back are pieces of a carbon atom. What else do you see?



Students: Oxygen.



Professor Michael McBride: Ah, there are three oxygen helices wrapped around the sodium and the carbon. And they wrote: "Note that this triangular arrangement of O3 has just been deduced by Bragg from his X-ray analysis of Calcite." We'll talk about Bragg and what he was doing next time. But they have experimental evidence that supports what they're reporting, independent experimental evidence. Here's their model of benzene, and they say about that, "Each of the three valencies of each Carbon are satisfied by Hydrogen, and the fourth valency, which some have postulated as going to the interior of the molecule, does actually do so." In fact, Annie Besant was the first woman to get a degree in Chemistry at the University of London; but that's another story. Her teacher was the guy who translated Das Kapital into English, and had a lot of other nefarious things that he did. Question six will have to do with that. Okay, so here's benzene and its resonance. These were structures that were first early proposed; we'll talk about that later on. But some people didn't like this idea of going back and forth between two things, and they thought that you just have the fourth valence of every carbon going into the middle, all satisfying one another. So this was what they said that their model confirmed, that the fourth valence of every carbon goes into the middle.



Okay, needless to say most people nowadays don't think too much of occult chemistry, even if you can still buy the book online. This portal you came through this morning. It says "Erected 1921". But here's the same place, in 1923, when the building was dedicated, and you notice those plaques weren't carved yet. It didn't say erected 1921. They were still finishing the building at the time the American Chemical Society had their national meeting here to celebrate the biggest academic building for chemistry in the world. Okay? So the slates were clean. Now if you go down into the crowd here, you see some interesting people. For example, there's G.N. Lewis with his Philippines' cigar in his hand. He always had a cigar; he smoked seventeen cigars a day. [Laughs] Groan.



Okay, but they were trying to decide how to decorate all these plaques around the building, and they wanted to put the names of distinguished American chemists. Had they done that, it would be pretty much a disaster because the people who were thought to be distinguished American chemists in those days were not so hot actually in retrospect. But they chose a wonderful group, and the idea came from Giuseppe Bruni from Bologna who had come to the meeting to speak. And he said, "Don't put living chemists there, put dead chemists." And that's what they did; in almost all cases they were dead. And it's a wonderful group; as we've already used Faraday over here. Okay, here's one of them that's over in that corner of the building. And van't Hoff, have you heard of van't Hoff? You'll have heard a lot of him by the end of this semester. And Gibbs?



Students: Yes.



Professor Michael McBride: Yes Gibbs, we'll talk about him too. And Mendeleev you've heard of. But I suspect that Crookes is not so familiar to you. This is Sir William Crookes. He was a Fellow of the Royal Society, and FTS. In 1861 he discovered the element thallium. He also developed the cathode ray tube, which he's holding there, which became the x-ray tube, and he invented the Cooke's radiometer. I bet every one of you has seen a Cooke's radiometer. Do you know? There's a picture of a Cooke's radiometer, to measure the intensity of light by how fast it spins. You've seen these things you buy in novelty stores that sit there and spin in the light. Okay? From 1913 to 1916 he was President of the Royal Society, the same thing that was founded back by Boyle -- remember? -- and Robert Hooke and those guys. In 1898 he was President of the British Association for the Advancement of Science, and he was also, in that year, President of the Society for Psychical Research. And FTS is Fellow of the Theosophical Society.



In his presidential lecture to the British Association for the Advancement of Science, one of the things he said was, "Telepathic research does not yet enlist the interest of the majority of my scientific brethren." But he's the guy who supplied the samples for Leadbeater and Besant to look at. He supplied them with lithium, chromium, selenium, titanium, vanadium, boron and beryllium samples so that they could draw these pictures of what they saw by clairvoyance. So you can read more on the Web if you want to about him, he's an interesting character. And there were a lot of semi-, well really serious scientists. Oliver Lodge was another one who invented radio and was head of the Physics Department at Bristol, I think [actually Liverpool and Birmingham], who was into communicating with the dead and stuff like that. So it wasn't obvious at the beginning who you should believe and who you shouldn't believe.



But to get back to the point, and forget clairvoyance for awhile, which I don't believe in, the question is, are molecules unobservably small for what Newton called "vulgar eyes"? Is there no way we can see something and measure it if it's that small? So consider water. A cubic centimeter of water is 1/18th of a mole, since the molecular weight is eighteen. That means it's 6/18 times 1023 molecules, in a cc of water. Now that's a really, really big number and it sounds like they must be just impossibly small, hopeless to try to see such a thing. But the difference is between the cube and the cube root, because when we measure distance we measure linear, not volume. So to get something about distances you take the cube root of the volume. So if we take the cube root of that, it's about three times 107. Now that's still a very big number, 30,000,000. Right? But it's not impossibly small. It's about -- 3 angstroms is this dimension, the size of a water molecule. Remember, we'll talk a lot about a carbon bond -- carbon-carbon bond, being about one and a half angstroms.



Okay, now think about 105. So this lecture room is about ten meters wide. One of my hairs is order of magnitude 100 microns in diameter, and that's a ratio of 105. Right? So 105 of my hairs sideways would go across this room. So that's a lot. Right? But it's not impossible to think of lining up hairs to go across the room, or putting them side by side. It'd take a while, but it's not impossible to think about that, a hair compared to the room. Right? But a molecule is about a nanometer, and a small atom is about 1/10 of a nanometer, or an angstrom. Right? And that ratio is 105. So the room to my hair is like my hair to a molecule, or even an atom, a big atom. So it's not impossibly small, it's just very, very small. In fact, a nucleus is 10 femtometers, which is another 105 down. So that means if the lecture room were an atom, the nucleus would be the diameter of a hair; very small, but not impossibly small to conceive of. Okay?



Now Newton, in Opticks, the book we looked at before, in 1717 wrote on page 369: "The thickness of the Plate where it appears very black, is three eights of the ten hundred thousandth part of an Inch." "Is" - that means he knew the size of it. Now how big is three-eighths of a ten-hundred-thousandth part of an inch? Well ten-hundred-thousand is a million; three-eights of a millionth of an inch is thirty water molecules. So somehow Newton was able to measure something, in 1717, that was the size of thirty water molecules. Now what tool could Newton possibly have had in 1717 that allowed him to measure something so very small? What?



Student: Glass ring.



Professor Michael McBride: Can't hear very well.



Student: Glass ring.



Professor Michael McBride: Glass springs?



Student: Ring, like you would --



Professor Michael McBride: That's right. Okay, so here it is, Newton's Rings, which really should be called Hooke's Rings, because he's the guy that in Micrographia published how they worked. But remember, they had different light theories. So here are two pieces of glass, two disks of glass. And I'll change the lights so we see light through them. Now I can't remember which is which, but let me guess that if I put this one over here, having turned it over -- okay. Oh I can see, yes you can see. See those patterns? You've seen things like that with microscope slides, colors. Okay? So these are two flats against one another, they get very close to one another. But if I turn the top one over, you see that it's a little bit -- let me -- so that's what we just saw. But notice that if I turn it over there's a little -- there turns out here at the edge to be a little bit of a gap, because the top one is just very slightly curved. And now let me see if I can see something. I can't see it yet, let me zoom in. We're going to need to focus on that. Ah, see that thing, see that.



[Technical adjustment]



There. Okay, so here's what you're seeing. In the middle is what's called Newton's Rings. They look like that. Different colors, right? And Newton could measure the thickness of the air gap that caused different colors. The colors repeat in higher orders. Okay? Now how could he know the distance? So he could associate every color with a distance. How could he know the distance? Because he could measure the diameter of the rings. And he said this: "Observation Six: The Diameter of the fixth Ring" (Right? One, two, three, four, five, six) "at the most lucid part of its Orbit was 58/100ths of an inch, and the Diameter of the Sphere on which the double convex Object-glass was ground was about 102 Feet; hence I gathered the thickness of the Air or Aereal Interval of the Glasses in that Ring." So in other words, here's the air gap he's trying to measure, which is the sixth ring out. He knows the diameter of that ring. He knows that they touch in the middle, the two glasses, and he knows that it's a sphere and it has a radius of fifty-one feet. So he can just use trigonometry to figure out that the air gap, if he knows that angle there, the air gap is 1.8 microns, right? So that's a lot bigger than an atom; but if you go into the first ring, or even closer, then you can measure it. Right? So all he needed was a spherical glass with a very large radius that he could measure his distances from. Okay, but here's a simpler measure of an even smaller distance, about 100 years later, and here's the guy who did it. You know who that is?



Students speak over one another]



Professor Michael McBride: Benjamin Franklin of course; we'll get back to the artist later on. So he published in the Philosophical Transactions of the Royal Society in 1774 -- the Society is now 110-years-old, 114-years-old -- this article on the stilling of waves by means of oil. Can you see what that means? What's that related to, a common saying?



Student: Separation?



Professor Michael McBride: Pouring oil on troubled waters; have you heard of that one? So the stilling of waves by means of oil. Extracted from sundry letters between Franklin and Brownrigg. So he wrote to Brownrigg, 1773: "I had, when a youth, read and smiled at Pliny's account of a practice among the seamen of his time, to still the waves in a storm by pouring oil into the sea; as well as the use made of oil by the divers…" Pearl divers would fill their mouth with oil and when they went down, if a breeze came up and ruffled the surface, which made it hard to see because the light wouldn't come through clearly, then they'd let oil out of their mouths that would float to the surface, stop the ripples, and they could see to get their pearls. This is what Pliny said. Right? But he said he smiled at that, what a quaint thing to think; it's like the occult chemists.



"I think that it has been of late too much the mode to slight the learning of the ancients. The learned, too, are apt to slight too much the knowledge of the vulgar. In 1757, being at sea in a fleet of ninety-six sail bound against Louisbourg," (off Cape Breton Island) "I observed the wakes of two of the ships to be remarkably smooth, while all the others were ruffled by the wind, which blew fresh. Being puzzled with the differing appearance, I at last pointed it out to our captain and asked him the meaning of it. 'The cooks,' said he, 'have I suppose just been emptying their greasy water through the scuppers, which has greased the sides of those ships a little.'" (So there's experimental evidence for what Pliny said.) "Recollecting what I had formerly read in Pliny, I resolved to make some experiments of the effect of oil on water when I should have the opportunity."



And when he had the opportunity was when he was in London. So here's London, and his experiment in 1762 was in that place, on Clapham Common in South London.



"At length being at Clapham, where there is on the common a large pond which I observed one day to be very rough with the wind, I fetched out a cruet of oil and dropped a little of it on the water. I saw it spread itself with surprising swiftness upon the surface; but the effect of smoothing the waves was not produced." (So the experiment was a failure.) "For I had applied it first on the leeward side of the pond where the waves were greatest; and the wind drove my oil back along the shore."



So it didn't spread on the surface. So what did he do?



Students: Go to the other side.



Professor Michael McBride: Go the other side, right? "I then went to the windward side where the waves began to form and there the oil, though not more than a teaspoon full, produced an instant calm over a space several yards square which spread amazingly and extended itself gradually till it reached the lee side, making all that quarter of the pond, perhaps half an acre, as smooth as a looking glass." So a teaspoon is five cubic centimeters, and he stilled half an acre, which is 2000 square meters, which means the thickness of this layer, to spread that many cubic centimeters over 2000 square meters, would be twenty-five angstroms, which is the length of a molecule of the fat that's in the oil. So Franklin had in fact measured the length of an oil molecule, in this way.



Now he didn't think he had measured it and didn't claim to have measured it. He said, "When put on the water it spreads instantly many feet around, becoming so thin as to produce the prismatic colours" (the Newton Ring's colors, right?) "for a considerable space, and beyond them so much thinner" (it became so thin that you didn't even get Newton's colors, right?) "so much thinner as to be invisible except in its effect of smoothing the waves at much greater distance." (So the technique was stilling the water; allowed you to measure how big the area was. Isn't that cool?) "It seems as if a mutual repulsion between the particles took place as soon as it touched the water." That is, they pressed one another apart and moved. So what he wouldn't have known was that they pushed one another apart but they stayed in contact with one another, to become a monomolecular thick layer over the water. So he didn't claim to have measured it, but indeed he did measure the size of an oil molecule. So molecules are very, very, very small, but they're not inconceivably small and there are ways you can measure them.



So we have this question, are there electron pairs? So let's try first feeling, with Scanning Probe Microscopy, or SPM. Can we feel individual molecules? Can we feel individual atoms? Can we see what bonds are by feeling them? So Scanning Tunneling Microscopy was invented in 1981 by these guys in Zurich, Gerd Binnig and Heinrich Rohrer. And here's the first publication. They worked at IBM and here's the first publication on the cover of their journal about a meeting about this new technique, Scanning Tunneling Microscopy. And they got the Nobel Prize within five years, or maybe it was even less; in very short order, 1986. Now how does that Scanning Tunneling Microscopy, which is one type of SPM scanning probe -- Scanning Tunneling is one way, and I'll mention that shortly. But first I'm going to talk about Atomic Force Microscopy. And for this you need a little chip, and I'll show you. The chip is in here, in this little bug box. Ah you can't see it here. Well there's the chip in the middle, right? And I'll pass it around so you can look at it. And if you tilt it so that the light from the ceiling glints off the gold, and you see it really shining, you may be able to see at the thin end these little v's; you can see here at the bottom little v's. The chip is 1.4 millimeters wide. It's a gold-coated silicon chip. And if you look at those with even higher magnification you see this -- and there's the size of a hair. You can actually see -- did you see them? They're not easy to see and maybe you will and maybe you won't, but they're there.



Okay, so there's the size of a hair. And in any given experiment you have a choice of five different cantilevers, as they're called, that you can use. Here's a higher resolution electron micrograph that shows the tip of one of these v's, and you see a little bulge here, at the bottom, and if you zoom in on that you see this; it's a little pyramid. And so at the bottom of that pyramid the radius of curvature of this tip is only about twenty nanometers, so about the size of twenty molecules. Right? It's round on the tip, it's not truly pointed and come to nothing; it's a little bit rounded. You can buy ones that are a third that size, their radius of curvature. Okay, and then you have this piece of glass. I said, "Do not touch" when you came in because that piece of glass costs $2000. Okay, so there's this piece of glass, and if I hold it up here there's a little chip on the bottom of the piece of glass. I know what I'm looking for. I don't know if you can see it or not. But anyhow it's held into this piece of stuff here. And those little tips are pointing up here, but in fact the thing is used upside down so they point down. So here's that thing mounted in an instrument, and you see there's a red light glowing in the glass, and the reason is it's being irradiated by -- so there's a blown up picture of one of these tips, pointing down. And laser light comes and bounces off the shiny gold on this tip, and gets reflected up to a detector. Now what would happen if you pushed up on the tip? It would deviate the reflection, right? It would go like this. And you have a very precise position detector for the laser light up where you see the light glowing. So you can tell the deviation of that tip, moving up and down, by a size that's less than an atom. Isn't that neat?



So now what you do in this is move this back and forth over the sample and watch the little needle go up and down by the reflection of the light. So if you had, if the surface was like this, they'll click, click, click, click, click, click, up, up, up, and just scan across. And it's sensitive, as I say, to less than one molecule change in height; although it doesn't feel an individual molecule because the tip, remember is twenty nanometers wide. So the width of the tip is like 100 or so atoms. Right? So it's touching a bunch of atoms at any one time. But if you come to a ledge, unlike a [smooth] crystal [face], it would be going at a certain height, touching a bunch, and then it would move up and touch a bunch more, and you'd be able to tell if it went up and down, if that distance was only one molecule; easy.



Here's, in fact, a picture taken by an undergraduate. In fact, it's a movie. So it's AFM traces. So what you do, it's like a TV where you go zoom, zoom, zoom, zoom, zoom, zoom, zoom, zoom, and then you go back to the top, zoom, zoom, zoom, zoom, zoom, say how high it is at every point along the scan, and then color code it to show how high. So the width of this is about 600 molecules, right? But the lines you see, which are ledges that go up and down; each of those ledges is one unit cell, 1.7 nanometers, one set of molecules. So these are -- and that, the thing at the top surface where it's flat, is absolutely flat, except for there's little holes there. And those holes were made by taking this tip and pressing down and scratching a little bit to knock some of the molecules away. So the larger pit is five nanometers deep, which means it's four layers of molecules that've been knocked off in there. And now this is underwater, or underwater in propanol. So the crystal is dissolving slowly. So those pits open up, as molecules come away. So watch. Right? Those are at one-minute intervals. So we're actually feeling individual molecular heights with this thing.



There's an interesting point here that I'm not going to dwell on, but the smallest one did not lose any molecules. If I back up, watch the smallest one. It always stays the same size as the others dissolve. That's an interesting puzzle isn't it? But at any rate, you can do it. And you can measure the rate at which different ledges dissolve. Some dissolve much faster than others. Here's scanning tunneling microscopy. This is done in a different way. It's not -- you don't reflect the laser light. What you do is detect the electrical conductivity through a layer. And if you have an atom there that is a good conductor of electricity, you get more current through and less and so on. So as you scan a real -- now this could be a really, really sharp tip where the tip is just one atom or a couple of atoms. So now as you go across you can feel things that are much smaller laterally. Okay? So this picture here, that thing right there, which is as you see a repeated pattern, that's one of those molecules, with 12 carbons and a bromine on the end. And here's a model that shows what the molecule looks like. The yellow is bromine, the brown are the two oxygens, and the green are carbon and the grey are hydrogens. They don't come colored that way, they come looking like this, as to conductivity. But you can see individual atoms. Right? Or one of the more dramatic early examples, in 1993, was at the IBM labs at Almaden in California, where they worked on iron atoms on a surface of copper, and they used this tip and changing the voltage to pick up atoms and move them where they wanted, then change the voltage and drop them there, then go back and get another one and do it. So they made this thing they called a quantum corral of whatever, fifty-eight or something iron atoms. Yes?



Student: Is that along the lines of when they moved the atoms --



Professor Michael McBride: Pardon me?



Student: Is this along the lines of when they moved the atoms to spell out IBM?



Professor Michael McBride: Oh yes, they also spelled out IBM. They knew which side their bread was buttered on too, just like the Royal Society did. Okay, so that's neat. So clearly you can see individual, or feel individual atoms, but you don't see the bonds. The bonds are smaller. It just looks like a bunch of balls, right? You don't see the electron pairs between them.



Here's SNOM. SNOM stands for Scanning Near-Field Optical Microscope. So this is in a sense seeing, but actually it's scanning. It's like feeling. So what you have here is a glass fiber that's drawn down to a very sharp point and coated with aluminum, and the hole at the bottom is just 100 nanometers; that's like 100 molecules, not really, really tiny like the scanning tunneling microscope. So what you do is you have something on the surface of a slide here that you want to probe with this, and you send light down the thing, and if there is a molecule where the light's coming out, through this tiny hole, if there's a molecule there, that will take the blue light and emit red light. Then you know when it's under the tip. Suppose there was just one molecule on the whole slide. You scan it around until suddenly red light comes out. Then you know that tip is pointed toward that particular molecule. Okay? So red light comes out and you have a detector that detects it, and you scan the sample back and forth, like a TV, and you find out where such molecules are. And here's a picture taken in that way. So this is a scale of microns, 2 x 2 microns; in fact it's a distorted scale, as I see here. Okay? And that arrow, doubled-headed arrow, is the wavelength of red light. So you're seeing things significantly smaller than the wavelength of the light, having done it in this way. You're not really looking with your eye, you're doing this scanning trick.



Okay, so scanning probe microscopies, like atomic force microscopy, scanning tunneling microscopy, scanning near-field optical microscopy, are really powerful, and you can see individual atoms. Right? The sharp points can resolve individual molecules, and even atoms, but not bonds. So it's not doing what we need to do for our purposes.



Now, that's -- I'm going to spend the last few minutes going over the problem that we were looking at with -- remember you were supposed to draw all the resonance structures for H, C, O, N isomers. I think you remember that. Okay, so let's just look at it. So what we're going to do is compare what we see with Lewis theory with things that have been calculated by reasonably good quantum mechanics; not the very, very highest level, but a pretty good level. Okay, so we want to try to make all the possible arrangements here. So we can put the three atoms, O, C, N, with any one of them in the middle, and then we can put hydrogen on either end. So then we'll do it with nitrogen in the middle and then with oxygen in the middle, so we'll have done all possibilities there. So here are two possibilities. Now are these good structures is what we have to decide? Okay, now it would be possible to shift electrons here and draw a different Lewis structure. Which of those two do you think is better?



Student: Probably the top one.



Professor Michael McBride: Why?



[Students speak over one another]



Professor Michael McBride: Okay, there's separation of charge on the bottom. Okay? So there's bad charge separation. The most electronegative atom, oxygen, in fact has a positive charge not a negative charge. Right? So that's -- not only is there a charge separation, it's in the wrong direction. But they're all octets that we've drawn in both structures. All atoms have octets. Now let's try the same trick over here. We can do it that way. How about that? Which of those is better? In both structures we could go through and count and see that they're all octets. Would you like me to do that, or is that something that's become second nature to you by now? Anybody want me to count them? Okay, you've got that. But now which of those two structures do you think is better?



[Students speak over one another]



Professor Michael McBride: So you say the top one's better, you say, because it doesn't have charge separation. Now, as compared to the case on the left, which do you think -- if you have to have charge separation, which one is better?



Student: The right.



Professor Michael McBride: The one on the right because at least you have it with the right atom, the oxygen being negative. You could've drawn the one on the top too, right? And that goes the other way. Notice that I've introduced something here that I haven't really talked about enough, the idea of curved arrows. Curved arrows show a shift of electrons, an electron pair. What would you draw if you wanted to show just one electron shifting, if you had to think up the notation?



Students: A dotted curve.



Professor Michael McBride: A dotted curve would be one possibility. Another possibility? Sophie?



Student: Half a head instead of a full head.



Professor Michael McBride: Ah, like a fishhook instead of the double arrow. And that's in fact what's used. If you want to shift just one electron you show a single barb, and if you do a pair of electrons you show a double barb, a curved arrow. So curved arrows don't show atoms moving. They don't show this atom goes to there. They show an electron pair moving, or sometimes a single electron moving, and you can use a double or a single barb to show which one. Okay, so get that straight, because that's often a problem. Okay, so the one on top is much worse than the one on the bottom. Okay, so there are our possibilities with C in the middle.



Now how about if you put N in the middle? Here are two structures with octets. Now -- oh pardon me, they don't all have octets, that one on the right has a sestet, because the carbon's making two bonds; so it gets three electrons out of those two bonds -- pardon me, it gets two electrons out of those two bonds, one from each bond, that's its share from the bookkeeping, plus an unshared pair. Right? That's only -- what's associated with it, it has only three… pardon me, I'm saying the wrong thing here. It has four electrons that it calls its own, from a bookkeeping point of view, one from each bond and two in the unshared pair. So it's not charged. It came in with four electrons, it's still got four electrons, but participates with only three pairs of electrons; so it's a sestet. So that's not so good on the right. The one on the left is not so good because it's charge-separated. At least it's on oxygen, the negative charge. We could draw this other one, which puts the negative charge on carbon. That's, we would say, not so good. Right? So that's a bad charge. Or we could do it that way. That looks like the best charge probably; better to put positive on carbon than on nitrogen. Okay, but that carbon has a sestet; the others are octet ones. Okay, we can do this other one over here. Now we've got an octet on the carbon, but we've got negative charge on the carbon. That's not such a great charge separation thing. Or we can do it this way, but again we got bad charge. Okay, so now if we put O in the middle we can do these structures, both of which you notice have -- what's bad about them?



[Students speak over one another]



Professor Michael McBride: Sestet there, sestet on the right too. There's a plus charge on oxygen; that's not so great. You can draw curved arrows and shift the electron pairs around. You've now got an octet on carbon but you got a sestet on nitrogen, and bad charge, negative on carbon; not so great. And over here you could do that. Sestet again, bad charge. Or you can make the three heavy atoms into a ring, and then you could put the hydrogen on any one of the -- coming out of any one of them. Now the one on the left, with my computer, I can't calculate it. I can't find an energy minimum that has that structure. So that one is very bad; which doesn't surprise anyone. It's got horrid angles for the bonds, although we haven't really said that we prefer one angle over another. But it also has bad charge separation. Okay? That one's a bad charge. So we could draw this other one here, but it's still got a positive on oxygen, a sestet. What do we have here? A sestet on carbon. Okay, so here are the bunch of isomers we've just gone through.



Now a couple of years ago, when I first talked about this in the class, and when these things had been accurately calc… or as accurately as people are likely to do, to calculate their true energy and structure -- I tested my colleagues in the department, asked them if they could rank these things as to what's the lowest and what's the highest energy and so on. And no one succeeded; in fact, only a few people get the -- know which the lowest one is, on the basis of their familiarity with Lewis theory. So the point is, don't be depressed when you are trying to do this. It takes a lot of lore, and even with all of the lore that's been gathered over many years, people can't use Lewis theory for this purpose because it's just not that good. Okay, but in fact the way I've drawn them there is their energy, according to these pretty good calculations. The very lowest energy is the C in the middle and H on the nitrogen, and then C in the middle and H on the oxygen. Now you can go back and look at what we said was good and bad about these things and convince yourself that "Ah! now you understand why those are on the bottom and these are on the top." But if it had come out differently you would have convinced yourself differently. Okay? So that's enough for today. Actually, there's a reference for this if you want to see it. The energies were published in the Journal of Chemical Physics in 2004.



[end of transcript]

Professor McBride introduces the theory behind light diffraction by charged particles and its application to the study of the electron distribution in molecules by x-ray diffraction. The roles of molecular pattern and crystal lattice repetition are illustrated by shining laser light through diffraction masks to generate patterns reminiscent of those encountered in X-ray studies of ordered solids.

Transcript

September 12, 2008



Professor Michael McBride: So as we ended last time, we said despite Earnshaw, which says you can't have these Lewis structures, might there really be shared-pair bonds and lone-pairs, and how do we know; we have to look, or feel. So last time we tried feeling with scanning probe microscopy -- AFM, STM, SNOM. They're really great. You can see atoms; you can see molecules, but you can't see bonds. Okay, so what's the key word here?



Students: Lux.



Professor Michael McBride: Lux, right? So maybe we'll see it, if we can't feel it. Now this is the entrance to this building, the old delivery entrance out on Prospect Street. And you know there's interesting stuff all over the building. Up here on top there's a fluted filter paper and a funnel, but hidden back in the shadows is something that's a little surprising to you, back there, which is a microscope. What science do you associate that with?



Students: Biology.



Professor Michael McBride: Biology, not chemistry, right? What's it doing? Well at least it's back in the shadows there. But maybe the eyeball up there says we can see the things we can't feel. Okay? This is a picture -- does anybody recognize this?



Student: Flea.



Professor Michael McBride: You know what it's from?



Student: Bubonic plague?



Professor Michael McBride: It's from Robert Hooke's great book, Micrographia, from 1665. It's a wonderful book. The page is about this big. And imagine the person who had fleas, seeing that for the first time. It's exquisite, this drawing. And he said in the book:



"But Nature is not to be limited by our narrow comprehension; future improvements of glasses may yet further enlighten our understanding, and ocular inspection may demonstrate that which as yet we may think too extravagant either to feign or to suppose."



So if he had a microscope that could do that in 1665, we must have a microscope now that's powerful enough to see bonds, right? And in fact we do. In fact, there's a brand new one that they promise will come online next Tuesday in a room about 100 yards over that way, and the trick it uses is the same one Newton used to measure the distance of the air gap for Newton's rings, or that Hooke explained by the pulses, right? And it's interference that comes from scattering. So you've seen oil on water, thin layers of oil on water, and you see rainbows in them. Why? Okay, so light comes in and scatters from two surfaces, that let's say are 200 nanometers apart. Okay? So the path difference between the one that reflects from the top and the one that reflects from the bottom, the difference is 400 nanometers; one goes 400 nanometers further than the other. Now suppose that happened to be one wavelength. Right? Then if that purple light that came in, one going further than the other, they would come out exactly one wave apart. So it would be as if it were a single wave and they would reinforce one another and you'd see the purple light strongly. Okay? But if the wavelength difference were half -- if the path difference were half the wavelength, like for red light --



Oh-oh. What do we do here, Elaine? Better get it quick, before something comes up. You're a good sport.



[Technical adjustments]



OK. Thank you.



Professor Michael McBride: If it's one wavelength path difference, then it reinforces and we get a nice strong wave coming out. Okay? But if it's half-a-wavelength difference, like for the red light, then they cancel. One is a maximum when the other is a minimum, and they're zero. Okay? So you get no scattering. So that's why you get different colors, because the oil is different thicknesses in different parts of the slick. Okay, so here's the new machine that's right over here that's supposed to come on -- new machines actually; there are two of them; it was a package deal. And that's Chris Incarvito who's the director of the Chemical Instrumentation Center and the proud owner of these two new machines. So the one he's looking at there is a user -- is to be operated by users. So it's just you sort of push buttons and you get where the atoms are in the molecule -- at least that's the hope -- and it costs about $200,000. Okay, so there's a little thread there that has a crystal on, but you need a magnifying glass to see it. You won't see it with your eyes; it's a tiny, tiny crystal. Okay? So here's an x-ray tube. X-rays come down the pipe there, hit the crystal and get scattered, and are detected by this CCD detector. Okay? And then from that information, those scattered rays, you get where the atoms are in the molecule; more precisely, as you'll see, where the electrons are in the molecule, or in the crystal.



The other machine that he's especially proud of costs $350K. And why is it more expensive? Because it collects more. Instead of using that little disc of a CCD, it has a curved image plate. Right? So when the x-rays come down here and hit the tiny crystal, many more of them get collected by the plate. So it's more efficient. So those will be coming soon. And if all you have to do is press a button and the structure comes out, that seems fine. But that's not the goal of this course, to press a button. We want to understand how such a thing works. Seeing individual molecules, atoms, and maybe bonds, there's a problem, and the problem is wavelength. Because you know in principle you can't resolve things that are closer together than the wavelength of light, and the wavelength of light we've been talking about here is 400-800 nanometers. Whereas what's the distance of a carbon-carbon bond? Anybody remember?



Student: 1.86.



Professor Michael McBride: One and a half angstroms, right? So three orders of magnitude smaller than the wavelength of light. So this is a problem, and it's why, as you'll see, you use x-rays, because they have short wavelengths. Now to understand this we'd have to know what light is. So what is light? You people have studied physics and so on. So what's light? Wilson?



Student: Who knows?



Professor Michael McBride: Who knows, okay.



Student: An electromagnetic wave.



Professor Michael McBride: It's an electromagnetic wave. Okay. Now I've seen -- indeed it is -- I've seen waves on the ocean, right? Does it mean that electromagnetic waves are like those waves? In what way is an electromagnetic wave a wave? What's wavy about it? Well you can make a graph that's a wave, that involves light. So here what we're going to plot is the force on a charge. The charge is at a particular position. We'll have the charge over there on the right; it's fixed there, and we're going to measure the force on it that'll make it accelerate up or down. Okay? Oops. It doesn't work quite the way it does on mine. Please let me know when that happens. No maybe it's going to work. [It does not.]



What that first thing did was went up and down, up and down, up and down. Okay? But if you plot it then as a function of time, it's a wave. Right? The field, at a particular point, as a function of time, is a wave. It goes up and down and up and down. Okay, now so we plot the field, or the force on a charge at one position as a function of time.



But there's another way of making a plot that also shows light as a wave, which is to show the field at different places at the same time. Right? So we're going to change it and now the horizontal axis is the position but it's at a particular instant. So as the thing goes along there are different forces at different positions, okay? And again, that will be a wave. So that's the sense in which light is a wave.



And what is it that we're plotting? We're plotting the electric field. So light is associated with an electric field that goes up and down. Right? And you can plot it in time or in space and get a wave. So that's the sense in which it's a wave. It's also, incidentally, a magnetic field; it's electromagnetic. Maxwell dealt with things like that. And there's a magnetic field perpendicular to the electric field, but we don't care about it, at least not until next semester when we talk about nuclear magnetic resonance, because the electric field is so much stronger in its effect on molecules. Now, its effect is on charges; on electrons, on protons, therefore on nuclei. Now, accelerated electrons scatter light. So here comes the light in. We'll see if this works. Okay, so it makes the electron go up and down, as the light goes by. But up and down moving of electrons, accelerating electrons, is what creates electric fields. Right? That's what an antenna is, is electrons moving back and forth.



So when the original light comes through and hits the electron and makes it vibrate up and down, that electron vibrating emits radiation, electromagnetic radiation, in all directions; all directions except this direction, and stronger the more you're perpendicular to the direction it's going up and down. But anyhow it scatters the light. So most of the light still goes through the direct beam, but a little bit of it gets scattered; much less would be scattered -- a single electron wouldn't scatter very much, you'd need lots of electrons to scatter a lot, and they have to be cooperating. Okay? Now you tell me, why are we interested in electrons scattering light? There are just as many protons in a sample as there are electrons. Why don't we worry about the protons scattering light?



Student: Electrons are a lot less massive.



Professor Michael McBride: So what?



Student: They move a lot more easily.



Professor Michael McBride: Ah, the electrons are what moves. You need the moving charge to generate the scattered light. The lightest positive charged things are -- of normal particles -- are a thousand times heavier than the electrons. So they don't move very much. It's the electrons that scatter the light. So with x-rays, you see electrons, not nuclei. Okay, they're too heavy. Now then, but you need a lot of electrons to scatter enough to see, and they have to cooperate. Right? So here we have ripples on a pond; you've all seen that kind of thing with a little bit of rain falling. And so suppose you had two spots that are generating ripples, two electrons let's say, and they give off circular waves here, in two dimensions, and they interfere with one another. But you can see a pattern is emerging, and when you're very, very far away from these, compared to the distance between the electrons, as indeed you are in these samples we're talking about where you have an infinitesimal crystal and the detector plate is a substantial distance away, right? The electrons are really close to one another. So you're way, way, way far away on the scale of the distance between the electrons. Okay?



And now you can see what pattern there is for high points and low points, for waves. Okay? Because you can see here that there's a pattern like that, that goes along the maxima. Everybody see that? And halfway in between those dashed yellow lines, there's no change at all; nothing's happening there. Right? And if you go out very, very far away, you can see that asymptotically these approach these straight lines that originate between the electrons, or on the crystal in a real case. So what you have coming out of the crystal are straight rays of light, x-ray light. Okay? And the angles at which they come out depend on where the electrons are. If the electrons are closer together or further apart or displaced in this direction, you'll get different patterns of the ripples. So somehow it must be possible, or there must be a connection between the positions of the electrons and where these rays are coming out. Right? And if you can go backwards, to go from where the rays come out, to where the electrons were, then you've solved the problem. Right?



You're not creating an image, the way you would with a normal microscope. You're trying to interpret the scattering. So the angular intensity distribution, at great distance, depends on the scatterer distribution at the origin; that is, in the crystal. Now, if you have normal light, and a lens like Hooke's microscope, then you can use these lenses to refocus it. So the same information is coming out. The sample there is emitting these rays, but a lens collects them and refocuses them to make an enlarged image that you then observe. The problem is that you can't do that with x-rays. People are making efforts now, with nanofabrication kind of things, to make things that will act like lenses for x-rays, and they've made some steps, but nothing like you would need to actually observe electrons in a crystal. Okay? Be sure to read -- there's a webpage -- have some of you read it so far? -- that has to do with what I'm lecturing on now, which will help you a lot I think; so look for that. So we're interested in seeing molecules, atoms, bonds, collectively, by x-ray crystallography. That is, we're not seeing the image of individual electrons, we're seeing the scattering that comes from all the electrons acting together. So we're not seeing them one at a time, we're seeing something collective about them. So that, for example, a real sample, or a sort of fake sample, of benzene, which had a bunch of -- six carbon atoms in a ring, might look like this at any given time. It's sort of regular but things are a little bit one way or the other, vibrating and so on.



Now you wouldn't see this in x-ray, because everything is cooperating in what you're getting. You'd see some sort of average of all of them, and it would be a little bit smeared because of that. Right? You don't see -- with scanning probe microscopy, with these sharp tips, you actually can feel individual atoms, and if one atom is someplace else, you'll see it, or feel it there. That's not true in x-ray. You see an averaged structure. And it's averaged in two ways. There's blurring, from motion and from defects. There's one benzene molecule missing there. It's time-averaged because it took you time to collect these, this information. With synchrotrons and really, really intense beams, people are trying to get faster and faster data collection, but still what you see is time-averaged on the scale of how fast atoms are moving in your sample. So it's time-averaged, what you get. And it's also space-averaged, right? It's as if you put them all on top of one another, that's what you would see, but some are displaced a little one way, the other, and so on. So it's a little fuzzy. Okay? And this is an advantage for scanning probe microscopy, which operates in real space, actually feels individual things.



Okay, so x-rays were discovered in 1895 by Röntgen, using Crookes' tube that we talked about. Okay? And he took a picture of his wife's hand there, Frau Röentgen's hand, in 1896. But what he sees is not a picture that you can blow up, like with a microscope, because it's just a shadow. All the bones do is stop the x-rays that are going through. So you don't get something that's enlarged. Right? You're not going to be able to use it the way you use medical x-rays. Okay. But in 1912 Max von Laue invented x-ray diffraction, which is this scattering and detect -- not trying to focus things, not using a shadow, but looking at these rays that come out and trying to figure out from them, something about what the atoms were. And that's his diffraction picture. That's what you'd see on that -- remember there was that round CCD plate that's on this new machine?



You might get a pattern that looked sort of like that. That was copper sulfate in 1912. But the real breakthrough by was William Lawrence Bragg, who was 22 years old. He had just graduated from Cambridge University when he determined the structure of a crystal using Laue's x-ray diffraction pattern. So he figured out how to go the other way, to go from the x-ray's scattered pattern to what the atoms were that were doing it. Right? And he did this in 1912 when he was 22-years-old, and he got the Nobel Prize in 1915. He's still the youngest Nobel Laureate. Right? His son gave me permission to use this picture. He's a very nice guy, lives in Cambridge. Okay, then of course we've come a long way from that. I think you probably recognize this picture, right? What is it?



[Students speak over one another]



Professor Michael McBride: It's the scattering from DNA. Is it a picture of DNA?



Students: No.



Professor Michael McBride: No it's not a double helix. What it is, is the way x-rays come out when they hit the double helix. Okay? And you have to figure backwards, and that's what Crick was able to do; and we'll discuss how he did that. Okay, so that's 1952, 40 years after Bragg's discovery. And then just in 2000, not so long ago, this has gone to the complete structure of the ribosome at 2.4 angstroms resolution, which was done here at Yale, in this building and the next building. Okay, and that's what it looks like. It has twenty-five nanometers across, 250 angstroms, as long as whatever number that is; lots of carbon-carbon bonds. And you see all those atoms. There are greater than 100,000 atoms, not counting the hydrogens. Incidentally, why is it easier to see other atoms than to see hydrogens?



Student: They have never counted them.



Professor Michael McBride: Because that's where the electrons are. Hydrogens come with only one electron. Other atoms come with lots of electrons. So it's hard to see hydrogens; much easier to see the other things. Okay, now what can electron diffraction show, x-ray diffraction show? That's what we want to know. Can it show molecules? Yes. Can it show atoms? Apparently; I just showed you a picture. But what we really want to know is whether it can show bonds and whether Lewis was right. So to understand this, we have to know how diffraction works. I mean, you could just be like the people that will go and punch the button on that machine, but that's not what I think you would be satisfied with. So I'll help you out. So like all light, x-rays are waves; they just are very short. So now I'm going to demonstrate with a machine here, which was designed for an overhead projector, but I believe it's going to work here.



Okay, so here are our waves. Okay, so here's a wave coming in. Okay? And when it gets to this position, it hits an electron here and an electron here; forget this one for now. Okay, everybody got me? At a given time, these two electrons are being pushed up; then they're going to be pushed down, as this wave passes by. Okay, now as those two get pushed up and down, they give off waves in almost all directions; all directions for our purposes. So they give off waves in all different directions. Right? Now notice the one, the part of that scattered x-ray that goes straight forward, in the same direction the original light came in, is bound to be in-phase with one another. Right? And we can test that with this line here, because if they're going right straight ahead, this one is a maximum when this one is a maximum. Everybody with me on that? And it'll keep that. So you'll get scattering by both of them cooperating, coming out of it straight ahead. But how about if it's at an angle? Okay, so I can push this up and change the angle. How about at that angle, how much light is going to be coming out from these two electrons?



Student: None.



Professor Michael McBride: None, they exactly cancel each other, right? But if I go to this angle, now they're just as strong as they were originally, right? And then it'll go weak, weak, weak, weaker, weakest, nothing; then stronger, stronger, stronger, stronger, very strong, weaker, weaker, weaker, weaker, weaker, zero. So there'll be a modulation -- as you go out at different angles here, going either up or down, it'll be strong, then nothing, then strong, then nothing, strong then nothing. Right? So now what will determine, what will determine how frequently these angles recur, at which reinforcement occurs? Is that clear, the question? What determines the angles? Yeah -- pardon me?



Student: The wavelength.



Professor Michael McBride: Ah, obviously the wavelength. What if the wavelength were very, very, very short?



Student: Lots of angles.



Professor Michael McBride: Then you'd get lots of them. You wouldn't have to do it very much before the next one came in, if they were very -- what else determines it, besides the wavelength. Yes John?



Student: The distance between the two of them.



Professor Michael McBride: The distance. That's what we're really interested in, is using this; knowing the wavelength of the x-rays, using this to measure distance. Okay? Now let's forget the one on the bottom here. Let's put this one in. Okay? Now -- oh okay, for reference let's look at the bottom a second. So the first reinforcement came here, between the top one and the bottom one. Everybody with me? How about for the top one and the middle one, where did the first one come? You had to go to a higher angle, there, to get reinforcement between the top and the middle. Right? So the closer things are together, the fewer the angles are. Everybody with me on that? Notice that's a reciprocal relationship. The closer things are in what we call 'real space,' the distance between the electrons, the further apart are the rays that come out, in angle. Okay? So it's backwards. Closer together, further apart, in what's called 'reciprocal space.'



Okay, now suppose you had a whole row of electrons that were evenly spaced. Okay? So here's the first, the second, the third. Now we'll take all three of them. Right? And what you notice is that as we go out it gets weaker, weaker, weaker, weaker, weaker, weaker, weaker, weaker, weaker. Now when we get to the angle here, where the first and the third were very strong, what do we see now? The one in the middle cancels the first one, and if it were a very long row of them, the fourth one would -- here the second one cancels the first one -- the fourth would cancel the third, the sixth would cancel the fifth, and so on, and you wouldn't get anything. Right? But then you'd get it again when you got to the second. The one that would've been the second angle, if it was just one and three, will be strong, because this one halfway in between will reinforce. Right? So closer together, further apart. And if you have a whole row of equally spaced things, they'll all be together, according to whatever the distance between successive ones is. Yes, Shai?



Student: What are the chances that electrons are going to be spaced evenly along --



Professor Michael McBride: Ah, what could cause electrons to be spaced exactly evenly?



Student: A crystal.



Professor Michael McBride: A crystal; x-ray crystallography, right? So that's why you use crystals. Okay, that's what we want to say here, I think.



[Technical adjustments]



Professor Michael McBride: Okay, so that's the wave machine. And if you want to do it in the privacy of your room, you can go to this website at Stonybrook and download something that allows you to run a Java applet that does sort of this kind of thing. This, if you blow it up in there, the waves look sort of funny there, at the atoms, and the reason is because you're measuring the phase perpendicular to different directions of the wave. That's just to help you out, if you try this and have -- and are confused by that. Okay, now there are -- so suppose you have just an arbitrary set of electrons, and the waves coming in and hitting them, what directions will you get reinforcement in? Well there are two directions that you're guaranteed, no matter what the spacing is, and that's this. Here the light comes in and goes out. So the direct beam, you have the same path, because one of them gets hit earlier, but has a longer path coming out, and the other one gets hit later but has a shorter path coming out. So they're guaranteed to be exactly in-phase if they're scattering straight ahead. Okay? But only a little of the light is getting scattered. Most of it is the direct beam. So you don't even notice the difference, from that. So that's not very exciting. But there's another angle at which you're sure that these two are going to be in-phase, and that's this angle. Now how do we get to that angle? I've lost some of the -- since you downloaded it there's some more stuff in here now. But -- so now I can't quite remember, let me just try.



Okay, so this blue line here is called the scattering vector, and that's how different the arrow coming out is from the arrow going in. Right? That's just a funny mathematical or geometric thing that people have defined; they call it the scattering vector. Right? Now this length is exactly the same as that length. This length is exactly the same as that length, and when you turn at a given angle, those two will -- the vector between those will be perpendicular to the direction they're going, and they'll be in-phase again. Right? And if you draw the line that connects the two points, notice the scattering vector is perpendicular to that line. So you say that this incoming wave was scattered perpendicular to this line. That's how a mirror works. Right? That angle is called the specular angle, because speculum is the Latin word for mirror. Right? So you're bound to get reinforcement at the specular angle, and you know that from having looked at mirrors.



Okay, now suppose there were another electron on that same line or plane. Okay? For the same reason it's guaranteed to be in-phase. So everything on that plane will scatter in-phase at that particular angle. They'll all reinforce one another. Great. So all electrons on a plane perpendicular to the scattering vector, scatter in-phase at the specular angle. Now suppose you have a whole bunch of electrons and you have to figure out how they're going to reinforce or cancel one another. There's a trick you can use. You see if you can discern a set of planes that are evenly spaced, that contain all, or almost all, of the electrons you're interested in. Okay, now if you look at this pattern, you can see that there's a set of planes, equally spaced, that passes through all of them. Can you perceive it? There.



Okay, there are three electrons on the first plane, four on the second, two on the third and one on the fourth. Okay? Now why is that handy? Okay, so there's the scattering vector, perpendicular to these planes, and all the electrons on any one plane will scatter in-phase with one another. Okay, so it's as if there were a single electron three times bigger, or four or two or one, on those successive planes, because they're all going to be in-phase with one another. So we could pretend they're all at one point, on any given plane. Okay? So now all we have to worry about is the phase relationship from one plane to the next. Got that? If they were in-phase from one plane to the next, as well as on the planes, then everything will be in-phase and you'll get really bright reflection coming out at that scattering angle. Okay? Now, so let's think -- so it's as if we had these collected electrons all lying on the scattering vector and we want to know whether they're in-phase with one another.



Okay, so here's light that comes in and out at a certain angle from the first one; three electrons worth of scattering from that, the ones that were on that plane -- okay? -- and four from the second plane. Now are the three and the four in-phase with one another? What condition would have to apply in order for those to be in-phase with one another? They have different path lengths, but if the paths differ by an integral number of wavelengths, then they'll be in-phase with one another and reinforce. Okay? So there's the path difference in red. If that happens to be -- suppose the wavelength of the light, and the angle of the scattering -- which also determines that path difference, as you can see -- suppose that the wavelength and the angle are such that that happens to be one wavelength? Okay, so those are in-phase with one another; all seven will be scattering together. Right? How about the next one, with two electrons, how about its phase? What was the condition we talked about, up here at the top?



Student: Evenly.



Professor Michael McBride: Evenly spaced. So what do you know about the next one?



Student: [inaudible]



Professor Michael McBride: It's going to be in-phase too, two waves behind the first one. Right? It'll be two wavelengths behind the first one, exactly, at some angle. Right? And the next one will be three wavelengths behind. So all these electrons are going to be scattering in-phase at that particular angle. So you get a really strong reflection coming out. Right? It'll be as if it were all the electrons working together. So the net in-phase scattering is as if there were ten electrons doing the scattering. That's great, right? Now suppose that the first path difference, instead -- suppose you had a different wavelength or a different angle. So the first, between the three and the four, suppose the difference in path was half a wavelength. Then how would it differ?



[Students speak over one another]



Professor Michael McBride: How about between the first and the second. Wilson?



Student: Cancel.



Student: The first and the second?



Professor Michael McBride: The first and the second, would they exactly cancel and get zero?



Student: There'd be one left.



Student: Zero.



Professor Michael McBride: Why?



Student: There's one --



Professor Michael McBride: Ah, there are three in the first and four in the second, right? So you got plus three but minus four. How about the next one? Speak up gang.



Students: Plus two.



Professor Michael McBride: Plus two. And the last one?



Students: Minus one.



Professor Michael McBride: Minus one. And what's the net scattering?



Student: Nothing.



Students: Zero.



Professor Michael McBride: Zero. Okay? So you can see how this is a neat trick to work. If you can see in the pattern a bunch of planes which would contain the electrons, then you can figure out in a particular direction, how strong the scattering would be. Now, we're going to do an experiment and it requires the room to be dark. And so I'm going to start turning the lights off and ask Filip to get into position to do stuff in the dark here. Okay, and turn this one off. Okay. Now I'll show you first what we're going to do. So pull out the little thing there. And there's a laser that's focused right here, but it's too bright, because the other things we're going to see are very dim, which is why we had to turn the lights off. So I'm going to put this black tube there, with a little hole in the end. Hopefully I can get it positioned so that most of the laser will go inside here and we won't see it very well. Let me get it positioned right. It's a moving target so it's not so easy.



Okay now, so here's the view from the ceiling. There's the screen, and this laser is coming from the back of the room and hitting the screen, right here. Okay? Now what Filip is going to do is put things called diffraction masks, which are just slides, thirty-five millimeter slides, and he's going to put them in the path of the light. And that's, the distance from Filip to here, is 10.6 meters; I measured it. Okay? And so put in the first slide please. And see this? Those are all deflections at different angles. And what's doing the scattering is a slide that looks like this. Okay? A jail window, right? Okay, so what it's giving is a row of dots. Now I'm going to ask Filip to do two things here. First, rotate the slide around this axis, so rotate it like this. What do you think will happen? You see, right? The row of spots is perpendicular to the direction of the lines. Okay? And now I'm going to ask him to do something else. I'm going to ask him to twist the slide like this, which changes the effective distance between these. And notice what happens as he makes them closer? Twisting makes them look closer together, right? What happens to the spots?



Student: Farther apart.



Professor Michael McBride: Does that surprise you?



Student: That's that reciprocal --



Professor Michael McBride: That's that reciprocal relationship. When things get closer together the angles get bigger. Okay, now things are going to get darker here. So I'm going to do some things. Turn that out and also get the room light. Well no, I want to show you something here first. Okay, so I'm going to show you what the masks are going to be and then we'll show you the effect from the masks. Okay, so the next -- so here's a question for you to work for homework. What is the spacing of the lines on Filip's slide? How far are the bars in the jail window apart on that slide, in order to give a spacing here -- oh I did the wrong, I was premature with that -- in order to get that 10.8 centimeter spacing here, at a distance of 10.6 meters, with 63 [ correction: 633] nanometer wavelength light? Okay? You can use that to find the distance between the bars on his slide, and I want you to do that, because if you can do it then you understand how this works. Okay, but the next one he's going to show -- not yet but he'll put it in -- is one that's a similar spacing, but of pairs of lines. There's a pattern being repeated, pairs of lines. Okay? And then he's going to go to a whole bunch of hexagons of dots, which we will call benzene, but it's not exactly like benzene would be, a benzene gas, because they're all oriented exactly the same way. Right? So it's oriented benzene; that's the next one we're going to look at. And then we're going to look at this one, which is also oriented benzene. Can you see the difference in the pattern between the one on the top and the one on the bottom? Those red lines are a hint.



Student: They're connected.



Professor Michael McBride: In the bottom they're all pairs of hexagons, that are distributed randomly, where in the top they're individual ones. Now then he's going to show you this one. How's that different?



Student: Quadrupled.



Professor Michael McBride: It's quartets of hexagons. And finally he's going to show you this one, which is a crystal of hexagons, and in a crystal they truly would be oriented. Okay? And then we're going to -- the pièce de résistance, at the end of this sound and light show, is going to be that. You know what that is? It's a light bulb filament, that he's going to hold up in the path of this; so a helix. Okay, now we're ready to do the trick. So we'll mute that and do this, and if you have a laptop open you'll need to close it because we got to -- it's not very bright, so we have to make it as dark as possible, and I have to find my way back there and close my laptop. Okay you ready Filip?



Teaching Fellow: Yes.



Professor Michael McBride: Okay, so first it's going to be pairs of jail bars. So how does it look different from what we saw before?



Student: Like triplets.



Student: Like different lights in there.



Professor Michael McBride: It's the same kind of dots but their intensity is modulated. They're not just evenly -- they're not just all the same or slowly dying away as you move out to right or left. They come in a pattern. Okay, now let's do the next one. This is the one that's benzene, but randomly, a random collection of benzene. So how would you describe that?



Student: A snowflake.



Professor Michael McBride: It looks like a snowflake. Oh, my laptop has come on. [Laughter] Go figure. I'm putting things to cover the lights up here. Okay, so it's a snowflake. Now we're going to look at pairs of benzenes. How is it different? It's still the snowflake.



[Students speak over one another]



Professor Michael McBride: But there are bars across it. Everybody see that?



Student: Yes.



Professor Michael McBride: Can we stop the green blink back there or put something across it? That's good. Okay, now we're going to do quartets of benzenes. How does that look? How would you describe that?



[Students speak over one another]



Professor Michael McBride: It's now got bars going in both directions, not just horizontaloid but also verticaloid; they're actually vertical I think, but not quite. Okay, now a lattice of benzenes. Isn't that great? So it takes the same -- it's the same underlying pattern, the snowflake, but it concentrates all the light that would've been spread out, into very, very fine points. Remember when it was just pairs, we put bars across, but it was the same amount of light coming out, so that it got focused sort of, and when we have a whole lattice it gets fabulously focused. I can see things quite far out from the middle here, because I'm so close to the screen. I'll hold my hand up but you can't see my hand. [Laughter] They go way out. So that's what a crystal is. The crystal takes the same underlying pattern, that comes from the molecule, and focuses it into intense points at many, many angles. So it's like looking at the scattering from a single molecule but looking at it through a pegboard. You know what a pegboard is? In a tool shop you used to put a pegboard up. It's a piece of masonite or something that had holes in it, regularly, and you could put little hooks in it to hang your hammer and your saw and whatnot. So it's as if you have, you're looking through such a pegboard at that underlying pattern. And now we're ready for the pièce de résistance, which is the light bulb filament, which is hard to see. This is why we really needed the lights out. We needed it for the snowflake too because it was so -- so what do you see for the light bulb filament?



Students: An X.



Professor Michael McBride: You see an X, right? Can people in the back see the X? And there are dots along the arms of the X. Okay? So that does it, you can open your laptops again now and we'll try to get some lights back on. Sorry for straining your eyes.



[Technical adjustments]



Professor Michael McBride: Okay, so there's the scattering from these things and we can -- now so let's try to understand it. So this was the slide with randomly positioned but oriented benzenes. It turns out that random positioning generates the same diffraction as a single pattern but gives it more intensity. If we'd just put one hexagon there you wouldn't have seen enough from it. Okay, so here's the pattern you got from isolated benzenes, which is that snowflake pattern. Right? And when we had a lattice made of those, you saw the same underlying pattern, but only at little -- only at regularly spaced spots that had to do with how far apart the hexagons were from one another, and much more intense, because all the light that would be scattered all over the screen here is focused in those little dots. Okay, now the thing that generated this snowflake was benzene, or what we're calling benzene, the hexagon. Okay, now can you see what we need in order to understand? What do you look for in order to see what directions will give you scattering? Ah sorry, I was having so much fun we went over. So we'll talk about this next week. Thanks.



[end of transcript]

Professor McBride uses a hexagonal "benzene" pattern and Franklin's X-ray pattern of DNA, to continue his discussion of X-ray crystallography by explaining how a diffraction pattern in "reciprocal space" relates to the distribution of electrons in molecules and to the repetition of molecules in a crystal lattice. He then uses electron difference density mapping to reveal bonds, and unshared electron pairs, and their shape, and to show that they are only one-twentieth as dense as would be expected for Lewis shared pairs. Anomalous difference density in the carbon-fluorine bond raises the course's second great question, "Compared to what?"

Transcript



September 15, 2008



Professor Michael McBride: So we want to understand diffraction patterns. It's clear that there's some relationship between the arrangement of the electrons that are doing the scattering and the pattern of the scattering you get. Right? At a particular angle there's a particular intensity, and it must depend on how those electrons are arranged. But how is the interesting question. So what we're going to do now is look just a little bit-- -- no one figures this out, in a real case; people have written fancy computer programs to do this. But so that it doesn't seem mysterious, I want you to get the idea of how it works. Okay, so the first slide we showed, or the third slide actually that we showed last time, was made of a bunch of these things we call benzenes, the collections of six dots, and they're all oriented the same way; that's not very reasonable except when they get regular in a crystal, which is where we're going. But anyhow, if you have a whole bunch of them, but randomly distributed, it turns out the intensity you get is stronger, proportional to the number of them that are doing the scattering. Right? But the pattern you get is the same you'd get from just one. So that's why we use this funny slide.



And this was the pattern that came out, this thing that looked a little bit like a snowflake, and I'd like you to understand why that hexagon of dots gives a snowflake. So we take the hexagon of dots, and what do we look for to see where scattering might be, to see--what's the--remember what the trick is? The trick is to look at things that lie on equally spaced planes. Try to choose a set of planes so that as many of the electrons as there are will fall on or near that set of evenly spaced planes. So And this is two dimensions, so we're looking for lines, or cross-sections of planes, right? So how can we draw lines on here, evenly spaced lines, so that all the dots will fall on them? Well we can do it this way, right? So they're evenly spaced, three of them, and all the electrons fall on or near them. Notice, they don't have to fall exactly on, because if they're displaced a little bit, the phase of the wave will change a little bit, compared to the others; but only a little bit. So it just has to be near, not exactly on. Okay, so there we have it. Now, so that means that you'll get scattering in that direction, where they will reinforce one another.



Let's go back to the display we were looking at before, here. Okay? We were talking about this just as class started. So a wave comes in, makes these things all move up and down, the electrons, here and here, at the same time. Actually, let's make all three of them vibrate for current purposes. Okay. Now we're interested at what angles light can come out and all these things will help each other out, reinforce one another. So as we go here and look at that vertical line, we see that they're not all in-phase at that angle. But as we increase the angle, when we get to say here, it looks not so bad. The top and the bottom are actually exactly in-phase, but the middle one is exactly out-of-phase, and it'll cancel out the top one. And if there were a whole row of these, the second one cancels the first, the fourth would cancel the third, the sixth would cancel the fifth, and so on; there wouldn't be anything coming out. And now we go a little bit away from that. And it's worth making a point here. Suppose that this were a very, very, very long line of scatterers, right? And the first two--let's get to this place here where they're all exactly in-phase; that'll obviously be a very strong angle. But suppose we looked at another angle that was close to that, but just a teeny bit off. So the first two are off by just 1° in-phase, and the next one's off by 2° in-phase; not much off, okay? So are we going to get pretty strong light coming out from that? Russell?



Student: If it's a very long string, you'll eventually get one that--



Professor Michael McBride: If it's a very long string, if the first one's 1° out, by the time we get to the 180th, it'll be 180° out, or exactly--so the 180th will cancel the first one, the 181st will cancel the second, the 182nd will cancel the third, and so on. Right? So it has to--if you have a really long string, it has to be exactly on, from one to the next, in order to get them so that you get something coming out. And what you get coming out then will be quite strong because it'll be the scattering from the whole bunch of them. Right? Where did you see that in last week's lecture, Friday's lecture?



Student: In the dots.



Professor Michael McBride: Right, you saw it that you got very fine dots when you had a really long row of things. Okay, so back to here. So there'll be things that are almost exactly in-phase, and it'll be a certain angle at which they'll be in-phase, and that angle is shown by the end of that arrow, and you get, therefore, a dot there, and also, by symmetry, the other five of the six dots; but you could draw the lines other angles the same way. Everybody with me on that? Okay, now so we look for electron density on or near evenly spaced planes. Now because it only has to be near, there are other directions we could draw it, so we get a set of planes where they're near but not on; like here, right? Now one is a little above the plane, two are a little below the plane--, right?-- Or vice-versa for the second line, right? But they're pretty close, and they're at a certain spacing. Now, that spacing is larger than the blue spacing. Does everybody see that? What does that mean about the angle at which the reflection will occur, or the scattering, where they'll be in phase?



Students: Smaller.



Professor Michael McBride: The angle will be smaller. So the part of the snowflake that comes from that is here, right? A smaller angle because it's a larger distance. And you can see immediately why that is, if we look at this thing, that if you look at the top to the bottom, where they're far apart, the lines are far apart, you get the first reflection there. But if the lines are closer together, like these two, then you have to go to a higher angle in order to get that first one, first coming into phase. Right? So that's that reciprocal relationship. Okay, so we get a set of six dots like that, and that's most of the pattern. But there are other places too you can see it. For example, you can take those lines. In that set of lines, every-- -- the center of every dot is on a line. There happens to be a line in the middle that doesn't have anything on it; but who cares? Right? The others are on evenly spaced lines. Right? So they'll all be in-phase. Where will that one come, that reflection, close in or far out?



Students: Far out.



Professor Michael McBride: Far out, and there you see it, way out at the edge. Can you see it there? Just a little bit with--I could turn the room lights down, but I don't think it's worth it. You can see that it's there. And so there are six of them out there as well. Okay, so you got the idea of how by looking for evenly spaced planes you can predict what the pattern will be for a certain arrangement of electrons. Now, notice those ones far out are very weak. That's because although the centers of the dots lie right on the lines, the lines are getting so closely spaced that there's part of the electron density that's not right on the lines, that in fact gets out to almost halfway between the lines. Right? And that part of the electron density will be canceling. So most of the electron density is all in-phase here, but some of it is out-of-phase, and as you get the lines closer and closer together, the fact that the electron distribution is finite, and not tiny points, means that things will get weaker and weaker as you go to higher angles. Okay, so there again we notice here that the closer spaced planes give higher angles. This is what we've just been talking about.



Now, when you take a slide with a regular lattice of these benzenes, then--remember, if you took just one, you get this snowflake pattern. If you take a whole row, across the top, like that-- -- oh pardon me, I meant to say that one will be very weak, you can hardly see it. What you need is a bunch of them to cooperate, to see something that you can see. Now first we could take the whole row along the top, and we're going to look only at scattering in the vertical direction; so I've narrowed in on the picture. But if we take that whole row along the top, remember, they'll all scatter in-phase at the specular angle, because--for scattering up and down. So now it gets more intense because they're all cooperating. For other directions they wouldn't all be in-phase from one hexagon to the next, but for scattering up and down, where they're all at exactly the same level, they'll all be exactly the same.



Okay, now suppose I add the next row. Will they be in-phase, for the same direction of scattering up and down, where that top row is all in-phase? So we would get the scattering shown here, right? Stronger than it was before because it's a bunch of these things now, right? But still not really strong. But let's add in the next row. Is that going to make them all stronger again? No, most of the time they'll be out-of-phase, from the next row, for scattering in this direction. Right? Most of the time they'll be out-of-phase, but for particular angles they'll be in-phase. So in fact what that will do, if we add the next row, we've made the picture brighter because we've got twice as many things scattering now, but there'll be bars across it, like that. It's always the same pattern, getting brighter and brighter as we add more of them, but at certain angles they're in-phase now and at certain angles they're out-of-phase, from one pattern to the next, or from one row to the next. Now, what's going to happen if we add the whole sheet of them? Can you see? Sam?



Student: Well you're only going to get very thin--



Professor Michael McBride: Ah, there'll only be really precise angles, at which they all interfere; otherwise the 180th will interfere with the first, the 181st with the second and so on. Right? So if we add the whole phase, the whole thing, then if we look at the pattern it's going to be brighter, right? But it'll be only dots, only at very precise angles. Right? And that's what we saw when we went to this final picture for scattering from a lattice. So what this means is that the lattice repeat concentrates the snowflake scattering into tightly focused spots, on a regular pattern. Right? So this is the diffraction from a 2-D lattice of benzenes, which is the same as the underlying snowflake that came from just one, but it's as if it were being viewed, as I said last time, through a pegboard. Right? All that scattering gets concentrated into a few points, very bright, and these regular points. Now some people seemed not to know what a pegboard was, so I took a picture of a pegboard over the weekend. How many people have never seen a pegboard before, just to calibrate me? Okay, most people have seen a pegboard. But you can put these hooks in it and then hang whatever you want on it, like tools or pictures or whatever you want to put on your wall. Okay, so it's like looking through the holes in that, at the snowflake pattern.



Now, there are two, quote, spaces that get talked about: direct space or real space; and diffraction space or reciprocal space. Okay? So the crystal is the real thing that you're interested in, and diffraction photo is in diffraction space; you see these dots that Lowy Laue saw, for example, for copper sulfate. Right? So in real space you have a unit cell structure; that is, the thing, the pattern that then gets repeated to make the crystal. But you have the pattern, right? And that gives rise to this fuzzy pattern, like a snowflake in reciprocal space. And then you have a crystal lattice, a regular repetition of these patterns, right? And that's as viewing through the holes in the pegboard. Okay? And remember that decreased spacing in real space corresponds to increased spacing in reciprocal space; which is of course why it's called "reciprocal". Okay, so that's what we want to understand about how it works. Right? There are patterns which give rise to patterns in reciprocal space, that fuzzy thing, and then if you have a repeating lattice of that, it gets concentrated into particular points; but it's the same underlying pattern. So there's the pattern, which is the structure of the molecule, and there's the lattice, which is the structure of the crystal.



Okay, now let's look at the light bulb filament, the last thing we looked at, and see why the scattering from that looks the way we do-- -- looks like this X with dots along it. Okay, so we take-- -- we want to find if there are electrons on this helix, right? We can find a set of evenly spaced, parallel planes like the yellow ones here. So we'll get scattering perpendicular to those planes. And sure enough there it is, that spot. But we could also take a set of planes, and it goes not only up but also down, both branches; both directions of that particular branch of the X, come from the same planes. Okay? But you could make planes that are twice as close to one another. Right? What will those give rise to?



Student: Further out.



Professor Michael McBride: Twice as far out; the sine is twice as big actually, not the actual spacing. But it's that one, the second one out along the branch. Okay? Now, will that be-- -- do you expect that to be brighter or not as bright as the first one?



Student: Not as bright.



Professor Michael McBride: Not--well it probably will be not as bright, and the reason is that for--there are two things that have to do with the brightness. One is how much density there is on the planes versus how much there is between the planes. Right? And we see here that--say take the original yellow one--there's a lot; all this stuff is nearly on the plane, and halfway between the plane there's not very much, just a thin cross-section of the wire. Okay? But as the planes get closer and closer, the fact that the wire has a finite thickness means that some parts of the wire, the same part of the wire, are out-of-phase compared to others. So because of the finite thickness you expect it to slowly fall off in intensity as you go out; but you'll then get the third one along the branch and so on. And you can do the same trick in the other direction. Right? So that's why it's an X. You can draw the analogous mirror image lines. Okay, now what does the X, the angle of the X tell you? The X could look like this, or it could look like this, right? What do you learn from that, the angle that the X makes?



Student: How tightly wound the helix is.



Professor Michael McBride: How tightly wound the helix is, because if the helix were very tight, then the planes would be like this, and if you grabbed the spring and stretched it out, they'd look like this. Right? So the angle of the X tells how tightly wound the helix is, tells the pitch of the helix. Okay? What does the spacing of the spots tell you? Did you do the homework I suggested for today, that you didn't have to hand in? Then you'll know what the spacing of the spots tells you. If you know the wavelength of the X-rays, or of the light in this case, what does the spacing of the spots tell you?



Student: The spacing of the plane that's--



Professor Michael McBride: It tells you what the actual distance is between successive turns of the helix, how far apart the planes are. So it tells the scale of the helix, right? The angle you get from the X, but how big the actual helix is you get from the spacing of those dots, if you know the wavelength and the distance that you're magnifying; in the case of the demonstration, the distance from the slide at the back of the room to the front of the room, for seeing this. And the spots weaken successively because of wire thickness, as we just explained, as you go on out. Now, that looks very much like Rosalind Franklin's Bb-DNA photograph from 1952. So we see the helix and we see how angled the successive turns of the helix are, what the pitch of the helix is, and from the spacing of the dots we get how big the actual helix is.



But there's something funny. The intensity doesn't just evenly decease as you go out. Instead of being strong, not quite as strong, less strong, weaker, very weak, as you go out, it goes weak, strong, strong, very weak, strong. Now why is that? Why do you get such a funny pattern of intensities? Let's look here at this thing again. Remember where, if we looked at the first, second, third and so on, we saw that they come in and out of phase. But let's suppose--so it would be, if you have things that are twice as close together here, as in the first case, what effect does that have on the actual pattern we see? The pattern we saw, if we had only every other one, was here the direct beam, first reflection, second reflection, third reflection, fourth reflection. How does it differ if we add one in between? Okay, here's the first reflection--can we see that, if we have all three. ? That's what would've been the first reflection if we didn't have the intermediate one. Do we see a reflection here, if we have a whole row of them? No, because every other one cancels out. So we won't see that one. How about the next one?



Student: It'll be very bright.



Professor Michael McBride: Yes, it'll be very bright, because there's more doing the scattering. How about if we go to the next one? No, nothing there, cancelled, right? How about the next one? Do you perceive a pattern emerging? What is it? What?



Student: It's successive; you see it, then you don't see it.



Professor Michael McBride: Can you say it in a more mathematical way? Suppose we numbered them. The one straight ahead we'll call zero; then there'll the first, one, two, three, four, five. What happens when we put one in between?



[Students speak over one another]



Professor Michael McBride: Is one there? This was one here, right? There was one for the top and bottom. Is that there?



Students: No.



Professor Michael McBride: How about two?



Students: Yes.



Professor Michael McBride: How about three?



Students: No.



Professor Michael McBride: No. How about four?



Students: Yes.



Professor Michael McBride: What's the--can you state it mathematically?



[Students speak over one another?]



Professor Michael McBride: Lexie, what do you say?



Student: [inaudible]



Professor Michael McBride: Can't hear very well.



Student: n+2?



Professor Michael McBride: Which ones disappear?



Student: Odd.



Student: The even.



Professor Michael McBride: The odd ones, one, three, five, seven, nine; the odd ones disappear but the even ones are still there, if we put an extra one in between. Now let's think about DNA. Here's a helix, right? But what do you say about DNA? Is it just a helix?



Students: Double helix.



Professor Michael McBride: It's a double helix, so there are two helices. Okay, so suppose we have--there are the successive turns and we get scattering at a certain angle--but suppose we make a double helix, now what'll happen? Lexie, you're going to help me again.



Student: Oh.



Professor Michael McBride: How different will it be if we have a second helix wound halfway between the first helix? What will happen to the--what we--from one helix we had one, two, three, four, five, six, seven, eight, nine angles, right? What happens now?



Student: You're missing even numbers.



Professor Michael McBride: Can't hear.



Student: You're missing even numbers.



Professor Michael McBride: Ah! All the odd ones disappear. Right? So what would you have expected to see in Rosalind Franklin's picture then? Right? It would be weak, strong, weak, strong, weak, strong, right? But generally getting weaker as you go out because of the finite thickness of the wire. Is that what we see? No, we saw--the first one is very weak, strong, strong, very weak, strong. Can you see any--how could that be? What is that telling you? Yes?



Student: They're not perfectly in-phase.



Professor Michael McBride: What do you mean they're not perfectly in-phase.



Student: They're not equally--



Professor Michael McBride: Ah. Maybe the second one isn't halfway between the turns of the first one. Right? What happens? So the planes are twice as close there and that would cancel every other reflection, all the odd ones. But suppose it's like that. Did you ever see a situation like that before where what repeats is not a line but a pair of lines? Angela?



Student: The second diffraction--.



Professor Michael McBride: The second diffraction pattern we looked at was of that sort, which showed the same dots but the intensity got modulated. Remember? Okay, so let's, instead of doing this where we have equally spaced ones, let's do one where they're not equally spaced, like this. Now--oops, it came loose here; here we go; oh no; there we go, okay. Now consider the top and the bottom one, which are in successive turns of the same helix, and this one is offset in the middle. But consider just the top and the bottom one. Okay, there's the first reflection, the second reflection, the third reflection, the fourth reflection, everybody see? Between the top and the bottom. Now tell me what happens if I add that other one offset this way. How about the first reflection there, is that going to be strong?



Students: No.



Professor Michael McBride: No, because this one is canceling it. And remember, there's also one like this below this one; remember, it's a really long row of these things. So it gets almost perfectly cancelled; not quite because it's not quite-- there would be exactly the minimum, and when the first and the third are in-phase, the second one is not exactly out-of-phase, but very close; very weak, that reflection's going to be. Now let's go to what would be the second reflection of the single helix there. How about there?



[Students speak over one another]



Professor Michael McBride: Yes, the second one now is helping out a little bit; not perfectly but it's--in fact, probably if we go just a little bit further maybe it'd be better. But anyhow there'd be a reflection there. Okay? Now how about if we go to the third? That's as good as the second one, right? And how about if we go to the fourth? Would the fourth--if they were evenly spaced, what would you expect for the fourth?



Student: Very strong.



Professor Michael McBride: Very strong, right? So let's go to the fourth; one, two, three, four. What should it be?



Students: Really weak.



Professor Michael McBride: Really weak. And how about the fifth?



Students: Strong.



Student: Really strong.



Professor Michael McBride: That's precisely what she saw, right? Weak, strong, strong, very weak, strong. So that's how much, this is how much it's offset from the center, in order to get that sequence of intensities. Everybody with me on this? Okay. So that's that repeated pair pattern that we saw before. And there you see that the outer coil of the DNA, which has phosphorous in it, with a lot of electrons, that's very much like this helix. Okay, much more electron density near these planes than between them, than halfway between. But there's also much more electron density on those, near those blue planes than in between, for the same reason, and that means that you get those spots on the side. That tells you the diameter of the helix. Okay, and the things top and bottom tell you the base stacking distance. Let me back up just a second to show you what we mean, right? So there are a bunch of electrons here, a bunch of electrons here, a bunch here, a bunch here, and that gives scattering in this direction. So all that information--the helix diameter, the fact that it's a helix, the fact that it's a double helix, the fact that it's an offset double helix, and the diameter of the cylinder, are all easily read out if you know what you're looking for; which Crick knew, because he had been studying helices of proteins, so he knew about--he knew from helix scattering.



Okay, so knowing the molecule's electron density, as we knew our artificial sort of benzene, or this DNA, it's easy to calculate the crystal's diffraction pattern, especially if you have a computer to help you out; but it's not an intellectual challenge, it's just work. Right? Using pretty heavy- duty math, people got the Nobel Prize for developing it; or a canned program, (that's what they got the Nobel Prize for), you can go the other way and go from the X-ray pattern to what it was that was causing it. Right? What we did was go from what's causing it to the pattern, but in fact you can go the other way now when you press a button on a machine. And you get out the electron density at every point in space. Okay, that's the yield from an X-ray structure. Why don't you get the nuclear positions, why do you get the electron positions?



[Students speak over one another]



Professor Michael McBride: Alex? Can't hear.



Student: It's too big to diffract.



Professor Michael McBride: Yes, it's too heavy, so it doesn't vibrate, the nuclei, so they don't function as an antenna, to give off the new waves the way the electrons do. Okay, so you get a result, which you can plot on a map. Now in the olden days when I started doing this kind of thing, the kind of map you got out was a big sheet of computer output paper, and it had numbers on it, and a given piece of paper was a given slice through the "unit cell', through the pattern, and numbers were printed out that said what was the electron density at that particular point on that level. So you had an x axis, y axis, and at this point it was seventy-five, forty-two, fifty-five, eighty-seven, or whatever the number is, and so on. And what you do is take a felt tip pen and connect dots of equal size, and you'd get a contour of that particular electron density. And here's where that was done for a particular compound.



This is taken from a book written at that time by Stout and Jensen. And you drew these by hands because you didn't have a computer that could draw--that could figure out where to draw it and so on. So here were atoms in a particular crystal structure. Then you had a whole bunch of sheets of paper for different slices, through the electron density; stack them up, right? Now sometimes the particular slice you're working in went right through the middle of an atom, so you got very high electron density. Sometime it just barely touched the atom, right? So you see here that that one, that atom, this particular slice, was very near the nucleus, so we got high electron density. But that one, the nucleus was not in this plane, so you just got a little glancing blow at it. Okay? But you could draw these things on acetate or plastic sheets and stack them up, right? So this was the model made for the first determination of the structure of penicillin; in fact, the potassium salt of penicillin. Where's the potassium atom? How can you distinguish potassium from nitrogen or carbon or hydrogen or oxygen?



Student: Electron density.



Professor Michael McBride: It'll have a lot more electron density. So do you see it? There's the potassium, lots of density. Okay, now here's a molecule that happened to be flat. So it was possible to slice through practically all the nuclei. It's not a very common type of molecule, but this one is, rubofusarin. Okay, now these contours are drawn at intervals of one electron per cubic angstrom. Right? So the first one is one electron per cubic angstrom, two electrons per cubic angstrom, three electrons per cubic angstrom and so on. Now there are both carbons and oxygens in this. How can you tell which ones are oxygens?



[Students speak over one another]



Professor Michael McBride: Yeah, Yoounjoou?



Student: They are more electronegative, so they should have more contour lines.



Professor Michael McBride: They should have more contour lines. So if we look here, we see that that one goes up to five electrons per cubic angstrom; one, two, three, four, five rings, right? That one goes up to seven electrons per cubic angstrom, and so do those. Right? So those are the oxygens. Okay? And there's a picture of the structure and you see indeed those atoms are oxygens. Now, what is our fundamental goal of looking at this stuff, what do we want to see?



Students: Bonds.



Professor Michael McBride: Bonds. Do you see bonds? What does it look like? It looks like just a bunch of balls, just atoms put there. Was Lewis right? Are there electron pairs halfway between the bonded atoms? No, it's just atoms. Okay. Why are there no hydrogens?



[Students speak over one another]



Professor Michael McBride: Because they have hardly any electrons compared to the others, right? So they're a very weak thing. Okay, now those are bonds. You can see them there. Where did those lines come from?



[Students speak over one another?]



Professor Michael McBride: Angela, what did you say?



Student: Someone drew them.



Professor Michael McBride: Yes, you drew them. How did you know where to draw the bonds?



Student: You connect the dots.



Professor Michael McBride: You connect the centers of the rings, right? Those are the bonds. Now how do you know what's a single bond, what's a double bond, what's a triple bond?



Student: More density?



Professor Michael McBride: You notice in the structure here, on the bottom right, there are double bonds and single bonds. How do I know which is which? Is there any way of knowing? Elizabeth?



Student: Maybe the distance between.



Professor Michael McBride: Ah, the distance! Double bonds should be shorter than single bonds. And, in fact, notice that here we have short, long and intermediate. Why intermediate?



[Students speak over one another]



Professor Michael McBride: Why do I draw a dotted line there?



Student: [inaudible]



Professor Michael McBride: Because we have this, look here; resonance, right? So there are one- and- a- half bonds. Right? Okay, but fundamentally what we see are spherical atoms, right? No bonds, no unshared pairs on oxygen, just atoms. So was Lewis way off base? Actually you can see the bonds with X-ray, but you don't look at total electron density the way we're looking at it. What you look at is difference density; you see how different is the picture I actually determine from what I would've expected if they were truly just spherical atoms, undistorted. How do you find--how would you make such a map? So you have these pieces of paper that say how much the electron density would be at all different points; experimentally, how much it is experimentally. Now I want to see how different is that from if I just had spheres for the atoms. What would I do? I have to find out what the spheres would look like, what density they would predict at all these different points, and then do what?



Student: Subtract.



Professor Michael McBride: Subtract, right? So that's the difference density. Sometimes it's called deformation because it's how much the electron distribution gets deformed when bonds get made. Okay, so you take the observed electron density, which is from X-ray, which is experimental, and you subtract atomic electron density, which you calculate somehow. You know what spherical atoms would look like; we'll get to that next, how you would know what that is. But anyhow, how do you know where to--when you're going to do the subtraction you have to make sure you position the atoms properly, to be subtracted, right? How do you know where to put them? How do you know where to put your spherical atoms so that you subtract it? Yes?



Student: On the center of the rings.



Professor Michael McBride: You put it on the center of those rings, right? Okay, so let's look at a case like this. This was done by Leiserowitz and Ziva Berkovitch-Yellin thirty-three years ago. So this is an interesting molecule because it has lots of different kinds of bonds in it: single bonds, double bonds--in fact three double bonds in a row--and also benzene rings that have these resonance bonds. Okay, so we're going to look first at one of the benzene rings. Okay? So this map that I'm showing you here is a difference density map. It's the total, minus what it would be for the atoms. But notice what was subtracted is just the carbon density; didn't subtract hydrogens. So let's look at the features that are big here. This is how the electron density changed in forming the molecule from atoms. Okay?



So first look up there. There are some very dense contours, right in the middle up there, and that's the hydrogen which was not subtracted, right? So it's the biggest game left in town, once you've subtracted the spherical carbons. Okay, so that's--and the total amount of intensity in that region is one electron, which is what you expect. Now the total amount of deformation density in this region, where there's a single carbon-carbon bond, is only 1/10th of an electron. What would Lewis have said? Two electrons, it's 1/10th of an electron. Okay, now let's slice that--and within the carbon-carbon bonds of the ring, the aromatic ones, it's bigger. Why bigger? Because it's partially double bond in the resonance structures, right? Okay, so it's 0.2 electrons for those ones in the ring. Now let's slice it and turn it so we can see its cross-section. And it's round; no surprise there. But let's slice the ones in the ring and turn them. It's oval, not round, the bonds are not round. Why?



Students: π bonds.



Professor Michael McBride: Ah, because they're partially made up of these p orbitals that point in and out of the ring, that distort it from being round. Okay, now let's look at the double bonds that are in the middle of this molecule, okay? First that one; we've already seen that bond before that was 0.1, had a round cross-section. But let's look at one of the double bonds. Not too surprisingly it has more electron deformation density in it; 0.2, right? And the one in the middle actually has 0.3. Now they're both double bonds, so probably they look similar, and in a sense they do. If we slice the first one, and turn it, it's oval for the same reason as before; there's a p orbital pointing in and out of the board, and if we look at it below, part of the bond is caused by overlapping, as we'll see, of those p orbitals. So it's got an oval, up and down shape. How about the next one? If we slice it and turn it, it's also oval in cross-section, but perpendicular. Right? What do you expect the next one would look like?



[Students speak over one another]



Professor Michael McBride: It's symmetrical; it'll look like the first one again. But here's something interesting. That cross-section of the middle one showed the deformation due to the pink but not the deformation showing--due to those neighboring blue ones. There must be something different about the interaction at the end and the interaction in the middle. Does everybody see that problem? You might think the middle one would be round or clover-leaf shaped, but it's not, it's distorted in the in-plane direction. But to understand this, you have to be patient and wait till next week, or later this week, when we start getting into quantum mechanics. But notice that it's as if the carbons formed bent bonds. So the bonds from those central carbons at the end go up and down, but the ones in the middle are in- plane. Now how many electrons are there in a bond? So here we're going to plot how many electrons vertically, and how long the bond is, which is related to whether it's single, double, triple. In fact, a single bond is there; a double bond and a triple bond have those distances. Now if you--and in fact you could also do a one-and-a- and a half bond, those bonds in the benzene ring. Now what would you expect this to be? How many electrons--if you were Lewis, how many electrons would you have in a single bond?



Students: Two.



Professor Michael McBride: Two. Double bond?



Students: Four



Professor Michael McBride: Four.



Students: Six.



Professor Michael McBride: Six.



Students: Three.



Professor Michael McBride: Three, okay one and a -and-a-half. Okay, so there's what you would expect for the plot, if Lewis was right:; two, four, six. Okay? Now we can take this molecule that we just looked at and see how much electron density there is in each of those bonds--and I've color-coded it--and it looks pretty darn good. Right? Hooray for Lewis. And in fact the people who did the work, Leiserowitz and Berkovitch-Yellin, looked at a bunch of other structures too and added those to the same plot, and they're not bad, except the scale is wrong. Instead of being two, four, six,, which I've faded out here, it's 0.1, 0.2, 0.3. So bonds are a very minor game in the distribution of electrons. Lewis, in a sense, was sort of right, there is electron density shifted between the atoms to hold them together, right? But much less than pairs of electrons. In fact, this amount of bonding density is about 1/20th of what Lewis might have said, but never did say, because he wasn't that naïve. Okay? So instead of getting two, you get a 20th of two; 0.1. Okay? So bonding density is about 1/20th of a Lewis. Incidentally, the unit "Lewis" is used only within this class. Okay?



So here's another example done by Dunitz in Switzerland and these people who were the--Schuyler Seiler and Schweizer, who were Swiss, who were the progeny of watchmakers and people that do very precise work; because to do this you have to have really precise experimental work, if you're going to subtract something from something else and have the difference be meaningful, when the numbers are almost precisely the same, right? Remember we looked at the total electron density. It looked just like spheres. So you have to be really good with this if you want to have this be meaningful; and those guys were really good. This was done, what, twenty-five years ago. So they took this molecule, which is highly symmetric, which is great. Remember it's got its resonance structure, so all the bonds around the middle are the same, top and bottom. Okay, so we need only look at a quarter of the molecule, or for convenience a little bit more than a quarter. So I'll blow it up here. And there is the electron difference density. So this is how the electrons shifted during formation of the bonds. Okay, so there's a carbon-carbon aromatic bond, a bond and a half, and if you turn it and look at it, it's distorted the way you expect, just as in the previous experiment we talked about. Here's another one, same deal. Now here is a single bond, carbon-carbon single bond. What's it going to look like in cross-section?



[Students speak over one another]



Professor Michael McBride: Round. Good. Now here is a carbon-nitrogen triple bond. This is something new. What's it going to look like in cross-section? Maybe like a clover-leaf, huh? Like that, it looks like that. It's round. It's not clover-leaf, and it's not a diamond, it's round in cross-section. So we'll address that later. What's that? That's not a bond.



Students: Unshared pair.



Professor Michael McBride: That's the unshared pair on Nitrogen. Now what's special here? What are we looking at? We're looking at the carbon-fluorine bond. Does it look like the carbon-carbon bonds, or carbon-nitrogen? How's it different? There's no there, there. Right? There aren't any electrons holding carbon to fluorine. What the heck is going on? Where is the C-F bond? This brings up the second great question in the course. What's the first question?



Students: How do you know?



Professor Michael McBride: How do you know, right? And the second question is illustrated by this film clip.



[Technical adjustments]



Come on baby. I worked hard on this.



Okay, I'll tell you what it says.



[Laughter]



[Technical Adjustments]



Professor Michael McBride: Okay, I can do it myself. (This is a disaster.)



"Catch some thieves Hobson, that's what we're here for."



"Very well sir."



Click, click, click, click, click, click, click, click, click, click. [Laughter]



"What do you think of him?"



"Compared with what sir?"



"Exactly."



Okay. [Laughter and applause] "What do you think of him?" "Compared with what sir?" "Exactly." That's the second question, compared with what? In anything quantitative, even many qualitative things, that's the question, and that's why we're in problems here looking for the C-F bond. We already saw this question when we talked about resonance. Remember, I asserted that when you have resonance the molecule is unusually stable. What should you have asked?



Student: Compared to what?



Professor Michael McBride: Compared to what? Right? In that case it turns out it's compared to observed or calculated energy. Compare the observed or calculated energy, (what really happens), to what you would expect if it were not resonant. And we'll talk later about how, on what basis, you might have an expectation. Now we're talking about difference electron density. So what do we compare?



[Students speak over one another]



Professor Michael McBride: We compare experimental electron density, (or it could be calculated, if we can use quantum mechanics to calculate it)--, but compare experimental electron density to what? What have we compared it to?



Student: [inaudible]



Professor Michael McBride: What it would be if it were just atoms. Right? But what kind of atoms, right? We compare the observed or calculated total electron density to the sum of electron densities for a set of undistorted atoms. Now what's an undistorted atom? Okay. Now, to avoid Pauli problems--and this isn't the time to discuss what those are, we need to subtract not an unbiased spherical fluorine atom, which if you count up how many electrons there are on fluorine, each quadrant of space around fluorine--of which there will be four-- each quadrant will have one and three three-quarters electrons in it. Right? You multiply that by four and you get seven electrons, which is the valence electrons of fluorine, right? But you can't have one and three three-quarters electrons from fluorine, and also an electron from carbon in the same place. That's too many electrons in the same place, right? So what you have to start with is not an undistorted spherical fluorine atom, but a fluorine atom that's prepared to make a bond, which has only one electron in the region where it's going to make a bond. So if you subtract a spherical one, you're subtracting too much from the region of the bond, right? So that's why you don't see any electron density in there. So, and if you want to understand this better than I've explained it, go on the Web and look at Professor Dunitz explaining it. We have a little movie of him talking about this particular problem. Okay? But at any rate the conclusion of all this is that bonding density is about 1/20th of a Lewis. All right? And it's time for me to quit. We'll go on with this next time.



[end of transcript]

After pointing out several discrepancies between electron difference density results and Lewis bonding theory, the course proceeds to quantum mechanics in search of a fundamental understanding of chemical bonding. The wave function ψ, which beginning students find confusing, was equally confusing to the physicists who created quantum mechanics. The Schrödinger equation reckons kinetic energy through the shape of ψ. When ψ curves toward zero, kinetic energy is positive; but when it curves away, kinetic energy is negative!

Transcript



September 17, 2008



Professor Michael McBride: Okay, so we just glimpsed this at the end last time. This is a crystal structure of a complicated molecule that was performed by these same Swiss folks that we've talked about, and notice how very precise it is. The bond distances between atoms are reported to ±1/2000th of an angstrom. The bonds are like one and a half angstroms. So it's like a part in a thousand, is the precision to which the positions of the atoms is known. Okay? But those positions are average positions, because the atoms are constantly in motion, vibrating. In fact, the typical vibration amplitude, which depends on -- an atom that's off on the end of something is more floppy than one that's held in by a lot of bonds in various directions, in a cage sort of thing. But typically they're vibrating by about 0.05 angstroms, which is twenty-five times as big as the precision to which the position of the average is known. Okay? So no molecule looks like that at an instant, the atoms are all displaced a little bit.



Now how big is that? Here, if you look at that yellow thing, when it shrinks down, that's how big it is, that's how big the vibration is. It's very small. But these are very precise measurements. Right? Now why did they do so precise measurements? Did they really care to know bond distances to that accuracy? Maybe for some purposes they did, but that wasn't the main reason they did the work very carefully. They did it carefully in order to get really precise positions for the average atom, so they could subtract spherical atoms and see the difference accurately. Okay? Because if you have the wrong position for the atom that you're subtracting, you get nonsense. Okay, and what this is going to reveal is some pathologies of bonding, from the point of view of Lewis concept of shared electrons. Okay, so here's a picture of this molecule. And remember, we had -- rubofusarin, which we looked at last time, had the great virtue that it was planar. So you could cut a slice that went through all the atoms. This molecule's definitely not planar, so you have to cut various slices to see different things. So first we'll cut a slice that goes through those ten atoms. Okay? And here is the difference electron density. What does the difference density show? Somebody? Yes Alex?



Student: It's the electron density minus the spherical --



Professor Michael McBride: It's the total electron density minus the atoms; that is, how the electron density shifted when the molecule was formed from the atoms. Okay, and here we see just exactly what we expected to see, that the electrons shifted in between the carbon atoms, the benzene ring, and other pairs of carbon atoms as well. It also shows the C-H bonds, because in this case the hydrogen atoms were subtracted. We showed one last time where the hydrogen atoms weren't subtracted. Okay, so this is -- there's nothing special here, everything looks the way you expect it to be; although it's really beautiful, as you would expect from these guys who do such great work. Now we'll do a different slice. This is sort of the plane of the screen which divides the molecule symmetrically down the middle, cuts through some bonds, cuts through some atoms and so on. So here's the difference density map that appears on that slice.



Now the colored atoms, on the right, are the positions of the atoms through which the plane slices, but the atoms are subtracted out; so what you see is the bonding in that plane. So you see those bonds, because both ends of the bond are in the plane, so the bonds are in the plane, and you see just what you expect to see. But there are other things you see as well. You see the C-H bonds, although they don't have nearly as much electron density as the C-C bonds did. Right? You also see that lump, which is the unshared pair on nitrogen. Right? But you see these two things, which are bonds, but they're cross-sections of bonds, because this particular plane cuts through the middle of those bonds. Everybody see that? Okay, so again that's nothing surprising. But here is something surprising. There's another bond through which that plane cuts, which is the one on the right, through those three-membered rings, right? And what do you notice about that bond?



Student: It isn't there.



Professor Michael McBride: It isn't there. There isn't any electron density for that bond. So it's a missing bond. This is what we'll refer to as pathological bonding, right? It's not what Lewis would have expected, maybe; we don't have Lewis to talk to, so we don't know what he would've thought about this particular molecule. So here's a third plane to slice, which goes through those three atoms, and here's the picture of it. And again, that bond is missing, that we saw before. Previously we looked at a cross-section. Here we're looking at a plane that contains the bond, and again there's no there there. Okay? But there's something else that's funny about this slice. Do you see what? What's funny about the bonds that you do see? Corey? Speak up so I can hear you.



Student: They're connected; they're not totally separate.



Professor Michael McBride: What do you mean they're connected separately?



Student: Usually you see separate electron densities, but they're connected.



Professor Michael McBride: Somebody say it in different words. I think you got the idea but I'm not sure everybody understood it. John, do you have an idea?



Student: The top one seems to be more dense than the bottom one.



Professor Michael McBride: One, two, three, four; one, two, three four five. That's true, it is a little more dense. That kind of thing could be experimental error, because even though this was done so precisely, you're subtracting two numbers that are very large, so that any error you make in the experimental one, or in positioning things for the theoretical position of the atoms, any error you make will really be amplified in a map like this. But it's true, you noticed. But there's something I think more interesting about those -- that. Yes John?



Student: The contour lines, they're connected, the contour lines between the top and the bottom bonds are connected. So maybe the electrons, maybe -- I don't know if they're connected.



Professor Michael McBride: Yes, they sort of overlap one another. But of course if they're sort of close to one another, that doesn't surprise you too much because as you go out and out and out, ultimately you'll get rings that do meet, if you go far enough down. Yes Chris?



Student: The center of density on the bonds doesn't intercept the lines connecting the atoms.



Professor Michael McBride: Ah. The bonds are not centered on the line that connects the nuclei. These bonds are bent. Okay? So again, pathological bonding; and in three weeks you'll understand this, from first principles, but you've got to be patient. Okay, so Lewis pairs and octets provide a pretty good bookkeeping device for keeping track of valence, but they're hopelessly crude when it comes to describing the actual electron distribution, which you can see experimentally here. There is electron sharing. There's a distortion of the spheres of electron density that are the atoms, but it's only about 5% as big as Lewis would've predicted, had he predicted that two electrons would be right between. Okay? And there are unshared pairs, as Lewis predicted. And again they're less -- but in this case they're even less than 5% of what Lewis would've predicted. But you can see them.



Now this raises the question, is there a better bond theory than Lewis theory, maybe even one that's quantitative, that would give you numbers for these things, rather than just say there are pairs here and there. Right? And the answer, thank goodness, is yes, there is a great theory for this, and what it is, is chemical quantum mechanics. Now you can study quantum mechanics in this department, you can study quantum mechanics in physics, you can probably study it other places, right? And different people use the same quantum mechanics but apply it to different problems. Right? So what we're going to discuss in this course is quantum mechanics as applied to bonding. So it'll be somewhat different in its flavor -- in fact, a lot different in its flavor -- from what you do in physics, or even what you do in physical chemistry, because we're more interested -- we're not so interested in getting numbers or solving mathematical problems, we're interested in getting insight to what's really involved in forming bonds. We want it to be rigorous but we don't need it to be numerical. Okay? So it'll be much more pictorial than numerical.



Okay, so it came with the Schrödinger wave equation that was discovered in, or invented perhaps we should say, in -- I don't know whether -- it's hard to know whether to say discovered or invented; I think invented is probably better -- in 1926. And here is Schrödinger the previous year, the sort of ninety-seven-pound weakling on the beach. Right? He's this guy back here with the glasses on. Okay? He was actually a well-known physicist but he hadn't done anything really earthshaking at all. He was at the University of Zurich. And Felix Bloch, who was a student then -- two years before he had come as an undergraduate to the University of Zurich to study engineering, and after a year and a half he decided he would do physics, which was completely impractical and not to the taste of his parents. But anyhow, as an undergraduate he went to these colloquia that the Physics Department had, and he wrote fifty years later -- see this was 1976, so it's the 50th anniversary of the discovery of, or invention, of quantum mechanics. So he said:



"At the end of a colloquium I heard Debye"(there's a picture of Debye)"say something like, 'Schrödinger, you're not working right now on very important problems anyway. Why don't you tell us something about that thesis of de Broglie?' So in one of the next colloquia, Schrödinger gave a beautifully clear account of how de Broglie associated a wave with a particle. When he had finished, Debye casually remarked that he thought this way of talking was rather childish. He had learned that, to deal properly with waves, one had to have a wave equation. It sounded rather trivial and did not seem to make a great impression, but Schrödinger evidently thought a bit more about the idea afterwards, and just a few weeks later he gave another talk in the colloquium, which he started by saying, 'My colleague Debye suggested that one should have a wave equation. Well I have found one.'"



And we write it now: HΨ=EΨ. He actually wrote it in different terms, but that's his way, the Schrödinger equation. The reason the one we write is a little different from his is he included time as a variable in his, whereas we're not interested in, for this purpose, in changes in time. We want to see how molecules are when they're just sitting there. We'll talk about time later. So within, what? Seven years, here's Schrödinger looking a good deal sharper. Right? And where is he standing? He's standing at the tram stop in Stockholm, where he's going to pick up his Nobel Prize for this. Right? And he's standing with Dirac, with whom he shared the Nobel Prize, and with Heisenberg, who got the Nobel Prize the previous year but hadn't collected it yet, so he came at the same time. Okay? So the Schrödinger equation is HΨ=EΨ. H and E you've seen, but Ψ may be new to you. It's a Greek letter. We can call it Sigh or P-sighi, some people call it Psee. Right? I'll probably call it Psi. Okay. And it's a wave function. Well what in the world is a wave function? Okay? So this is a stumbling block for people that come into the field, and it's not just a stumbling block for you, it was a stumbling block for the greatest minds there were at the time.



So, for example, this is five years later in Leipzig, and it's the research group of Werner Heisenberg, who's sitting there in the front, the guy that -- this was about the time he was being nominated or selected for the Nobel Prize. Right? So he's there with his research group, and right behind him is seated Felix Bloch, who himself got the Nobel Prize for discovering NMR in 1952. So he's quite a young guy here, and he's with these other -- there's a guy who became famous at Oxford and another one who became the head of the Physics Department at MIT. Bloch was at Stanford. So these guys know they're pretty hot stuff, so they're looking right into the camera, to record themselves for posterity, as part of this distinguished group; except for Bloch. What's he thinking about? [Laughter] What in the world is Ψ? Right? Now, in fact, in that same year, it was in January that Schrödinger announced the wave equation, and Ψ. Right?



And that summer these smart guys, who were hanging around Zurich at that time, theoretical physicists, the young guys went out on an excursion, on the lake of Zurich, and they made up doggerel rhymes for fun about different things that were going on, and the one that was made up by Bloch and Erich Hückel, whom we'll talk about next semester, was about Ψ. "Gar Manches rechnet Erwin schon, Mit seiner Wellenfuction. Nur wissen möcht man gerne wohl, Was man sich dabei vorstell'n soll." Which means: "Erwin with his Psi can do calculations, quite a few. We only wish that we could glean an inkling of what Psi could mean." Right? You can do calculations with it, but what is it? -- was the question. Okay? And it wasn't just these young guys who were confused. Even Schrödinger was never comfortable with what Ψ really means. Now if we're lucky, this'll play this time. So this is a lecture by Schrödinger, "What is Matter", from 1952.



[Short film clip is played]



Schrödinger's voice: "etwa so wie Cervantes einmal den Sancho Panza, sein liebes Eselchen auf dem er zu reiten pflegte, verlieren läßt. Aber ein paar Kapitel später hat der Autor das vergessen und das gute Tier ist wieder da. Nun werden sie mich vielleicht zuletzt fragen, ja was sind denn nun aber wirklich diese Korpuskeln, diese Atome - Moleküle. Ehrlich müßte ich darauf bekennen, ich weiß es sowenig, als ich weiß, wo Sancho Panzas zweites Eselchen hergekommen ist."



Professor Michael McBride: Right? So twenty-six years later Schrödinger still didn't really know what Ψ was. Okay? So don't be depressed when it seems a little curious what Ψ might be. Okay? First we'll -- like Schrödinger and like these other guys -- first we'll learn how to find Ψ and use it, and then later we'll learn what it means. Okay? So Ψ is a function, a wave function. What do you want to know, from what I've shown here? What is a function?



Student: A relationship.



Professor Michael McBride: Like sine is a function; what does that mean? Yes? I can't hear very well.



Student: You put an input into a function and you get an output.



Professor Michael McBride: Yes, it's like a little machine. You put a number in, or maybe several numbers, and a number comes out. Right? That's what the function does; okay, you put in ninety degrees and sin says one. Okay? So what do you want to know about Ψ first?



Student: What does it do?



Professor Michael McBride: What's it a function of? What are the things you have to put in, in order to get a number out? Okay? So it's different from the name. The wave functions have names. That's not what they're a function of. Right? You can have sine, sine2, cosine. Those are different functions, but they can be functions of the same thing, an angle. Right? So we're interested in what's it a function of; not what the function is but what's it a function of? So you can have different Ψs. They have names and quantum numbers give them their names. For example, you can have n, l, and m. You've seen those before, n, l, m, to name wave functions. Those are just their names. It's not what they're a function of. Or you can have 1s or 3dxy, or σ, or π, or π*. Those are all names of functions. Right? But they're not what it's a function of. What it's a function of is the position of a particle, or a set of particles. It's a function of position, and it's also a function of time, and sometimes of spin; some particles have spin and it could be a function of that too. But you'll be happy to know that for purposes of this course we're not so interested in time and spin. So for our purposes it's just a function of position. So if you have N particles, how many positions do you have to specify to know where they all are? How many numbers do you need? You need x, y, z for every particle. Right? So you need 3N arguments for Ψ. So Ψ is a function that when you tell it where all these positions are, it gives you a number. Now curiously enough, the number can be positive, it can be zero, it can be negative, it can even be complex, right, although we won't talk about cases where it's complex. The physicists will tell you about those, or physical chemists. Okay? And sometimes it can be as many as 4N+1 arguments. How could it be 4N+1?



Student: [inaudible]



Professor Michael McBride: Because if each particle also had a spin, then it would be x, y, z and spin; that'd be four. And if time is also included, it's plus one. Okay, so how are we going to go through this? First we'll try -- this is unfamiliar territory, I bet, to every one of you. Okay? So first we're going to talk about just one particle and one dimension, so the function is fairly simple. Okay? And then we'll go on to three dimensions, but still one particle, the electron in an atom; so a one-electron atom, but now three dimensions, so it's more complicated. Then we'll go on to atoms that have several electrons. So you have now more than three variables, because you have at least two electrons; that would be six variables that you have to put into the function to get a number out. Then we'll go into molecules -- that is, more than one atom -- and what bonding is. And then finally we get to the payoff for organic chemistry, which is talking about what makes a group a functional group and what does it mean to be functional, what makes it reactive? That's where we're heading. But first we have to understand what quantum mechanics is.



So here's the Schrödinger equation, ΗΨ=EΨ, and we're talking about the time-independent Schrödinger equation, so time is not a variable, and that means what we're talking about is stationary states. We don't mean that the atoms aren't moving, but just that they're in a cloud and we're going to find how is the cloud distributed. If a molecule reacts, the electrons shift their clouds and so on; it changes. We're not interested in reaction now, we're just interested in understanding the cloud that's sitting there, not changing in time. Okay, now the right part of the equation is E times Ψ, right? And E will turn out to be the energy of the system; maybe you won't be surprised at that. So that's quite simple. What's the left? It looks like H times Ψ. If that were true, what could you do to simplify things? Knock out Ψ. But HΨ is not H times Ψ. H is sort of recipe for doing something with Ψ; we'll see that right away. So you can't just cancel out the Ψ, unfortunately. Okay, so ΗΨ=EΨ. Oops sorry, what did I do, there we go. Now we can divide, you can divide the right by Ψ, and since it was E times Ψ, the Ψ goes away. But when you divide on the left, you don't cancel the Ψs, because the top doesn't mean multiplication.



Now I already told you the right side of this equation is the total energy. So when you see a system, what does the total energy consist of? Potential energy and kinetic energy. So somehow this part on the left, ΗΨ/Ψ, must be kinetic energy plus potential energy. That recipe, H, must somehow tell you how to work with Ψ in order to get something which, divided by Ψ, gives kinetic energy plus potential energy. So there are two parts of it. There's the part that's potential energy, of the recipe, and there's the part that's kinetic energy. Now, the potential energy part is in fact easy because it's given to you. Right? What's Ψ a function of?



Students: Position.



Professor Michael McBride: Position of the particles. Now if you know the charges of the particles, and their positions, and know Couloumb's Law, then you know the potential energy, if Couloumb's Law is right. Is everybody with me on that? If you know there's a unit positive charge here, a unit negative charge here, another unit positive charge here and a unit negative charge here, or something like that, you -- it might be complicated, you might have to write an Excel program or something to do it -- but you could calculate the distances and the charges and so on, and what the energy is, due to that. So that part is really given to you, once you know what system you're dealing with, the recipe for finding the potential energy. So that part of HΨ/Ψ is no problem at all. But hold your breath on kinetic energy. Sam?



Student: Didn't we just throw away an equation? There was an adjusted Couloumb's Law equation.



Professor Michael McBride: Yes, that was wrong. That was three years earlier, remember? 1923 Thomson proposed that. But it was wrong. This is what was right.



Student: How did they prove it wrong?



Professor Michael McBride: How did what?



Student: Did they prove it wrong or just --



Professor Michael McBride: Yes, they proved this right, that Couloumb's Law held, because it agreed with a whole lot of spectroscopic evidence that had been collected about atomic spectra, and then everything else that's tried with it works too. So we believe it now. So how do you handle kinetic energy? Well that's an old one, you did that already in high school, right? Forget kinetic energy, here it is. It's some constant, which will get the units right, depending on what energy units you want, times the sum over all the particles of the kinetic energy of each particle. So if you know the kinetic energy of this particle, kinetic energy of this particle, this particle, this particle; you add them all up and you get the total kinetic energy. No problem there, right? Now what is the kinetic energy that you're summing up over each particle? It's ½mv². Has everybody seen that before? Okay, so that's the sum of classical kinetic energy over all the particles of interest in the problem, and the constant is just some number you put in to get the right units for your energy, depending on whether you use feet per second or meters per year or whatever, for the velocity.



Okay, but it turned out that although this was fine for our great-grandparents, it's not right when you start dealing with tiny things. Right? Here's what kinetic energy really is. It's a constant. This is the thing that gets it in the right units: (h2)/8(π2) times a sum over all the particles -- it's looking promising, right? -- of one over the mass -- not the mass mv², but one over the mass of each particle -- and here's where we get it -- [Students react] -- times second derivatives of a wave function. That's weird. I mean, at least it has twos in it, like v², right? [Laughter] That's something. And in fact it's not completely coincidental that it has twos in it. There was an analogy that was being followed that allowed them to formulate this. And you divide it by the number Ψ. So that's a pretty complicated thing. So if we want to get our heads around it, we'd better simplify it. And oh also there's a minus sign; it's minus, the constant is negative that you use. Okay, now let's simplify it by using just one particle, so we don't have to sum over a bunch of particles, and we'll use just one dimension, x; forget y and z. Okay? So now we see something simpler. So it's a negative constant times one over the mass of the particle, times the second derivative of the function, the wave function, divided by Ψ. That's kinetic energy really, not ½mv². Okay? Or here it is, written just a little differently. So there's a constant, C, over the mass, right? And then we have the important part, is the second derivative. Does everybody know that the second derivative is a curvature of a function. Right? What's the first derivative?



Students: Slope.



Professor Michael McBride: Slope, and the second derivative is how curved it is. It can be curving down, that's negative curvature; or curving up, that's positive curvature. So it can be positive or negative; it can be zero if the line is straight. Okay. So note that the kinetic energy involves the shape of Ψ, how curved it is, not just what the value of Ψ is; although it involves that too. Maybe it's not too early to point out something interesting about this. So suppose that's the kinetic energy. What would happen if you multiplied Ψ by two? Obviously the denominator would get twice as large, if you made Ψ twice as large. What would happen to the curvature? What happens to the slope? Suppose you have a function and you make it twice as big and look at the slope at a particular point? How does the slope change if you've stretched the paper on which you drew it?



Student: It's sharper.



Professor Michael McBride: The slope will double, right, if you double the scale. How about the curvature, the second derivative? Does it go up by four times? No it doesn't go up by four times, it goes up by twice. So what would happen to the kinetic energy there if we doubled the size of Ψ every place?



Student: Stay the same.



Professor Michael McBride: It would stay the same. The kinetic energy doesn't depend on how you scale Ψ, it only depends on its shape, how curved it is. Everybody see the idea? Curvature divided by the value. Okay, now solving a quantum problem. So if you're in a course and you're studying quantum mechanics, you get problems to solve. A problem means you're given something, you have to find something. You're given a set of particles, like a certain nuclei of given mass and charge, and a certain number of electrons; that's what you're given. Okay, the masses of the particles and the potential law. When you're given the charge, and you know Couloumb's Law, then you know how to calculate the potential energy; remember that's part of it. Okay? So that part's easy, okay? Now what do you need to find if you have a problem to solve? Oh, for example, you can have one particle in one dimension; so it could be one atomic mass unit is the weight of the particle, and Hooke's Law could be the potential, right? It doesn't have to be realistic, it could be Hooke's Law, it could be a particle held by a spring, to find a Ψ. You want to find the shape of this function, which is a function of what?



Student: Positions of the particles.



Professor Michael McBride: Positions of the particles, and if you're higher, further on than we are, time as well; maybe spin even. But that function has to be such that HΨ/Ψ is the total energy, and the total energy is the same, no matter where the particle is, right, because the potential energy and the kinetic energy cancel out. It's like a ball rolling back and forth. The total energy is constant but it goes back and forth between potential and kinetic energy, right? Same thing here. No matter where the particles are, you have to get the same energy. So Ψ has to be such that when you calculate the kinetic energy for it, changes in that kinetic energy, in different positions, exactly compensate for the changes in potential energy. When you've got that, then you've got a correct Ψ; maybe. It's also important that Ψ remain finite, that it not go to infinity. And if you're a real mathematician, it has to be single valued; you can't have two values for the same positions. It has to be continuous, you can't get a sudden break in Ψ. And Ψ² has to be integrable; you have to be able to tell how much area is under Ψ², and you'll see why shortly. But basically what you need is to find a Ψ such that the changes in kinetic energy compensate changes in potential energy.



Now what's coming? Let's just rehearse what we did before. So first there'll be one particle in one dimension; then it'll be one-electron atoms, so one particle in three dimensions; then it will be many electrons and the idea of what orbitals are; and then it'll be molecules and bonds; and finally functional groups and reactivity. Okay, but you'll be happy to hear that by a week from Friday we'll only get through one-electron atoms. So don't worry about the rest of the stuff now. But do read the parts on the webpage that have to do with what's going to be on the exam. Okay, so normally you're given a problem, the mass and the charges -- that is, that potential energy as a function of position -- and you need to find Ψ. But at first we're going to try it a different way. We're going to play Jeopardy and we're going to start with the answer and find out what the question was. Okay? So suppose that Ψ is the sine of x; this is one particle in one dimension, the position of the particle, and the function of Ψ is sine. If you know Ψ, what can you figure out? We've just been talking about it. What can you use Ψ to find?



Student: Kinetic energy.



Professor Michael McBride: Kinetic energy. How do you find it? So we can get the kinetic energy, which is minus a constant over the mass times the curvature of Ψ divided by Ψ at any given position. Right? And once we know how the kinetic energy varies with position, then we know how the potential energy varies with position, because it's just the opposite, in order that the sum be constant. Right? So once we know the kinetic energy, then we know what the potential energy was, which was what the problem was at the beginning. Okay? So suppose our answer is sin(x). What is the curvature of sin(x), the second derivative?



[Students speak over one another]



Professor Michael McBride: It's -sin(x). Okay, so what is the kinetic energy?



Student: C/m.



Professor Michael McBride: C/m. Does it depend on where you are, on the value of x? No, it's always C/m. So what was the potential energy? How did the potential energy vary with position?



Student: It doesn't.



Professor Michael McBride: The potential energy doesn't vary with position. So sin(x) is a solution for what? A particle that's not being influenced by anything else; so its potential energy doesn't change with the position, it's a particle in free space. Okay? So the potential energy is independent of x. Constant potential energy, it's a particle in free space. Now, suppose we take a different one, sin(ax). How does sin(ax) look different from sin(x)? Suppose it's sin(2x). Here's sin(x). How does sin(2x) look? Right? It's shorter wavelength. Okay? Now so we need to figure out -- so it's a shortened wave, if a > 1. Okay, now what's the curvature? Russell?



Student: It's -a² times sin(ax).



Professor Michael McBride: It's -a² times sin(ax). Right? The a comes out, that constant, each time you take a derivative. So now what does the kinetic energy look like? It's a² times the same thing. Okay? So again, the potential energy is constant. Right? It doesn't change with position. But what is different? It has higher kinetic energy if it's a shorter wavelength. And notice that the kinetic energy is proportional to one over the wavelength squared, right?; a²; a shortens the wave, it's proportional to a², one over the wavelength squared. Okay. Now let's take another function, exponential, so e^x. What's the second derivative of ex? Pardon me?



Student: e^x.



Professor Michael McBride: e^x. What's the 18th derivative of e^x?



Students: e^x.



Professor Michael McBride: Okay, good. So it's e^x. So what's this situation, what's the kinetic energy?



Student: -C/m.



Professor Michael McBride: -C/m. Negative kinetic energy. Your great-grandparents didn't get that. You can have kinetic energy that's less than zero. What does that mean? It means the total energy is lower than the potential energy. Pause a minute just to let that sink in. The total energy is lower than the potential energy. The difference is negative. Okay? So the kinetic energy, if that's the difference, between potential and total, is negative. You never get that for ½m(v²). Yes?



Student: Does this violate that Ψ has to remain finite?



Professor Michael McBride: Does it violate what?



Student: That Ψ has to remain finite?



Professor Michael McBride: No. You'll see in a second. Okay, so anyhow the constant potential energy is greater than the total energy for that. Now, how about if it were minus exponential, e-x? Now what would it be? It would be the same deal again, it would still be -C/m, and again it would be a constant potential energy greater than the total energy. This is not just a mathematical curiosity, it actually happens for every atom in you, or in me. Every atom has the electrons spend some of their time in regions where they have negative kinetic energy. It's not just something weird that never happens. And it happens at large distance from the nuclei where 1/r -- that's Couloumb's Law -- where it stops changing very much. When you get far enough, 1/r gets really tiny and it's essentially zero, it doesn't change anymore. Right? Then you have this situation in any real atom. So let's look at getting the potential energy from the shape of Ψ via the kinetic energy. Okay, so here's a map of Ψ, or a plot of Ψ, it could be positive, negative, zero -- as a function of the one-dimension x, wherever the particle is. Okay? Now let's suppose that that is our wave function, sometimes positive, sometimes zero, sometimes negative. Okay? And let's look at different positions and see what the kinetic energy is, and then we'll be able to figure out, since the total will be constant, what the potential energy is. Okay? So we'll try to find out what was the potential energy that gave this as a solution? This is again the Jeopardy approach. Okay? Okay, so the curvature minus -- remember it's a negative constant -- minus the curvature over the amplitude could be positive -- that's going to be the kinetic energy; it could be positive, it could be zero, it could be negative, or it could be that we can't tell by looking at the graph. So let's look at different positions on the graph and see what it says. First look at that position. What is the kinetic energy there? Positive, negative, zero? Ryan, why don't you help me out?



Student: No



Professor Michael McBride: Well no, you can help me out. [Laughter] Look, so what do you need to know? You need to know -- here's the complicated thing you have to figure out. What is minus the curvature divided by the amplitude at this point? Is it positive, negative or zero? So what's the curvature at that point? Is it curving up or down at that point? No idea. Anybody got an idea? Keith? Kevin?



Student: It looks like a saddle point so it's probably zero.



Professor Michael McBride: It's not a saddle point. What do you call it?



Students: Inflection points.



Professor Michael McBride: A saddle point's for three dimensions. In this it's what? Inflection point. It's flat there. It's curving one way on one side, the other way on the other side. So it's got zero curvature there; okay, zero curvature. Now Ryan, can you tell me anything about that, if the curvature is zero?



Student: Zero.



Professor Michael McBride: Ah ha.



[Laughter]



Professor Michael McBride: Not bad. So that one we'll color grey for zero. The kinetic energy at that point is zero, if that's the wave function. Now let's take another point. Who's going to help me with this one? How about the curvature at this point right here?



Student: Negative.



[Students speak over one another]



Professor Michael McBride: It's actually -- I choose a point that's not curved.



Student: Ah.



Professor Michael McBride: It's straight right there. I assure you that's true. So I bet Ryan can help me again on that one. How about it?



Student: Zero.



Professor Michael McBride: Ah ha. So we'll make that one grey too. Now I'll go to someone else. How about there? What's the curvature at that point do you think? Shai?



Student: It looks straight, zero curvature.



Professor Michael McBride: It looks straight, zero curvature. So does that mean that this value is zero?



Student: Not necessarily, because the amplitude --



Professor Michael McBride: Ah, the amplitude is zero there too. So really you can't be sure. Right? So that one we're going to have to leave questionable, that's a question mark. How about out here? Not curved. So what's the kinetic energy? Josh?



Student: Questionable.



Professor Michael McBride: Questionable, right? Because the amplitude is zero again; zero in the numerator, also zero in the denominator; we really don't know. Okay, how about here? Tyler, what do you say? Is it curved there?



Student: Yes.



Professor Michael McBride: Curving up or down?



Student: Down.



Professor Michael McBride: So negative, the curvature is negative. The value of Ψ?



Student: Positive.



Professor Michael McBride: Positive. The energy, kinetic energy?



Student: Positive.



Professor Michael McBride: Positive. Okay, so we can make that one green. Okay, here's another one. Who's going to help me here? Kate?



Student: Yes.



Professor Michael McBride: Okay, so how about the curvature; curving up, curving down?



Student: It's curving down, that's negative.



Professor Michael McBride: Yes. Amplitude?



Student: Zero. So it should be green.



Professor Michael McBride: Ah, green again. Okay. How about here? Ah, now how about the curvature? Seth?



Student: I don't know.



Professor Michael McBride: Which way is it curving at this point here?



Student: Curving up.



Professor Michael McBride: Curving up. So the curvature is --



Student: Positive.



Professor Michael McBride: Positive.



Student: The amplitude is negative, so it's positive.



Professor Michael McBride: Yeah. So what color would we make it? Green again. Okay, so if you're -- you can have -- be curving down or curving up and still be positive; curving down if you're above the baseline, curving up if you're below the baseline. Right? So as long as you're curving toward the baseline, towards Ψ=0, the kinetic energy is positive. How about here? Zack? Which way is it curving? Curving up or curving down?



Student: It should be curving up.



Professor Michael McBride: Curving up, curvature is positive. The value?



Student: Positive.



Professor Michael McBride: Positive.



Student: I guess it'll be negative.



Professor Michael McBride: So it's negative kinetic energy there. Make that one whatever that pinkish color is. Okay? Here's another one, how about there? Alex? Which way is it curving at the new place? Here?



Student: Curving down.



Professor Michael McBride: Curving down, negative curvature.



Student: Negative amplitude.



Professor Michael McBride: Negative amplitude.



Student: Negative kinetic energy.



Professor Michael McBride: Negative kinetic energy; pink again. Is that enough? Oh, there's one more, here, the one right here. Okay?



Student: Negative.



Professor Michael McBride: Pardon me?



Student: Negative.



Professor Michael McBride: Negative, because it's -- how did you do it so quick? We didn't have to go through curvature.



Student: It goes away from the line.



Professor Michael McBride: Because it's curving away from the baseline, negative. Okay, pink. Okay, curving away from Ψ=0 means that the kinetic energy is negative. So now we know at all these positions whether the kinetic energy is positive, negative or zero, although there are a few that we aren't certain about. Right? So here's the potential energy that will do that. If you have this line for the total energy, right? Then here and here you have zero. Right? Also, incidentally, here and here, you have zero kinetic energy. With me? Okay. So no curvature, right? At these green places, the total energy is higher than the potential energy. So the kinetic energy is positive. Okay? At these places, the potential energy is higher than the total energy. So the kinetic energy is negative and the thing is curving away from the baseline. Right? And now we know something about this point. If the potential energy is a continuous kind of thing, then, although we couldn't tell by looking at the wave function, it's curving away from the baseline, but very slightly, right? It's negative kinetic energy there, and also on the right here is negative kinetic energy. And here we know, just by continuity, that at this point it must've been positive kinetic energy, even though we couldn't tell it by looking at the curve. There must be an inflection point when you go through zero, otherwise you'd get a discontinuity in the potential energy. Okay, so that one was green. Okay, now I have to stop.



[end of transcript]

Professor McBride expands on the recently introduced concept of the wave function by illustrating the relationship of the magnitude of the curvature of the wave function to the kinetic energy of the system, as well as the relationship of the square of the wave function to the electron probability density. The requirement that the wave function not diverge in areas of negative kinetic energy leads to only certain energies being allowed, a property which is explored for the harmonic oscillator, Morse potential, and the Columbic potential. Consideration of the influence of mass reveals an "isotope effect" on dynamics, on the energy, vibration frequency, and length of bonds.

Transcript



September 19, 2008



Professor Michael McBride: Okay, so last time we saw the jeopardy approach to solving the Schrödinger equation, which is to take the answer and find out what the problem was, and we saw that for a constant potential you got a sine wave. You could choose any length of sine wave you wanted, any wave period you wanted. It just turned out to be different energies and it had a relationship between the wavelength and the energy, the kinetic energy, or the total energy in that case. And we saw an exponential is -- corresponds to -- negative kinetic energy, which not very many people in the class liked, and not very many people ever like it, but suck it up because that's the way kinetic energy is for small particles. If you don't like to call it kinetic energy because you're so wedded to the fact that for big things it's ½mv², then just consider it the difference between total energy and potential energy. Okay? But anyhow it's kinetic energy.



Okay, so we looked at the jeopardy approach, to start with the answer and get the question. But we'd a lot rather do it the other direction, to start with the problem and find out what the answer is. And you can do that by guessing the energy. If you have a sufficiently simple system, that is, one particle in one dimension; and this is a great way to get the idea of what's special about quantum mechanics, make it as simple as possible and then we'll go to realistic systems. Okay, so we can rearrange the Schrödinger equation. Remember, this is the Schrödinger equation: HΨ=EΨ; divide through by Ψ and split the HΨ/Ψ part into the potential energy part, V, which is a nothing, it's given to you, and the kinetic energy part, which is a negative constant that involves the inverse mass times the curvature of Ψ over Ψ. So it has to do with the shape of Ψ. And remember, if you multiply Ψ by a constant, you multiply the curvature by the same constant, so the kinetic energy stays the same. So it's only what we say, the shape, not really the amplitude, because that scales with the curvature.



Okay, but now we can rearrange this, this way. So we have a formula here for the curvature of Ψ. Now if you have a formula for the curvature, if you know how a curve is going to change, how a line is going to change its curvature [correction: slope], then you have a recipe for drawing it, if you know how to start. If you start with a certain height, and a certain slope -- suppose you know that, and you know the curvature, at that point, then you know if you take a little step in that direction what the next slope is going to be. Okay? Because that's how curved it is. And if you recalculate and know the curvature at that point, then you know what the curvature at the next point is going to be, and the next and the next and the next, and you can use this to just draw a curve that's bound to satisfy the Schrödinger equation because you're using the Schrödinger equation to generate it. Does everybody see this? This is really important. Do you want me to say it again, or you got it now? If you have a formula for the curvature -- that's how fast the slope is changing -- and if you know the initial height and the initial slope, then if you know the direction of the slope you know where the next point is going to be. So you go to that -- just think a tiny -- now this gets a problem in a computer. If the thing is really going like this and you take a finite step, the slope will have changed during your step and you won't get it just right. So you'll have to take smaller and smaller steps so that it doesn't change during the step. Chenyu?



Student: [inaudible]



Professor Michael McBride: Can't hear very well.



Student: Are you saying that the curvature is a second derivative?



Professor Michael McBride: Yes, the curvature is a second derivative, yes.



Student: How do we start with the correct slope?



Professor Michael McBride: We'll get to that. But if you have the initial slope, and the initial height, then you use the formula for the curvature to trace out the curve. Everybody got that now? Okay, so that's great, and it would be tedious to do this by hand, but that's what this little program, Erwin Meets Goldilocks, does is start the curve and draw it out, according to what -- if you know everything else in the equation. Okay, so, and remember it curves away from zero when the potential energy is greater than the total energy, and it curves towards zero when the potential energy is less than the total energy. Right? So the potential energy being greater than the total energy is a bad situation intuitively, right, it means the kinetic energy is negative, and it curves away from the baseline. And what's bad about that, or dangerous about that? Yes, Kevin? Pardon me?



Student: It's not bounded.



Professor Michael McBride: Yes, it'll keep going until it gets to infinity, if it's curving away. Okay, and that's not allowed, it has to not go to infinity. Okay, so now do we know everything else? Well we know m, the mass. We know the constant, C, and the V is given to us, the potential energy changes with position. Right? But we know how it changes with position; it's Coulomb's law typically, or we can do some other law if we want to. So this recipe then, you know everything on the right, except the total energy. Ψ you don't care about because you can scale it, you can make it anything you want, call it one. Okay? Then we can double it if we want to or whatever. So Ψ we know; m we know; C we know; V we know. We're going to guess E. So that means if we know the initial slope -- and for the problems we work on the initial slope will be zero, and you'll see why -- then we can use this formula to trace out the curve. Okay? And that's what the computer does for you. So people were worried last night about math. Don't worry about math. Let the computer calculate how these slopes change and so on. Right? All you have to do is think about the curvature. Okay? So you just have to think about the graph.



Okay, so we're going to talk now about nodes and quantization, in one dimension, using this program Erwin Meets Goldilocks. And there's a problem set for Monday, as you know, where you do this, and the answers are on the Wiki, but don't look at them before you've given it a good try. Okay, so let's -- there are simpler things to start with, like a constant potential energy or running into a wall or into a pillow more like. But the harmonic oscillator, or Hooke's law, is a very popular one to do; so we're going to do that. Okay, so here we're looking at the potential energy as a function of distance. And also on the -- so you see the distance goes from zero to two and a half angstroms and the potential energy goes from zero to 100 kilocalories per mole. You can change that if you want to on the computer. Okay? And the horizontal black line is a completely different plot. It's a plot of Ψ. That's the plot on top of this one where we're going to trace out the wavefunction. Okay? And that's zero of the wavefunction; here's zero of energy. Okay? And the reason we put them on top of one another is so that it's easy to see the influence of one on the other, of the wavefunction on the potential energy and vice-versa, and because they're functions of the same thing. What's Ψ a function of? What's it a function of?



Students: Position.



Professor Michael McBride: It's a function of the position. Position of what? The one particle we're talking about. Okay? And the potential energy is also a function of the position. Okay, so the way we're going to do this then is to guess an energy, let the computer trace out this curve, and see whether it's okay. What will tell us whether it's okay? How will we know if it's okay? It's got to be continuous. We're not going to have a problem with that because the computer is tracing it out this way; it's not going to be discontinuous. But the main thing is that it doesn't diverge, that it doesn't go to infinity. Okay, so let's just guess twenty-one kcal/mole. Why not? Okay, now you can already tell me something about what the curve is going to look like. Notice that at those two distances, between these two distances, the total energy is higher than the potential energy. What's the curve going to look like in that region? The kinetic energy is going to be positive, right? What will the curve look like? Pat?



Student: Spike up in between them.



Professor Michael McBride: Can't hear very well.



Student: It's going to spike up in between those two lines.



Professor Michael McBride: It may go up, it may go down. Yes? Pardon me? It's not -- would it be a sine curve? Does anybody think it wouldn't be a sine curve? Why not Russell?



Student: It does sinusoidal behavior when the potential is constant.



Professor Michael McBride: That's when the potential is constant. It's not constant through here. Right? So it won't be an exact sine curve. If this isn't very high, it might be close to a sine curve because it's sort of constant-ish, if this is a very small distance. But what characteristic must it have?



Student: It'll curve towards zero.



Professor Michael McBride: Say it again.



Student: It'll curve towards zero.



Professor Michael McBride: It'll curve towards zero. So it might -- it'll look something like this. Always -- it's curving and going up and down, maybe going high, but it's always curving to go back toward the baseline. It's never going to go to infinity. Okay? Now, there's a problem, because if you go the other way, the rest of the graph, the kinetic energy is negative because the potential's higher than the total. So in that region there's great danger that the line is going to go to infinity, because it's going to be curving away from the baseline. Okay? Now how could you possibly avoid that? How could you have it such that the curvature divided by the value of the wavefunction gets very big and the curvature away from the baseline, but still it doesn't go to infinity? How can the kinetic energy be big and negative, the curvature divided by the value is enormous, but it doesn't go to infinity? How can that be?



Student: Do we have to bound it?



Professor Michael McBride: Pardon me?



Student: Bound.



Professor Michael McBride: Bound?



Student: Yeah bound what we're looking at?



Professor Michael McBride: No, it's not a question of bound. This thing is, the particle, if it has this total energy, is bounded, it can't go all the way to the left or the right, that's true enough. Okay, but that's not the secret of how the wavefunction avoids going to infinity. Yeah?



Student: Make it a constant.



Professor Michael McBride: Pardon me?



Student: Make it a constant.



Professor Michael McBride: We can't -- make what a constant?



Student: Like -- another one a little bit.



Professor Michael McBride: Okay. Russell?



Student: You'd have very little curvature.



Professor Michael McBride: But -- you say very little curvature. But the kinetic energy is enormous. The curvature divided by the value of the wavefunction has to be enormous. Mike?



Student: You could make the value of the wave function go to zero.



Professor Michael McBride: Ah, if the value of the wavefunction is almost zero, then you can have an enormous -- pardon me, wait a minute -- then you can have a tiny numerator, hardly any curvature at all, and still the quotient can be big. Right? The kinetic energy can be enormous, and negative, if the denominator is near zero. So that means that if you're in those red regions, you know what the wavefunction has to look like. What does it have to look like? It has to get down near zero and stay there so that it doesn't have the curvature that'll -- enough curvature to pull it away. So it's got to look like that, on the left. Right? Because out here, that parabola is going way up; it's enormous, curvature divided by wavefunction, but if the wavefunction is essentially zero, still the curvature can be essentially zero. Okay? And the same on the right. So we know that the curve is going to look like this, something like that. Okay? And all we have to do is connect the pieces. Okay, so let's try it. We guessed twenty-one kcal/mole. This we did in our head, just by thinking about it. But let's let the program do it, and start at the left. And now do you see why it has to start with zero slope? Because if it's out in this region and has a finite slope, or an appreciable slope, and it's curving away, you're in trouble, because it's going to go to infinity. So we're going to use Erwin Meets Goldilocks, this program, and we're going to start at -- here goes the curve, and there's the part where the kinetic energy is negative, right? It hasn't gone to infinity because it started low enough and with low enough slope out at the left. Okay, now what's going to happen when it crosses that line?



[Students speak over one another]



Professor Michael McBride: What?



Student: The curvature changes.



Student: Now the curvature is going to be toward the baseline, as we guessed it would be; so like that. And now what's going to happen? Now it's going to start curving away from the baseline again. Uh oh, you don't win them all, right? Okay, now did we have too much energy or too little energy here? Notice that -- what did we want it to do on the right? On the left is where we started, that looks great, but on the right we're having trouble. What do you need to do in the middle in order for it to behave okay on the right? It has to not curve so much toward the baseline. Does everybody see what I'm saying? If it curved less, then it could join that other line we had originally. Okay, so we'll give it a little less energy. Let's try twenty. That's too hot, too much energy. Okay, so we'll guess twenty kcal/mole. Okay, here we go. Start out; almost exactly the same, it's nearly the same energy, so it doesn't surprise you that it looks about the same. And then it's going to curve in here, the way we expected, but a little bit less than it did before. Right? So now we're on the way to success. So what was that one?



Student: Too cold.



Professor Michael McBride: Too cold. And what do you need?



[Students speak over one another]



Professor Michael McBride: Someplace in between these, there's going to be one that's right. Okay? So it's between twenty and twenty-one. Now should we guess halfway between twenty and twenty-one, is that the reasonable thing to do? Which original guess was better? Which guess was better originally, and why? The red one or the blue one, which guess was better?



Student: The red one.



Professor Michael McBride: Why was the red one better?



Student: Because it curved close to the baseline.



Professor Michael McBride: It lasted longer before it went to infinity. So it'll be closer to the red than to the blue. And if you fiddle around -- and that's what I want you to do, because this is the way I think it gets into your head is by guessing numbers and saying "where should I make the next guess?" and so on. After you've done this a certain number of times -- but enough times that it really gets into your bones and your brain -- after you've done it enough times, then you can use the function that says "Solve" and it will do that guessing game for you. But do it yourself at the beginning. Okay, and it turns out that after you've done that, you get 20.74 da-da-da-da-da, like a lot of decimal points, decimal positions, and you get that one, just right. Okay? So that's the only math you need to do. Is it too hot or too cold or just right? Okay? And you have to know why it has to be that way, because of this curvature thing. Okay, so there we have it, 20.74 kcal/mole is a satisfactory energy, and it comes with this wavefunction, whatever that's good for. Okay? But the energy is what we were looking at, what energies are allowed for this system? Now, could there be a lower energy allowed for this system? Could you imagine a less curved wavefunction that would also have this property of not going to infinity? Can somebody guess what it would look like? Nick?



Student: You'd have one that would just go up.



Professor Michael McBride: Ah, it could just go up and then come back to the baseline -- it has to come back to the baseline -- but it could just go up rather than curving so much in the middle. Okay, so we guess a lower one. Could there be a lower energy Ψ? Yes indeed, there could be one at 4.15 -- it's actually .149-something-or-another, to a lot of decimal points. And there it is, exactly what Nick expected. Okay, so you got this. Could there be a still lower one?



Student: Yes, well in between the two.



Professor Michael McBride: Not in between. Can there be one lower than the blue?



Student: [inaudible]



Professor Michael McBride: Can you imagine a way to get it even lower than the blue, an even lower curvature than the blue? Yes, Angela? Pardon me?



Student: What if it curved, like the one --



Professor Michael McBride: Can't hear very well.



Student: If the blue one is going up, then there could be one going down.



Professor Michael McBride: Ah, there could be an upside-down blue one; absolutely true, there could be an upside-down blue one, but it's the same thing, because it's the blue one multiplied by minus one. Remember, you can multiply it by any number and it's still the same thing. Right? So it's exactly the same energy. It's not a different wavefunction, it's just multiplied by a constant. But you're right, that also does the trick, but it turns out to be the same thing. Can you imagine anything lower in energy than this? It's impossible, because if it had less curvature in the middle, it's bound to go to infinity when it got back in the other region. Okay, so that's -- the blue is the lowest one. Now could there be one in between these two?



Student: Yes.



Professor Michael McBride: How? It looks like -- how could you do it? Yes, Kevin?



Student: One peak is above the value and one peak --



Professor Michael McBride: Ah, one peak above and one below. Right? Could there be an energy in between? Yes indeed, there could be that one. Now could there be others in between? What happens as you add curvature? You have these nodes. A node is where the wavefunction is zero. It's not a node out here, where it's almost zero. A node is when it's exactly zero, and it has to be zero because it's going from being positive to being negative; obviously it has to go through zero. Out here, and out there, it's very, very, very small but it's not zero; that's not a node. A node is where it's zero because it's changing sign. Okay, now nodes are related to energy. Why? Notice that the lowest energy one, blue, had no node, the next one, the purple, had one node. The next one, the violet or whatever color it is, had two nodes. Why is there that relationship?



Student: I was noticing inflection points.



Professor Michael McBride: They are inflection points, but why do they get higher in energy when you have more nodes? Sam?



Student: Higher frequency.



Professor Michael McBride: Say it in terms of energy. Some people would say frequency is energy, that's true, but I want you to say it terms we've been talking about. How do you get more nodes? Yeah?



Student: A higher absolute value of curvature.



Professor Michael McBride: Yes, if the curvature gets bigger, you get more nodes, and curvature is the energy, the kinetic energy. Right? So more curvature, more nodes, more energy; or more energy, more curvature, more nodes. So we have zero nodes, one node, two nodes. Those are the lowest three you can get. And then there would be three nodes, four nodes, five nodes, six nodes, as many nodes as you want. Okay, now this is great. You can do all sorts of neat tricks for Monday, and what I want you to do with this program is just to fiddle around with it and make discoveries for yourself. And I'll show you some of the kinds of discoveries that you can do in the next slides here. But I warn you that if you have more dimensions, or more particles, things quickly become impossible. The only thing you can do this particular approach with is one particle and one dimension, because if you have more, then you don't know which curvature to change. You know the curvature -- remember, it's a sum of a bunch of curvatures, and knowing the kinetic energy, the difference between total and potential, doesn't tell you which curvature to change. If there's only one then you know that that's the one you have to do. So it's going to be a heck of a lot of work to do this for more complicated systems. Right? And that's not our purpose now. Our purpose is to understand how quantum mechanics works with this simple system. But the question is, is it going to be worth our effort to learn how it works if we're unable to do anything with it? Or another way of putting it is, "What's the reward, what are we going to get out of this? Is it worth spending some time on?"



Well, here's the reward for finding Ψ. [Laughter] The knowledge of everything. Okay? For example, you know what energies are allowed to a system; you know what the structures can be; you know how it behaves dynamically, how it moves; you know how bonds work -- that's what we're really after, remember? And you know what makes for reactivity. So there's a pretty good pot of gold at the end of this. So fasten your seatbelts and put your nose to the grindstone and all sorts of metaphors, and we'll do this for the next couple of weeks. Okay? First we'll look at allowed energies and structure. Okay now, of course, we already did this, right? No nodes, one node, two nodes. But what were the energies that were allowed? Okay, so here's a whole bunch of them. No nodes, one node, two nodes, three nodes, up to seven nodes or is it seven? Yes seven nodes. Now here are the energies: 4.15, 12.44, 20.74, 29.04, 37.34. Do you see any pattern to the energies? You might read the title if you have trouble seeing it. They're evenly spaced; after the first one they're evenly spaced. How about the first one compared with zero? How high is the first one if you know this spacing?



Students: Half.



Professor Michael McBride: It's half. The first one above zero is half as big as the spacing of the others. Right? So the energy is some constant, and that constant is what? 8.3 roughly, times an integer minus a half. Why do you have integers? Because nodes come in integers; you can't have half a node, right? It's zero nodes, one, two, three, four and so on. Okay? So that's pretty great. We know for springs, or harmonic oscillators, or an atom stuck on a bond, as long as the potential -- it doesn't vibrate so much that the potential deviates from looking like a parabola; at the bottom of the curve it looks like a parabola. So this is the way atoms should behave when they're attached by bonds -- approximately. Okay, now remember that doggerel poem from 1926: "We only wish that we could glean an inkling of what Ψ could mean." What have we used Ψ for so far?



Student: [inaudible]



Professor Michael McBride: We used it -- its not going to infinity -- to find out when we guessed the energy right. It was just a tool for telling us whether we had the right energy. But does Ψ really mean something, besides just being a handy tool for this purpose? And it was later, six months later, in 1926, that it was suggested what the meaning of it was. And it was in this paper by Max Born that the -- Ψ doesn't mean anything, but Ψ² is probability density, or proportional to probability density. If you scale it right -- remember you can scale Ψ by -- multiply it by any number you want. If you scale it right, it is the probability density of the particle. Now here's what he wrote in German, and what it says is: "If one wishes to translate this result into physical terms, only one interpretation is possible, Ψ signifies the probability [of the structure]." But there's a footnote there, because he made a mistake, and he corrected it after the thing was already set in type. So he put a footnote here which says: "A correction added in proof: more careful consideration shows that the probably is proportional to the square of Ψ." Why couldn't it be -- why couldn't the probability be proportional to Ψ? Because Ψ can be negative! You can't have a negative… Well that's one reason. There were other reasons too. I don't know how embarrassed he was by this, but anyhow -- but since nobody else knew what Ψ was at all, I think he could be proud nonetheless, right? But anyhow, it's the square of Ψ, not Ψ itself, that gives probability density.



Now, six months later Albert Einstein wrote a letter to Max Born where he said something you've probably heard: "But an inner voice tells me that this is not the real thing. Theory yields a great deal, but it brings us no nearer to the secret of the Old One. Anyway I am convinced that He does not play dice." Right? So it's not probabilities. There's this fundamental determinism, Einstein thinks. But it turned out that in this respect Einstein was wrong. Right? It is probability.



Now, it's not probability, it's probability density, and we have to think a little bit about this, and we can think of it in -- by analogy, with mass density. Suppose you had a flask and it had mercury and water and oil in it, right? And you're going to ask the question, if the total mask in the flask is one kilogram, what fraction of that, or how much weight, is exactly one centimeter from the bottom? How many grams, or what fraction of the total do you think, would be exactly one centimeter from the bottom? Anybody got a guess? 5%, 1%, ½%, 10%?



Student: 12%.



Professor Michael McBride: 12%?



Student: Three.



Professor Michael McBride: There'll be -- this'll be an auction, right? Do I hear higher than 12%? Do I hear lower than 12%? Who's bidding? Yes, what do you say?



Student: Zero.



Professor Michael McBride: Zero?



Student: It's an exact value.



Professor Michael McBride: Why zero? Well obviously it has to do -- different heights would have different densities, right? So it'd be a lot more in mercury than it would if I drew the line in oil or in water, right? But you say zero. Why?



Student: Because it's an exact value. It won't have --



Professor Michael McBride: Right. A plane, or a line, or a dot doesn't have any volume. Right? You need a volume -- it's zero, you're right, you need a volume. You have to multiply density, grams per cubic centimeter by volume, how many cubic centimeters, in order to get mass. And the same is true of -- so that in this case the real question would be how much is between one centimeter and 1.01 centimeters, or something like that? You could ask that question, but you can't say how much is exactly one centimeter? So you have to multiply by volume, multiply density by volume, to get the amount. Right? And so Ψ² is probability density. It's how probable it is to be in a small volume, a unit volume, around the point in question. So you go to some point in question, square Ψ, find the density, multiply it by the volume over which Ψ is constant -- because it doesn't vary real rapidly, so you could choose a really small one -- multiply, and that's the probability of being in that little block. Right?



Now, remember we said you could multiply, you could scale Ψ by any constant you wanted to, even minus one or -Ψ or whatever. Okay? Is there a good value, a preferred value to scale it by? You could scale it by such a number that when it's squared and you've multiplied by volume over every place, sum up all these little boxes, you get one. Why do you want to get one? Because the total probability is one that it's someplace, right? So that process is called normalization. We don't usually worry about that, right? We usually just take whatever numbers we want, that's convenient to plot on the graph, and know that we have to scale it such that if we want to calculate actual probability density that's going to sum to one. Okay, so here's an example where above we show two of these curves and below we show squares of those, not properly scaled together. There's, I think, a little more area in the red than in the blue, although maybe I'm bad. They should have the same area, if the thing were normalized. Okay, so that shows what it is. Now there's an interesting thing about that. If you have a pendulum, or a thing on a spring, and it's going back and forth, where does it spend most of its time?



Student: Extremes.



Professor Michael McBride: At the extremes, because it's going so slow there. Right? In the blue one where does it spend most of its time? Where does it spend most of its time? Where is it most dense if it's the blue wavefunction, the lowest wavefunction, the zero-node wavefunction? Obviously it spends its time in the middle, not at the extremes. But notice if you get to the seventh wavefunction; one, two, three, four, five, six, seven; the seventh wavefunction is going to have seven humps -- that then it's most probable at the extremes, and as it gets -- as you get higher and higher it gets more and more focused at the extremes, and that's why when you get to something that's really heavy it behaves the way you expect a pendulum or a thing on a spring to behave. Okay? This is not normalized. But notice something else that's really valuable here, really crucial, which is that this line is where the potential and the total are the same for the red curve. Right? So here is zero kinetic energy. In here the kinetic energy is positive, out here it's negative, and the thing can be out here. Right? So the thing exists in a region of negative kinetic energy. There's nothing special about that. Okay? It's just that that's the way things work. Kinetic energy can be negative. And in the case of the blue one, here was the point where the kinetic energy was zero, and all this out here is probability in a region of negative kinetic energy.



Now, what if we do it with a Morse potential? Notice if we look at the first one with a Morse potential, it looks the same as it would in a Hooke's-law harmonic oscillator. Does that surprise you? No, because the potential looks that way, pretty much. Right? But as you go higher and higher and get here to whatever that is, the twelfth or something like that, you can see that it's very different and it's spending a lot of time far out. It's shifting its average to the right. But that doesn't surprise you because the well moves out to the right, as you move up, in the Morse potential, as you move to higher energy, because low kinetic energy, which is what you have as you move to the right, means low curvature. Why does that make it high probability? Because as long as you have high curvature-over-wavefunction, whenever the wavefunction peeks up, it gets pulled down again; it can't stay up. But if you have very low curvature toward the baseline, very low kinetic energy, then it can go up and you don't mind, because the kinetic -- you don't need much curvature when the wavefunction is big. Okay, so it's like that, and then out in here it's pretty much like an exponential decay. Does that surprise you? Kate you look like it didn't surprise you.



Student: No it didn't.



Professor Michael McBride: Why not? When do you expect the wavefunction to be an exponential decay, under what condition?



Student: When the potential energy --



Professor Michael McBride: Can't hear very well.



Student: When the potential energy is constant.



Professor Michael McBride: Ah, when the potential energy is constant, and what else to be exponential? The sine wave is also constant.



Student: Right, but it has to be --



Professor Michael McBride: Right? What makes it exponential, rather than a sine wave? Yes?



Student: Negative kinetic energy.



Professor Michael McBride: Ah, the kinetic energy is negative, the potential's higher than the total. But that's exactly the case on the right, because there's the potential, there's the total, and the kinetic energy is going down; it's negative, the difference between the two. Okay? Now, here are a whole bunch of these for the harmonic -- for the Morse potential. How does it look different from Hooke's law? Does the spacing of energies look the same as it did in Hooke's law? Again, it's zero nodes, one, two, three, four and so on. But how is it different? Yes?



Student: [inaudible]



Professor Michael McBride: Ah, the spacing gets closer as you go up. So real bonds, which have a potential that looks more like this than like a spring, will have their energy levels get closer as you get more and more vibrational energy. Okay, they're not evenly spaced as they are for Hooke's law. As the energy increases, along with the number of nodes, the well widens more than it would for Hooke's-law parabola. So the wavelengths become longer, those nodes are spaced over a bigger distance, and the energies are lower than expected for Hooke's law, because you don't have as much curvature. You have the same number of nodes -- like the eighteenth level has the same number of nodes as it would in Hooke's law, but they're spaced over a wider distance, so there's less curvature, so the energy doesn't go up as much. Okay now, this is interesting. Look if you have the blue energy. Right? Now the wavefunction looks like this. It's no longer bounded. Does it surprise you that the wavefunction goes on out to infinity when the total energy is higher than that potential energy at the right? No. It means you have enough energy to break the bond. So the particle could just keep going. Okay, now what does this look like? In this region it looks like it does in a Hooke's law or something like that, more or less. What does it look like out at the right? It's a sine wave. Why? Sam, does that surprise you that it looks like a sine wave?



Student: Because the potential is just --



Professor Michael McBride: Ah, the potential is constant, effectively constant out there, and lower than the total. So it's a constant. It's about a sine wave because you have a constant positive kinetic energy when the total energy is above that dissociation limit. So there's the potential, there's the total, and now the kinetic energy is positive and about constant, so it's approximately a sine wave. Okay, now Coulombic potential is a more complicated one. The reason it's complicated is the energy goes to infinity when r becomes zero; when the positive and the negative are on top of one another it becomes infinitely favorable. So it's hard to do this numerically, and the program will screw up sometimes when you try to do Coulombic potential. And it's one-dimensional, and it turns out, curiously enough, that the system is simpler in three dimensions than in one dimension, for a Coulombic potential. Don't worry about the technical aspects of that. But at any rate, this is using the program for that, and you can see that the lowest -- that you have very, very high curvature when you get near zero because you have such high kinetic energy when the particles get very close to one another, and because the Erwin program is approximate, taking finite steps, you get in trouble often. Okay, but anyhow you get the idea that again as you get higher energy, you get waves that spread out. And they spread out a lot for the same reason they did in Morse potential, although here we're using a much lighter mass, we're using an electron. Right? But it spreads out much more because the well spreads out so much. Okay? So higher levels spread way out, and now the sequence of energies -- here's the lowest, the next and the next. You have a big spacing first and then it gets much, much smaller. It turns out that the energy is some constant divided by n². So the number one level has an energy of -k, the next one is -¼k. The blue line is only a fourth as far down as the green, from zero, which is up at the top. Well zero is that grey line at the top, because it keeps going out to zero far out there. And the next one is 1/9th of the way from the top to the lowest level.



So the reward for finding Ψ was "the knowledge of everything." We've seen the allowed energy and structure. Now we're going to see another case where we get structure and we also see something about dynamics that's interesting, about the motion of the atoms. So here's -- suppose this is a proton attached by a bond to something that's heavy. So the heavy thing doesn't move, the hydrogen does the moving. Okay? And here's its wavefunction. This is the bond distance, the minimum energy. If you stretch the bond the energy goes up; if you compress the bond the energy goes up. And that's the wavefunction for the position of the hydrogen atom. Right? Now suppose we increase the mass and instead of talking about a hydrogen atom we talked about a carbon atom which had say fourteen, depending on which isotope we're using -- suppose we went to fourteen. How is that going to change things, if we have mass -- can we use the same function, if we use mass of fourteen? Notice that we're using the same potential. So the kinetic energy at any point is going to be the same for the two. How does mass enter into kinetic energy? Yes?



Student: Would it be more concentrated?



Professor Michael McBride: Can't hear.



Student: Would it be more concentrated?



Professor Michael McBride: Why do you say it would be more concentrated? What would make it more concentrated?



Student: Higher nuclear charge, so the electrons would want to--;



Professor Michael McBride: Yes, but you're not thinking about quantum mechanics. You're thinking that the higher -- well no, wait a second, we're not talking nuclear charge here. All we're talking is mass. It's held by a spring, we're thinking, okay? So electrostatics doesn't come into it, it's a spring that's stretching. Elizabeth?



Student: In that equation that you showed us before, the curvature is proportional to mass?



Professor Michael McBride: Yes, the kinetic energy has one over mass in it. So if you increase the mass, and keep the same wave, what are you going to do to the energy?



Student: Decrease.



Professor Michael McBride: You're going to decrease the energy. But it's the same energy as it was for hydrogen. You can't just decrease the energy, at a given position, if you know the total. Okay, so you want it to be -- to get the same energy it has to be more curved. Okay, everybody with me? To have the same energy, if you have a bigger mass in the denominator, you're going to have more curvature, fourteen times as much curvature, to compensate for the bigger mass. So it's going to curve more toward the baseline. Okay? So if we used that same energy and had the higher mass, so we got more curvature for the same energy, then it's going to curve a lot more, it's going to go through and go up to infinity, and it's even higher energy than the second wave would be. It's curved way too much. So how would you find the lowest energy, what would you do in your guessing? If you wanted to find the lowest energy what would you do? You'd lower your guess to make it less curved toward the baseline. Okay? And if you do that, so lower the energy, from that, and there we've got the answer. Right? So hydrogen has a much higher vibrational energy than carbon does for the same bond, to get the lowest energy. I think this will be clear to you when you have time to think about it. If not, ask questions. Okay? Now what if you had U-235 or a marble at the end of the same spring? What do you think its wavefunction would look like, the lowest energy wavefunction?



Student: A spike.



Professor Michael McBride: It's just a spike. The heavier it is, the narrower it gets. Right? So if you have something like a baseball held by a bond, it's going to sit right there. You're not going to be able to see that it's spread out. But electrons are so much lighter than nuclei that electrons really spread out. Okay, now so there are the probability distributions for hydrogen and for carbon, held by a single bond. Why is their average not in the same position? See what I mean? Elizabeth? Pardon?



Student: Acceleration has to do with mass.



Professor Michael McBride: No, no. It has to do -- yes?



Student: Because the electrons are not attracted equally.



Professor Michael McBride: No, no, it's -- remember, electrons aren't entering into this. We're just supposing that we have a nucleus held by a spring and that spring is the strength of a single-bond. Okay? What is the potential for a bond? Is it Hooke's law? What's a better potential for a bond than Hooke's law?



Students: Morse potential.



Professor Michael McBride: The Morse potential, which broadens out, remember, as it goes up. So since hydrogen goes up more, it samples further out to the right, so it's displaced to the right from the carbon. Okay? Now so let's -- here they are at their minimum energy position, and they vibrate. The higher energy is shifted to the right in the unsymmetrical Morse potential. And here's the half-maximum of the -- so it spends -- it's half as probable -- the probability density here is half what it is there. So when you say, how far does it spread out? This thing goes out to infinity. But we can talk about how wide it is by saying how wide it is at half-height. So the hydrogen at half-height is from here to here, but the carbon at half-height is only from there to there; that is, if you think of these things vibrating, the hydrogen has a bigger amplitude than carbon, because it's lighter. So the hydrogen vibrates by 0.1 angstrom, 9% of the distance. The carbon vibrates only half as much, 3% of the carbon-X distance, just because of the mass difference. And remember we said before, last time, that typically the atoms are vibrating by about 0.05 angstroms in the crystal. That's where this came from, it's this quantum mechanical thing. That's the least that they can -- and when they're in their lowest possible energy state they vibrate by that much, right? And that's how much it is, that little yellow dot after it shrank. So we've seen allowed energies and structure. We've seen dynamics, the vibration. Now, next time, we're going to look at dynamics, and also the payoff, bonding, how we understand bonding in one dimension.



[end of transcript]

After showing how a double-minimum potential generates one-dimensional bonding, Professor McBride moves on to multi-dimensional wave functions. Solving Schrödinger's three-dimensional differential equation might have been daunting, but it was not, because the necessary formulas had been worked out more than a century earlier in connection with acoustics. Acoustical "Chladni" figures show how nodal patterns relate to frequencies. The analogy is pursued by studying the form of wave functions for "hydrogen-like" one-electron atoms. Removing normalizing constants from the formulas for familiar orbitals reveals the underlying simplicity of their shapes.

 

Transcript



September 22, 2008



Professor Michael McBride: Okay, so we have the promise of the knowledge of everything, at least everything within a certain category, and we saw how we could see quantized energy last time, and structure, and structure-related dynamics, how nuclei vibrate and how it's different for heavier atoms. Today we're going to look again at a different kind of dynamics, and also at bonding. So in the one-dimensional world this is the payoff. We have to go to the three-dimensional world.



Okay, so you did problem sets for today and you're now familiar with things like this. What do you know about the blue curve and the red curve here? Anybody got a suggestion of what to do, or what they tell you about what the true solution is? Yes Angela?



Student: It looks like the true solution is somewhere between them.



Professor Michael McBride: I can't hear very well.



Student: It looks like the true solution will be somewhere between them, because it looks like --



Professor Michael McBride: Yes, the true solution should be between them. The red one is too hot and the blue one is too cold. Right? So there'll be, in between them, there'll be one that'll just come back down to the baseline for Hooke's law. Now the problem is that this isn't Hooke's law, because I didn't show you the whole picture. It's in fact a double minimum. Right? And those are both true solutions. So the correct single minimum energy would lie between those two, but for the double minimum there's one, there's a solution that's lower in energy and a solution that's higher in energy. Now what would happen if we moved the wells further apart? Here you see the separation of the two wells as 0.6 angstroms. Suppose we moved it apart to be 1.3 angstroms. Right? Now we have blue and red solutions. But now you notice the blue solution (or the red solution, they're the same in the left), they're both the same as the single minimum. So if you get the wells far enough apart they look like single minima with respect to their wavefunction. Okay?



Now, closer wells. If we move the wells closer together, we get a lower minimum energy than for the single well solution. Okay? And we get a higher energy for the next higher one. But they both look like the single minimum -- but they both started when the wells were far apart, looking like the single minimum solution, one with no nodes, one with one node. When the wells were far apart they had the same energy as the single minimum, but when the wells come together, the one with no nodes has less curvature, less kinetic energy and is lower energy than the single minimum, and the other one has one node, more curvature than the single minimum, and a higher energy; that is to say, there's a splitting between the energy of these two. If the wells are far apart, they're the same as the single minimum, both of them, the no-nodes and the one-node solution, are the same as the single minimum, but if you bring them together the no-nodes is lower in energy and the one-node is higher in energy.



So the energies split about the original single minimum structure. Now that's really important because that's what makes bonds. Right? So if we have two atoms or one-dimensional wells far apart, say A and B, and we get a solution in A and a solution in B, and if we want the solution for the whole thing, we just put them together that way, no nodes. Or we could flip B upside down, multiply it by minus one, add them together then and we get one node. But they have the same energy as the single ones would do. Right? But if we bring them together the energy gets lower, less curvature. Right? And that consists, or gives rise, to a stabilization of the particle. The particle is more stable when the wells are closer together. Right? What would you call that?



Student: A bond.



Professor Michael McBride: That's a bond. The energy is lower when they're close together. So that holds A and B together because the energy of the particle is lower when the wells are close together. So that's bonding. And part and parcel of this, the flipside of this, is that you have this wavefunction, which is higher. What will we call that?



Student: Antibonding.



Professor Michael McBride: Antibonding. So you get a combination of bonding and an antibonding combination because the kinetic energy changes when the wells come closer together. You get one that's less curved, one that's more curved. Okay, now that's bonding. How about dynamics? Well, suppose we have this double minimum and suppose we put a particle in and stop it there so it has zero -- that's our zero of energy -- and then we're going to let it go. What will it do if it's classical? It'll roll back and forth, right? Bingo, let it go. Roll across, to the same height, and roll back again, and then roll across again to the same height. With no friction it goes on forever, right? But here's what happens special, in real quantum mechanical systems -- oops, it went too far -- and now it's going to roll back and forth on the right; but every once in awhile it goes across, from one well to the other. Now, this is called tunneling, for obvious reasons. Right? But the word tunneling I hate. It's one of my pet peeves. It's misleading and mischievous, because it suggests there's something weird about the potential energy, that you can tunnel through and have a potential energy lower than you would've guessed from the potential energy curve. But that's not right. What happens is when it's in the middle you have negative kinetic energy. You have the potential energy that the curve shows, but the kinetic energy is negative. The potential energy can be higher than the total. And this happens in every bond and wavefunction, as you saw in your problem set, that the waves always go out into that forbidden region of negative kinetic energy where they curve away from the baseline. So there's the forbidden region, out to the left, to the right, and also in the middle for that energy. Right? And what really happens is that you go to negative kinetic energy and high positive energy to get over that hump, and then you get across to the other side.



Now how often does that happen? It turns out that the time to get from one well to the other, which I won't be able to prove to you because it requires time-dependent quantum mechanics, and we're talking about time-independent quantum mechanics, but let me just tell you that the answer is, if you know the energy difference between the blue and the red, that the rate of getting across is 5*10-14 seconds, divided by whatever that energy is, expressed in kilocalories/mole. Right? This is just an assertion, and it's based on time-dependent quantum mechanics. It's true, but I'm not telling you why. So you won't be satisfied, I hope. Okay? So the energy difference here is 1.4 kilocalories/mole. So how rapidly does a particle then get from -- if it started out in the left, how long do you have to wait, on average, until it's in the right one? You have to wait 5*10-14 divided by 1.4, which means about 4*10-14 seconds it takes to, quote, "tunnel"; although it really doesn't bore through, it goes over with negative kinetic energy. Okay, so there's something else about dynamics. And we're hoping soon to get to reactivity, but that'll come after the exam, after we talk about atoms and molecules.



So now it looks like we're in a pretty powerful position, at least with respect to one dimension, because with this Erwin program we can find satisfactory wavefunctions for any complicated potential. We could make it anything we would like and just follow the recipe, find out allowed energies, shapes of wavefunctions, probability density, rates of tunneling, all this kind of stuff, and we can rank all the wavefunctions by their energy, or their curvature - the number of nodes they have. Here's an unsatisfactory wavefunction. What do you notice that's bad about it, just to show that you've learned something? Yes?



Student: It doesn't start at the baseline.



Professor Michael McBride: It doesn't start at the baseline, on the left. How about on the right, does it look okay on the right?



[Students speak over one another]



Professor Michael McBride: Lucas?



Student: It tends to be going down.



Professor Michael McBride: Ah, it crosses the baseline, in the forbidden region. If it crosses the baseline while curving away, it's bound to go to negative infinity, or positive infinity if it crosses in the other direction. Right? And on the left, because it didn't get to the baseline when it became flat, it's going to keep curving and go up to infinity. So you've internalized a lot of this stuff now. Congratulations on doing well on the problem set. Okay? So that was a bad one. Okay, so it looks like we're in pretty good shape. We even can handle multiple minima and understand tunneling and so on. The unfortunate thing is this curve tracing recipe doesn't work if you have more dimensions. Remember, the reason it would work is if you knew the potential and the total energy, then you knew the kinetic energy, and you could assign it to curvature. Right? But if you have two dimensions, then there are two different curvatures you could assign it to, and you don't know how much to put in each of them. So it's no longer such a clear-cut problem if you have more dimensions, especially 3n dimensions. So when there are many curvatures, it's not clear how to partition the kinetic energy.



But Schrödinger had no trouble. The Schrödinger equation is what's called a differential equation, it involves derivatives. Right? And how many people have had differential equations? Okay, maybe a fifth of you I guess, or fewer. Right? So the rest of you are feeling, "Gee I wish I knew differential equations because then I could do this." But you don't have to, because people already have done it. You don't have to develop this from ground zero, right? And Schrödinger didn't solve the differential equations either. Right? He knew what the solutions were, because people had been studying them forever, and the reason they studied it was because of studying acoustic waves, and light also. And Chladni is the guy who originated this.



And here we have a book by Chladni called Acoustics. I got it out of the library. This is in fact the new, unchanged edition from 1830. The original edition was something like 1805 or something like that; actually we'll see it here I think. There's Chladni, Ernst Florens Friedrich Chladni, and this is the title page of that book, Acoustics; the original one was 1803, ours is 1830. Okay, now what he did was take plates of different -- he also did violin strings and timpani heads and things like that. He was interested in all kinds of acoustics. But the particular experiment we're interested in was he would take a plate and suspend it in the middle and touch it in different places, while he was bowing it with a violin bow to make it vibrate. Okay? So you bow it in different places, and he put sand on the surface of this thing to see what patterns would be formed when the thing vibrated -- this is why violins have the shape they have, so that they'll vibrate in certain ways when you bow to make them vibrate -- and the sand collects where the plate isn't vibrating, because when it's vibrating it shakes it away. So you get a picture of the pattern of vibration.



Now, I'm going to try to show this to you. But it might not work, and just in case I can go to the Web. So what we have here is a -- I can't play a violin, but as you know I can play a bell. In fact, I can play a bell in several ways; not only that, [Bell sound] I can do it this way too. [Bell sound] Now, that's not a very pleasing sound. Why? Why is it -- what's noise? I'm not very good at playing the bell. [Sound] Why does dry ice make a bell ring? This was discovered in London by an ice-cream vendor who had a bell on his cart, and he asked a physics teacher, a woman in London, why his bell rang when the dry-ice from his ice-cream hit it. [Sound] See, when you touch metal like that, it gives heat to the dry ice, which causes it to give off the pressure of gas, but when it gives off the pressure and the metal moves away, then heat's not being transferred, so it stops. So you get rapid pulses of CO2, which make the thing ring. But depending on how you touch it, you get noise, because there are many different patterns of vibration at different frequencies that are being generated. If I could do it just right, I would generate just one pure tone and then you'd say, "What a great player of bells he is."



Okay, now I'm going to try this, and I may fail in the same way, to do an experiment like Chladni's. But instead of using a violin bow -- I'm going to put sand on here -- instead of using a violin bow I'm going to use this piece of dry ice. Okay? And I have to -- if I just touch it, it won't do anything because I'll be touching sand and I won't get heat transfer. So I have to brush a little bit to get a clean part of the brass, and then let's see what happens here. [Sound] Now I'm doing the same thing I did with the bell, right? It's just noise. But let me try someplace else here. This is an empirical science. [Amazed sounds and laughter] Try something else here. Spread some around here. If you're lucky, this is really great. Hear that pure one? Maybe it's not sharp enough. We don't want to spend too much time doing this. [Sound] I want to try one more and then we're going to -- you can go to the website and see a much better once. Once I played just the lost chord, a really beautiful tone; and it's on, you can see it on the Web. [Sound] I'm trying to find where and how hard to touch so that I get a really pure tone. [Sound] Well, you can get patterns that way.



Okay, you can get much prettier patterns and prettier tones than I'm able to coax out of the thing right now, and you can look at it on the Web to see it. [projection adjustments] Okay, so these are some crude Chladni figures that I've made in the past, three rings; two rings with a line through the middle, vertically. That's the one we just made now, I think, isn't it? Isn't it the same number? A ring and three diameters. And there's a circle and four diameters. I actually made that in class last year. Okay, these are ones that Chladni did back in the late eighteenth century, taken from that book, right? And here, the first ones are just diameters, and the next ones are rings, together with a certain number of diameters, or two or more rings. Now let's see what that picture means. Below you see a vertical diameter and two rings. On top you see color coded as to whether the plane, at an instant of vibration, whether it's toward you or away from you, as it's vibrating; that's the pattern. Because those lines don't move, but either side of the line, one side moves up, the other side moves down. Okay? Or you could look at it this way. The dotted patterns and the circles and the line don't move as the rest of the thing deforms. This would be like a drum head on a timpani. Got the idea? So that's the pattern of motion.



Now Chladni was interested in sound, right? So he found what pitch each of these different patterns corresponded to. So this table shows the number of diameters that are nodes, the number of circular nodes, and what pitch it corresponds to, and the little lines may mean it's a little sharp or a little flat, something like that. For example, the one with two diameters and no rings is the pitch C. Got the idea? So then he tried to figure out if there's a mathematical relationship among these things. These were forty-seven patterns that he observed there. So here he says: "The pitch relationships agree approximately with the squares of the following numbers." So how many diameters, how many circles, and what he sees is the frequency is roughly the number of diameter nodes, plus twice the number of circular nodes, squared. This is his empirical observation. So, for example, you could have two diameters or one circle, both would give the number two, which you then square, to get something proportional to the frequency. Right? So there are two different ways to get the same pitch -- that's what's crucial -- either rings or lines. Or you can have a combination of rings and lines. So for higher frequencies you can get more and more combinations that give the same pitch. For example, the number eight, you can have four circles, three circles and two lines; two circles and four lines; a circle and six lines; or eight lines. All of them give the number eight and all give the same pitch, at least approximately. Okay? So this is what he -- but the lesson we want to take is that there are different ways of getting the same frequency, by combinations of these nodes.



Okay, now Chladni didn't solve his problem mathematically about how plates or strings or whatever vibrate. But there were great mathematicians working on this, like Bernoulli, Lagrange, Euler, who not only made a Communist Germany stamp, he also made a Swiss ten-franc note. Okay? Okay, so the Ψs for one-electron atoms -- now we're going from one dimension into three dimensions, for a real atom, electron and an atom -- so for one-electron atom there's going to be three variables, x, y, z for the electron, and the solutions, the waves we're going to get, involve what are called spherical harmonics, and they're 3D analogues of Chladni's 2D figures. These mathematicians could work in three dimensions as well as two. Okay, so a 3-dimensional H-atom wavefunction, Ψ, can be written as a product of three functions, an R function, a Θ function, and a Φ function. And these were available from other old-time mathematicians. The R function is called the Associated Laguerre Function, and it's named for Edmond Laguerre. And the Θ function is called the normalized Associated Legendre Polynomial, after Adrien-Marie Legendre. Okay? Now here are the solutions. Schrödinger didn't find these, he just looked them up. These guys had already done it, from acoustics, for 3-dimensional things vibrating. Right? So here's a table that you can use, the same way Schrödinger would've used it, to find out what wavefunctions for electrons in one-electron atoms actually look like. In all your books you've seen pictures of these things but you've never seen -- my guess is -- you've never seen the real thing, and you should wonder, "Are these pictures that people have shown me right?"



[projection adjustments]



Now how do you understand? This looks pretty complicated, this table, and in a sense it is. But if you look at it the right way it's really pretty simple. It relates to the position of an electron, shown on the right there, relative to the nucleus. And there's only one electron; it's a one-electron atom, otherwise we have more dimensions. But the nucleus can have any charge you want it to. So it's a one-electron atom of any nuclear charge, and we have the coordinates, x, y, z, and we have a potential. What's the potential law that's given to us, whose law?



Student: Coulomb's.



Professor Michael McBride: Coulomb's law. So it's 1/r2. And how do you write r2?; r2 is x2 + y2 + z2. Right, everybody know that? Everybody know that? Good, nod, yes; good, okay we're on the same page here. Okay, but that's a pretty complicated function, right? So the equations that involve that are going to be really, really complicated. But there's a clever way to get around this. Do you know what that way is? What's the clever way of getting around this? The clever way of getting around it is to change the coordinates of the system that you're using to one that's more natural for the problem. And the one that's natural for the problem is spherical polar coordinates. So the three dimensions are r, the distance, and θ, the angle down from the z axis, and φ, the angle rotated around from the x axis.



Student: Isn't it the other way around?



Professor Michael McBride: Well you can define it any way you want to. This is the way it's defined for these purposes. Okay? Now, so that simplifies the expression for potential enormously. Instead of being 1/√x2 + y2/c2, it's 1/r, or some constant over r. Okay? And what that means is -- and this is the mathematical payoff -- that the wavefunction can be written as a product of three functions, each of which is the function of only one variable. Instead of having x2 and y2 z2 all mixed together in this complicated function, you have a function only of r, a function only of θ, and a function only of φ, and you multiply them together and you get the solution to the Schrödinger equation for a one-electron atom. Now, that -- this table -- gives you those functions, the R function, the Θ function, and the Φ function. It's like a restaurant where you get to choose one from column A, one from column B, one from column C, right? Here's your appetizer, your main course and your dessert, right? So you choose one of these, one of these, one of these, multiply them together, and you have the true function. Right?



Now they look very, very complicated. You can't choose any combination you want to. Once you've chosen one -- once you've chosen your appetizer -- you can only have certain main courses, and once you've chosen that, you can only have certain desserts, to get solutions. Right? This is what these mathematicians had all figured out. Okay? And so how do we go about doing it? Well we can name the Ψ we're going to get by a nickname, like 1s or 2px or something like that, but you can also name it by the numbers that name these individual functions, and those are n, l and m, which I think you've encountered before probably. So let's make a try at it. No, actually notice something about the complicated form of the functions. First, every one of them has Z/a0 to the 3/2 power. Z is the nuclear charge; a0 is a unit of length of about half an angstrom about. Okay, so to the 3/2 power; that's pretty weird. But remember what you're going to do with the wavefunction? What do you do with wavefunctions, other than find energy?



Students: Probability.



Professor Michael McBride: Get probability density, and what do you do to do that Sam?



Student: Square it.



Professor Michael McBride: You square it. So when you square it, it's going to be (Z/a0)3, when you square it. Everybody with me on that? Okay, so what that means is you're going to get units that are Z3 per unit volume; because remember a0 is a distance. So a03 is a volume. So it's going to get a number -- the nuclear charge is a number Z, the atomic number is a number -- so it's going to get a number per unit volume, which is the right units to have for probability density. What is it per unit volume? Okay? So this is just the bit that scales it right so that you get the right density and the right units. So you can drop that out, that's not something very fundamentally interesting. Okay, let's try making up a 1s orbital. Where do you start? Can somebody tell me where to choose? It's not hard. They're labeled with red, right? If you want 1s, you choose the top one. Okay? And now if you've done that, n is one and l is zero. So when you come to the next one, l has to be zero. Now m can be either zero or one; or no, actually wait a second, if it's 1 -- no, if l is zero, then m has to be zero, it says in the second column. Does everybody see that? See up at the top? It says if l is zero, then m can be zero; lm. Okay? For m, pardon me, to be non-zero, like here, you have to have an l that's greater than zero. Okay? So to get a solution, if you chose n one and l had to be zero -- because remember that's the only choice here, you can't have a higher l -- then m is also going to be zero. So you've got to take this times this times this. Okay? Now how complicated are those functions? Okay, √2/2, that's just a constant, that's a nothing function. Or 1/√(2π), big deal. Why do you have these constants in there? Why do you care what constant you multiply the wavefunction by?



Student: No idea.



Professor Michael McBride: For what purpose?



Student: Normalization.



Professor Michael McBride: For normalization, right? If you want to get -- but if all you want is the shape, forget constants. So what's the real working part of the 1s wavefunction? What's the part that varies? We have here the (Z/a0)3/2; forget that; two, forget that, √2/2 forget that, in fact the 2s would cancel; √2 here would cancel that √2; so it's 1/√π times this, times what?; e^(-ρ/2). That's the real working part of it. Right? When you square it, what are you going to get, for e ^(-ρ/2)? What's e^(-ρ/2) squared? Yes Alex?



Student: e-ρ.



Professor Michael McBride: e-ρ. Okay, so the probability density is going to be e-ρ, times a constant. Okay, so here we got it. It's a constant times e^(-ρ/2); when we square it, it's e-ρ. Now, there's something -- there are two things that are interesting here. One, we wanted to get a function of r, θ and φ, but we don't have r in it. Right? We have ρ instead. Why do we substitute r by ρ? There's a good reason for that. ρ is defined -- which is, indeed, the Greek version of r is ρ -- it's defined as r times a constant. The constant is twice the atomic number, divided by n, the quantum level n, times that distance a0. And why do you do it that way? It's because if you do it that way, then the same ρ works, no matter what the nuclear charge is, and no matter what n is. So you get the same -- so it makes the table very compact, because you use the same thing no matter what atom you're dealing with, if you work in units of ρ rather than r; that is scale r, to take into account the fact that you can have various nuclear charges. Okay, allows using the same e ^(-ρ/2) for any nuclear charge (Z) and any n. And notice, every one of these has e^(-ρ/2) on it. Now, why does that make sense? Why is it no surprise that these things have e-r? Have we ever seen that before, e-x as a wavefunction? What kind of wavefunction -- what does that mean when you have e-r?



[Students speak over one another]



Professor Michael McBride: That's a constant negative kinetic energy, constant negative kinetic energy, which is going to be the situation for any nucleus and an electron. Once the electron gets pretty far away, the energy stops changing. Right? So the potential energy is constant, the kinetic energy is constant. Right? If the electron is bound to the nucleus, that means it can't just fly off to infinity, right? So it's below the ultimate potential energy. Then you have constant negative kinetic energy, when you're far from the nucleus, right? So e-ρ . So it doesn't surprise you that every one of these has e-ρ in it.



Okay, now let's just look at this scaling quickly. Okay, so here's e-ρ and ρ -- as a function of ρ. Okay? Now suppose -- so we can rearrange this. r is that. If we wanted to make a new plot, which has the horizontal axis being r, being distance instead of ρ. Okay now, so r for a hydrogen atom, the 1s orbital of a hydrogen atom, is 0.53 -- so n is one, a0 is half an angstrom, Z is one; so it's .53/2 times ρ. So we could just put a new scale on there. So here's half an angstrom for a hydrogen, here's one angstrom for a hydrogen. Everybody with me on this so far, how I did that? Right? I just went through and found out for any given ρ here, that I already had, one, two, three, four, five, e-ρ. For any give ρ, what would the r be if I were talking about a hydrogen atom? Okay?



And I just put a new scale on there. Now if I were talking about a carbon atom, which has a nuclear charge of plus six, then this number Z is going to be much bigger, it's going to be six times as big. What effect will that have on the scale? So suppose we're at five, for ρ. Instead of dividing by one, for Z here, I'm going to be dividing by six. So I'm going to get much shorter distances, real distances, rather than ρ. Right? Does that surprise you -- that the function is going to squash in if I have a plus six nuclear charge? Does it surprise you or not? It's what you expect. A bigger charge is going to suck the electrons in more. Okay? So if I do it for carbon, I get .53/12 instead of .53/2, and the scale is going to look like that. Instead of this being 0.5 here for hydrogen, it's 0.1, right? So then if I squash it in so that they're on the same scale -- those are different angstrom scales for carbon and hydrogen -- if I make them the same scale, the bigger nuclear charge sucks the 1s function in by a factor of six, and it looks like that. But now, of course, if I want the real, the normalized function, it's got to be higher, so that the total area is the same. I have the squared function here, e-ρ; remember that the wavefunction was e ^(-ρ/2); the density is e-ρ.



To get the same thing I'm going to have to multiply that by six. So it's going to look like that. So that's how the radial distribution, probability density distribution, looks different, for a one-electron atom with a plus one or a plus six nuclear charge. Carbon holds its core electrons in much more tightly than hydrogen does; no big surprise there. Okay. So here's something to think about. How would it differ if instead of talking about the 1s orbital of carbon, I was talking about the 2s orbital of carbon? Right? Now, ρ is going to be different, and here, instead of one, it's going to be twp. So it's going to change the scale by a factor of two. So the 2s are further out. Okay, so for Wednesday I want you to do these problems -- you can do them in groups if you want to -- some about Chladni Figures, and then some things about energy and some atomic orbital problems. Okay?



Now we already looked at the 1s. Let's look at some other atomic orbitals. Incidentally, this function, the Coulombic function, is simpler in three dimensions than it is in one dimension. It was that complicated thing in one-dimension, remember that had the cusp on it. It was a really complicated function, but it's really simple in three dimensions, it's just exponential in r. How does it vary with θ? As you come down from the axis, how does the wavefunction vary, the 1s? What does it say? How does it depend on θ?



Student: It doesn't.



Professor Michael McBride: It doesn't. How does it depend on φ?



Student: It doesn't.



Professor Michael McBride: It doesn't. So what shape does it have?



Student: A sphere.



Professor Michael McBride: It's spherically symmetric, it depends only on r. And the dependence on r is as easy as pie, it's just e-r (or ρ -- it depends on what units you measure r in). Okay, now let's look at 2s. Now tell me what to do. How do I write a 2s function from this table? First I have to choose the appetizer. Where do I go?



Student: Second one.



Professor Michael McBride: Second one down is 2s. Okay? So I take this one, and now 2s, that means l is zero. So where do I go; what direction do I go? Here. And then what direction do I go?



Student: Straight across.



Professor Michael McBride: Straight across, because m is zero. Okay. So here's -- there's 2s; multiply those together. Now what's interesting, there are a bunch of constants that are going to give me -- make it normalized. So that's a constant, that's a constant, just like they were before; this is a constant, that's a constant. But this part is interesting: (2-ρ) times e^(-ρ/2). That's the function that we're interested in, (2-ρ) times e ^(-ρ/2). That has an interesting thing. What happens when ρ has the value two? It has the e ^(-ρ/2); that's just decaying, all of them have that. But how about the 2-ρ? That gives an interesting thing. What's its value when ρ is two?



Student: Zero.



Professor Michael McBride: Zero. So there's a node. What shape is the node? Is it a point, is it a line, is it a wiggly thing?



Student: A sphere.



Professor Michael McBride: It's a sphere, it's spherically symmetric again. So there's a spherical node. So inside -- and at a node, the wavefunction changes sign. So inside it's one sign, outside it's the opposite sign. Okay? Now how about a 2pz orbital? So where are we going to go? 2p, we'll start with this, and for pz, we do this one, and for z we do this one. So it's this times this times this. I think I have that circled here; let's see, yeah. Is this one interesting?



Student: No.



Professor Michael McBride: No. Is this one interesting?



Student: Yes.



Professor Michael McBride: What part of it is interesting?



Student: Cosine.



Professor Michael McBride: cos(θ) is interesting, right? That's real variation. And what part's interesting here? It's ρ times e^(-ρ/2). Okay, so it's some constant times ρ times the cosine of θ times e^(-ρ/2). You always have the e ^(-ρ/2); forget that. It just means it decays as you go out. Now ρ times cos(θ). What does that mean?; ρ, which is this distance, times the cos(θ), what is that? Russell? It's z. So I can simplify this a lot, if I mix my metaphors between polar and Cartesian coordinates, it's z times e^(-ρ/2). That's the 2pz orbital. Now, can you guess what the 2px and the 2py look like? Can somebody guess the 2py? Josh?



Student: You replace x with --



Professor Michael McBride: Replace what?



Student: Replace z with x.



Professor Michael McBride: Replace z with x, you get the 2px orbital; replace z with y, you get the 2py orbital. Pretty straightforward functions. Now how about -- you've seen pictures of p orbitals. Now that you know what the functions look like, it's interesting to go back to the pictures you've seen and see what they mean. How do they relate to the actual formula? So you could plot -- remember, it's three parts, an R function, a Θ function, a Φ function. We could look at each of those functions separately, if we wanted to. Okay? We've been talking about the R function, how it would behave. But how would the Θ function behave in this, cosine θ? So we'll do a polar plot that shows the value of cosine θ as a function of θ. Everybody with me? Okay, so what is it when θ is zero; what's cosine of θ?



Student: One.



Professor Michael McBride: One. Okay, so there it is, and we'll put a point there, okay? Now let's go to + and -30°. Cosine is 0.86, and we'll put points there. And let's go to 45°, it's the √½, 0.71, put points there. At 60°, it's ½; put points there. What is it if it's 90°? If it's 90°, the cosine is zero. So we got a point at the origin. Right? So you can see what the shape is. What if you go the other direction, what if you go beyond 90°, what happens to the cosine?



Student: It's negative.



Professor Michael McBride: Changes sign. So you've got another circle that's negative on the other side. So there's a p orbital. But I'm not showing the whole p orbital, I'm just showing how the angular part varies. So you've seen pictures like that of p orbitals, which is the Θ function. Or I could square it, in order to look at probability density, as a function of θ, and it would look like that. So you've seen pictures more or less like that. Or I could make a picture like that, which is a contour plot showing how big it is, on a slice, through the nucleus. Right? So it's ρ times e ^(-ρ/2) times cos(θ); that multiplied by a constant is the wavefunction. Okay, so you can see it's positive up at the top, negative at the bottom. And let's look at it a little bit. And, notice it doesn't depend on φ, the angle of rotation around here. So we can take that picture and rotate it to give a three-dimensional dumbbell. Okay? We're just looking at one slice here, in order to be explicit. Okay, now so there's a certain point there that has a certain ρ and a certain θ, and therefore it has a certain value. Now, let's find the maximum, of that function. Okay? So first it depends on cos(θ). What will θ be in order for this to be a maximum?



Student: Zero.



Professor Michael McBride: Zero. So it has to be along the axis. Okay? Now, so θ has to be equal to zero. Okay, now how do we find where it's going to be maximum with respect to ρ? We take the derivative with respect to ρ, set it equal to zero. So that's this, or simplified that, or simplified that; ρ=2. So there's the maximum. And if we put the constants in, we could find the true probability density anywhere; just plug in θ, plug in ρ, and square it, multiply it by the constant. That's the electron density at any given, at the point you're talking about.



Now, or you could have a computer do it for you, and that's where we're indebted to Dean Dauger, who's a physicist. He went to Occidental College and actually wrote this program while he was there, but then he got a Ph.D. at UCLA in Physics, and he wrote this for physicists, but we'll use it. Okay, here's a picture of Dean Dauger. He also juggles. And here's a picture of him at an Apple Developer Conference, and he's juggling, the guy on the left there. He also drops one of the pins, he's not perfect. Okay. So this is his program. He's at the Apple Developer Conference. It works only on an Apple, so if you don't have an Apple, borrow somebody's Apple and look at the program, because it's really fun. This is the screen of the program, and I don't have time to take you through it now, but you should try it. So look at the PowerPoint of this on the next few slides to learn what all the different information is that's being given here, and the kinds of questions you can ask. There's a 1s orbital, right? And it's as if the thing were really -- the electron were really colored, and it's a picture of how it would actually look if you could see electron density. But all, of course, he's done is used fancy computer graphics to plot the square of the wavefunction. Right? But it's a really nice program. We'll talk about this a little more next time.



[end of transcript]

In discussions of the Schrödinger equation thus far, the systems described were either one-dimensional or involved a single electron. After discussing how increased nuclear charge affects the energies of one-electron atoms and then discussing hybridization, this lecture finally addresses the simple fact that multi-electron systems cannot be properly described in terms of one-electron orbitals.

Transcript



September 24, 2008



Professor Michael McBride: So back to business. We're in a really great position here because we've gone from one dimension to three dimensions, to get something real, a real atom. And if you take that table from last time, that we got from the old-time mathematicians, and plug in according to what you can use together, you get the exact wave function that's a solution of the Schrödinger equation for the hydrogen atom, or any atom that has only one electron. It can have any nuclear charge you want, right? So it's the real thing. People have shown you pictures, which are some kinds of approximations of this, but you have the equation, the actual formula, for the true wave function, and no one has a better one than you have. That's the real McCoy. Okay?



And you can check these pictures that people have drawn to see if they're realistic graphs, or whether they're just some kind of artist's impression, or somebody who doesn't know anything's impression; maybe he's not even an artist. Right? So you can actually do these. And many of them are very simple functions, like e-r, right? You can hardly get easier than that. Okay, but you can also draw pictures that look just like what you could see if the electron had a color to it. They're done by ray tracing where the computer pretends it's your eye and sees how much density it would go through, if it went through this point or that point or that point, along a certain ray from your eye, and puts a color here that's proportional to how much electron density it sees. It's not the wave function; it's the square of the wave function, the electron density, that Atom-in-a-Box plots. Okay? So that's what it would really look like, if you could see it and if there weren't problems with wavelength and photons knocking electrons to kingdom come and so on.



That's what it would look like, right? But it'll be colored; it'll be colored according to the sign the wave function has. It's not the wave function; it's the square of the wave function. So it would be positive. But it's colored to say the wave function in this region would be positive or in this region negative. And I'll show it to you in just a second, but first let's -- or a few minutes -- but first let's describe it. So there's lots of information on the screen that you get with Atom-in-a-Box. How many of you have tried it? Okay, all of you should. Okay? Okay, first you have the Schrödinger equation. That always appears and it's sort of silly but if you need the Schrödinger equation for reference, there it is. Then up at the top is the particular function that's being plotted, and it's given as n, l and m, and also the nickname 1s, for this particular one, which is 1,0,0 for n, l and m. And then there's the formula that that particular wave function has, and you know what it is, it's a bunch of constants, which we can forget, but it's e-r. He's not using ρ here, notice. He uses actual r and puts in explicitly the Z and the ao, and the two is cancelled. Okay? And then he says what the energy would be, in units of electron volts. An electron volt is 23.06 times as big as a kilocalorie, which is what we usually talk in. Now, you can look at these pictures and ask questions, some fairly simple questions, like where is the electron density highest? The nucleus is in the center of this picture, not surprisingly. So this is not a hard question. Where's the electron density highest?



[Students speak over one another]



Professor Michael McBride: In the middle, right? No big surprise there. Okay, how do you know? You take the wave function, you square it and find out where it has a maximum, and e-distance has a maximum when the distance is zero; the value is one. It falls off from there. Okay, so you just, to find where the density is highest, you just square the wave function and maximize it. Okay, now but here's a slightly different question: What is the most likely distance? Can you see why that's a different question than where -- you might think the most likely distance is wherever the density is highest, but that's not quite true. Do you see why?



Student: [Inaudible]



Professor Michael McBride: Because what's most-- yes?



Student: It could be moving very quickly in the area.



Professor Michael McBride: No, it doesn't have to do with motion. Yes Lucas?



Student: There's almost no volume in the center.



Professor Michael McBride: There's almost no volume in the center. If you take--say if you talk about a point, then there's no volume at all. Okay? But take a little box and put it in the center, or actually a sphere of a certain radius, a small radius, 0.0001 Å say, and put it in the center, and see how much density is inside. Okay? But then you could look at one angstrom out, and now you're looking at a little spherical shell that's 0.0001 Å thick. Is everybody clear? So there could be different directions in which you have this electron density, which is at that particular distance from the nucleus. And the further out you go, the more volume this shell has. It's always the same thickness, but as you move out there's more stuff in it. You see what I mean? More volume in it. What's it proportional to? As you increase r, how does the volume of the shell increase? How does the surface of the sphere increase as you go out?



Student: r3.



Professor Michael McBride: No, volume is r3.



Student: r2.



Professor Michael McBride: It goes with r2. The surface goes with r2. Right? So you have to multiply the probability density by r2, to see at what radius you have the most stuff. Is that clear to everyone? Because you want to count everything that's at that radius, not only in that direction but also in that direction, that direction, that direction, and so on, and the sphere gets bigger as you go out. So you'll notice that what's given in this plot here on the bottom right is r2 times the wave function squared. Okay? So that'll tell you what's the most likely distance. Okay? So you have to do surface weighting by r2, and it's not at zero, it's at the unit that we use to measure distance in, a0, which turns out to be half an angstrom, 0.53 Å. Okay, now here's a 2s orbital. Right? Here we'll draw two different shells of the same thickness, right? We'll number them one and two, and we're going to ask the question, which one has higher density? [Technical adjustments of room lighting] Okay, which shell has higher density, one or two?



Student: One.



Professor Michael McBride: One, clearly, okay? Now, here's a different question -- okay, it shows one -- which shell contains more electron density, more stuff? Which one has more probability?



Students: Two.



Professor Michael McBride: It's not obvious, because one has higher density but two is much bigger. Okay? And in fact, the two has about three times the radius of one. Right? So how much more volume does that shell have?



Student: Nine times.



Professor Michael McBride: About nine times more volume; not of the sphere but of the shell. Okay? So it's about nine times the volume. So it turns out that actually two has more stuff, and you see that in this plot. Right? The red for two is bigger than the green for one. Right? So depending on what question you ask, you have to decide whether you have to weight it by r2 to get the answer you need. Okay, now here's another way of looking at a wave function. This happens to be the -- it's actually a 3D wave function; as you see, not from here, but from here, it's 3D. This is the picture in three dimensions, using that ray tracing, the way we talk about. But you can do it, if you click "Slice" down here, then you get a slice through it. Okay? So that's 2-dimensional, more like the contour plots we've been using. It's just a slice through the 3D thing. So this is a 2D, not a projection, but an actual slice of it, and with this slider here you can control where the slice is. It's in the middle there; so it goes through the nucleus. But if you want, you can slide it down and see what the slice would be if it were in front of the screen that you sliced this picture. Everybody got it? So you can look at different levels that way, if you want to, or you can pretend you're looking at the whole thing. So it can be near or far.



Okay, now let's look at these two different, the 2D and the 3D pictures, and look at the pattern of nodes, the shape and the energy of the 3d orbital. You see here that there are diagonal nodes in this 3d orbital. Okay? So that's 3d. Let's look at this one. Now we still have those two diagonal nodes. In fact, it turns out that the nodes are actually cones. This is a slice through a cone, on the top, and on the bottom. Okay? So the nodal surface -- bear in mind that in one dimension a node is a point; in two dimensions a node is a line; in three dimensions a node is a surface. Right? So we're slicing it and see an intersection of the conical surface here, but the actual node is a cone. Okay, but is that the only node that you see in that picture on the right?



Student: There's a circular--



Professor Michael McBride: Ah-ha, Angela says she sees a circle too. Right? So there are two kinds of nodes in this orbital. There's both a conical node, which we had in the previous one; the 3d orbital had a conical node. That's what makes it a d orbital, having a conical node. But this one also has a spherical node, which is a circle when we slice it. Okay? So is this higher energy or lower energy than the one we looked at before?



Students: Higher.



Professor Michael McBride: How do you know it's higher?



Student: Nodes.



Student: More nodes.



Professor Michael McBride: It has more nodes. Did you ever see a situation before where you can have two different kinds of nodes? Here you have both, in this section, both a circle and lines. You ever see anything like that before?



Student: Chladni plates.



Professor Michael McBride: Right. Anyhow it's higher energy, it's 4d, not 3d, as you said, and it's like this, right? You have both circular and diameter nodes, but this is three-dimensional so you have both sphere and cone. So can you guess what a 5d orbital would look like? We looked at a 3d orbital. This is a 4d orbital. How about a 5d orbital? Lucas?



Student: It would have a third layer of nodes.



Professor Michael McBride: Not a third layer of nodes. It would have how many more nodes?



Student: One.



Professor Michael McBride: One more node.



Student: A circle, a sphere.



Professor Michael McBride: And it would be -- in the section it would be another circle; or actually it's another sphere of a different diameter. Okay? So you see how you can combine cones in this case--or it can be planes too; you have planes for p orbitals--you can combine spheres with planes, or cones, and you can combine them in different ways to get the same energy. For example, you can have a sphere and a cone that has a certain energy--that's 4d, the one we're looking at here--or you could have two sets of -- forget the sphere but have another cone. Does that remind you of anything? And get the same energy. Or you can have a plane and a cone, or you could have a sphere and a plane, or you can have a sphere and a cone -- I said that already -- or you could have two spheres, or you could have two planes. [Correction: a conical surface is in fact a double cone and counts as is equivalent to two spheres (3d ≈ 3s) or a sphere and a plane (3d ≈ 3p).] All those are two nodes--does everybody see what I'm saying--and they all have the same energy. Does that remind you of anything? It's not exactly the same, but it should remind you of something. That's what Chladni saw; remember, you could get different combinations, in his case, of circles and lines, and get the same frequency. Right? So that's why the 2s and the 2p have the same energy, or the 3s, the 3p and 3d have the same energy. Okay, now let's talk about scaling -- H-like. What do we mean by H-like?



Student: One electron.



Professor Michael McBride: It's one electron; any nuclear charge, but one electron. Why incidentally do we want it to be one electron? Zack?



Student: Repulsion.



Professor Michael McBride: That will turn out to be the case, but even if that weren't the case, there's a good reason to start with one electron. Yeah?



Student: You get a Coulombic interaction between the electrons.



Professor Michael McBride: Yeah, it's true that we don't want -- well we'll get into big trouble when we get Coulomb interaction between electrons. It's absolutely right. But we want to walk before we try to run; work with three variables, not six variables, which you'd have with two electrons, or 3n if you had n electrons. Right? So we want simple functions at the beginning. So that's what an H-like wave function is, just one electron, just three variables; ρ, θ, φ are the convenient ones. Okay, but now we want to see how we scale it as we change the nuclear charge. We already started talking about that last time when we talked about ρ, but we're going to talk about size, also about electron density, and also about energy and how the energies of these one-electron atoms change as you change the nuclear charge. Okay, this is what we talked about last time, that if you make the table up, the formulas, in ρ, instead of in r, then you could use it for any hydrogen-like atom, because the distance scale gets multiplied by the nuclear charge; that is to say, if you have a given ρ -- that's a given size of the function we're talking about -- a given ρ, then you have that value at shorter r, if you have a higher Z; a tradeoff between Z and r. If you double Z, you get the same value at half the r. Does that make sense, that as you get Z bigger, the distances get smaller? Does that make sense? Why?



[Students speak over one another]



Professor Michael McBride: Because the nucleus pulls the electron in more. And notice that it's linear, right? If you double the nuclear charge, you shrink by a factor of two, in the distance. Okay? So increasing Z shrinks the wave function, makes r smaller for the same ρ; that is, the same height of the wave function, or the same position in the shape of the wave function, because you have to renormalize if you shrink the wave function. Okay, so let's compare hydrogen, carbon+6, and potassium+19. Right? So the distance scale will go 1, 1/6th, 1/19th. So this is what it looks like. If a hydrogen atom is that big, this is how big the carbon one-electron atom is, and that's how big the potassium one-electron is, 1/19th as the linear distance that's in hydrogen. Now, how much does the electron density change? I think that's the next thing; yeah, scaling electron density with Z. You can see that just from what we just talked about. If you shrink the distance by a factor of two, how much do you change the density? Factor of two, does the density get twice as high if you make the distance half as long? Ryan?



Student: 23.



Professor Michael McBride: It's got to be 23, because you shrink it this way and this way and also this way. So it goes up with Z3, the density. That means if you have a hydrogen atom, the electrons are every which-a-way, all over, but if you have even a carbon atom, six times the nuclear charge for the same 1s orbital, the density is 63 as big. What's 63? Thirty-six -- ~forty five -- over 200, right? So you have 200 times the electron density for the 1s electron in carbon, compared to the 1s electron in hydrogen. Where does that crop up in experiment -- that you have so much higher electron density for carbon than you do for hydrogen -- and other things are even more so, like potassium -- where does it crop up? Yeah?



Student: The contour plots.



Professor Michael McBride: Of what?



Student: Of the molecules, the bond lengths.



Professor Michael McBride: What experiment?



Student: I forget.



Professor Michael McBride: Close. What?



Student: I forget.



Professor Michael McBride: You can do it by calculation and you'd see much higher density. That's what we're talking about is theory, but where do you see it experimentally? Lucas?



Student: Refraction difference diagrams.



Professor Michael McBride: Not to do -- well you can do it in the difference, but if you look at the total electron density you don't see hydrogens, unless you look -- work -- really hard, because the density of electrons, which is what you're seeing, is so much higher for other atoms. Okay, so the normalization has to be one. So if you shrink it, and that's where that comes in. We talked about that Z/ao to the 3/2 that gets squared before, and so that the Ψ2 is proportional to Z3, and the density of the 1s electron, as you go across these atoms, goes from one to 216 to 7000; potassium is 7000 times higher. Remember where we saw that in X-ray, what a very dark atom that was. Okay, so it helps X-rays find heavier atoms more easily.



Now how about the kinetic energy? Right? So Ψ is a function of r, some kind of function of r, but it includes a Z; Z times r, because ρ is Z times r. So now, how does that influence the derivative, right? If you have a function of Zr, the derivative of it is Z times the function of Z, by the derivative of -- oh what's it?; blah-blah. I got that wrong, or do I have it right?



Student: That's right.



Professor Michael McBride: That's right, yeah I got it right. Okay, now about how about the second derivative?



Student: Z2.



Professor Michael McBride: It's going to have a Z2 in it, right? So the curvature is going to be Z2 times what the curvature was before, but the value of the wave function is the same at some -- the corresponding ρ. So what's going to happen to the kinetic energy? If you have a bigger nuclear charge, what's going to happen to the kinetic energy at any point, at a corresponding point? Right? It's going to be proportional to Z2, the curvature divided by the value of the wave function. Right? So the kinetic energy is going to really zoom up for a hydrogen-like atom, when you get a heavier atom. Right?



So let's summarize what we did so far. So distance will shrink according to Z, electron density will go up according to Z3, and the kinetic energy will go with Z2. Okay?



Now how about the potential energy, how will it change as we change the nuclear charge? Well if we change the nuclear charge; Coulomb's law includes the nuclear charge, right? So at the same distance you'll have Z times whatever it was for hydrogen, much higher, or indeed lower potential energy, more negative, further from zero. Okay, but at a fixed-- that's-- so at a fixed distance the potential energy will be proportional to Z; but the distance shrinks, right? So if you look at corresponding positions in the wave function, or you look at the average potential energy, it's going to go-- it's going to also be, take into account that the distance gets smaller, r gets smaller, because the thing got shrunken by a factor of Z. So what happens to the average potential energy? How does it scale with Z? Maria, you got an idea? At a given distance, the energy, according to Coulomb's law, will be multiplied by Z, because you've got a bigger nuclear charge, but every distance gets shrunk by a factor of Z. Right? So if you average everything, what are you going to get? There are two factors of Z that are increasing (or decreasing, making more favorable) the energy, the potential energy. Got that? Two factors. So what's it going to be?



Student: Z2.



Professor Michael McBride: Z2, right? So the potential energy--what did the kinetic energy go with? Remember how the kinetic energy scaled?



Student: Z2.



Professor Michael McBride: Z2. How does the potential energy scale?



Student: Z2.



Professor Michael McBride: How about the average potential energy, if I double the nuclear charge? How much lower is the -- obviously if I increase the charge, the average energy of the electrons is going to get lower, but for two reasons, because the nuclear charge is bigger, and because the electron is twice as close. Okay? So how does it scale, the potential energy average? Z2. So the kinetic energy scales with Z2, the potential energy scales with Z2. How about the total energy? This is a hard one.



Student: Z2.



Professor Michael McBride: Z2, right? Okay, so the potential energy goes as Z2, and the total energy, here's a big surprise, it goes with Z2 too. It also happens to be 1/(n2). Right? So the energy gets -- yeah, so as you go up in n, you get less stable, right? Now R turns out to be about 300 kilocalories/mole, if you want to really get what the energies are. And here is a plot of the energies. Here's the 1s, 2s or 2p, 3s or 3p or 3d, 4s, p, d, f and so on, right? But the distance from zero, the amount of stabilization, is proportional to 1/(n2); and we saw the same thing if we did-- in the one-dimensional Coulomb case, the energy was also 1/(n2). Okay. Notice it's independent of l and m; 3s is the same as 3p is the same as 3d. Does that square with what you have learned before?



Student: No.



Professor Michael McBride: What did you learn before? Lexy, did you learn anything about that before, about the energies of 3s, 3p, 3d? No. Anybody? Shai?



Student: s, p, d.



Professor Michael McBride: Yes, s is lower than p is lower than d. Now how can both these things be true, that you heard something that was true, and I'm telling you they're all the same?



Student: There are more electrons.



Professor Michael McBride: Ah-ha! What you heard about was real atoms, things that have many electrons. I'm telling you about if the atom has only one electron, then s and p and d are the same. Okay? But it'll be different when you have many electrons, and we'll see why shortly. Okay, because this is one-electron atoms. Okay, so we've -- this just summarizes -- the size goes with 1/Z, the electron density is proportional to Z3, and the energy goes with Z2, whether you're talking potential, kinetic or total energy. Okay, and it's also 1/n2. And the size happens to be n/Z as well. Okay, now here's a really pretty one, and to do this, to see this right, you want to see the real thing. So notice that this is a 2p orbital; n, l, m are 2,1,1. Everybody see that? I want to show you the real program that does that. [Technical adjustments to show 3d orbital in Atom-in-a-Box program] Okay, now what I want for the display -- this display is just as if electrons were all white, okay? But we want to color them according to the color of the -- [Technical Adjustments of Atom-in-a-Box format] -- we want to color them with the phase, plus, minus. That's 3d. The phase keeps changing. That has to do with time dependence, which we're not talking about. Okay? Now, that's a 3d orbital. We want to have -- what did we have before? We had 2p, right? So I'll make this instead of a three, a two. So there's a 2p orbital, a particular 2p orbital, 2,1,0. Which one is that? Notice you have a coordinate system next door there, and I can rotate this thing, I can grab it and rotate it; the coordinate system rotates. See that?



Student: Cool!



Professor Michael McBride: So I can look at it end-on if I want to, or look at it from the side. Which p orbital is that?



Students: 2pz.



Professor Michael McBride: How do you know?



[Students speak over one another]



Professor Michael McBride: Points along the z axis. Okay? Now, let's look at another 2p orbital, because we can make m one instead of zero. So the 2,1,0 is 2pz. Now what's 2,1,1? Let me look at it this way. That's time-dependence, right, and that's weird. And physicists love this sort of thing, because they deal with atoms in isolation, and atoms in isolation have angular momentum, things going around. But atoms, when they're in molecules, when they're with other atoms, or in other things that distort it, like in an electric field, they don't behave that way. So chemists use a different way of doing it, right? It's not that physicists are right and chemists are wrong, or that chemists are right and physicists are wrong, it's if you're dealing with isolated atoms, you use one thing; if you're dealing with atoms interacting with other atoms you should use something else. And the way you do this, dealing with something else -- I'll show you here. You do what's called superposition; that is, you add two functions in order to get a new function, and the two functions I'm going to add, when I click "Superposition" here, are 2,1,1, and the other one I'm going to use is 2,1,-1. And now what's that?



Students: 2py.



Professor Michael McBride: That's the 2py orbital. Okay, does everybody see what I did? So I can combine two of the physicist's orbitals to get the chemist's 2py orbital. Dean Dauger, who wrote this, is a physicist, so he writes physicist programs, right? But we can trick it to be chemistry by adding two of his together, or by subtracting them; and when you subtract them you have to put an i in -- imaginary. So I'm not going to do it right now. You can do it yourself. You can see down here, you can do the phase and so on; the phase ofp makes it imaginary. I won't do that now. We'll go back to the PowerPoint. [Technical Adjustment to change program] So superposition of that kind, when you add two functions together to get a new function, for whatever purpose, is a kind of what's called hybridization. Now there are other kinds of -- have you heard about hybrid orbitals? Right. What that is is adding two orbitals together to get a new orbital. And let's look about when that would be. As we said, this is the physicist's 2p, m=1, with orbital angular momentum, and it has complex numbers involved. And the chemist's 2p is that sum. Okay.



Now multiplying and adding wave functions. We've already seen a case where you multiply functions. Do you remember where we've seen things be multiplied?



[Students speak over one another]



Professor Michael McBride: Alison, do you remember? No.



Student: You multiplied the wave functions.



Professor Michael McBride: Multiplied which wave functions? Let me tell you one place we already multiplied was squaring a wave function is a kind of multiplication to get probability. But I want a different one.



Student: When we, for different variables --



Professor Michael McBride: Ah, for separating the variables, r, θ and φ; we had an R function, a Θ function, a Φ function, and we multiplied these pieces together to get the total wave function for the electron, the three-dimensional wave function. So that's one place you multiply, to get a one-electron wave function, which is -- we call an orbital. An orbital is a one-electron wave function; that's what an orbital is. Now, you can add orbitals to create a hybrid orbital; that's what we just did, we took two physicist's orbitals, added them together, and got a new orbital. Okay? 2py + 2pz is a hybrid orbital. An orbital is a one-electron wave function. What's it a function of? What variables do you put in, in order to get a value out? How many variables? Sophie?



Student: Three.



Professor Michael McBride: Three. What are they?



Student: Those.



Professor Michael McBride: The position of that electron, the one electron. If it's a one-electron orbital, you can write the orbital as many pieces -- here are two pieces, right? But it still is a function of just the position of one electron; it's not a function of six, or eighteen, or whatever number of variables, it's just a function of -- it's an orbital, a one-electron function, so it's a position of one electron that you plug in. Now, you can change orientation of orbitals by hybridization. Let's look at that. So here's the 2pz orbital. I've rotated the coordinate system so z is horizontal. Okay? And here's the 2py orbital, which we already made by adding two physicist's orbitals together. Okay? Now what happens -- and notice the 2py , surprise, is pointed along the y axis, 2pz was pointed along the z axis. What if I add the two together? What if I add 2pz and 2py? Now let's go back and look at 2pz a second. Notice it's red on the left, blue on the right; say positive numbers on the left, negative numbers on the right -- everybody with me? -- because it assigns numbers to different points in space, positive numbers on the left, negative on the right, let's say. Now this one assigns positive numbers on the top, negative numbers on the bottom. Suppose I add together, what am I going to get? Andrew?



Student: You get diagonal.



Professor Michael McBride: Ah, they should reinforce in the top left where they're both positive, and in the bottom right where they're both negative, but in the bottom left and the top right they should cancel. Okay? So if we add them together to make a weighted sum -- we don't have to add them 50:50, but we'll start by adding 50:50. Okay? So if we add them, a 50:50 mixture will give that, just what you said. Okay, so we changed the orientation of the p orbital by adding two. Okay, now suppose we didn't use 50%, suppose we used another mixture, suppose we used 75%, right? Or there's 50%, or 25%. Okay? So you can get any orientation you want by mixing -- within that plane, within the yz plane, by adjusting the ratio. Okay, there's 2py.



Now, what would happen with a 2s orbital if you put it in an electric field? You're changing the rules, you're changing the potential, because now the electron has -- doesn't just -- its energy doesn't depend just on its distance from the nucleus, it also depends on where it is in the field. So we're changing the definition of the potential energy. So it's a different quantum mechanical problem. But I don't want you to think about quantum mechanics, I just want you to think about what would happen if you put an electric field on. What do you think?



Student: It shouldn't --



Professor Michael McBride: Would it move the atom?



Student: Yes.



Professor Michael McBride: No, because the atom is neutral, overall. We'll do a hydrogen atom here. The atom's neutral, so it won't move. But what will happen?



[Students speak over one another]



Professor Michael McBride: The electrons will move one way and the nucleus will move the other, and if you've preserved the center of mass, then the nucleus will hardly move at all, and the electrons will do most of the moving. Okay. So you can change the shape by hybridization. We can get something that will work in an electric field, describe an atom in an electric field, by mixing the ones without electric field. Now let's look how we do that. We make an spn hybrid orbital. Have you ever seen that?; n is a number. Can you give me an example? two, sp2 or?



Student: sp3.



Professor Michael McBride: sp3 or sp, where n is one, right? Or sp0. Have you ever seen that?



Student: No.



Professor Michael McBride: sp0 is just s. This is how much p there is in it, after you square it. That is, it's a weighted sum, this thing you're using. You have different values of a and b, and it's weighted such that -- so when you square it, you get a2, a fraction of 2s squared; that's the distribution 2s would have; b2 of the way py would be when squared. But you also get a cross-term, 2ab times 2s times 2p. Okay, now so n, the number you put in to talk about sp2, sp3 and so on, is the square of b2/a2. So if you have an sp hybrid, it means equal parts of s and p, and if you have sp3 , it means three times as much p as s, after squaring. Okay, so it's what fraction you have of something that looks like 2s squared and 2p squared; and then there's also that part. Okay, so now let's just look at some examples here. Suppose n is zero; then you have just pure s. Okay? Now suppose you put an electric field that wants to shift the electrons. It would push a positive charge to the right but it will push electrons to the left. Okay, so let's -- here we went to 0.02, just a tiny amount of p. But see how it -- here's back to s, and there's sp0.02. It shifted the electrons to the left. What if I go to a higher value? 0.04, 0.06, 0.09, 0.11, 0.18, 0.25, 0.33. So what I'm doing is turning up a dial to make the electron -- electric field stronger -- and the electron is shifting, by changing the hybridization. Okay, 0.5, 1. There's an sp hybrid. Now what'll happen if I keep on going, make it sp2, for example. What do you think will happen? Guess.



Student: It'll get more and more to the right.



Professor Michael McBride: It'll keep going to the right. Wrong! Watch this. That's the maximum extension is sp. That's the most you can extend. Now watch what happens if I make it bigger; sp2, sp3, sp4, sp9, sp24, pure p. Right? Because both pure s and pure p are symmetrical. The 50:50 mixture is the one that shifts the most to the left.



Okay, now we're going to close here with trying to go beyond what we've already seen. What we've seen is one-electron atoms, hydrogen-like atoms. So we find that we can multiply pieces, R, Θ, Φ, to create this one-electron wave function, the orbital, and we can furthermore add these things to make hybrids that shift things around in useful ways. We can add orbitals to make an sp hybrid, for example, as what would happen in an electric field. But it's still an orbital, so it allows adjusting to new situations -- for example, an electric field -- while preserving the virtues of real solutions of the nuclear potential, because you never put on a field as strong as the field felt by the electron near a nucleus. Adding the external field by turning up your dial is just a small change. So you want things to look pretty much like what they looked originally, with the nucleus, just a little change, and you can do that by hybridization. Okay, the big deal is the nuclear potential at small distances.



Now, let's try to go to a two-electron wave function. Okay? So what's it a function of? What's a two-electron wave function going to be a function of, how many variables?



Students: Six.



Professor Michael McBride: Right, six, right? x,y,z for one electron, and x,y,z for the other electron, or r, θ, φ of electron one, and r, θ, φ of electron two, and they're on the same nucleus, so you measure the distances from the same point, but they're two different distances, one to electron one, another to electron two. They could be the same, but they don't need to be, they could be different. Okay? Now wouldn't it be neat -- that's the question mark -- if we could get that by multiplying two one-electron wave functions? Because we have one-electron wave functions cold; we look in the table and we've got the exact thing. If we could take two of those and multiply them together, so orbital A for electron one times orbital B for electron two -- remember, what these things are is just numbers. If you put in some value of position of electron one and some value of position of electron two, this first thing will give you a number, the second thing will give you a number. If you could multiply those two numbers together and get the number for the total wave function, that would be really great, because then you'd know how to do it. The question is, can you do that? Remember, an orbital is a one-electron wave function. So can we multiply orbitals in order to get a many-electron wave function? If we could -- and then, of course, we would square it to get probability density. Now it would be a different probability. In the first case it was probability of electron one being here. Now it's the joint probability of electron one is here, and electron two at the same time is here; that's the probability when you have six variables. Okay? Wouldn't that be great? It's just paradise, right?



Because if we had many electrons we could just multiply all those orbitals together, a certain orbital for electron one, another orbital for electron two, another one for electron three. We know how to write those; multiply them together, we get the thing. We find the total energy is whatever energy electron one had, plus whatever energy electron two had. That's great. We find the total electron density. We know how to find the electron density of each individual electron. So the total electron density should be just the sum of those. So if we were interested in the total electron density at this particular point, we'd see what the electron density of electron one is at that point, and what the electron density of electron two is at that point, and we'd add them together and that would give the total electron density, at that point. So the total electron density is a function of x, y, z. It's whatever this is, evaluated at x, y, z, plus whatever this, plus all the others. Absolutely wonderful. Right? And then you could write things. So the whole is the sum of the parts. Right? And you could write things like this at the top, that Neon has the configuration: two electrons in 1s, two electrons in 2s, two electrons in 2px, 2py. Right? Things like that. You've seen things like that. So those are the orbitals in which electrons are, to describe the neon atom. Right? Now, this is a problem in joint probability. So it's like tossing two coins to try to get to two heads. So I don't know if I can do these at the same time, but I have two quarters here. Need your back of your hand. Okay, so I got tails. What did you get?



Student: Heads.



Professor Michael McBride: How often?



Student: Half the time.



Professor Michael McBride: Half the time. Actually if I get tails and you get heads, that's a quarter of the time.



Student: Yes, that's a quarter of the time.



Professor Michael McBride: But a quarter of the time you'll get tails and I'll get heads. So heads/tails is half the time. How about heads/heads, how often?



Student: Heads/heads. Still a quarter.



Professor Michael McBride: A quarter of the time. Now I'll bet you, I'll bet you that I can get head/head half the time, actually doing the experiment.



Student: Yes.



Professor Michael McBride: You want to take the bet? I would advise against it.



[Laughter]



Student: I won't take it then.



Professor Michael McBride: Okay, here's what I'm going to do. Okay?



[Professor McBride tapes the quarters together]



Professor Michael McBride: Two tails, I lose.



[Laughter]



Professor Michael McBride: Whoops.



[Laughter]



Professor Michael McBride: Two tails, I lost again. But I think you can see that if I do it a lot of times I'm going to win half the time. Why? Why was the rule wrong?



Student: They're not independent.



Professor Michael McBride: Because they weren't independent, right? If they're independent, the probability of both A and B, that is, the probability that A will be here and at the same time B will be here, is the product of the probabilities; but if you do this, that's not true anymore. Right? So if you have a two-electron wave function, you square it to get the joint probability, that electron one will be here and electron two will be here at the same time. So is that valid, to multiply -- here I'm multiplying the individual probabilities that electron A is at a certain position, electron B is at another position. Do I get the joint probability that way? Well under what conditions do I get the joint probability that way?



Students: If they're independent.



Professor Michael McBride: If they're independent. Now, here's the big question. Are they independent?



Students: No.



Professor Michael McBride: Why not?



[Students speak over one another]



Professor Michael McBride: Because of Coulomb's law, they repel one another. There's no way they can be independent, they repel one another. So orbitals, one-electron wave functions, to express many-electron wave functions, cannot be right. Orbitals are a fiction, except for one-electron problems. Right? hydrogen-like atoms fine; we've got hydrogen atoms cold. But if you got more than one electron, orbitals won't work. But still we're going to use them.



[Laughter]



Professor Michael McBride: Right? And that's what you're going to find out after the exam. Okay, good luck on -- so that doesn't work -- good luck on the exam.



[end of transcript]

The lecture opens with tricks ("Z-effective" and "Self Consistent Field") that allow one to correct approximately for the error in using orbitals that is due to electron repulsion. This error is hidden by naming it "correlation energy." Professor McBride introduces molecules by modifying J.J. Thomson's Plum-Pudding model of the atom to rationalize the form of molecular orbitals. There is a close analogy in form between the molecular orbitals of CH4 and NH3 and the atomic orbitals of neon, which has the same number of protons and neutrons. The underlying form due to kinetic energy is distorted by pulling protons out of the Ne nucleus to play the role of H atoms.

Transcript



September 29, 2008



Professor Michael McBride: Okay, so what's coming for the next exam? Well we've been looking at atoms, and at the idea of orbitals for many-electron atoms, which we showed last time is wrong. So today we want to recover from the orbital approximation. Okay? Then we're going to look at molecules. And first we're going to look at them in a very unconventional way, from the point of view of molecules as plum-puddings, what's called the "United Atom" limit, to apply what we know from atoms to molecules. But then we're going to look at a very, very different way of looking at it, to try to understand bonds in terms of what are called linear combinations -- that means weighted sums -- of atomic orbitals to make molecular orbitals; but we're going to try understanding bonds.



And then these terms that don't mean much to you now but will mean a lot later on: "energy-match" and "overlap." And then we're going to get to reality -- all this stuff is theory -- but we're going to look at something real: at XH3 molecules, with different atoms for X, at their structure and at their dynamics, and how that ties in with our understanding of bonding. And then we'll go on to reactivity, which is of course the main goal. We'll talk about HOMOs and LUMOs, and you'll see what those are, and how to recognize functional groups and their reactivity; and that's the real prize, is to be able to -- you've been memorizing functional groups for this previous exam, but what I want you to be able to do is look at a molecule and recognize when it has a functional group, even if you've never seen it before, and how it might react, what it would react with. And then we're going to see how organic chemistry really developed, the real thing; how it developed from the time of Lavoisier.



Okay, so that's what we're going to. And here's where we were last time. We were wondering, when it comes to a two-electron wave function, might it be possible to write it as a product of one-electron wave functions? Because we know how to write 1-electron wave functions; we've got that table, we can write the real thing, at least for an atom. So if it were possible to write the six-variable, two-electron wave function as a product of one-electron wave functions, that would be a fantastic simplification, and all we'd have to do is square and we'd get the joint probability; not really that you care very much about the joint probability. I suspect you haven't stayed up late nights worrying about joint probability. But anyhow, if you want to handle things with more than one electron, you have to have a two-electron wave function. And if we could do it from one-electron wave functions, then we're really in a good position. But we ended the first quarter of the semester on a downer, the idea that there's no way electrons can be independent. They repel one another, so orbitals are fundamentally wrong. Okay? So forget that. And here's our paradise; it's gone. Okay? But are there tricks that would allow us to salvage orbitals, to use them, even though we know they're wrong?



Okay, well the first, the simple-minded trick, is Z-effective. We talked about how you could scale the one-electron wave functions, the atomic orbitals, for nuclear charge; if you increase the nuclear charge they contract; and we saw what different properties are proportional to. So if you have other electrons in the atom, it could be that they could be sort of approximated, as if they were just reducing the nuclear charge, right? There's a certain repulsion, as well as an attraction taking place, if there are other electrons there. But maybe we can just reduce the attraction, right? And things will be sort of -- at least corrected in the right direction. Right? So pretend that the other electrons, or some fraction of them, are concentrated at the nucleus. So from the point of view of an electron you're thinking about, they just reduce the nuclear charge; and that's a problem we know how to handle, what happens if you change the nuclear charge. Okay? So might there be an effective nuclear charge that we could use? So we pretend the other electrons just reduce the nuclear charge for the electron we're interested in, the orbital, the one-electron wave function that we want to find. So we're going to just to try to find a one-electron wave function in a many-electron problem. Okay? And this is what we know, that 1s looks like that and ρ scales with Z. So if I just reduce Z a little bit, maybe I get something at least that's corrected in the right direction.



Okay now, these guys, Clemente and Raimondi, who worked at IBM in the early days, when IBM was the only place that had computers that were as powerful as your laptop, or almost as powerful as your laptop, they did things like this, where they did good quality calculations, better than this, and then tried to match the wave functions they got, by adjusting Z, so they looked like the wave functions that you get with better approximations. Right? So they got a best fit to better calculation. So they found that helium -- which has two electrons; so it has a nuclear charge of two; Z=2 -- you get electrons distributed, more or less right, if you pretend that the nuclear charge is 1.69, instead of 2. Okay? So each electron sort of what's called 'screens' the electron from the nucleus. Right? Okay, now in the case of carbon, which has Z=6, you can also use Zeff, but it's different depending on which electron you're talking about. Why would that be so? Why would it be that you'd use a different screening constant, or a different effective Z, if you're talking about the 1s orbital, than if you're talking about the 2s orbital say? Why would it be different, how much you want to reduce the nuclear charge? Sherwin?



Student: [Inaudible].



Professor Michael McBride: Yes, the electrons that are outside, you don't care about. It's the electrons that are between you and the nucleus that are screening you from the nucleus. Right? So a 1s is already really close to the nucleus. There's very little other electron density that's inside it. Right? That's why in helium it wasn't screened very much. But in the case of carbon, the 1s electrons, which are way down near the nucleus, have an effective charge that's almost 6; 5.67. But the 2s have 3.22. Right? The 1s electrons are all down inside and making the nuclear charge appear small. Right? And to a certain extent the other, the 2p electrons, are also inside; to a certain extent, but not as much inside as the 1s is. So 3.22. And the Zeff for 2p is different than for 2s; it's 3.14. What do you notice about those two numbers? You might think they'd be more or less the same; and they are more or less the same. But 2s is slightly less screened. It sees more of the nucleus than the 2p does. Okay? So it looks like 2s gets more down inside than 2p does. Does that surprise you? Russell?



Student: No, because the 2p orbital dumbbell doesn't have anything at all.



Professor Michael McBride: Yes, the dumbbell doesn't have anything at the nucleus. Right? It's got a node at the nucleus. Whereas 2s has that first little core, down in -- the stuff that's inside its first spherical node is down quite close to the nucleus. Now we could look at that. There's what 2s looks like, and here's the probability distribution that we talked about: r² times the radial function squared. So this stuff that's in here is inside. If we compare it with the 2p orbital, you see that this part is way down inside. It's not being screened by the 2p electrons. Right? In fact, we could also look at the 1s. This is how the 1s is distributed. Right? So the 1s will screen this. And of course it screens this part of the 2s, and it screens this of the 2p. But it's not really trivial to look at this and figure out which one would be more screened, the 2s or the 2p, because this part is further in than this part, but this part is way down inside. So it's a balancing act as to exactly which one would be bigger, and they're not very different. But as it turned out, according to Clemente and Raimondi and their calculations, the Zeff that you should use for 2s is a little bit bigger than the one you should use for 2p. Kevin?



Student: Why are there two peaks for 2s instead of 1?



Professor Michael McBride: Because there's a radial node; remember 2-ρ, the function is 2-ρ. So here's ao; 2 ao, when the distance is two; two minus two is zero; there's a node there. Okay? And it's been squared of course. The wave function changes sign, but here we square it. Any other questions about this? Okay, now let's look at sodium. So sodium has a nuclear charge of eleven, and you won't be surprised to see that Zeff for the 1s electrons of sodium is 10.63; almost 11, they are very little screened. 2s is 6.57. 2p is 6.8. But 3s is only 2 and a half, because it's further out still. So there are more electrons inside, hiding the nucleus. Everybody got the idea of what's going on here? You'll notice one thing sort of funny here. Do you notice what?



[Students speak over one another]



Professor Michael McBride: John?



Student: The 2p is higher than --



Professor Michael McBride: Ah, it turns around! This time it's vice-versa. Right? So it's a balancing act, and who knows? And there's nothing fundamental about this. This was just adjusted by Clemente and Raimondi, in order to get shapes that looked pretty much like better quality shapes. Okay, so it's a very subtle thing. But this is very crude. Right? Nobody argues that it's the last word. So isn't there a better way to do it? And there is a better way to do it, to get back to orbitals, even for many-electron problems, and that's called Self-Consistent Field, or SCF. And it's a recipe for calculating orbitals; for calculating better orbitals than you would get with an effective-Z. So first you go through all the electrons in an atom, or a molecule, and you find an approximate form of the orbitals; for example, you could use Zeff, to get something that's sort of approximate. Okay? So you have approximate wave functions, square them; you have approximate distributions for all the electrons. Now with Zeff what we were pretending was that a certain fraction of the other electrons are on the nucleus, and the rest of them we forget; that's pretty crude. Okay?



Now what we're going to do is pretend we know, at least approximately, how the other electrons are distributed in a cloud; the other electrons. We're interested in one electron, an orbital. Right? Now how will knowing how the other electrons are distributed help us find the orbital we're interested in, the one-electron wave function? What do you need in order to solve a quantum mechanical problem? You need the mass of the electron -- that's easy. What else do you need? The potential law; what its energy is, its potential energy at different positions. But if you know where the nucleus is, or nuclei, and you know the cloud of the other electrons, and assume they're just static clouds, then you can -- it's laborious, you need a computer to do it -- but you can calculate the potential that the electron you're interested in would have, at different positions, if there were this fixed cloud of other electrons and the nuclei. So you have the potential law, which means you can find your one-electron orbital. Okay? Does everybody see the strategy here? So we're going to do it one electron at a time. So we fix all the other electrons, all but one, from some crummy estimate, like Zeff or something like that, calculate the potential for the one electron we're interested in, and then we get -- use that potential to calculate an orbital for that one electron that's better than if we had used Zeff. Right? Because it's not just putting a certain fraction of the other electrons at the nucleus, it's treating them as a cloud. Okay? So now we have a much better guess for that one electron than we would've had earlier with Zeff, or whatever type of guess. What do you do next?



Student: Do it for every orbital.



Professor Michael McBride: Dana?



Student: You do it for every other orbital, in sequence.



Professor Michael McBride: Ah, one at a time. You now know how that one is distributed in the cloud, much better. Right? And you take all the others but one, fix them in their approximate clouds. Now you calculate a potential for the second electron, and do that trick. Okay, so we repeat steps two and three to improve the orbital for another electron. Then what do we do? Will, what would you do, at this point?



Student: See how it changes shape?



Professor Michael McBride: Well yes, it'll change shape. That second electron will have a better shape now. So you've got that. You've got a good shape for the first electron. Now you've got an improved shape for the second electron. What do you do?



Student: Start from step one or the first electron.



Professor Michael McBride: Or you go, if there's -- what if there are three electrons?



Student: Do all three again.



Professor Michael McBride: No, you do the third one.



Student: Oh yes, definitely.



Professor Michael McBride: Right? And then the fourth, fifth, sixth, until you get through all the electrons. And then?



Student: Start over.



Professor Michael McBride: Start over. So you improve all the orbitals, one by one, and then you cycle back to improve the first one again; go through them all. Then what do you do?



[Students speak over one another]



Professor Michael McBride: Start again. Then what do you do? This is why it's good to have a computer, right? Then what do you do? When do you stop?



[Students speak over one another]



Student: When they stop changing.



Professor Michael McBride: They'll stop changing; at least -- it'll be like Erwin meets Goldilocks where it's out in eighth decimal place that things are changing. So then you know you've got it as close as is reasonable to go. So you stop. And what do you call it, when you get to that situation when you stop? The system is self-consistent. Right? That's why it's a self-consistent field. Okay? So you quit when the orbital steps -- shapes stop changing. So now you have the right wave functions, the right orbitals. So now we've got the real thing. Right, or wrong? What could be wrong? We've got it self-consistent. Where's the weakness in the assumption on which we've been doing this? Kevin?



Student: Well you're assuming fixed positions for the other ones.



Professor Michael McBride: I'm assuming what?



Student: You're assuming fixed positions for the other ones.



Professor Michael McBride: You're assuming that all those other electrons, except the one you're working on, are fixed in clouds. But they're not fixed. Right? The electrons can move around. So depending on where your electron happens to be, those other electrons may change their shape, at any given instant. Okay, it's still wrong, because real electrons are not fixed in clouds; they keep out of each other's way by correlating their motion. Right? They don't move independently so that this one is a cloud and when this moves around it, nothing happens. As this one moves, when it gets near this place, this one gets away. Okay? So they keep out of other ways by correlating their motion. So the true energy must be lower, more favorable, than you calculate by self-consistent field, because the orbitals are able to get away from where you think they should be. Is everybody clear on why the limit goes in that direction; why the true energy is lower? Because the electrons are smarter than you are. They know to get out of each other's way; or at least than you pretended. Okay? So what do you do? You hide the residual error so that people won't embarrass you by saying, "What the heck were you thinking?" Right? You say that when you go to the Hartree-Fock limit -- which is a fancy name, named after two people that thought of doing this self-consistent field thing, or thought of a method for doing it -- when you get to the Hartree-Fock limit, that's a self-consistent field, but there's going to be an error because of correlation. So you give it a fancy name so people will think, "ah, what in the heck is that?"; they must really know a lot. Right?



[Laughter]



Professor Michael McBride: So what do you call the error? Do you call it error? No way. Sophie, do you know what you call it?



Student: Correlation energy.



Professor Michael McBride: Correlation energy. Right? There is no such thing as correlation energy. That's not a fundamental energy. It's not like Coulomb's Law or gravity or something like that. Right? It's just the error you make when you do self-consistent field, that the true energy is lower. Now if you want to measure correlation energy, what do you have to know?



Student: True energy.



Professor Michael McBride: You got to know the true energy, so that you know how bad your estimate is. Right? And where do you get the correct energy, or the true electron density that you have an error in?



Student: Experiment.



Professor Michael McBride: You get it from experiment. Or you get it from some whopping calculation that's not as simple as SCF, Okay? For example, there's a thing called "configuration interaction", which we won't know anything about, and don't need to. Right? But it's a much more complicated calculation. It takes into account the fact that electrons keep out of each other's way. Or there's another one called "density functional theory", which is also approximate. They're all approximations. You can't solve the real equation. But these are better approximations than just self-consistent field. Okay? But they're hard to think about, because they involve so much manipulation that it's hard to reason about them. Self-consistent field is easier to understand. Okay? So if we're really lucky though, correlation energy might be negligible; it might be so small we don't care about it, in which case we're golden, for practical purposes. We can use orbitals, self-consistent field orbitals, treat things as if they were independent electrons. Okay? And then we're in business. Okay? So we should think about the magnitude of energies and whether we care about correlation energy, this error that is orbital theory. What do I mean by saying orbital theory? What's an orbital?



[Students speak over one another]



Professor Michael McBride: A one-electron wave function. But we're trying to understand many-electron problems. Can we analyze many-electron problems as sums of one-electrons, as sums of things that come from orbitals? Is the whole equal to the sum of the parts, the parts being orbitals? We know that orbitals have to be fundamentally wrong, but if the error is -- if the correlation energy is really small, we don't care. So should we care about the error in orbital theory? So here is a scale, a logarithmic scale, of energy changes that occur when things happen, and how big are these energy changes. Let's start at the beginning. So we take a bunch of neutrons and protons and bring them together to make a nucleus, and energy comes out. Right? And the amount of energy that is given off when a C-12 nucleus is formed is 2*109 kilocalories/mol. How do I know? Because I look at the mass, the rest mass of the proton and the neutron, and I look at the mass of C-12, and mass was lost when it came together. Right? And the amount of mass lost, E=mc², is 0.1 atomic mass units, which is 2*109 kilocalories/mol. So that's a lot of energy. Okay? Now so we got C+6. Now we're going to put two electrons on it, the 1s electrons of carbon. Right? Bingo! And that gives us 2*104 kilocalories/mol, given off when that happens. Okay?



Now that is our old, familiar friend. What do you notice about the ratio of these energies? 105, right? So it's 105 smaller, like the ratio of my hair to the width of the room. Right? So much, much smaller energy involved in putting electrons into the 1s shell of carbon, than in putting the carbon nucleus together. Okay? Then we're going to put the four valence electrons onto carbon, right? The 2s and 2p. And with that we get another 3*103. So it's an order of magnitude less. The 1s electrons are bound much more strongly than the 2s and 2p; and you know that, from this scaling, Z²/n². Okay? Because you could use Zeff to guess how the 1s's are affecting the energies -- the nuclear charge for the 2s's. So you could lower the nuclear charge from six to four, because there are already two electrons way down in there; and then you have that n² as well. So you could scale the energy and find that it's an order of magnitude less. Okay.



So then what are we going to do next, now that we have atoms? Going to put them together to make bonds. Okay. So we make four single bonds from carbon, but they're to other carbons. So for this one carbon we should count only half of each energy, right? Because of half of it we'll assign to the other carbon. So half of four single bonds -- a single bond is order of magnitude 100 kilocalories/mol. So that's about 200 kilocalories/mol. So another order of magnitude down, the energy in making bonds. Okay?



And then we can have non-bonded contacts. Now there are different kinds of non-bonded interactions between molecules. But typically, to be worth talking about, they're in the range of one to twenty kilocalories/mol; so another order of magnitude down, or more. Okay, and the weakest of all, the weakest bond known, or interaction known, attractive interaction, is between two helium atoms. It turns out that two helium atoms, although they don't form a bond, are attractive; you know, things at long distance are attractive and then they become repulsive. So the minimum energy distance is fifty-two angstroms -- right? -- thirty-some times as long as a normal bond. And the depth of that well, how favorable is that energy, is 2*10-6 kilocalories/mol; so that's nothing.



But all these things we're looking at down below here are all based on Coulomb's Law. The first one was not. Right? The nuclear binding energy is not Coulomb's Law. But all these others are Coulomb's Law. But they're all 105, up to 1015, times weaker than what goes on in the nucleus. And that means if you made any error, at all, in the energy of the nucleus, it would completely wipe out anything that had to do with Coulomb's energy. Right? Because it's so much bigger. So an infinitesimal error in nuclear energy, or change in nuclear energy during a reaction, would completely wipe out everything that we're talking about. Right? So this sounds like we would really worry about it, except -- so a 0.001% change in nuclear energy would overwhelm everything Coulombic. Right? But fortunately nuclear energy doesn't care about chemistry. If you change from one arrangement of atoms to another, the nuclear energy doesn't change at all. Right? So why is that good? You can just cancel it out. The starting material and products of any reaction have exactly the same nuclear energy. So if you're not a physicist, you can just forget nuclear energy, even though it's so enormous. Okay? Is that clear? So forget nuclear energy, we don't have to worry about it. All we have to worry about are these Coulomb things.



Now, so black that out. Okay. Now here's how big correlation energy is. It depends on what problem you're dealing with, how big the molecule is, what kind of atoms there are in the molecule. The error you make in SCF is different, for different cases. But for the kind of cases we're interested in, normal organic molecules, it's of the order of 100 kilocalories/mol. Now that's an error. Do you care about that error? The error in nuclear energy, an error, would have been enormous and completely wiped anything out. Is this enormous? Do we care about an error of 100 kilocalories/mol? Do we? Would you care about an error of -- Devin, what do you say? Would you care if you made an error of 100 kilocalories/mol? How big is that in the scale of things we're talking about?



Student: Doesn't seem too big.



Professor Michael McBride: Compared to what?



Student: That's the question.



[Laughter]



Professor Michael McBride: Yes. What are we interested in?



Student: Bonds.



Professor Michael McBride: Bonds. How big is it compared to bonds?



Student: Fifty times as big.



Professor Michael McBride: It's as big as a bond. A bond is 100 kilocalories/mol. That means that we're -- that's a disaster, right? -- unless what? The nuclear would've been a disaster too. Why isn't it? Errors in nuclear energy are not a disaster. Why?



[Students speak over one another]



Professor Michael McBride: Because they don't change. The starting material and the product have the same. So the same thing would be true here. If correlation energy didn't change, from one arrangement of atoms to another arrangement of the same atoms, then it would just cancel out and you wouldn't care. Everybody got that idea? Okay, so the correlation energy is about equal to the magnitude of a bond. But how big are changes in correlation energy, when you change the arrangement of atoms? Well changes tend to be about 10 to 15% as big as bond energy, or 10 to 15% as big as correlation energy. So correlation energy does change. You make different errors for different arrangements of the same atoms. But those errors are about 10 to 15% as big as a bond. Now do you care about correlation? You don't care as much. So you'll get approximate ideas. If you don't care within -- if you're satisfied with getting about 80% of the right answer, then you don't care. Okay? So that implies that orbital theory is fine, as long as what you're interested in is answering qualitative, rather than fine quantitative questions. Okay?



So if you want to get numbers really right, and the magnitude of the number really makes a difference to you, a few percent change, then you have to do something better than use orbitals. Okay? But if you just want to understand why bonds work, then orbitals are fine. Okay? Because the changes in correlation energy aren't that big.



So, but for these properties, for non-bonded contacts, especially for helium-helium, the correlation is the only game in town often. That's what holds helium atoms together. Helium atoms are nuclei, positively charged; electrons, negatively charged. So nucleus repels nucleus; electrons repel electrons; nucleus attracts electrons; nucleus attracts electrons. Right? At first, at fifty-three angstroms, or fifty-two angstroms, far apart, those essentially cancel. Right? But the motion of the electron around this nucleus correlates with the motion of the electron around this nucleus. They tend to be in-phase with one another. So at any given time you have plus, minus, plus, minus, and they attract one another. So precisely what holds helium to helium is correlation. So if what you're interested in is bonding, then use orbitals, fine. But if you're interested in non-bonded interactions, then correlation can be a big problem. Okay? But we're talking about bonds now, not correlation.



Okay, so orbitals can't be true. That's clear, because electrons influence one another, they repel one another. If you have just one electron, fine; more electrons, orbitals can't be true. But still we'll use them, to understand bonding and structure and energy and reactivity. And we know we won't get precise values, we'll be off by one to ten kilocalories/mol, but we'll get insight that's very useful. Okay. Now what gives atomic orbitals their shape? Why does this particular orbital have a node, for example? We know Coulomb's Law tends to attract the electrons to the nucleus. Why does it have a node and spread out? Because of kinetic energy, right? This curvature of wave functions that's required to solve Schrödinger. So it's kinetic energy that gives them their shape. Or if you double the nuclear charge, the thing gets half as big. That's Coulomb's Law, sucking it in. Right? So the potential energy scales the radius through the formula for ρ; we've seen that. But the kinetic energy is what creates nodes. So the 2s has that spherical node. Or this orbital here has a conical, two cones, and a spherical node; it's the 4d orbital. So kinetic energy is what creates the shapes. The charge just scales things in and out.



Now if we use orbitals, how should we calculate the total electron density? Well we have two one-electron wave functions. We know the density of electron one, at this position, by squaring its wave function. We know the density of electron two at that position, by squaring its wave function. How do we get the total electron density at that same position? How would you get it? You know how much of electron one is there, its probability density. You know two. How do you get the total? Ilana? If you know how much of electron one is there, and you know how much of electron two is there, how much total electron density is there? How would you get it? Can't hear very well.



Student: You'd just add them together.



Professor Michael McBride: Yes, add them together. A whole is the sum of its parts. Okay? So the total density is the sum of those two squared wave functions; just right. But notice it's a sum; it's not a product. This is not a question of joint probability. It's not what's the probability that electron one and electron two are there at the same time? That's not the question. The question is, what's the total probability of finding any electron there? Okay? So it's a sum, it's not a product. Okay? Now we had this question you looked at before: How lumpy is the nitrogen atom? This picture is taken from a recent organic text and they had a fancy graphic program or an artist or something that drew something that looks very realistic. Right? But we can check it because we know the formulas. And if we want to get the total electron density, we add the electron density of the electron that's in the px and the one that's in the py and the one that's in the pz orbital. So we square them, and we get this, and we sum it up to get the total electron density; and it's some constant times x²+y²+z² times e-ρ ; after we've squared. Right? And how can you simplify that? What's x²+y²+z²? It's r². Okay? So how does it depend on θ? How does it depend on φ? Elizabeth?



Student: It doesn't.



Professor Michael McBride: It doesn't depend. What does it look like?



Student: A sphere.



Professor Michael McBride: It's a sphere. It's a ball. It doesn't look like this thing. So it's spherical. So forget, for all the elegance of that picture, forget it; they're not showing you the truth. Okay, or this problem we had before of looking at the cross-section of the CN Triple Bond, where we sliced it and turned it and thought it might look like a cloverleaf, or maybe some kind of diamond shape or something. But in fact it's round. Why? Because (2px)² +(2py)² depends on x²+y², which is r², in two dimensions. Right? So it's cylindrical, doesn't depend on the angle around the bond axis.



Okay, so this is what we've seen. We've seen three-dimensional reality, hydrogen-like atoms. We talked about hybridization. We saw that orbitals are fundamentally wrong but that we can recover from the orbital approximation and use it, if we -- for example, with self-consistent field; as long as we're not interested in getting the very finest energy but can be satisfied with approximation. Now we're going to move on to something more interesting, which is molecules. We're going to look first at Plum-Pudding molecular orbitals, and then understanding bonds, and overlap, and energy-match.



Now there are lots of different ways of looking at the electron distribution. You know this story, presumably, about different blind men doing experiments on an elephant and getting completely different ideas. The same is true of the electrons in a molecule. So here's the electron density in a hydrogen molecule. It's calculated, but there are ways to get that kind of information experimentally as well. So, and you know how contours work; it's a lot more dense near the origin, near the nuclei. So which contour do we want to look at in order to understand it? Well we can choose our contour, and we get different pictures, different understanding, depending on which contour we choose. For example, we could choose, way down, a very high electron density. Right? And then we see just a set of two atoms; it just looks like two atoms. Right? We don't see the fact that it's a molecule. Where have we done this before?



Student: [Inaudible].



Professor Michael McBride: Andrew? Shai? Pardon me?



Student: Difference density.



Professor Michael McBride: Not difference density.



Student: Well, we were subtracting to get the difference density.



Professor Michael McBride: Yes. We looked at total electron density; it just looked like atoms. We had to do the difference in order to see that it was not just atoms. It was essentially just atoms. So if you look at high density, you see just a set of atoms; no excitement there. So that's the molecule as a set of atoms, just a set of atoms. Okay?



Or we could take a somewhat lower electron density contour and we'd see that the atoms are a little bit distorted. Or we could do a difference map, as Shai says, in order to see that they're a little bit distorted, because bonding distorts the shape of the atoms a little bit. But first -- and we'll, this is what we're going to look at shortly -- but first I want to get some insight by looking at the very lowest electron density. So that would be molecules formed from a set of atoms, rather than as a set of atoms. Right? There are little changes because of the bonds.



But how about if we look way out there? Now, if you don't look really close, it looks spherical, it looks just like an atom, or almost like an atom. And if you went even further out, it would get spherical, for all you could tell. Okay? So that's the molecule looking like an atom, with whatever number of electrons the molecule has. So that's the molecule as one atom. The only difference is that the nucleus, which for an atom would be in the middle, has been split, to give two nuclei, which of course distorts the shape of the electrons a little bit, if you move -- cut the nucleus in two and split it out. So we want to look at a few molecules, from this point of view, as single atoms -- because we know about atoms now -- but distorted by the fact that the nuclei has been split apart. So this is nuclei embedded in a cloud of electrons.



Now what gave those electrons their shape, the shape of the cloud in which this thing is embedded; what gave them their shape? We just talked about this a short time ago. What gives orbitals their shape? Dana? Can't hear very well.



Student: A desire to be as far apart as possible?



Professor Michael McBride: That's potential energy, right. But that's not what gives -- that's not what creates nodes. That just causes things to spread out. What gives them their characteristic shape of nodes, planar nodes, spherical nodes and so on? Elizabeth?



Student: Kinetic energy.



Professor Michael McBride: It's the kinetic energy. And the same kinetic energy considerations will apply in a molecule as in an atom. It's always curvature of the wave function divided by the wave function. So the electrons are dispersed by electron repulsion, and noded, given nodes, by the kinetic energy; and the kinetic energy also causes them to spread out, as we've seen before. Okay, but you see then that. Although Thomson was wrong about the atom, it wasn't a plum pudding of nuclei [correction: electrons] embedded in a cloud of positive charge. But a molecule is a plum pudding! It's nuclei embedded in cloud of negative charge, but that cloud of negative charge is given its form by kinetic energy largely; also potential energy, as you say. So it's backwards from what Thomson thought. But his idea wasn't a silly one, it just happened to be wrong, for atoms. So here's a piece of plum pudding; which you know is like fruitcake. So how do the plums distort the pudding? Right? We know what atoms would look like now. Right? But if you split the nucleus into several nuclei, and moved them around as the plums in the pudding, they'll change the shape of the electrons. How? That's what we want to look at today.



Okay, so first we're going to look at methane and ammonia, and we want to understand them visually. Why do the orbitals in these molecules have the shapes they do? Okay, so there are four pairs of valence electrons -- there are also two core electrons, 1s electrons on carbon and on nitrogen. So we want to compare the molecular orbitals to the atomic orbitals of neon, which has the same number of electrons; four electron pairs with n=2 in the neon atom. So there are the same number of electrons as neon. So they should, at a big distance, if you look at very low contours, it should look like a Neon atom. But what does it look like as you get closer? Okay, so here's a 1s orbital of the Neon atom. Right? And here's the 1s orbital of methane; it's that net that's shown, the red net. Everybody see it? And here's the -- the contour level that we're drawing is where the orbital, you know, which level are we choosing? We're choosing 0.001 electrons/cubic angstrom is the level we've chosen to draw, of the onion. Right? And here it is for nitrogen. The blue one there is the lowest molecular orbital. But they're just little spheres, right? They're like the 1s of neon. Okay, so the core orbitals are like the 1s of the carbon or nitrogen atom. They're tightly held, they aren't distorted very much because the nuclei have split apart, right? And they're boring. So forget them, we won't talk about them anymore. We'll focus on the valence orbitals where the bonding action will occur. Now, there are eight valence electrons. That means two to an orbital. So there'll be four molecular orbitals that have electrons in them, in methane. And they are arranged in energy this way. There's a lowest energy, and then there are three that have exactly the same energy. Do you remember what we call those when several orbitals have the same energy?



Students: Degenerate.



Professor Michael McBride: Degenerate. Okay? So there are three degenerate molecular orbitals for CH4 that have exactly the same energy. In the case of ammonia it's a little different. There's one lowest one, and then there are three, but one of them is higher than the other two; there are only two degenerate ones there. Incidentally, what does that remind you of, to have one that's a little bit lower and then three that are the same energy? Have you ever seen that before?



Student: 2p and --



Professor Michael McBride: Ah! 2s and 2p, right? There's one 2s and three 2p's, in an atom. Right? Here's a molecule and there's one that's lower and then three that are equivalent. Okay, now there's the lowest one, and it's a 2s orbital. It's the neon's 2s orbital that's been distorted by taking four protons out of the nucleus and pulling them out to be where the hydrogens are. Okay? So you can see how this has been distorted, by distorting the electron cloud into the direction where the protons have been pulled out. Do you see that? Okay, now you don't see the -- you have to remove the ball there, that's the carbon or the nitrogen, to see that little spherical node. It's way down near the origin that made it a 2s orbital. Right? That's the spherical node. And it doesn't look like a sphere because of the algorithm the computer uses to draw this, connecting dots with straight lines. Right? But it is more or less spherical; a little bit distorted by the fact that you've pulled these things out. Okay, now what do these molecular -- these are molecular orbitals. But that, you see, is a 2px orbital, that's been a little bit distorted. Right?



Notice that the difference between the CH4 and the NH3, there was a proton that went up in CH4, which pulled them up. But if you didn't have that, it didn't go up as high. Right? But then essentially this is the 2px orbital of the molecule, distorted from the atom by pulling protons out of the nucleus. Okay? Or that one, what's that one? It's 2py, but you have to rotate it to see it. If we rotate it 90 degrees here, you can see that it's a 2py orbital. And again, the CH4 is distorted at the top because two protons came out and stretched it out. Same thing in nitrogen. Now we go to the third of these 2p orbitals, which is different in nitrogen; for the nitrogen case, higher in energy. There they are. Why is that orbital higher in energy than the others in Nitrogen? Why isn't it so good? Because you have electrons up in that red region that don't have a proton there; it's not being stabilized. Here, the electron density that went up is around a proton. There the top lobe of the p orbital doesn't have a proton stabilizing it. So it's higher in energy. And it turns out to be more reactive. What do you call it?



[Students speak over one another]



Professor Michael McBride: That's the unshared pair. The high-energy reactive electrons are the ones that don't have a proton stabilizing them there. Now there are also vacant orbitals. Right? Remember, you could have any number of orbitals. We're looking at the ones that you can make from the 2s and 2p. But we have a number of atomic orbitals that we can use. We're looking at the lowest set, the ones that come just from 1s orbitals of Hydrogen, and 2s and 2p orbitals of carbon or nitrogen. But there are eight such orbitals: four, 2s and three 2p's, on the central atom; four on the hydrogen for CH4. So there'll be a number of other combinations you can make from mixing those, which won't have electrons to go in them. But let's look at them. So here are the next four orbitals made from the valence level orbitals of the atoms involved. So what you see on CH4 that this is the 3s orbital. It has a radial node. Right? Or pardon me, yes, a radial or a spherical node here, that surrounds the sphere. So it's red sign inside, blue sign outside.



The same thing is true of the NH3, but you have to rotate in order to see it. That's the lowest vacant, the lowest unoccupied molecular orbital. And if we rotate it here, you can see the inner part. And bear in mind that there are two spherical nodes. There's one that's really tiny, that's inside anything we can see here. Okay, then there's this orbital, which is the 3dx²-y², looked at from a funny angle where it's not very clear what it is. But if we rotate all these by 90°, you can see that it's like that cross, or whatever you call it, of the 3dx²-y². But why doesn't it look just like that? Why are they distorted? Why does it have this funny U-shape here. Instead of this lobe, this lobe, here and here, the lobes moved up there. Why? Angela? Pardon me?



Student: There are protons.



Professor Michael McBride: That's where the protons pulled them, the potential energy. But the fundamental shape, before it got distorted by the potential energy of where the protons are, the fundamental shape was from the kinetic energy. It was like the 3dx²-y². Or how about this one? There you can see -- but it's clearer still if we rotate it -- that again it's that X-shaped thing. But the ones on top have been pulled out by the protons. Or this one. That's the one that has a doughnut around the middle. But the doughnut has been pulled down by the three hydrogens that aren't on the z axis. Or there it is, rotated. Okay? Now I hoped to get -- let's just start ethane and methanol, and then we can get to more interesting things in it next time. It has a lot more pairs of electrons, ethane and methanol. So we can compare those MOs to the AOs of Argon, which has the same number of electron pairs. Actually I should quit now and let you get to your next appointment.



[end of transcript]

This lecture begins by applying the united-atom "plum-pudding" view of molecular orbitals, introduced in the previous lecture, to more complex molecules. It then introduces the more utilitarian concept of localized pairwise bonding between atoms. Formulating an atom-pair molecular orbital as the sum of atomic orbitals creates an electron difference density through the cross product that enters upon squaring a sum. This "overlap" term is the key to bonding. The hydrogen molecule is used to illustrate how close a simple sum of atomic orbitals comes to matching reality, especially when the atomic orbitals are allowed to hybridize.

Transcript



October 1, 2008



Professor Michael McBride: Okay, let's get started. So last time we looked at methane and ammonia, and saw something interesting about molecular orbitals: we could give them the same name as atomic orbitals; that is, if we, especially if we look at a low electron density contour of the molecular orbitals, we see that what it looks like is an atom where the nucleus is split into pieces. Right? Now that splitting into pieces changes the potential energy for the electron. So you expect that to distort the orbital. The electrons will move in the directions that pieces of the nucleus have gone. But then, in addition to potential energy controlling the shape of orbitals, kinetic energy also does. And that's the same thing as it is in an atom. You have a thing with no nodes, a thing with one node -- and when you have one node you can have either a spherical node, or a planar node, and there can be three planar nodes; so a 2s and three 2p's. Exactly the same considerations apply in a molecular orbital as in an atomic orbital. You have the kinetic energy, which comes with curvature, which comes with nodes.



Now, as you split the nucleus up and pull pieces in different directions, it doesn't have the same symmetry it had when it was all together in the nucleus, a spherical kind of symmetry. So the nodes get distorted. But still you can see them there. And we saw them last time and went through all the occupied and vacant valance orbitals of ammonia and methane, and saw how they looked like atomic orbitals. This, not surprisingly, because it's so fundamental, the potential energy and the kinetic energy, applies to every system. It applies to you, viewed as a single atom, right? With a zillion electrons. Okay? But pieces have moved around, so the orbitals change. We'll look at two more complicated cases and then we'll get on to a different way of looking at bonding. So we'll look at ethane and methanol. And we use -- I didn't tell you last time, explicitly, where I got the molecular orbitals from. I got them from my laptop. There's a program -- the particular one is called Spartan, that I use -- and it calculates what molecular orbitals look like, using approximate molecular orbital (that is, Schrödinger equation kind of) theory.



Okay, so let's just look at the ones of ethane and methanol. Now both of these have seven pairs of valence electrons. There are also core electrons, and if we were looking at the orbitals for all the electrons, we'd include those. And exactly how we're going to count those -- you could do it one way or the other -- whether you consider the core electrons, the 1s electrons to just be part of the nucleus and then treat the rest as the electrons you're interested in; or whether you want to count the core electrons too. You can do it either way, but what you analogize to what depends on whether you count them as part of the nuclei. So anyhow, we're going to compare these molecular orbitals to the atomic orbitals of argon, which has also seven electron pairs. Okay, so there's the 2s orbital. I'm going to start with that, because I'm going to pretend that that red part is the core -- that there's a 1s which is core electrons. But in this case -- here's my pedantic note on this subject which I just added. So if you have -- before we had just one heavy atom, carbon in methane, nitrogen in ammonia. Right? So there was one 1s orbital. Right?



Now we have two heavy atoms, carbon and carbon in methane, carbon and oxygen in methanol. So there are two heavy atoms and therefore two boring core orbitals. So for purposes of making analogies, we'll use the atomic 1s orbital, of the atom that we're analogizing things to, to stand for all the molecular core orbitals. You can do it any way you want to. We're not really interested in that. We're interested in the valence orbitals. So whether I start with the 1s or the 2s depends on how I'm handling the core electrons; it's not a big deal. Anyhow, let's pretend it's the 2s of argon here, and we're going to compare it with this lowest valence level molecular orbital of ethane.



Now we can -- as I do this, on the left side of the pictures I'm going to show one view, and then, because it's more complicated than methane and ammonia were, I'm going to also show a picture rotated by ninety degrees. So I'm going to rotate around that axis, and on the right I'll show a different view of the same orbital. Okay, so that's the lowest valence level molecular orbital of ethane. And it's not very exciting; it's just a distorted sphere. And you can see the way in which it's distorted. It's distorted vertically, because the two carbons are pulled apart; so it's got sort of a narrow waist to it. And then it's pulled out, where each of the protons left the middle atom to come out and be hydrogens. Okay? Now if we look at methanol, it'll be a little different. Can you anticipate how it might be different if, on the bottom we'll have CH3 again, but on the top, instead of having CH3, we're going to have OH. How do you think it might be different, in the way it looks?



Student: It might be skewed more towards the OH because oxygen is --



Professor Michael McBride: Okay, so oxygen has a bigger nuclear charge. So that's going to pull the lowest energy orbital toward the oxygen. That'll be one thing. We expect it to be bigger, top-heavy. Okay? What else, about how the top will look? How will it be different from if it were a CH3? Yes, Alex? I can't hear very well.



Student: Oxygen has lone pairs.



Professor Michael McBride: Oxygen has lone pairs. Now, how is that going to change things? Or another way of saying the same thing is it has only one hydrogen up there. How's that going to change it, do you think? You have to speak up.



Student: It's not going to be symmetrical.



Professor Michael McBride: It's not going to be symmetrical. Which way is it going to be distorted to be unsymmetrical? You have to speak up.



Student: Towards the electron pair.



Professor Michael McBride: Toward the electron pair did you say? I couldn't hear.



Student: Away from hydrogen.



Professor Michael McBride: Away from hydrogen? Well how about -- the proton goes out; does it bring electrons with it, or does it repel electrons, the proton? Speak up.



Student: It brings them.



Professor Michael McBride: It pulls, so it should distort toward the hydrogen. So here's what it looks like. Right? It's top-heavy, as Angela said, and it's distorted out toward the hydrogen; there are no protons pulling it to the top left. And you see the same thing end-on there, on the right. Okay, this is the next orbital. What does that look like? Obviously you can peek in the middle and see. It's obviously a 2pz orbital, with the node across the middle. Okay? In both cases. Now the ethane case is symmetrical. The other case is unsymmetrical. Why is it unsymmetrical? Because the first orbital pulled electrons to the top. Right? So the next- in the next orbital, electrons aren't going to want to be at the top so much, because they're going to be repelled by the other electrons; so there'll be more toward the bottom, because the first ones went to the top. Okay? Okay, then here's the next orbital. You can see there are energies marching up here. The lowest one was s, then 2pz, now 2px, if we define the horizontal axis here on the left as the x axis. So again, it's what you expect, and it's pulled out, stretched vertically from being a dumbbell by where the nuclei went. And here's 2py, which we can see more clearly on the right, perpendicular to the 2px.



Okay? And notice that it's obviously so, that the methanol orbital will be less symmetric, in all cases, than the one for ethane. But still we can recognize the nodes, because they must go that way. It must be no nodes, one node of three different kinds, and so on. Okay, now this is 3s. You notice it has -- the node that we don't see, the one that's down near the nuclei, in fact two nodes down near the nuclei, one for each of the heavy atoms. But then this now has another spherical node, or approximately spherical node. So we have that extra red lump in the middle on the top, or blue, in the middle. Just to review, what's the difference between red and blue? On the top, the computer decided to draw it with red in the middle. On the bottom it decided to draw it with blue in the middle. What do those colors mean? Yes, Cathy?



Student: Positive and negative signs.



Professor Michael McBride: The mean positive and negative. So which one is right? Should it be positive in the middle or negative in the middle?



Student: Positive.



Professor Michael McBride: Positive is a guess. Wilson, what do you say?



Student: It doesn't matter.



Professor Michael McBride: Why not?



Student: Because it's not really positive or negative; it's just kind of a phase of it.



Professor Michael McBride: Yes, it's the sign of the wave function. But you can multiply the wave function by minus one, any constant, and it's just as good as it was before. So it's arbitrary, and the computer was arbitrary in choosing the colors. I think there's actually a function that I could've changed it, if I'd thought to do so, so they'd be the same. But actually it tells a story if I leave them different. Okay, so here's the next one. Now this is -- if you look on the left now, it's more clear where the nodes are, that that's a dxz orbital. The name 'xz' means that the product of x and z appears in the wave function. Right? So when both x and z are positive, then the product is positive; on the top right, red. Right? When x is negative, and z is positive, the top left, it's negative, the product of them. So that's why it has the name xz.



Okay, and there's dyz, which you see on the right, turned ninety degrees. And here's the dz2, which is that thing that has a doughnut that goes around the middle. Right? It's hard to see the doughnut. Can you see? It's blue on top. You can easily see the red on the top left, which is what's blue in the middle; the sign has changed. Right? But the doughnut is highly distorted, because as you go around, first a proton on the top pulls it up, then a proton on the bottom pulls it down, then up, down. So it's like a crown, the doughnut has been made into a crown around the end, by the protons pulling in that way. Okay. Then here's the 3pz. So it has the horizontal nodal plane, but also it has a spherical node, which you can see in either picture really. Okay?



Then here's the 3p y orbital, which again has that spherical node, but now the planar node, or approximately planar node, is vertical instead of horizontal. So say on the top right there you have red on the right and blue on the left. There's a vertical node over which it changes sign, going right to left. Or here's the 3px. Or here's the 3dxy; which you don't see so well here; well, but if you turned it down you would. And here's the 3dx^2-y^2; which again, to see it well, you'd have to turn it. And here's the 4f orbital. So you can see, especially say in the top ones, compared with the atomic orbital, that it's exactly the same general pattern of nodes, slightly distorted.



And incidentally, remember all the n equals whatever, n=3; all the orbitals had the same energy before. Now they don't have all the same energy. Notice that they all have different energies. None of them are degenerate. Why? Because if you broke the nucleus apart, how stable a thing that has that general shape, like a dumbbell or a cloverleaf or something, how stable the electrons are in those lumps depends on whether a proton got pulled into the lump. If there happens to be a proton in the lump, where kinetic energy, the node pattern, wants it to be, then that'll be unusually stable. If the protons have been pulled someplace where there is a node, because of kinetic energy, then it won't be stabilized. So that breaks the degeneracy of these different patterns that have the same number of nodes. It depends on where the -- how the potential energy changed.



Okay, now just finally and very quickly, I want to look at 1-flouoroethanol, which is a molecule that would have, I think, no stability at all, but you can put it into the -- as a practical matter, you're never going to put it in a bottle, but you can easily put it into the computer and calculate what its molecular orbitals would look like. And I did it to show you something that's very, very unsymmetrical, and has atoms of very different nuclear charge. Okay? So what will the very, very, very lowest orbital look like, do you think, for this thing? It has a fluorine, an oxygen, two carbons and five hydrogens. So what do you think? If you were an electron, and you had that set of nuclei, arranged this way, where would you want to go? Elizabeth?



Student: Very biased to the left, especially around the fluorine.



Professor Michael McBride: Yes. Now what would the very, very, very lowest one? It should be more toward the left, and really close to the fluorine. It would be a 1s orbital, and mostly on fluorine, because that's where most of the protons are, the most concentrated protons. So let's look. There's a smaller scale ball model, so that we can see really small orbitals, and there that's, you're absolutely right, it's essentially the 1s orbital of fluorine, is the very lowest orbital. What would be next? Suppose this one, you came up and this seat was already taken, now where would you go, the next electron? Zack?



Student: Oxygen.



Professor Michael McBride: To the oxygen, because that's the next highest concentration of protons. And next? Where will you go next? Steve, what do you say? The 1s of fluorine is taken. The 1s of oxygen is taken. Where do we want to go next? Pardon me?



Student: One of the carbons.



Professor Michael McBride: Ah, which carbon?



Student: The left carbon.



Professor Michael McBride: Why?



Student: Because it's closer to the fluorine than the oxygen.



Professor Michael McBride: And what does the proximity have to do with it? The proximity means that the electrons that are on that atom will have been drawn toward -- and we'll talk about this more later -- will have been drawn toward the fluorine and the oxygen. Therefore that atom will have fewer electrons, or lower electron density. Right? Therefore, it's a better place for other electrons to go. Okay? So you'd expect the next one to be the middle carbon. Right! And finally, of course, it's going to be the second carbon, the one that's remote from the electronegative, high nuclear charge atoms. Okay, now we've done all the 1s's. So we can look at what's interesting, the valence orbitals, the ones that are going to be involved in bonding; these don't have anything to do with bonding. Okay, so there's the first one. Russell, tell me something about the shape of this? Does it surprise you?



Student: No. It's highly distorted towards the fluorine.



Professor Michael McBride: Yes, it has no nodes; except it actually has little nodes down near the nucleus, because it's actually more like a 2s orbital in that respect, because of the core electrons. So it has tiny nodes that we don't see around the nuclei. But it has no nodes within the valence; big orbitals that we're looking at. So it's like an s orbital. Right? And it's big where the nuclear charge is big, as you say. Okay? What's the next one going to look like? Any ideas? Pardon me?



Student: Towards the oxygen.



Professor Michael McBride: Toward the oxygen. Will it also have no nodes? No. The next highest orbital has to have a node. Where do you think the node will be? What orbital will it look like, sort of? It'll be highly distorted. But what -- if this is the s orbital, right? What's going to be next? Elizabeth?



Student: It's as if it's switched but with a planar node in between.



Professor Michael McBride: Oh let's see. Right! So the first one was -- you were right Russell that this one should be big on oxygen. But notice it's like a p orbital, it has that horizontal node, because it's higher kinetic energy. Okay? And then this one, that's on the carbons, right? Mostly. But it's a py, it has a vertical node. So it's negative say on the fluorine and oxygen and positive on the carbons. Or if we rotate this one around a horizontal axis to look at it from the side, it looks like that, and you can see that it's a py kind of orbital. Elizabeth?



Student: I'm just asking a point of clarification. These aren't technically p orbitals, right? We're just saying they're analogous to them.



Professor Michael McBride: They're analogous to p orbitals; because it has to be the same. The orbitals must go in order of the number of nodes. Right? So the same thing is in an atom or in a molecule. You go from no nodes to one node. And there are three ways of getting one node: spherical or distorted sphere, planes, or distorted planes if the nuclei are pulled around; and so on. So they're really very much like the atomic orbitals. Okay, or this one. Now that's an interesting one because that's like a hybrid orbital. It looks like the orbital up here. It's a mixture of s with p. Did everybody see how that is? It's a big sort of blue lobe on the top and a small red one on the bottom. And also the next one is the other one, the hybrid that points the other direction, another combination of the s and the p, that points down. Okay? And then this one looks like dxy again. Right? It's sort of a cloverleaf with two nodes. Okay, so that's all I want to do with that, and we won't go any further with it. This is just an interesting way of looking at how -- that molecular orbitals are really just like atomic orbitals, and have energies for the same reason, except the potential energy gets screwed up by breaking the nucleus and pulling pieces around, but in an understandable way, and the nodes get distorted because of this.



Okay now, now we're getting into really -- we just looked at this view, at the single united atom view. But the other view is the one that's going to be more generalizable, and that's the one where we looked at bonding. Right? So you have to probe a little harder to get a qualitative understanding of what chemical bonds are. And that's what we're going to do now by choosing a higher contour with which to look at a molecule. Now, true molecular orbitals, to the extent that orbitals are true all together -- why aren't they true all together; why aren't orbitals true all together? Yes, Alex?



Student: Because you have multiple electrons.



Professor Michael McBride: Because you have many electrons; you can't have independent electrons, you can't have orbitals. But we're approximating things by orbitals, trying to take electron interaction into account in a sort of a left-handed way by Self-Consistent Field, or something like that. Because it's much easier if we can divide the whole into a bunch of parts, each of which we can understand. So, to the extent that molecular orbitals are true -- the kinds of things I've just been showing you, calculated with my laptop -- they extend over the whole molecule; they're not local. Right? Except like the 1s of fluorine was local, but mostly they go over the whole molecule. But bonds are thought of, and have always been thought of, as interactions between pairs of atoms. So we want to divide things completely differently and look at the bonds now, at pairwise LCAO molecular orbitals. Now what's an LCAO? It's a sum, or a linear combination. Right? A weighted sum of atomic orbitals. So here's an example. Ψ, which is an orbital -- what's it a function of?



Student: Position.



Professor Michael McBride: Position of what?



Student: One electron.



Professor Michael McBride: One electron. So it's a function of x1,y1,z1; we're talking just about electron one. So the wave function for electron one we say is 1/√2 times one atomic orbital plus another atomic orbital. Right? Now, have you ever seen adding orbitals like that before? That's what hybridization is; we added s and p. But this is different, because when we added s and p before, they were on the same nucleus, and we did it to get a new orbital for that particular nucleus for that atom; to distort it one way or the other, for example, or to rotate a p orbital. Right? But this is very different, because we're adding orbitals that are on different nuclei: A, nucleus A, and nucleus B. See the difference? Adding is just -- wave functions are numbers, we just add the numbers. But in the first case, hybridization, those two functions were on the same nucleus. Now they're on different nuclei, what we're adding together. Okay, now why is it sensible to think that you might get pairwise molecular orbitals that can be expressed like this? How do you interpret an orbital? Corey? What good is an orbital? What do you use it for?



Student: It's a one-electron wave function.



Professor Michael McBride: Well what do you use it for?



Student: For probability.



Professor Michael McBride: And how do you get probability? From the wave function -- if you have the wave function, how do you get the probability density?



Student: You square it.



Professor Michael McBride: You square it. So do you see why we have a 1/√2 in this? Because when we square it, that's going to be half, and we're going to get half of atomic orbital A squared, and of atomic orbital B squared; so it's a half of each of them. That's why we have 1/√2. So let's go on with this. Suppose we have a hydrogen molecule, and suppose that the nuclei are at a great distance from one another. So far they don't interact, or negligible interaction; they're very far apart. What would you expect the lowest energy, one-electron wave function to look like? One possibility is that the electron could sit exactly halfway between the two nuclei? Right? Is that a reasonable place, is that the low energy place for it to be? Lucas, what do you say?



Student: No, because the added probability density there is not the greatest.



Professor Michael McBride: Why not?



Student: It needs to be one or the other.



Professor Michael McBride: Why shouldn't the electrons sit -- if the two nuclei are this far apart; and I don't mean two angstroms apart, I mean two meters apart -- there's a proton here and a proton here. Is the electron most likely to be here, halfway between?



Student: No.



Professor Michael McBride: Where would it be most likely?



Student: Probably nearer to one of the two atoms.



Professor Michael McBride: And which of the two?



[Students speak over one another]



Professor Michael McBride: Suppose you took the long view. Suppose you averaged it over eighteen zillion millennia -- time averaged. Sometimes it would be near this one, sometimes it would be near this one. Okay? What would it be if you took a really long view?



Student: Both.



Professor Michael McBride: And half here, and half here. So the wave function, when you square it, you want it to be half looking like this atom, and half looking like this atom. Then you see the time average. Right? It might take a long time to achieve that average, because it'd take a long time for the electron to tunnel two meters, right? But in the very, very long time it would look like that. So we know what we want it to look like. What we want is that the probability density, the square of this one-electron wave function, should look half of the time like the atomic orbital A squared and half of the time like atomic orbital B squared. So on time average it's half of one and half of the other. Everybody with me? So that's what the -- yes, Nate?



Student: Why isn't there a 2A -- or a 2AB?



Professor Michael McBride: Oh, because I'm telling you what it looks like. It's got to look half like this and half like this. Right? So the square of it has to be this. So all we have to do to find the wave function is what, if we know what its square is? All we got to do is take the square root and we've got the wave function. Bingo! There's the square root. Right? So that is a reasonable way to write the wave function, 1/√2 (AOA + AOB). Claire, you have a question.



Student: From my small understanding, with the math that I've got -- and this may be wrong, correct me if I'm wrong. But don't you, if you have a bracket and two things inside it, you square outside of it, don't you have four things that come out of it.



Student: [Inaudible].



Student: You can't just put the square --



Professor Michael McBride: You want that?



Student: Yes.



[Laughter]



Professor Michael McBride: Okay, now Claire you're going to help me out. How big is -- that's a number. It's the product of -- atomic orbital A assigns numbers everywhere in space, everywhere in space. Atomic orbital B assigns numbers everywhere in space. So at some point in space atomic orbital B assigns a number and atomic orbital A assigns a number; and A times B is the product of those two numbers. How big is that product? How big is atomic orbital A here? This, a meter away from the proton?



Student: Not very big.



Professor Michael McBride: And how big is atomic orbital B there?



Student: Not very big.



Professor Michael McBride: Now how big is atomic orbital A here?



Student: Very big.



Professor Michael McBride: Okay. So now, how big an error do we make if we neglect A times B? Where do we make an error? Do we make an error here?



Student: No.



Professor Michael McBride: Do we make an error here?



Student: No.



Professor Michael McBride: Do we make an error here?



Student: Yes.



Professor Michael McBride: No! We make an error; yes, sure enough, we make an error. How big is the error?



Student: Not very big.



Professor Michael McBride: Ah, it's negligible, because it's at great distance. Okay, so at great distance we can forget that. So now it's easier to take the square root, right? Okay? The old fox up here, huh? [Laughter] Okay, now your problem is what happens if H₂ is at the bonding distance? What if they're only an Angstrom apart? Now there should be a problem, because A times B is not going to be negligible everywhere now. Okay, so now that's going to come back. So now we got an error, right? Or do we? Let's think what that does. Okay, so we if approximate the molecular orbital as the sum of atomic orbitals, this way, then it looks very good near the nuclei. Because near A it looks like A; near B it looks like B. And if we want to square it to find the electron density, we do this. But if we then subtract, what the atoms would give for electron density. Now what does this remind you of, where we look at the total electron density and subtract the atoms?



Student: Difference density.



Professor Michael McBride: So we're actually looking for the difference density. Everybody with me on this? So we're going to subtract the atoms, which is ½A² and ½B². Right? If we subtract, what do we get? What do we get for a difference density? We get A times B. So we get the difference electron density, which is due to overlap. And what do I mean by overlap? I mean that is only important in regions where both of them have a finite value. Right? The product of A and B is negligible if A is zero; it's negligible if B is very, very small. Right? So it's only where they overlap, the two functions have simultaneous values, that you care. Okay, now that thing, the thing that's the bonding, the difference density, is really a byproduct -- that's a little bit of a pun because it's a product -- but it's a byproduct of squaring the sum -- of what Claire didn't like about it. So the very thing you didn't like is what's going to give rise to bonding density. Isn't that neat?



Okay, but notice that here we're multiplying A times B. But this is a completely different instance of multiplying from what we had before. Right? Before we multiplied two orbitals to try to get a two-electron wave function. This has nothing to do with this, because both of these are functions of the same electron; it's like one electron that we're squaring here. So this A times B, this product, this overlap, comes from the squaring. It was when we squared it that we got that. Okay? Now, because we have this extra term, we have not only ½(A²+B²), which would -- what's the probability of A² summed over all space, or integrated, if it's normalized?



Students: One.



Professor Michael McBride: And this one?



Students: One.



Professor Michael McBride: And what's this whole quantity?



Students: One.



Professor Michael McBride: One, because of the half. But actually we've got it bigger than that, because we added the overlap term to it. Right? So it's actually not going to be half; it'll have to be something a little less than half, so that it'll sum up to one, if we want to normalize it. Okay? So there's going to be those less than halves there. Okay. Now what does that do? That shifts electron density, right? We're taking electron density away from where the nuclei are, from A² and B², and where's the electron density going? Because we're using less than half to begin with, we're taking electron density away. There's going to be -- we're subtracting more than was there at the beginning. Right? Which means that we're going to have negative electron density in the difference map. Electrons are going away from the atoms. Where are they going to?



Student: Into it.



Professor Michael McBride: Into the region where there's overlap. Right? So they go away from there, into the overlap region. So this is just like what we were seeing with X-ray. Okay, so that overlap, the A times B term, is what creates bonding. And we've seen this before. Remember when you have wells far apart, the wave function is the sum of the two; we saw this in one-dimension. Right? But if they come close together, you get a wave function that looks like that, which we looked before, just from the point of view of the energy, and saw that that would stabilize the particle, because it's got less curvature, less kinetic energy. Right? But also the electron density grows in the middle. Right?



So from the point of view of the electron distribution, that was the glue holding the atoms together. So it's held together, both because the energy goes down and because you put this glue in the middle, which is what causes the energy to go down. Okay? So that's bonding. And remember we also had this. So as the energy went up in the middle one, the energy is lower [correction: higher] here than it was in the atoms apart. So the nuclei push one another apart now, without the glue in the middle, and that was anti-bonding. So we've seen it before in one-dimension, but it's true in three-dimensions as well.



Now let's think about this again. So here's atom A. Now where is the square of that function significant? Is it significant there? No. Is it significant there? Yes, it's a little bit significant at least. How about there? A little bit. Right? Okay. Now suppose we have another atomic orbital there. Now, where is the product significant, of A times B? Okay? So is the product A times B significant there? No. Is it significant there? Nick, what do you say?



Student: No.



Professor Michael McBride: Why not? Speak up.



Student: It's very small near A.



Professor Michael McBride: What's very small?



Student: The value of A.



Professor Michael McBride: No, no. The value of A there is something -- we said that before, when we were looking only at A -- it's not very big but there's a significant value. But how about the product? Josh?



Student: The value of B is very small.



Professor Michael McBride: The value of B is very small there. So the product is going to be very small. How about there?



Student: There's going to be a change.



Professor Michael McBride: Ah, now they're equal; halfway between they're equal. So both of them are a little bit small, but their product is still significant. Right? Only in this region, where they overlap, is that product significant. Okay. So at the center, notice that the number ΨA assigns and the number ΨB assigns are the same number. So 2(ΨA ΨB) is as large as (ΨA)²+(ΨB)²; because ΨA times ΨB is the same as (ΨA)². So the electron density is nearly doubled in the middle from what it would've been if it had just been two atoms. So that's the region of significant overlap, and that's what we care about. So the overlap integral, summing this product -- or integrating it -- over all that space, that's a certain density. Right? We squared in order to get that. Right? That's part of the density. So we sum that, the density that comes from that product, over all space, and that's called the overlap integral. If the atoms are very far apart, the overlap integral is essentially zero. If the atoms are close together, the overlap orbital will be finite, and the better the -- the more the orbitals overlap, the bigger the overlap integral, obviously. And that measures the net change that arises on bonding, the difference density, as we've just seen.



Now let's look at some theoretical examples here. So let's look at the total electron density as calculated for two -- adding two 1s orbitals of hydrogen at the appropriate distance for H₂. This was this was done forty years ago and published in the Israel Journal of Chemistry. Okay now, and here's -- so on the left we have the total electron density that you'd calculate from that. A very simple thing: 1/√2(A+B). And you square it, and you get the density, and that's the density, contoured at 0.025 electrons/cubic ao, the unit of distance. Okay, now, on the right is the difference density. So from that thing on the left we've subtracted the atomic orbital, the atomic electron densities. And you see exactly what you expect. It builds up in the middle, where there's overlap, and at the expense of the atoms. Yes Russell?



Student: Shouldn't it be H₂+ ion?



Professor Michael McBride: Oh, I think it's the H₂ molecule, I'm sure it's the H₂. They aren't so fantastically different, because the two electrons are in the same orbital; two electrons can be in the same orbital. So one'll be twice as big as the other. Qualitatively they'll look very similar. And I think this is the H₂ molecule; but it might be the ion, I'm not sure. Okay, at any rate the contours on the right are much smaller. Remember, difference density is much smaller than total density. So what you see is that it's contoured at 0.004. So you have one, two, three, four, five contours. So you get up, in the middle, to 0.02 electrons per cubic angstrom [correction: ao]. That's how much bonding has changed things, at the maximum. Okay, now the energy that's calculated, with this very, very, very crude wave function, just one half -- 1/√2 times the sum of the two atomic orbitals. The energy, you calculate that, is 92.9% of the true energy. That's pretty darn good, right? But almost all of that energy that you calculate was already present in the separate atoms. We're not interested in the energy of the separate atoms, we're interested in how much it changes when you make a bond, which is a small difference between large numbers. So it turns out that although we're within 7% of the true total energy, this simple model only calculates about 50% of the change in energy that came from putting them together, right, which is much smaller.



Okay, so high accuracy is required to calculate a correct value of the bond energy. This simple thing won't do it. Well it's in the right direction, you're halfway there, so it's a pretty good start. Right? But to do the difference, as in the same way you needed high precision to do X-ray difference maps, you need better orbitals than this, if you want to calculate good bond energies. So you need to make the orbitals better. Okay? Now -- so but already we can take heart that the very crudest model shows most, 52%, of the energy of the bond, and it shows the electron density building up by 0.02 electrons per cubic bohr radius. And what we saw qualitatively was there was a shift from the atom to the bond, of electron density. Okay, now we can adjust the molecular orbital to get a better approximation of the true thing.



How will we know when we've adjusted it and it's gotten better? If we adjust it and get a lower, calculate a lower average energy -- I should've said a lower average energy, because if we don't have a true wave function we get different values for the total energy; the kinetic won't exactly offset the potential as you move from place to place. But if you get the lowest average energy, then that is, by definition almost, more realistic, because you can easily prove that the true energy is the lowest possible energy; that makes a certain amount of sense. The lowest possible calculated energy is the true energy. So if you change your wave function and get a lower average energy, you're closer to the truth. Okay? That's called the Variational Principle. Okay, so here we've changed the form of the molecular orbital. And how did we change it? What does a 1s orbital look like? Kate, do you remember the form for a 1s atomic orbital? I can't hear.



Student: A sphere.



Professor Michael McBride: Angularly it's a sphere. How does it change as you go out, do you remember? How does it depend on ρ? The same way they all depend on ρ. Anybody remember?



[Students speak over one another]



Professor Michael McBride: e-ρ. So it falls off exponentially at a certain rate. And that rate, how fast it falls off, is determined by the nuclear charge. Okay? Now one way to change that shape would be change how fast it falls off. Right? It wouldn't be correct for the atom anymore. We've already got the correct solution for the atom. But we could change the shape of the thing by changing how fast that exponent falls off. And we could vary that, in the molecule, using one half -- 1/√2(A+B). But those A+B are no longer true atomic functions, they're a little fatter, a little skinnier. Okay? And we can change how fat or skinny it is, until we get the lowest molecular energy. See, that's a way you can vary it and find the best value. And that's what was done here to optimize the exponent. And now you get a total electron density that looks essentially the same as it did before. And if you look at the difference density, how has it changed, if you do this? First, notice that the energy got lower. We're now to 73% of the lowering of the bond energy. So the total energy's gotten lower, it's better.



And how has the electron density changed? It got higher in the middle. Because what we did was spread the exponent out a little bit, so you had more overlap in the middle. Okay? So this wouldn't have been good for the single atom, to spread it out, but it gives a better function for the molecule. And it's still very, very simple. And what you see it did is it increases the bonding density and the bonding strength. You get a larger shift from the atoms to the bond. Now, how else could you change the shape of the atomic orbital in order to increase the overlap; some way other than making a single exponential and having it get fatter or thinner? Can you think of some other way? Here you have an atomic orbital, a sphere, and you want to change its shape so that it overlaps better over here. Right? How could you change the shape of an atomic orbital, without doing really gross damage to it, making it a cube or something like that, or a line? How could you change it so it looked pretty much still like an atom did, has a lot of the virtues of the atom, but is shifted over here? Sam? I can't hear.



Student: Can't you just allow the electrons to shift?



Professor Michael McBride: Yes, how am I going to write a function that allows the electrons to shift in the direction I want them to? Lexy?



Student: You could hybridize it.



Professor Michael McBride: Hybridize it! We could hybridize to shift the electrons. So that's the next one. So here we're going to -- instead of this we're going to hybridize it. Now, this particular calculation did the hybridization and also did a little self-consistent field calculation. And the hybridization left it 96.7% 1s. So essentially it's still a normal 1s orbital. But they added .6% of 2s, which expanded it a little bit, because 2s is -- goes further out than 1s. And they added 2.7% of 2p. Now why was 2p much more helpful than 2s? Lucas?



Student: It has those lobes.



Professor Michael McBride: Yes, so it takes density from one side and shifts it to the other. In fact, this is precisely what we saw before. Notice what it did was -- how much it increased the density in the middle. And notice now, it's not taking electron density away from the nuclei, which was a good place for electrons to be. Where does it take it away from?



Student: The left one.



Professor Michael McBride: Out beyond the nuclei, and shift it to the middle. So even before, when it was an atom -- here's the nucleus, a certain distance out here, and a certain distance out here, were the same in energy, for that atom. Now we've taken it from a place which is -- out here, and put it here. Right? Great idea. Well done Lexy. So there's what it looked like. Remember, if it's 100 percent at 1s, it looks like that, and if you change it to be hybridized that way, with Atom-in-a-Box, it looks like this. Right? It's not much shift, but it shifts from the left to the right, and gives better overlap. Okay, but it requires the atom to be a little bit less happy as an atom, because it's partly 2s and 2p now. The electrons are further from the nucleus, but you make up for that by having a better bond. Okay, now notice that didn't change the energy very much. It went from 73% to 76%. Right? We were already about the right energy, but it changed the density a lot, to be what we want it to be. Okay, so there's a much bigger shift, and it's now from beyond the nucleus into the bond. Right? And now we're going to do the last thing. We've done SCF already, but now you have to do a higher level calculation that'll do correlation, take the correlation of the electrons into account. And if you add some correlation calculation to this, you now get 90% of the bond. If you did complete correlation, you'd get 100% of the bond. What does it mean to do a calculation with complete correlation? It means you don't have an error anymore, you've used a really good calculation. So already at this level, that was done fifty years ago almost, you get 90% of the bond. And the bond density notice, what about the electron density?



Student: No change.



Professor Michael McBride: It hasn't really changed. All that happened was that the electrons kept apart from one another, but the average density was the same. So already, with just that hybridization, we got very close to the truth in electron distribution, and three-quarters of the way to the truth in how strong a bond is. So not a bad approximation. So hybridization, to give better overlap, is a great thing. So the density wasn't changed, but you got a much better energy. And how so? Because the electrons kept apart from one another, when you allow correlation.



Okay, so here's a pairwise atomic orbital: < 1/√2(A+B); and you can use hybridized orbitals. And the virtues are it's very easy to formulate and to understand. And it looks like atoms, especially when you get down near the nuclei. And you don't want that to change because that's the main event for electrons. You get much more energy forming atoms, as we saw before, than you do making new bonds, once you have the atoms already. Okay. It builds up electron density between the nuclei, through overlap, which is the source of bonding. It smoothes Ψ, to lower the kinetic energy. And then there's a pedantic footnote here that actually there's a thing called the Virial Theorem, which I'm not going to stress you with, but that little bit there, it happens to be true. But still we have the proper understanding of what's going on. Hybridizing AOs provides flexibility that gives you better overlap. And if you use all the H-like Atomic Orbitals, you have perfect flexibility, you can make any shape you want.



Okay, but we're going to keep it simple, use only 2s and 2p orbitals to hybridize, because that'll get you most of the way there and it's much simpler, rather than to try to mix 5f orbitals into it also. Okay, so that's great. And when we square it, we get this, which has the overlap part that helps us out. We have the atoms, plus the bond, which is the overlap, that product part. But we could've done the same thing to get the same product, as far as the atoms go, if we'd used a minus sign instead of a plus sign in combining things; although then we would've changed the sign of the overlap thing. It would become minus. So we would change it from being less than to being greater than, in order to have it be normalized at the bottom. Right? And that's the anti-bond. So we get both the bond and the anti-bond by doing this. And now we're going to go on, next time, to overlap, which we've already introduced, and also the concept of energy-match. And when you put these two together, you'll really understand bonding.



[end of transcript]

Professor McBride uses this lecture to show that covalent bonding depends primarily on two factors: orbital overlap and energy-match. First he discusses how overlap depends on hybridization; then how bond strength depends on the number of shared electrons. In this way quantum mechanics shows that Coulomb's law answers Newton's query about what "makes the Particles of Bodies stick together by very strong Attractions." Energy mismatch between the constituent orbitals is shown to weaken the influence of their overlap. The predictions of this theory are confirmed experimentally by measuring the bond strengths of H-H and H-F during heterolysis and homolysis.

Transcript

October 3, 2008



Professor Michael McBride: Okay, let's get started. The stuff we're doing these few days are really the focus of the first half of the semester. So pay attention and think about it. Okay, so the topics today are Overlap and Energy-Match, as determinants of how strong bonds are. So we'll start with overlap. We saw last time that overlap is what creates the difference density, what focuses some small amount of the electron density between the atoms and serves to hold them together. The overlap integral is the total amount of that product of A and B. Remember, you square (A+B) and you get A² + B² + 2AB. That 2AB term is the overlap. It's called overlap because it has values when both functions have appreciable magnitude, at the same point in space; they overlap. And summing that over all space, or integrating it, gives what's called the overlap integral. It'll depend on the distance. Obviously if they're very far apart, they don't give simultaneous values to the same point in space, because it's far from either one or the other, or from both. So it depends on distance. It also depends on hybridization, and that's what we're going to talk about first.



Okay, first let's just look for scale. Here are the 2s orbitals of carbon atom, and you can work out, with your formulas, that the diameter of the spherical node is 0.7 angstroms. Right? That means we have a distance scale here. Now, the distance between two carbons is roughly 1.4, 1.5, when they're forming a bond from one another -- between one another. Okay. So that means we can use that 0.7 angstrom scale to see how far we have to slide these together, in order to get them to bonded distance. If we superimposed the X's -- remember those diameters are each 0.7 -- so if we slide them together until we superimpose the X's, then they're 1.4 angstroms apart. So that's about the distance that these two orbitals are, in the proper scale, when they're in a carbon-carbon bond. Now, how big is the overlap integral, the sum of the product of A times B, over all space? Now I want you to guess, guess how big that integral is. And we have a hint. What would the overlap integral be if we moved them until they superimposed on one another exactly? That's the maximum we could get, if the distance were zero. So if pulled them all together, then A and B would be measured from the same center. So it would be the same thing as A². Okay? So what would the integral of A² be, over all space?



Student: One.



Professor Michael McBride: One; that's normalization. Okay, so the maximum value you can get when they're right on top of one another -- negatives multiply negatives to get positive; positive areas, multiply positive areas to get positive. Everything is optimum, is going to be one. Now I want somebody to give me an idea, a guess, of how big it's going to be here, or at least some considerations that would go into that. The maximum value you can get is one. How big do you think it would be here? What's something that would hurt it; that is, make it less than one? Russell?



Student: When there's very little overlap on the opposite sides.



Professor Michael McBride: Yeah, out beyond the red areas, there's very little overlap altogether. So you lose whatever overlap you'd get from there, in doing the normalization. Now, in the middle, right in the middle, you have blue on top of blue. That's good. But notice there's also a node that's going to interfere with some of the wave functions' overlapping, and there's going to be a little bit of blue on top of red. Everybody got that? So I just want people to guess. Okay, I know what I'll do. We'll do a poll. This is a democracy after all; we're about to get into November. So I'm going to start with 0.1, as what you think it would be. Right? The maximum, if you put them right on top of one another, is one. So I'm going to start with 0.1, and move up and see hands.



Okay, how many people think it's 0.1? 0.2? 0.3? 0.4? 0.5? 0.6? 7, 8, 9, one? Obviously not one. Now, I think, my sense of this is it peaked between 0.3 and 0.4, but there were some votes for 0.2, and I would've voted for 0.2, because it looks like you've lost an awful lot here. But in fact I personally am surprised that it's pretty big. It's 0.41. It's almost half as big as it would be if they were right on top of one another, despite the fact that you're losing great parts of it, that they don't overlap at all, and there's negative overlap where red's on top of blue. Still the fact is it's 0.41; so pretty big. Now that's going to depend on distance obviously. Right? So here's how it depends on distance. There's that point; at 1.4 angstroms it's 0.41. And I've put on here for reference the distance of a carbon-carbon single bond, the distance of a carbon-carbon double bond, and a carbon-carbon triple bond. And zero, notice, would be way down, out in the hallway someplace. Right? We're only looking at the region that's carbon-carbon bonds -- or maybe just about to the far wall. Okay? Here's half an angstrom; we got to go two more halves. Right? Okay, so here's how -- how do you think -- what should the line look like? Should it be sine wave? Should it be increasing as we go to the left? Should it be decreasing as we go to the left? What do you think? Some guess.



[Students speak over one another]



Professor Michael McBride: Should probably decrease. Because we know it's going to approach one. Right? And there it is. Yes, it decreases -- pardon me, it increases as you go to the left, and ultimately it'll go up to one, when you get out to the hallway. Okay? So that's s with s. Now let's look at some other orbitals that might overlap, and see how big their overlap integrals would be. Okay, suppose you have s on one, overlapping with p on the other. Now that p is called, for this purpose, π; π is the Greek version of p; in fact, in Greek you pronounce it 'pee,' not 'pie'. Right? So that's the Greek letter p. Am I right about that?



Student: It depends on classical or modern Greek.



Professor Michael McBride: Okay, which way is it?



Student: It'd be 'pee' in modern Greek; it'd be 'pie' in classical.



Professor Michael McBride: Uh-huh, okay, good. So anyhow, 'pie.'



[Laughter]



Professor Michael McBride: Now, so that symbol is used to talk about molecular orbitals, or parts of molecular orbitals; where p is used to talk about atomic orbitals. They correspond to one another. Now what is it that characterizes a p atomic orbital? Has a nodal plane. And in the same way a π orbital has a nodal plane that contains the nuclei that are in question. Right? So you see, if those two came together there'd be a nodal plane in the orbital on the right that contains the two nuclei. Okay? So that's called a π orbital. Okay? Now, but notice the contributions to the overlap. Below that plane, you're going to have blue overlapping with blue. So that's going to be a positive contribution. But above, you're going to have blue overlapping with red; that's going to be a negative contribution, and it's symmetrical. So what will the total be? What will the total overlap integral be; the sum of all the overlaps?



Student: One.



Professor Michael McBride: One?



[Students speak over one another]



Professor Michael McBride: It'll be zero, because for every contribution on the bottom that's positive, there'll be one on the top that's the same value and is negative. So it's going to be zero. So the one on the left is called σ. Now σ is the Greek equivalence of 's.' Right? So it means, for a molecular orbital, what s meant for an atomic orbital; there are no nodes that contain the nucleus. Okay? Okay, so if you have σ overlapping with a π, there's going to be this symmetry that causes a cancellation. So you can never get any overlap, by overlapping a σ orbital with a π orbital. Does that make sense to everyone? It does make sense, doesn't make sense? If you have one that doesn't have a horizontal node, that's the same top and bottom, and one that does have a node that's plus on the top, minus on the bottom, then you're going to get this cancellation everywhere, and zero. And that's called ‘orthogonal'. Okay, so you won't get any net interaction there. Okay, but if you turn the p orbital so that it is also σ -- notice it has a plane, a nodal plane, that contains that nucleus, but it doesn't contain the other nucleus you're interested in. So to call σ and π, we need a plane that contains both nuclei. That'll create this symmetry that cancels things out. Okay, so this one will give overlap. Now what would you expect for the trend of this one, as you go from great distance to small distance? If they're very, very far apart, how big will the overlap integral be; like ten meters apart?



Student: Zero.



Professor Michael McBride: Zero. Okay? Now, as you bring it together, as they start coming together, will you get positive or negative overlap?



Students: Positive.



Professor Michael McBride: Positive; right, the blue will overlap with the blue. And you keep coming, it'll get bigger and bigger. And then what will happen?



Student: [Inaudible].



Professor Michael McBride: Then the red is going to start coming in and -- well the blue is going to start, the blue on the right, from the p orbital, will start -- and going to overlap that red part on the left. And then, as it slides on over, till they get exactly on top of one another -- when they get exactly on top of one another, what will it be?



Students: Zero.



Professor Michael McBride: Zero, because it'll be canceling right to left. Okay, so here's a plot of that, for s-p σ overlap. Right? It's almost the same as s with s. In fact, it's a little bigger at great distance, a little smaller at small distance; except, at very small distance what will happen?



[Students speak over one another]



Professor Michael McBride: When you approach zero? It'll go to zero. Whereas the other one went to one. Right? The s with s went to one. So, but in the region we're interested in, the carbon-carbon bond distances, it's about the same. Okay, now how about pσ with pσ? Well we won't go through an elaborate guessing game. If they're very far apart it'll be zero. How about if they were on top of one another? It's obviously going to grow as blue gets on top of blue, and then as they get really close, the blue up from the right will be on top of the red, and the red on top of the blue. Right? So what will it approach? Zack, what do you say?



Student: Zero.



Professor Michael McBride: Approach zero? What will it be when they're right on top of one another? So red on blue here; blue on red here. Will the overlap be zero?



Student: Negative one.



Professor Michael McBride: Pardon me?



Student: Negative one.



Professor Michael McBride: It'll be minus one, because it would be like the orbital on top of itself -- that would be one -- except you've changed the sign, of one of them. Right? So every contribution will be negative. So that one starts there, and it's heading toward minus one. So it's going down. Okay? So that's pσ with pσ. Can you think of any other way we should talk about it? We've talked about s with s. We saw that s with pπ was orthogonal; s with pσ we've got there; pσ with pσ. Any others we need to think about?



Student: pσ with pπ?



Professor Michael McBride: How about pσ with pπ? Can anybody guess that one? Pσ is the same sign, top and bottom; pπ is opposite signs, top and bottom.



Student: So it's zero.



Professor Michael McBride: That's going to be zero. There'll be orthogonal. So have we done it now, or is there anything else to think about?



Student: π, π.



Professor Michael McBride: We could do pπ with pπ. Okay, so there's pπ with pπ. Okay? Now at great distance they'll be nothing. We'll bring them together, they'll begin to overlap. What will it approach as we go to zero?



Student: Zero.



Professor Michael McBride: Zero? If they're on top of one another, the product will be zero?



[Students speak over one another]



Professor Michael McBride: It'll be one, right? It'll be the orbital with itself, normalized. So that'll be one. Right? But now which do you think is going to be bigger, pσ with pσ, or pπ with pπ? Any guesses? Well, I'll give you the answer. Right? pπ with pπ is approaching one. At big distance, it's smaller than σ. Right? But at small distance, it must get much bigger, and in this region it's crossing; so it's bigger. Okay, so there are the kinds of overlaps of simple 2p orbitals. Now, there's a curiosity here. Over most of this range, more than half of the range, 2s overlaps with 2pσ better than either 2s with 2s or 2pσ with 2pσ. That just seems curious to me. You'd think that one of them would have a shape that gives better overlap. Right? So two of those together would do the best job. But it's not that way. It's that an s with a p is the best, over most of this range. I just think that's curious, and it has an important implication, which you'll see right now. What if we use hybrid orbitals, that are partly s and partly p? Now, since this one is partly s, it's s plus a certain amount of p. This one's s with a certain amount of p. The overlap will be s here with s here; s here with p here; s here with p here; and p here with p here. Right? There'll be four contributions: s with s; p with p; s with p; p with s. Right? So if we took sp3, for example -- and we can make, remember, four such orbitals -- then you get that. It's better than any of the pure hybrids. Right? And the reason is that it has s with s, and p with p; and also s with p, that comes in as a bonus, twice. Okay, now how about if we use, instead of sp3, how about if we use sp²? Remember what happens as we approach sp? The orbitals extend more. So what do you expect?



Student: More overlap.



Professor Michael McBride: More overlap. So there's sp² with sp², a little better. How about sp with sp? Better still. Could we get any better? How about s²p with s²p, or sp½ with sp^½? Right? It's a little better at short distance, and a little not so good at long distance. So it optimizes around sp hybridization. That gives the best overlap. Why do we think now you want good overlap? Why would it be good to have good overlap?



Student: Keep the bonds strong.



Professor Michael McBride: Because that's what's building up the difference density, as we saw last time. Right? That's putting the glue in the orbital. Okay, so there's an interesting observation here, that hybrids overlap about twice as well as pure orbitals. Right? All the hybrids are roughly twice as good as the pure ones. Okay? And they're not very, very different from one another. But sp is about the best, over much of this range. The trouble with sp is you can only make two of them, because each involves 50% of s, and you only have one s to go in it; you have one s and three p's. So if you want to make just one orbital, fine, use sp, because you'll get good -- only one bond from the carbon; fine use sp and get the best overlap. Or maybe even s²p, depending on the distance. Right? But if you want to make more bonds, more than two bonds, then you're going to have to cut back on the s in each bond. Because you only have 100% to use, you can't use more than 50% in more than two bonds you're making. Okay. And what you see here is that sp3 isn't much worse than sp². So why not make four -- right? -- because you can. Okay, anyhow, that's the hybridization of carbon. So, and the reason they do this, as I already said, is because they allow nearly full measure of s with p overlap, plus s with s, and p with p, when you have mixtures. So that all depends on the fact that s-pσ is so good in its overlap. Now, so sp gives the best overlap, but only allows two orbitals, with 50% in each; sp3 gives four, and nearly as much overlap. Okay?



Now we're going to look at the influence of overlap on molecular orbital energy. But we're going to use Erwin Meets Goldilocks, just for familiarity. So we're working here in just one-dimension, to get the idea of it. And we'll also then think in more general terms. And we've already done this, so it's partly review. So we'll assume that there's perfect energy match; that is, the two atoms we're talking about. Of course, atoms would be Coulombic potentials, not harmonic oscillator, Hooke's Law potentials. Right? But the idea of a double-minimum is the same. So we'll suppose that the atoms are equivalent. So the bottom of the well is the same on both sides. And guess an energy. And lo and behold we were right; we got the solution, and you've seen that one. Okay? But you can also get that one, one with one node. Right? And on the far left you can see that they have the same curvature-over-amplitude. So they have essentially the same energy. If the green energy is right, the red energy is the same deal. Right? They're ‘degenerate' orbitals, because they're far apart. Okay? No significant energy difference.



Now, suppose we increase the overlap. What will happen if we shorten the distance? You did this on a problem set with Erwin. Right? So you bring them closer together, and you have the one with no nodes, and the one with one node. And now there's a little -- right halfway between there's more electron density in one than there is in the other. Right? The antibonding one has zero and the bonding one has some buildup of electron density in between. But the energy hasn't changed visibly on this scale that we're using. It looks just the same. Okay? Now we increase it more and now there's a big difference in the middle. Right? And now the energies are beginning to change. The unfavorable orbital is going up and the favorable bonding orbital is going down in energy. And if we get it still closer together, we can see that there's now a big difference between them. Okay? So that increasing overlap creates the splitting that we talk about.



This is just review, as you know, but it's worth -- it's so important that it's good to mention it again. Now that overlap holds the atoms together. And this again is a picture we saw before, except we're going to do it with writing-in the equations. Remember, a good guess of the form of the molecular orbital is 1/√2 (A+B), or 1/√2 (A-B). And that will give two new orbitals, one better and one worse. And, in fact, if you think about normalization, it should be a little less than 1/√2, in the one that's going to have +2AB in it, when you square it, and a little less [correction: more] than 1/√2 when it's going to have a minus sign for the 2AB term. Okay? Now, because of that ‘a little bit greater' and ‘a little bit less', the energies aren't quite, as we saw them before, equally split, up and down. The one that goes down doesn't go down quite as much, because this is less than, and the one that goes up, goes up a little more because it's greater than. The mathematical requirement for that isn't immediately obvious to you, but it's true.



Okay, so there are the new orbitals you get when you bring the atoms together. Okay? And if you had more overlap, there'd be a bigger difference. So that's why overlap is important. The more overlap, the more splitting. We just saw that in one-dimension with Erwin Meets Goldilocks. Now, how many electrons do we have to put in these orbitals? That's going to make a big difference. So suppose we have just one electron for these two orbitals. When we bring the two wells together, when we bring the two atoms together, that electron will go down in energy, to the new molecular orbital. So that's going to be good. And if we tried to pull them apart, the electron would have to go back up in energy, and that would require energy, and that would oppose the breaking. So that's the bond strength. Right? You have to put an electron up that much. Now suppose there were two electrons. Now a second one goes down the same way. So it's even stronger. It won't, in truth, be quite twice as strong. Do you see why? Yes?



Student: Electron repulsion.



Professor Michael McBride: Because the electrons will repel one another. We haven't taken that into account when we think about the energies here. But it'll be certainly stronger, considerably stronger. Okay. Now suppose we had three electrons. What's going to happen?



Student: [inaudible].



Professor Michael McBride: One of them is going to have to go up instead of down. And, in fact, it will go up further than either of the others went down, even neglecting electron repulsion, right? Because of that 'greater than/less than' thing that shifted the pair up a little bit. Okay? So that's going to be even weaker than having one electron to hold them together, when you have three. How about if you have four?



Student: No more bonds.



Professor Michael McBride: Wilson?



Student: They'll be antibonding, so --



Professor Michael McBride: Yeah, you're now definitely losing, because the two that go up, go up more than the two that go down, go down. So that's bad. So we can summarize that here, the effect on the bond. If you have one electron, it's bonding; two electrons is strongly bonding; three electrons is weakly bonding, probably; and four electrons will definitely be antibonding. So that's why helium bounces off helium, until it gets fifty-two angstroms apart, and then there's this very, very, very weak bond, that we talked about, due to the correlation of the electrons. Right? But as soon as you get overlap with helium, it's repulsive, when you start to split these levels. Okay, now let's look at the other end. Why doesn't increasing overlap, as you bring things closer together, why doesn't that make these plum-puddings collapse? The electrons are getting more and more stable as you increase the overlap by bringing things closer together. So why doesn't it just collapse? In fact, the electron -- if a hydrogen molecule collapsed to helium -- it would be a funny helium because it would have no neutrons -- but if it collapsed, the electrons, indeed, would become 55% more stable. The electrons in helium are lower in energy than the electrons in H₂. Right? So that would tend to pull them together, all the way, to distance zero. Why don't they? Elizabeth?



Student: The protons repel.



Professor Michael McBride: Because the protons will repel one another. That depends on 1/r. Right? And that increases much more dramatically. In fact, the amount by which the electrons become more stable is 650 kcal/mole, or would become more stable. Right? But the nuclear repulsion increases by that much by the time they get to 0.3 angstroms, let alone the last little bit, when 1/r gets enormous. Okay? So it's not worth it; unless you have glue to hold the neutrons together. Right? So on the sun you can take deuterium and fuse it into -- to make a helium -- you have neutrons there to hold the protons together -- and then you get 200 million kcal/mole; which is, of course, nuclear energy, not electronic energy. It is possible, but not under the chemical circumstances that we're talking about. Okay, so here's the Morse potential. Yes, Lucas?



Student: Have we been taking proton-proton repulsion into account before, with all our energies?



Professor Michael McBride: No, or only sort of subliminally. Right? Because if you take -- at first glance there's going to be nuclear-nuclear repulsion, and electron-electron repulsion, between the atoms; nuclear-electron attraction; nuclear-electron attraction. So to first approximation, those will all cancel out. Right? But then you get -- on top of that you have a little change from the fact that the nuclear-nuclear goes as 1/r and gets quite big, when they get really close, and that the electrons are coming down because of overlap. And we've been focusing on this overlap thing that causes the bonding, but obviously things aren't going to collapse because of the nuclear part. But we haven't been talking about it explicitly; you're right. Okay, so anyhow this is the form of the bonding potential, the Morse potential, which was just cobbled together artificially to make it convenient to calculate things and to have reasonable properties for a bond. But now we understand why it should look like that. This first part, the attractive part, is because the electron pair becomes more stable with increasing overlap, as the atoms come together. And then the nuclear repulsion becomes dominant, when it becomes too close. Right? So all this is Coulomb's Law. There's nothing special. There's not ‘correlation energy', which is some completely new physical phenomenon. Right? It's all Coulombic. Correlation energy is just, remember, an excuse for the error you're making, or a way to hide it.



Okay, so it all comes from Coulomb's Law for potential energy. But the funny thing was the kinetic energy of the electrons, according to Schrödinger. Right? Because as you make things -- as you concentrate things, you change the curvature as well. So you have to take that into account. But this curve provides the potential for studying molecular vibration; that is, once you know the form of this curve, then you can use quantum mechanics for the motion of the atoms -- not of the electrons but of the atoms -- and figure out how atoms vibrate; which we've already done, when we were doing Erwin Meets Goldilocks. Okay, but here, finally we understand the atom-atom force law. Right? That's what we've been doing the whole time, for four weeks or five weeks or whatever it's been, is trying to understand where -- what the force law is. Wouldn't Newton be happy? Remember, he said, "There are therefore agents in Nature able to make particles of bodies stick together by very strong attractions, and it's the business of Experimental Philosophy to find them out." And now we've found it out. Now, can we use this understanding then to gain greater purchase on chemistry?



So we've looked at overlap. But there's one other thing we have to look at, in order to apply it, and that's what we call energy-match. I don't think anybody else talks about energy-match, but we do. Okay, so here's what we talked about already. Right? And we have this splitting, which is about proportional to overlap. It's not numerically strictly proportional to overlap, but it certainly is close to proportional to overlap. Okay, now what would happen if the two atoms you started with were not the same, if one was a better place for electrons to be than the other; like there for example? So why would you -- remember what happened is you mix these. You get A+B and A-B. But why would you put any A together with B, if B is already much better than A? If they're equal, you don't care where it is, then fine, use a 50/50 mixture. But if one's better than the other, why use any of the inferior one? Why not just use the one that's good?



So we already actually saw this in the case of the 1s atomic orbitals, which did remain pure and unmixed, when we made the molecular orbitals of that funny molecule last time, the fluoroethanol. Remember, it looked like that. One of them was just an s orbital on F; another one was an s orbital on O; s orbital on one carbon, s orbital on the other carbon. The didn't mix. Right? So the compact 1s core orbitals did remain pure and unmixed, but the valence level atomic orbitals were heavily mixed. For example, the next one was biggest on fluorine, but also substantial on oxygen, and carbon even got into the act a little bit. Right? Now why is it that the core orbitals didn't mix and the valence orbitals did mix? What's different? The core orbitals are very small, very compact. These orbitals are bigger. So what happens? They overlap. Right? So it's overlap that causes things to mix. If there's no overlap they don't mix. If they overlap they mix.



So why use any of an inferior orbital? Well suppose the energy of A is much higher, less favorable, than that of B. Right? So now we make a sum, a weighted sum, a parts of A, b parts of B, and we square it. So we get a² parts of the A density, b² parts of the B density, and also this overlap term, 2ab. Now, can you profit from shifting electron density toward the inter-nuclear AB region, without paying too much of the high energy cost of A? That is, A² is higher energy than B². Right? The electron density around A is higher energy than the electron density about B. And if you mix A with B, you're going to get a certain amount of that bad, or less good, distribution, A². But at the same time you're going to get some of AB, which puts stuff in the middle, at the expense of outside. That's good. Right? At a certain distance from the nuclei, you'd rather be between the two nuclei than out beyond one of them. So as you increase a, you increase -- oops sorry [technical glitch] -- you increase this bad part, but you also increase that good part. Can it be worthwhile to make a non-0? Which one is going to help?



But notice this. If you put only a small amount of A in, then the amount of A² probability density, this bit here, is a² parts of that; really, really tiny, if a is tiny. But the amount of ab you put in is a times b, not a². Right? So it's much bigger. So you get this good stuff, without getting very much of this bad stuff. Right? For example, suppose you used, a was 0.03 and b was 0.98. So you square them, to find out how much -- [technical adjustment] -- you square them, and you find out that the amount of a² you get is only 0.001. The amount of b² is 0.96. And the amount of 2ab is 0.06. So you're getting sixty times as much of ab as you get of a². So that's why it can be useful to put a little bit in. And if you're pedantic, you can look at the footnote there.



So let's look at Erwin Meets Goldilocks in the case where the two wells are not matched, where the energies are a little bit different; because that's what we're talking about. So here are non-degenerate atomic orbitals. So the one on the left is a little bit lower than the one on the right, but only a teeny bit. Okay? Now we're going to look at the wave functions here. So there's the lowest energy wave function. Right? And notice it's the sum, it's a weighted sum, with no nodes. But there's almost no mixing at all, just a really, really, really tiny amount of A in it. So essentially it's just the solution you'd get in the left well. Right? There'd be one with one node as well. What will it look like, can you guess? This is the one with zero nodes. What's the wave function with one node look like? Any guesses? Yes, Josh?



Student: It'll just have a node in the middle of that first potential and then be straight on the second one.



Professor Michael McBride: And be what?



Student: And be sort of straight on the second one.



Professor Michael McBride: Right. So the first one, the lowest one, is the one on the left. The second one, the next highest energy orbital, is the one on the right, with just a little bit of the one on the left in it; negligible amount of mixing. Right? And their energies, you might think their energies would be the same, because they're not mixing. But the green one is in this well, and the red one is in that well, and that well's higher. Okay? So it's got to be higher in energy. That spitting is due only to the original offset between the wells. There's no shift of it. Right? It's just you have A or you have B, and you have what you would expect for them. Now suppose we increase the overlap between these. What do you think is going to happen? You should get a little more mixing because AB is going to count for more, if you have overlap. Okay, so here they are close together. You still don't have very much mixing, just a little bit, and you haven't seen any appreciable change in the energies. You still have that same splitting that's due to the original offset. Okay?



Now we'll bring them still closer together. And now it's getting worthwhile to put it in the wrong well, because the AB term is so important. Right? So the antibonding energy is rising, the bonding energy is falling -- that's why they're antibonding and bonding -- and if we bring them still closer together, we get that. Right? Which, if you looked at it casually, they look like a single well. But notice that they're still quite unsymmetrical, they're still biased. The one without any nodes is still mostly in the left well, and the one with a node is mostly in the right well. But they're heavily mixed. Right? So what's happening is this increasing overlap, as you bring them together, is fighting the effect of the difference in the two wells. So when we had the core electrons in fluoroethanol, if we had tiny orbitals that didn't overlap, then they stayed pure. Right? But once we went to the valence level orbitals, that were much bigger, so that they overlapped, then they started mixing, like this.



Okay, now what if one partner is lower in energy than A? That's the case we're dealing with. But what will these ultimate energies be, when you have this competition between overlap, which is trying to mix them, and energy mismatch, which is trying to keep them separate from one another? So here's what you get. You get again splitting, but not as much shift as you got before. The lower level will look mostly like C, because C is better than A. The antibonding will look mostly like A, with just a little bit of C in it. And if they had had the same energy, you'd get a big energy shift because of overlap. But now you get a smaller energy shift, because the mismatch is fighting that. Right? You don't want to -- you lose as you try to mix, because of the difference between A and C, in this case. Now, so that one looks mostly like C, both in shape and in energy. That one looks mostly like A, both in shape and energy. But it's a little worse than A, and the first one is a little better than C.



Now how much smaller is that bonding shift? When they were mismatched, they didn't shift as much with the same overlap. How much less? Exactly where does it end up? Well that's not so crucially important to you, but there's a neat trick you can do to see that. Suppose we have that much energy mismatch. Okay? And the red dot is halfway between B and C, or between A and C, since A and B are the same energy. Now, so when they were perfectly matched, A and B perfectly matched in energy, that's how much splitting you got from that amount of overlap. So these are the two factors that go into it: how much splitting you get when they have the same energy, and how different the original energies are. How can you put those together, to find out how much shift you actually get? Well what we do is slide that over, and then bend down the blue one, so it's perpendicular to that, and make a rectangle around it, and draw the diagonal. And that diagonal will be the new energy difference between the new levels. Okay? You say, why do I make this construction? It's because there's a quadratic formula that comes in. So the diagonal is the square root of the squares of the two sides, the sum of the squares of the two sides. Okay, so we can rotate that; put a tack in there and rotate it. And that's going to be the new energy difference. Okay? And it's bigger than the original energy difference, because the diagonal has to be bigger than one of the sides. But because of the normalization, because one of them is a little less than -- one of them is a little smaller and one's a little greater -- that was the less than/greater than √2, 1/√2 thing -- it's going to shift up a little bit, both levels will shift up a little bit, like that. Right? I'll do it again so you see it. Both will shift up. So there are the new levels. That's how you do it. Okay?



Now, so for a given overlap, the bonding shift, how much the electrons changed when the atoms came together, is reduced if the energies aren't well matched, if you have the same overlap. You don't get as strong a bond, from that point of view. The energies didn't shift down as much, or up as much. But still A+C will be lower in energy than the original one was, A+B. Okay? And we had that construction, which shows us an interesting thing, which is the splitting is not very sensitive to the one of these contributions that's smaller. There are two contributions: the overlap part, that's the black arrow -- or pardon me, that's the blue arrow -- and the energy difference, the energy mismatch, which is the black arrow. But if one of those is very small, like here, the mismatch, which is black, is very big, and the overlap, the blue one, is very small. But you can see that the length of the diagonal is not going to be very sensitive to what the overlap is. You could increase that blue overlap, the width of the rectangle, which wouldn't change the diagonal very much. So it's sensitive to the one that's bigger. Okay, don't worry too much about that part. I just think it's fun.



Okay, so we can generalize from this. Mixing two overlapping orbitals gives one composite orbital that's lower in energy than either of the parents, and one that's higher in energy than either of the parents. Okay? The lower energy combination looks -- that is, its shape -- is most like the lower energy parent; and the same is true for energy. Right? And the higher one looks like the higher parent. For a given overlap, increasing energy mismatch decreases the amount of mixing, and decreases the magnitude of the energy shifts. Now what does this have to do with bonding? Okay, the AC electrons are clearly lower in energy. But that's not really what we're interested in. We're interested in the change of energy that comes when you break the bond, or make the bond. So which bond is stronger, AB or AC? How many think AB is stronger, given this scheme? How many think AC is stronger? It's clearly got lower energy electrons, AC. But this is a classic example of "compared to what?" Okay? How are we going to break the bond? Suppose what we're going to do is break it, to put two electrons in C, because that's the lower energy atom; or in the other case, to put two electrons in B. Okay? So both electrons go the same way. Can you see anything bad about that? Why might it be better to put one electron each way? Kevin?



Student: Well in terms of spin, they have to be --



Professor Michael McBride: Not spin. You can put two electrons in an orbital with opposite spin. So spin's not a problem, if you only have two. If you had three it would be a problem, but with two, spin is not a problem. But there's an obvious problem with having both the electrons go to one. Dana?



Student: They'll repel each other.



Professor Michael McBride: They'll repel each other. If you could put them on opposite ones, so much the better. Right? But it would be possible to do this, and it might plausible in the case of AC, where C is lower in energy than A; although the repulsion might make it better to go one on C and one on A. Okay, but anyhow, suppose you break it this way. Which one would be easier to break? Which requires less energy, red or blue?



Students: Red.



Professor Michael McBride: Red requires less energy. Okay, so AB is stronger, if you're forming A+ B-, if both the electrons are going the same way. Right? So mismatch aids heterolysis. Heterolysis, (uneven breaking) is to put them both on the same side, rather than one on each side (that's called homolysis, as you'll see in a second). So if you put them both on one side, then it's easier to break if they're mismatched. The bond is stronger if they're well matched. Okay? That's because that. But now suppose you do homolysis and put one on each. Now which one's easier?



Student: The blue one.



Professor Michael McBride: Now the blue is less energy than the red. Okay? So now the AC is going to be stronger. So mismatch hinders homolysis. Right? So you can't really say which bond is stronger, unless you say, "Compared to what -- how are you breaking it apart?" Okay?



Is all this true? So I've been weaving this fairytale for you, for the last couple of weeks, but is it true? Well, we can check it with experiment. For example, compare HH with HF. HH has perfect matching between the two atoms. In HF, there's a much bigger nuclear charge on F; electrons in the valence orbital are lower in energy on F than they are on H. Okay, so if we look at the σ orbital, the valence σ orbital, it's symmetrical in the case of H₂ and quite unsymmetrical in the case of HF. It's big on F and small on H, just as we were speaking of. That's the σ, big on F. Right?



But there's also an antibonding orbital, σ*, which has a node. So that one is the contrary; it's big on H, small on F. This is just what we were talking about a few minutes ago. And the electrons, of course, are only in the σ, not in the σ*; there are only two electrons and that one's empty. And the * means that it's antibonding. Right? So it has the node between the nuclei, planar in the symmetrical case and bent in the case of HF; but it's still the node. Now, how about the experiment? Heterolysis, where you break, in order to get H+ H-, in one case, or H+ F- in the other case. In the case of HH, it costs 400 kilocalories to break the bond. In the case of HF, it costs only 373. Incidentally, this is a lot of energy, to break this bond, because you're putting both the electrons the same place. It can happen in solution, but it's not so easy in the gas phase, which is what we're talking about. Okay, so indeed we were right, that mismatch weakens the bond; for heterolysis, the HF bond is weaker.



That's why you call it hydrofluoric acid, because it can easily break off an H. You don't call H₂ hydrogenic acid or something like that. Right? But if you break it into two atoms, two H atoms or an H atom and an F atom, then it costs only 104 for HH, but 136 for HF. So the HF bond is stronger, if you're breaking to atoms. So that's the first verification, experimental verification, of what I've been talking to you about. And we're going to go on next time to talk about XH3, which will have lots of experimental data that will show that this stuff is sensible.



[end of transcript]

This lecture brings experiment to bear on the previous theoretical discussion of bonding by focusing on hybridization of the central atom in three XH₃ molecules. Because independent electron pairs must not overlap, hybridization can be related to molecular structure by a simple equation. The "Umbrella Vibration" and the associated rehybridization of the central atom is used to illustrate how a competition between strong bonds and stable atoms works to create differences in molecular structure that discriminate between bonding models. Infrared and electron spin resonance experiments confirm our understanding of the determinants of molecular structure.

Transcript



October 6, 2008



Professor Michael McBride: Okay, so we're in the payoff period of having spent all this time on quantum mechanics, and we want to see first whether what we've done is realistic or not. So last time we looked at H-H and H-F, in terms of what the bond strengths were, and we saw that indeed one of them was more stable for breaking into ions and the other more stable with respect to breaking into atoms. But here's a more extensive reality check today, which involves a lot more experimental evidence. And it has to do with XH₃; that is, the molecules BH₃, CH₃ and NH₃. And we're going to look at their structure, and how they move, and see how that checks out with what we expect theoretically. So first, what do we expect theoretically? Well BH₃, CH₃, NH₃ are the same, except boron has three valence electrons, carbon has four, and nitrogen has five. Now each of them uses three of their valence electrons to share with hydrogens to make the three XH₃ bonds. But boron then has no leftover electrons, carbon has one and nitrogen has two. So we want to see what the implications of this are going to be for the structure and dynamic behavior of these molecules, and then we'll check it against reality. First we need to look at something that shows that the hybridization that you use to make bonds should be related to structure.



Now suppose there is one of these atoms that uses an sp hybrid orbital, that looks like that. You see the little dot where the nucleus is. And suppose it happens also to use some other hybrid orbital to make another bond. Now, you want to have, because of the Pauli Principle -- and I'm sorry, we don't have time to go into this now -- the orbitals that you're using must not overlap one another. Because if there's net overlap between two orbitals then -- and you have electrons in both of them -- those electrons are, to a certain extent, the same thing, if the orbitals overlap. If they don't overlap, then they're independent. So to have only two electrons per orbital, the orbitals must not overlap. So we can use that as a criterion to say something about what other orbitals could be used at certain angles. For example, suppose the other hybrid orbital that this atom was using was sp³. Now you'll notice that -- remember -- let's suppose the green and the blue are positive, and the red and the yellow are negative. Now how much overlap is there? Well there'll be positive overlap where green overlaps blue, and there'll be positive overlap where red overlaps yellow. But there'll be negative overlap where green overlaps red, or where yellow overlaps -- no, where blue overlaps red, or where green overlaps yellow. Okay? Is that right? Yeah, where blue overlaps red, or green overlaps yellow; that'll be negative. And this, there'll be some angle at which those things exactly cancel out and you get no net overlap, no overlap integral. And we're interested in knowing what that angle would be; that is, how should hybridization be related to angles, be related to structure? Well, so the question is, what's that angle? And there's a nice formula, that isn't too hard to work out, why it should be that way, that relates spm and spn, the values m and n, to the angle that should be between them, in order to have no overlap. So let's use that formula and look at several mn's, and what the angles is. Let's look at sp with sp; sp² with sp²; sp³, sp³; and sp-anything with spinfinity. What's spinfinity? That says the ratio of p to s is infinite. So it's a pure p. So what angle would be pure p orbital make with any hybrid? Okay, so those are the things we're going to talk about. So first let's try the first one. You can do the math for me. Okay, so it's sp1, sp1; m and n are one. So the denominator's the √1, that's one. The cosine of the angle is minus one. What's the angle? What angle has a cosine of minus one?



Student: 180.



Professor Michael McBride: 180. Okay, good. So that would be linear. And you knew that, I bet, already. Didn't they teach you that in AP Chemistry, that if you have two sp orbitals, they're linear, or if you have bonds that point opposite direction you use sp hybrids for them. If they didn't, that's true anyhow. Okay? Now how about sp², sp²? So we're going to have m and n are both two; √4 is two. So it's the angle whose cosine is minus one half. Anybody know what that is? I don't -- pardon me? Sixty? That's +½. It's 120, to be -½; it has to be greater than ninety to be negative. So that's going to be 120º, and that's trigonal. So sp² hybrids make an angle of 120º. Now how about sp³ hybrids? That's √9; so minus a third. That one probably isn't stored in your head, but it's an interesting angle. It's 109.5º, which is the angle in a tetrahedron. Right? So that methane, which makes sp³ bonds at the carbon, has tetrahedral angles, 109.5º. Now, how about sp-anything, with n=infinity? What's the denominator?



Students: Infinity.



Professor Michael McBride: It's infinity. So what's the angle whose cosine is…



Student: 90º



Professor Michael McBride: Okay, 90º. Right? So that means -- that is what makes things σ and π. Right? A p orbital, with any other hybrid, has to be perpendicular to it. So you won't -- that's what we talked about, σ and π, last time. So the p orbital is π, and it's perpendicular to anything else, 90º, no matter what other hybrid. Okay, so now how about this particular one we have here where it's sp³ and sp¹? That turns out to 125.3º. So that's what the angle should be, if the carbon were making one bond with sp and one with sp³. Right? But the important thing here is that's there's a relationship between hybridization of two orbitals on the same atom and what the angle should be between them. Yes, Angela?



Student: When would you have an sp³ and an sp hybrid; like be at angles?



Professor Michael McBride: Well we're going to have to decide what determines the hybridization. Right? And it could be different for two bonds from the same atom, and we'll see examples of that. So that will become answered soon. It's if they're bonding to different kinds of things, they may have different hybridizations; and we'll see an example of that, several examples. So we want to optimize the hybridization, that is to get -- yes, Lucas?



Student: Does different weightings of the hybridized orbitals change the angle, degrees?



Professor Michael McBride: Yes, there are two ways. You could change the angle by moving the nuclei around, for some other reason, whatever it is, and that would then change the hybridization of the carbon that's making the bonds. So it works both ways. If you change the hybridization, you change the angle; if you change the angle, you change the hybridization. Okay, now we want to optimize hybridization. What do mean, optimize? We mean we want to get the lowest possible energy. We want to choose the hybridization that'll give the lowest hybridization [correction: energy], and therefore a particular geometry, for XH₃, certain angles. Okay, now let's think about that. We're going to have one of the valence electrons of the X atom in each of three bonding atomic orbitals, bonding to the hydrogen, and whatever's left over goes into the fourth atomic orbital, and we want to minimize the energy. Okay? Now let's first look at it from the point of view of the bonds. We want to make the strongest possible bonds. We want to maximize the overlap in the bonds. What hybrid gives the best overlap? sp. But we have to make three bonds; sp you can only make two bonds. So what's the best you can use if you want to make three bonds?



Student: sp².



Professor Michael McBride: sp². So no matter whether it's BH₃, CH₃ or NH₃, all of them will overlap best in making three bonds, if you use sp². So the bonds cry out, "Use sp²"; that'll maximize the overlap and minimize the energy of the electrons in the bonds. Okay? But that's not the only game in town, because there's also the other electron that's on the atom, if there is another electron. Okay? Because carbon has one and nitrogen has two other electrons. So they should get a say in it too. So the X atom, which has to accommodate whatever's left over, says, "Okay, go ahead, make three bonds, but get the lowest energy for the other electron, that I'm the only one responsible for." Right? So what's the lowest energy you could make for a pair of electrons, or for a single electron; what's the lowest energy valence orbital?



Students: s.



Professor Michael McBride: s. So the atom says, "Okay, make your three bonds, but make sure you don't let s orbital go begging." Right? Use the maximum you can of s orbital, because otherwise you're using high-energy atomic orbitals when you might not need to. Now sometimes these would agree and sometimes they would fight with one another, the bonds and the atom. Okay? So from the point of view of the atom, boron, with only three valence electrons, all being used in bonds, doesn't have anything else. So it would certainly want to use three sp² and leave the p orbital vacant. That gets the maximum utility out of the s; puts the s where the electrons are. So that's what boron says, use three sp²'s, for the bonds. What does carbon say -- no, pardon me, what does nitrogen say? Nitrogen, if it wanted to be the best atom, would put two electrons, the unshared pair, into the s orbital, and one into each of the remaining three p orbitals. Right? That gets the maximum use of out of s. You don't want to waste s by putting it where there's only one electron on the nitrogen atom. You want to use it where there are two electrons. Okay, so nitrogen says, "Well, I'd prefer to use three p orbitals to make my bonds, and put two electrons into the s orbital." Okay?



Now, what does carbon say? Carbon says, "Look, I got four electrons; three of them are going to be in some kind of hybrid, that's going to make the bonds, and one of them's going to be something else. But if I take the s out of one place, I'm going to put it into the other one. Over all the four, I'm going to use up all the s character. There's one electron in each of them. Who cares?" Right? So carbon really doesn't care, from the point of view of the carbon atom, about hybridization. It'll care for making -- for the bonds, but it doesn't care for the unshared electron. Okay, but now we have to reconcile these two considerations. So the boron will make the best bonds if it uses sp² and it'll make the best atom if it uses sp². So great, they agree. So boron should really want to use sp² hybrids for its bonds. Remember, those are the ones that form 120º. So it'll be the pattern of a Mercedes star. Yes, Yoonjoo?



Student: Can you explain again why we use three p instead of like, the hybrid orbitals for nitrogen?



Professor Michael McBride: If you're just looking at a nitrogen atom, it's going to have two electrons in one of its orbitals, and one in each of three other orbitals. Right? So those two electrons, that orbital that holds two electrons, should be as low as possible in energy; and that's s. So put two in s and one in each of the three p's, and that'll give the lowest energy atom, with one electron in each of the four orbitals -- pardon me, with two in one and three in the other three; one in each of the other three. Okay? Okay, so but anyhow, with BH₃ they agree, from both points of view: the strongest bonds and the best atom. So it'll certainly want to be flat, and 120º bond angles. Now how about CH₃? CH₃, from the point of view of the bonds, wants to be sp², and the atom doesn't care. So fine, be sp², be flat, be planar, with 120º bond angles. But it won't care quite as much as boron does. So carbon says it'll prefer to be planar, but not as strongly; so it'll be easier to bend it, because it doesn't have this consideration of leaving a p orbital vacant. Okay?



And now about nitrogen? Now you got a fight on your hands, right? Because the bonds want to be flat and 120º bond angles, but the atom wants to use p orbitals, 90º angles for the bonds; so like the corner of a cube would be the nitrogen, the bonds would be going out at 90º angles. So here there's going to have to be some compromise. Okay? So NH₃, since it must compromise, won't be either perfectly flat or 90º angles, but something in between. It'll be pyramidal. Okay? Now are the predictions clear; and the reason for them? Yoonjoo, do you see now, why it's the way it is? Don't let me bully you into it, if you don't really understand. Dana?



Student: Can you explain the difference in what the atoms and the bonds want for nitrogen again?



Professor Michael McBride: Yes, what we want to do is get the lowest possible energy. The lowest energy for the bonds will be the best overlap. You're making three bonds. So sp² will give the lowest energy, as far as the bonds go. But there also might be -- and that's the only game in town for boron. Right? But in carbon you have another electron you have to accommodate, and there's a question of what energy it will have, even though it's not being shared. And in nitrogen there are two of those other electrons. Now, the best orbital of an atom, for an electron to be in, the lowest energy, is s. And in nitrogen there are going to be two electrons that want to be in an s orbital. If they do that, they use up the s orbital, and the only thing you have left over to make the bonds is the p orbital. So now you've got a fight on your hands. Are you going to have sharp angles and have the unshared pair in a low-energy, s orbital but weak bonds, or are you going to have strong bonds, sp², but put the unpaired electrons in a p orbital? Okay, Kevin, did you have a question or did that answer it? Okay. Yes? I can't hear very well.



Student: Is there such a thing as an sp³ orbital?



Professor Michael McBride: Yes. What does sp³ mean? It means a hybrid orbital, a mixture of s and p, with three times as much p as s. Okay? You could have that. You could have three sp³ bonds, for nitrogen, for example, and then an sp³ hybrid, in which the unshared pair lives. Okay? That would be a possibility, sure. And that's the kind of thing I'm talking about here, by saying it would be intermediate, between being 90º bond angles and 120º bond angles, flat. Any more questions? Yes?



Student: Why are we concerned about the atomic orbitals rather than molecular orbitals?



Professor Michael McBride: Ah, I'm going to get to that just shortly here. So, because we're talking about bonds. Bonds are local. Bonds don't go over the whole molecule. Molecular orbitals go over the whole molecule. And we've been talking thus far, mostly, in terms of molecular orbitals, but now we're shifting to talk about bonds, because what we wanted to do was talk about bonds. And I'll do a little about the philosophy of that shortly. Okay? So here are the predictions. BH₃ should strongly want to be planar and be hard to bend. CH₃ should want to be planar but be easy to bend. And NH₃ should be easy to distort out of plane, so that it's not flat; in fact, not only should it be easy to distort, it should prefer to distort in its lowest energy form.



Okay, so now let's see if that's true. Is it true that that unshared pair will compete for s character, and therefore make nitrogen pyramidal? Now, how do we know? There's our prediction: distortion from the plane will weaken the bond and deprive the electrons of s character; so it doesn't want to bend. Here, distortion from the plane weakens the bonds, but it shifts s character to the lone electron, where at least it's not wasted, because there is at least an electron there. And in NH₃, it weakens the bonds but it shifts s character from single electrons to a pair of electrons; which is something good. Okay, so are these predictions true? Well, we need experiment. How about X-ray for seeing the structure of BH₃, CH₃, NH₃? How about NH₃? Have you ever seen NH₃? Have you ever smelled NH₃? It's ammonia. So how about doing X-ray on it?



Student: No.



Professor Michael McBride: Why not? Shai?



Student: Because you can only use X-ray to see electrons --



Professor Michael McBride: Can't hear very well.



Student: You can only use X-ray to see electrons; those are the ones that are going to move when you ---



Professor Michael McBride: No, that's not the main problem. Yes?



Student: You can't make it a crystal.



Professor Michael McBride: You have to have a crystal to do X-ray crystallography, and NH₃ is a gas. Now you might say, okay, we'll cool it way down to liquid nitrogen temperature and then maybe it'll crystallize. It'll be difficult to grow a crystal of ammonia, but it's conceivable we might be able to grow a crystal of ammonia. Okay? Can we use the same trick for CH₃ and BH₃? That one's a gas. Okay? CH₃, unfortunately, if you get two of them together they react, to get C₂H₆. So you're never going to be able to cool them down and get a crystal. And the same thing is true of BH₃. It forms B₂H₆. Right? So forget crystals for this. We can't use X-ray crystallography, the technique we've used so far to do this. But there are other techniques for finding structure, and the one we're going to talk about -- actually we're going to talk about two -- but we're going to talk about, not X-ray, we're going to talk about IR -- we're going to talk about spectroscopy, infrared spectroscopy, and electron spin resonance spectroscopy. Okay? We're going to talk more about infrared spectroscopy next semester, but I'll show you just enough to get through the question that we're interested in now. Does that remind you of anything that moves like this in a high wind?



Student: A weathervane.



Professor Michael McBride: It's like an umbrella, right? In a high wind it blows inside out. Okay, so this is sometimes called the "umbrella" vibration of XH₃. And you can treat it with Erwin Meets Goldilocks, as if it's one-dimension; just how much motion of all these atoms, how much motion. You know what the masses of the atoms are. You know the kind of motion you're interested in. It's all coordinated. So you're changing really just one-dimension, in moving all these things in a coordinated way back and forth. Right? So you can treat it as an Erwin problem, with a fictitious mass that reflects the amount of motion of each of the four atoms, while that's happening. Okay, so here's BH₃ doing that out-of-plane-bending vibration, the umbrella vibration, and CH₃ doing the same thing with its electron. Now, it turns out that BH₃ vibrates 34-trillion times a second. Okay? And that's called 34.2 terahertz; that's how often it vibrates. CH₃ vibrates 18.7 terahertz; so only half as fast as BH₃. Does this make sense? Suppose the springs that you're bending when you do this vibration -- suppose it's springs -- which set of springs is stiffer? The BH₃ or the -- does a stiff spring make a vibration faster or slower?



Student: Faster.



Professor Michael McBride: Faster. So it says the springs are stiffer in BH₃. It prefers to be planar more strongly than CH₃ does. Right? Now, you might ask -- what might you ask at this stage, when I tell you that it vibrates that fast?



[Students speak over one another]



Professor Michael McBride: No, that's not the question I want.



Student: How can you tell?



Professor Michael McBride: How do you know? How do I know that it vibrates at that frequency? Well, so it would be a stiffer spring, but how do I know? Well, when it vibrates, the electric charge is moving back and forth. Right? So the direction of the -- there's a dipole there, +/-, and it's moving back and forth. And what does that create, when you have charges moving back and forth? Electromagnetic radiation. Right? So electromagnetic radiation is going to come out, as this thing vibrates, like that. And it comes out with a shorter wavelength for BH₃ than for CH₃. Right? And it turns out that the wavelength is such that for BH₃ there are 1100 vibrations, waves, in a centimeter; short wavelengths. And it's longer wavelengths, only 600 waves in a centimeter for CH₃. Now, we can also -- so different light would come out, infrared light, of these two; or different colors of light, infrared light, get absorbed. That's how you actually measure it. And remember, the frequency of light has to do with the energies of the photon. So in, if you take the quantum mechanical view that you go from one quantized state to another quantized state, when you absorb, or emit, light, then we know, from this frequency, that in one case the two energy levels you're talking about are separated by 3.26 kcal/mole, and in the other case by 1.73 kcal/mole; only about half as much. Is everybody on board with this so far? So it's from the absorption of infrared light that we know how fast they vibrate, and therefore how stiff the springs are. Okay?



Now just how stiff are they? Well we can use Erwin for this point of view. We have the amount of deformation, which is our one-dimension, as it goes back and forth, and we can assume that it's something like Hooke's Law -- that's an assumption -- the distortion, these springs. And then we can adjust the potential, how sharp the parabola is, of potential energy, until we get two energies that are spaced like that. Does everybody see what we're doing? So here we do it. On the left, we get these lowest two energy levels and we -- there's the parabola that does the trick. So that's a strong, a stiff spring, strong planar preference. And in the case of CH₃, to get that lower energy separation, you need a broader parabola. It's easier to bend; the springs are softer. Okay? Everybody see how that worked? Okay, so infrared spectroscopy then tells us that first, that it prefers to be planar, and second, that it's stiffer for boron than it is for carbon; just as we predicted. Okay? Now how about for NH₃? For NH₃, you see two absorptions in the infrared, and they're very near one another; 968 and 932 wavenumbers; that is, waves per centimeter. If we translate that, it means that there are two energy levels -- there's a low energy level -- and then there are two that are very close to one another, in order to have these two different colors of infrared light being absorbed. How can we get this? How can we get -- this is not Hooke's Law. Remember what happens with Hooke's Law in the energy spacing? What happens? They're even. Remember? So this isn't anything like Hooke's Law. Where do you get two that are very similar to one another? Do you remember?



Students: Double minimum.



Professor Michael McBride: What?



Students: Double minimum.



Professor Michael McBride: A double minimum gives that. So NH₃ must be a double minimum, like this. Right? And those must -- the blue and the red one, must be the one with three nodes and the one with two nodes. There's also one with zero nodes, and one with one node. But those are both down at the bottom. Right? They have the same energy, because there's not enough overlap between the two to make them split. Right? So the IR spectroscopy of ammonia tells you it's a double minimum, and also, in order to get that spacing, tells you how big the barrier is. It costs three kcal/mole to make NH₃ planar. It prefers to be pyramidal. Okay? And that splitting, that "tunnel" splitting, that we talked about before, for those two levels, is thirty-seven wavenumbers. Remember, there's also the two that are down at the bottom, with zero and one node, and they're separated by only one wavenumber; they're very close to one another. But we can translate that into the tunneling frequency, when you're down in the lowest energy levels. How fast does it go back and forth? And you remember how to do that. If one wavenumber is two small calories/mole, not kilocalories, and we use that thing that we talked about in lecture nine, about how to go from splitting to tunneling rate; and it turns out that in the ground state, in the lowest energy state, ammonia is flipping back and forth 1011 times a second. And in fact for a while that was used as a fundamental clock; here was something that you really knew the frequency of. Hooke, when he was trying to devise a clock that could tell where you were in longitude, could've used ammonia, if he'd had the technology to do it, which would be an absolute frequency. Okay, so IR shows -- completely confirms what we said before, that BH₃ is stiff and flat; CH₃ is easily bent and flat; and NH₃ is not bent, but can tunnel back and forth, by the umbrella motion.



Okay, so now the other technique I want to tell you about is electron spin resonance spectroscopy. And it turns out -- and we don't have time to go into this -- well you know already that electrons are magnetic, and some nuclei are magnetic, and you can get a magnetic interaction between the nucleus and the electron, which influences how hard it is to change the electron from being pointing one way to being pointing another way. That's a kind of spectroscopy; see what color of light you need to make an electron flip from being pointed one way to the other way. But it's influenced by the magnet, which is the nucleus. But there's a neat thing about it, which is that influence can occur only when the electron is inside the nucleus. Where, in a 1s orbital, is the highest electron density? Remember the formula for a 1s electron? What's the atomic orbital? It's a constant times what?



[Students speak over one another]



Professor Michael McBride: e-ρ. Where's it biggest?



Student: At zero.



Professor Michael McBride: Inside the nucleus is the most dense part of the electron. Right? So when the electron is inside the nucleus it feels this effect. When it's not inside the nucleus it does not feel the effect, for the purpose of this spectroscopy. Okay? So we can measure how much s character there is in a hybrid orbital that holds an electron by seeing how strongly the nucleus interacts with the electron; because it happens only to the extent that it's s. Right? So we could measure the percent of s character by measuring the light that's absorbed by the electron. Christopher?



Student: How could the electron ever actually be inside the nucleus? Wouldn't that have two particles occupying the same space?



Professor Michael McBride: But they're different particles. And in fact it's a little more complicated than that, because what's the kinetic energy, if the distance is zero? Coulombic energy is 1/r. Suppose r is zero. What's the kinetic energy?



Student: It's infinite.



Professor Michael McBride: Infinite. Right? Now, that means the electron is really zinging, which means it becomes relativistic and its mass changes and all sorts of wild things occur. So it's actually that which causes this interaction. So we'll leave that across the way here in Sloane Physics Lab. Okay? And take it from me, that the amount of this splitting that you see in the ESR spectrum depends on how much s character there is on the orbital. Okay? And you can worry about whether it's precisely inside the nucleus, or just so heavy and fast that it's behaving weirdly. Okay? But anyhow, that's the trick. Okay, so the singly-occupied molecular orbital, the one where there's only one electron, in CH₃, will give rise to electron spin resonance, and the magnitude of this effect will tell you how much s orbital there is. Okay, so let's look at it. And this is what I just said. There's a line separation due to magnetic interaction, and it occurs only if the electron spends time on the nucleus, which happens only for an s orbital; so we can measure how much s orbital.



Okay, so here's a picture of planar CH₃, and the orbital, the singly occupied molecular orbital, SOMO, that contains the electron. How much time is it spending on the nucleus? Zero; there's a node at the nucleus, it's in a p orbital. Now suppose we bend it, so it looks like that; or we could look inside and it looks like that. Now it's got some s character, in that orbital. Right? And if we make little boxes and blow them up to see what it looks like down near the nucleus, one of them has no density at the nucleus and the other has electron density at the nucleus. So you'll see magnetic interaction, in this case. If it's bent you'll see magnetic interaction, and the more bent, the smaller that angle from 120 -- remember, the smaller the angle, so it'll go down toward ninety; the more s character, there'll be more of it. Okay, so here's what you see for CH₃. Bear in mind it has to be magnetic C, it has to be C-13 H₃, not C-12, in order to see the spitting. But at any rate, the amount of this effect is thirty-eight gauss, which are the units it can be measured in. And that implies that, on average, it's 2% s. Now, remember we already said that its lowest energy form is flat, with no s. But it's constantly undergoing this umbrella vibration. So it's spending time with different amounts of s character, not zero, as it vibrates. And the average is 2%. Now suppose I made the hydrogens heavier, by making them deuterium. So I make the atoms heavier. Do you remember how that changes the shape of the wavefunction, if I make things heavier? Here it is for H is moving. What if it were deuteriums that are moving, how will the thing change?



Student: [inaudible]



Professor Michael McBride: It'll get narrower. So as it gets narrower, that means it's not bending as much, on average. And what should happen to the thirty-eight gauss, if it doesn't bend as much?



Student: [inaudible]



Professor Michael McBride: I'll let you think about it a second. It's only when it's bent that it has s character and you see this effect. If it were rigidly flat, the splitting would be zero; thirty-eight would be zero. What would happen if I change H to D, so that it doesn't bend as much, on average? Sophie, what would you say?



Student: [inaudible]



Professor Michael McBride: Can't hear.



Student: If it was more rigid it would --



Professor Michael McBride: If it were more rigid it would -- how big would the effect be? Bigger or smaller, the gauss?



Student: Smaller.



Professor Michael McBride: Pardon me?



Student: Smaller.



Professor Michael McBride: Thirty-six. Right? So again it confirms what we're saying, that it prefers to be flat. It's only when it bends that you get this effect, and it bends less with heavier atoms than with lighter atoms. Wonderful. So again it confirms what we said. There's what CD₃ looks like. It's a little bit narrower and therefore a little bit less of this magnetic interaction. So this is a structural isotope effect. CH3 spends more time more bent than CD₃, and thus uses more s character, and thus gives a bigger magnetic interaction. Okay, so again we've confirmed what we predicted. Now here's one that we didn't work on in predicting, is CF₃. Now what's changing H to F going to do? Well, it's going to be heavier. So the same deal that would happen with deuterium should happen here. But there might be more to it than that, because now we have more protons and a bigger nuclear charge of the atoms that were H before. So they're going to attract the electrons in the bond more. So, but there's another thing too. Because the fluorines have a partial negative charge, and are bigger, they have more electrons altogether, they'll repel one another. Now what would -- if the things attached to the central carbon get bigger and repel one another, what effect should that have on it being bent? Should it make it easier or harder to bend?



Student: Harder.



Professor Michael McBride: If these things are big here? Easier to bend from flat, or harder, if these things are --



Students: Harder.



Professor Michael McBride: Harder, right? So valence electron pair repulsion, and all that, should say that if CH₃ is flat, what about CF₃?



Student: It's flatter.



Professor Michael McBride: It should really be flat, harder to bend, if these things are bigger. Okay. But there's another point of view that you can take, which is that fluorine has the lion's share of the bonding electron pair. The bonding electron pair, in each of the C-F bonds, is spending more time near fluorine than near carbon. So carbon, being a cagy sort of character, says, "Why should I waste my valuable s character trying to stabilize these electrons that are spending their time on fluorine anyway?" Right? "Why don't I use it to stabilize that other electron, that's only on me and not on fluorine?" Got the idea? So if carbon uses its s character to go to the odd electron, forgetting the ones that are mostly on fluorine anyway, then that would make it more bent. That would favor bending, if you put the s character in the odd electron rather than in the bonds. That'll favor using more p character in the bonds and therefore more bending.



So this is exactly the kind of thing that Bacon would have liked. Right? This is a crucial experiment. There are two opposite predictions, and you can do an experiment and find out which one is true. Is it that fluorine is bigger and negatively charged, so it makes it more flat? Or is it that the carbon says, "I want to put my electron, the only electron I'm really solely responsible for, in the orbital with the most s?" Right? So this is a real horserace and you place your bets. Okay? Now how many people think it's -- we'll guess ahead of time -- how many people think that fluorines are bigger and negatively charged, so it will be harder to bend away from planarity? And how many people think that it's the odd electron, that you want to put s character in? Well this is very close. Okay? So we'll have to look at experiment. Okay, so this is the singly-occupied molecular orbital of CF₃. It's mostly on the central carbon, but a little bit on the fluorines. And if you look in the middle, you see that there's the s character on the carbon. And if we measure the splitting, it's 271 gauss. What was it for methane?



Students: Thirty-eight.



Professor Michael McBride: Thirty-eight. This is ten times almost, or whatever number of times, maybe like eight times more s character than there was in methane. What's the conclusion? It's much more bent, much more of the time, than CH₃ is. Right? Okay, so that means it's 20% s. That means that it's sp⁴ in that, almost sp³. Right? So it's almost perfectly pyramidal, like methane would be. Right? So it's the character of that odd electron, wanting to be s, that's controlling the geometry. Okay, now we're going to get to what Christopher wanted, which is why the heck are we talking about bonds when we've been talking about molecular orbitals all the time? Well one reason is we wanted to talk about bonds. That's how we got into this whole thing, asking what bonds are. Okay? But there are three different operators here, and they have different motivation and methods. First there's the molecule deciding what to be; and then there's the computer doing quantum mechanics, deciding what the molecule should do; and then there's you, trying to understand all this. And these are different. And let's see the way in which they're different.



The experimental molecule, all it wants to know -- it doesn't care about Schrödinger -- all it wants to do is get the lowest possible energy. So it minimized the total energy, kinetic plus Coulomb energies, of the electrons and the nuclei, by just settling down into the most comfortable, lowest energy arrangement. Okay? That's what the molecule does, it just gets happy. Okay? And once it does that, then it has a certain structure, a certain arrangement of the nuclei and of the electron clouds. It has a total electron density, which you can measure by X-ray. It has certain energies that we showed measuring by IR; although you can measure them other ways as well. It has nuclear electron density, that we just measured by ESR. It has dipole moment, how much there's a separation of positive and negative, that you can -- that if it vibrates gives off IR and so on. So there are these real things that come from the molecule just doing what comes naturally. So it doesn't care about Schrödinger.



Now you have a computer that comes along, armed with Schrödinger's equation, and it knows it can't possibly solve the true Schrödinger equation, it's too complicated. So it'll do some approximation and try to get an approximate value. So it'll minimize the total energy, using the Schrödinger equation, with some realistic constraints; for example, using orbitals, or using configuration interaction, or using density functional theory, or some approximation. Okay? So it'll use like a limited set of atomic orbitals, only say 2s and -- it won't use the 5p orbital or the 5g orbital or something like that. Use Self-Consistent Field, or some correlation, or delocalized molecular orbitals come out. And its goal is to get a useful prediction of the properties; maybe the electron density, maybe the energy, maybe the vibration frequency, maybe absorption of, what colors of light get absorbed and so on. So the goal of the computer is to use any tricks it can get around -- it can do, and it doesn't care how hard it works -- it's not something you have to be able to do on a piece of paper -- just to get something useful. And it will be validated, the approximations you made, by experiment. If it agrees with experiment, then you can believe the computer. But that doesn't mean you understand what the computer did. Right? It's a really complicated calculation.



And that's different from what you want to do, because your goal is to understand structure and reactivity, with the simplest realistic model; not Schrödinger's fundamental equation, because you don't have the brain power to manipulate numbers that way and do all these integrals and so on; leave that to the computer. Your goal is to try to look at things and understand: ah! that should be reactive with this; that bond should be very stiff; that molecular structure should be very stiff because the, quote, "bonds" should be stiff. Okay? That's what you want to do is to develop an intuition that allows you to understand structure and reactivity. And you want the simplest, obviously, but realistic model. And that, to be both simple and realistic, may not be possible. Right? There's a certain tension there. Okay?



So what kind of things you'll talk about? You'll talk about localized bonds. You won't talk about these big things that are complicated and go over the whole molecule. It's interesting to see, for example, that those are analogous to atomic orbitals and that kinetic energy plays a part in them, but that doesn't empower you to predict the properties of things. Okay? So you want to talk about localized bonds, maybe pretend they're springs or something like that. And what makes a spring stiff or not stiff? And that, Christopher, is why we have been talking about springs and things here, rather than talking about molecular orbitals. So you talk about -- these are Lewis's ideas -- bonds, lone-pairs, hybridization. We talk about energy-match and overlap. And we'll be talking soon about HOMOs and LUMOs -- highest occupied molecular orbital, lowest unoccupied molecular orbital -- and what a very important thing those are.



So we're going to look at just little bits of things, to try to find out what's crucial and will allow us to understand. We want them to be realistic. We want them fundamentally to be based on something true, like quantum mechanics, not just on springs. But we have to make it simple enough to understand. So we're aiming for qualitative insight. And we'll know when we're right if we agree with experiment, and also if we agree with what the computer predicts; even though the computer doesn't know why it predicted it, it just grinds through and gets you a number. But we want to understand why the computer gets that number. Okay. So let me just do one thing quickly here. Pathological Bonding. Or let me let you go to your next class; we'll do Pathological Bonding next time.



[end of transcript]

Professor McBride begins by using previous examples of "pathological" bonding and the BH3 molecule to illustrate how a chemist's use of localized bonds, vacant atomic orbitals, and unshared pairs to understand molecules compares with views based on the molecule's own total electron density or on computational molecular orbitals. This lecture then focuses on understanding reactivity in terms of the overlap of singly-occupied molecular orbitals (SOMOs) and, more commonly, of an unusually high-energy highest occupied molecular orbital (HOMO) with an unusually low-energy lowest unoccupied molecular orbital (LUMO). This is shown to be a generalization of the traditional concepts of acid and base. Criteria for assessing reactivity are outlined and illustrated.

Transcript



October 8, 2008



Professor Michael McBride: Okay, so now we have a number of different perspectives we can take on understanding what holds molecules together; that is, on bonding. We can look at the real molecule, for example, with X-ray. Or we can do a computer calculation, and if we do a careful enough computer calculation, we can get something that we believe is pretty close to reality, like the total electron density I showed you. Right? And the goal of the computer is to approximate the Schrödinger equation. You can't do it absolutely, but you can, depending on how much computer power you want to throw at it, you can do it close enough for most purposes. Okay. But you're neither the molecule -- well you are a molecule -- but you're neither the kind of molecule we're studying, nor are you a computer. You're a chemist, and your goal is to understand things; let the computer do the heavy lifting mathematically and you understand it. Okay, so let's see if we can understand some of the things that we didn't understand when we were looking at the molecule really, with X-ray diffraction. Right? When we were looking, we looked at this molecule and said that the bonding, from the Lewis point of view, was pathological. And you remember how we saw that. We looked at a cross-section through those three atoms, and it looked like that in the electron difference density. And you remember what's funny about it? What's pathological? Dana?



Student: The bonds appear to be bent.



Professor Michael McBride: Some of the bonds are bent, and what else?



Student: And there's another bond that's missing.



Professor Michael McBride: And one bond is missing. So that bond isn't there in the electron density, the difference density. And those are bent. They don't lie on the line between the nuclei. Now, can we understand that, from the point of view of molecular orbitals, of Schrödinger's equation; why it should look like that in this particular case? So would a computer's molecular orbitals provide understanding? So I was just showing you, as the class assembled, how you get these molecular orbitals, and I was showing you some of them. There's the framework of the molecule superimposed on this map, and that triangle on top there, on the right, is what we're looking at. And so we could look at the thirty-second of the thirty-three occupied orbitals, the next to the highest occupied orbital, and it looks like that. That's the one we were just looking at, before we went on to the PowerPoint. And you see there is electron density in those -- the wavefunction has big value in the region where those bonds are bent. But there are thirty-three of these occupied orbitals, and they put electron density every which way, and it's very difficult to understand why this is the way it is, why it's bent. Even if you grant -- which is not easy to grant -- that this is the proper orbital to look at, and that indeed the electron density of the red and blue, big red and blue lobes, are not on the line between; we don't know why they are. Right? And there's so much other electron density in the molecule, with all complicated nodes, that it's very difficult to know where to look. Right? So the answer is no, it doesn't provide understanding. It gives far too complicated an answer. It doesn't tell you why. Even if you believe that it gives you the right numbers, it doesn't tell you why. Okay? But analysis in terms of pairwise -- that is, don't go over the whole molecule, just look at two atoms adjacent to one another, making a bond, so just a very local view -- pairwise bonding overlap of hybrid atomic orbitals explains these two pathologies. So let's look.



First, I will assert that the best overlap you can get between hybrid orbitals on these carbons -- if the nuclei are arranged in an equilateral triangle, so 60º angles -- the best hybrids you can get point at the angles shown by the red arrows here. They do it in order to get the best overlap. Now this seems funny, because the angle between the red arrows on the right carbon is greater than 90º. Why do I say it's funny that it's greater than -- you're not going to be able to make it 60º because two orbitals that made a 60º angle would not be independent orbitals. Obviously if they were right on top of one another, they'd be the same orbital. But you can't say they're not the same orbital if they're just a little bit away, because still they're mostly the same orbital. It's only once they get to 90º that they can be different. What would the orbitals be that would be independent of one another at 90º?



Students: p orbitals.



Professor Michael McBride: p orbitals, right? So 90º is the smallest angle you could get. Why don't you use 90º? Why do you use greater than 90º? Okay, that's the question. Why not use p orbitals and make the smallest angle you can? Does anybody see why that is? Okay here, there's a p orbital that points at 90º. You see its axis, the yellow axis, is a little bit inside the red arrow, and the blue axis, which would be another p orbital, perpendicular to that one, is a little inside the other red arrow; they make an angle of 90º. Got it? Now why don't the atoms use p orbitals to make the smallest possible angle? Can anybody see why that won't give the best overlap? How could you change the orbital so as to get better overlap?



[Students speak over one another]



Professor Michael McBride: Claire?



Student: The best overlap is between an s and a p orbital. So if it's two p orbitals, you don't have the best overlap.



Professor Michael McBride: Remember, that depends on the distance. And they're very close to one another. And it turns out it might be that at this angle -- if you used the s on one of the carbons, and the p on another -- it might be that you would get better overlap. I don't know that it's true, but it's not implausible. But what would be wrong about using this p orbital from that carbon, and on the top one of the triangle using an s, which might give better overlap? Can you see what's bad about that? If you use all the s of that top carbon there, you can't use any of it to make any of the other bonds from that carbon. Right? All you have left over on that one is p's, which give crummy overlap. So even though you might be able make this single overlap better, you couldn't make all the bonds strong from that top carbon. So you wouldn't get the best overall overlap. So that's a good suggestion. But there's another way to do it, which is to use a hybrid orbital. Now here we've put in a little bit less than 20% of s, to give the ratio sp⁴.1. Right? But notice that it really distorted; it made it much larger in the region where we're interested in it overlapping. Right? In the process, the axis moved out, to close to where the red arrow is. Right? But even though the bonds didn't point -- let's see, like this -- even though the bonds didn't point at one another, by extending they overlap more. Got the idea? So the hybridization, by shifting it to one side, gives better overlap, even though the angle might not be quite as good. So you have bent bonds, paradoxically enough, in order to get the best overlap. Isn't that funny? You'd think the best overlap would be when they'd point at one another. But you get it better when they make an angle because they stretch out.



Okay, so that explains the bent bonds. Now how about the missing bond? Well again we can see that in hybridization. Because what's left over, after you make those sp⁴.1's, three of them, pointing up from the bottom carbon, after you make three sp⁴.1's, what you have left over is an sp⁴.1's. And it points down, not up. If it pointed up, if the big blue lobe were up and the red down, then it would be part of those other three orbitals. The only way you can make it point up is to make the others point down. Right? And then they would have crummy overlap. So there's very poor overlap between the tiny lobes of these two. So there's not enough overlap to get a bond. You don't stabilize the electrons because you don't have overlap. Right? And rehybridizing this one, to strengthen this particular bond, by making it point up, would drastically weaken all the others, making them point down. Right? So the best overall overlap, for all the pairs that are involved in these bonds, is to have one bond not there, or to have very, very poor overlap so that the other three are very good. Sam?



Student: So when you hybridized the other two and you get the sp^4.1's, shouldn't -- and I understand that shifts the electrons towards the other atom, but how does that shifting distort the angle?



Professor Michael McBride: Remember the angle between hybrids depends on m and n; spm and spn, we did that last time. If it's pure p, it's 90º. If it's sp³,-sp³, it's 109.5. If it's sp²,-sp² it's 120; and so on. If it's sp^4.1, it makes that certain angle that's shown there bigger than 90º. Okay? And the only reason for that is so that these hybrids will be different from one another; they won't be the same. Right? They won't overlap each other on the same atom, because to the extent they do, they're the same, and so they can't at all. They have to be orthogonal, if they're on the same atom.



Okay, now three views of BH₃. First nature; the total electron density. I'm going over the same thing again and again here, with different examples. First look at total electron density; nature. Then we'll look at molecular orbitals; that's what a computer does. And then we'll look at bonds from hybrid atomic orbitals; and that's what we do, as chemists.



Okay, first we'll look at the electron cloud of BH₃, calculated here, using that same program I just showed you. So we can look at the successive layers of the onion, the high electron density, lower, lower, lower, the various contours. So this lowest one is mostly 1s of boron, but a little bit of the 1s's of hydrogen too. And one of them doesn't even show up, because it's so small. All the hydrogens should be the same, but the graphing capability of the computer, for something that has as few points as this, the way it draws surfaces, doesn't draw them smooth. Right? It connects points, so it doesn't even show one of those. Nobody would ever make such a plot. Okay? That's 0.3 electrons per cubic angstrom. So suppose we go to 0.15 electrons per cubic angstrom, half that density. Now we're beginning to get smoother surfaces. And it's on the boron, it's 1s of the boron, and a little bit of the hydrogens as well. Right? And now we're going to lower the electron density, 0.05. And now there's an interesting feature you can see here. You see that dimple? Why is that? Why is there not as much electron density, at this level, around the boron? We're out in the valence-shell now; not 1s anymore, because the 1s is really down close, high density. It's because the hydrogen atoms are pulling the electron density away from the valence atoms [correction: orbitals] of Boron.



Often you talk about this in terms of electronegativity. But why is something electronegative? It's because of what the effective nuclear charge is. The effective nuclear charge of hydrogen, for the valence level, is one. Right? But for boron it should be higher. Boron is the fifth element. But it's got the 1s electrons shielding the nucleus. So it's not as high an effective nuclear charge as hydrogen does. So when you put the orbitals together, hydrogen is lower than boron, and the electrons, in those bonds, are mostly on the hydrogen. So you see that dimple in the middle, because there's not much valence electron density on the boron.



Now we go to 0.02 electrons per cubic angstrom; and finally to 0.002 electrons per cubic angstrom, which is defined as (for purposes of drawing what the van der Waals surface is, that is) how close molecules tend to get to one another at normal pressure in a liquid or in a crystal. Right? And that's, it's just found empirically, just by testing and measuring things, that if you draw a surface at 0.002 electrons per cubic Angstrom, you have about how close things would come to one another when you bump these surfaces into one another. Okay, so that's the van der Waals surface; that's a definition, right? And we showed once before electrostatic potential, where you show that same surface but color it, to show how the electrons are distributed, and the protons in the molecule. So where would a positive charge on the surface be happy? Where it's negative. That's red. And where would it be unhappy, the positive charge, because the molecule is positive there? That's the blue. So you can see what it is. It's red where the hydrogens are, because they're pulling the electrons, and it's blue where the boron is, because it's giving up electrons. Okay, so there's what the real molecule looks like.



Of course I didn't do it by an experimental measurement from BH₃, because you can't crystallize BH₃. I did it by a computer calculation. But I believe that what it gives is what you would see in an experiment, if you could do the experiment. Okay, so that's the total electron density. Now let's look at how the computer breaks that up into molecular orbitals in a calculation like this. So it looks at -- it makes symmetrical molecular orbitals, like Chladni things, with nodes; we've seen lots of those. So here's the framework, BH₃, and we're going to look at the molecular orbitals the computer calculates. And we're going to make those molecular orbitals by adding up atomic orbitals, linear combination of atomic orbitals.



And which atomic orbitals will we use? The best would be if we used every atomic orbital. Then we could get perfect flexibility, but the calculation would take too long. So we use just a minimal number of atomic orbitals; namely the 1s orbital on boron, and also the valence orbitals, 2s, 2px, 2py, 2pz, and also the 1s orbitals of hydrogen. So there are one, two, three, four, five, six, seven, eight, there are eight atomic orbitals we're going to use, to make up our molecular orbitals. How many independent molecular orbitals can we make, if we start with eight atomic orbitals? We can make eight molecular orbitals, the same number you started with. If you tried to make more, they wouldn't be independent of one another. But you can make -- from eight, you make eight. Okay, so these are going to be the energies that'll turn out of the eight molecular orbitals that we're going to get, that the computer is going to get, starting from that. If you added more atomic orbitals, if you let it use 3s, 3p, 3d, and so on, then you'd get more molecular orbitals, but they'd be higher in energy, and less relevant. Okay? In fact, how many of these are relevant? How do we know how many we care about, the energy of? Russell?



Student: The total number of electrons.



Professor Michael McBride: How many electrons. We can put two electrons into each. Once we've got it filled up, we don't really care what other energies there could be; at least not for this purpose. Okay, well, it turns out that boron has five electrons, hydrogen has three; eight altogether. So there'll be eight electrons, four pairs, and that means four occupied orbitals. We don't care so much about the others, right? So those are the Occupied Molecular Orbitals, OMO. Okay? And then there'll be, also from this set of atomic orbitals, there'll be four Unoccupied Molecular Orbitals, UMOs. Okay, now those two are called -- nobody talks about OMO and UMO, but they talk about HOMO -- that's very important and we'll be talking about that -- that's the Highest Occupied Molecular Orbital. And also they're very interested in the Lowest Unoccupied Molecular Orbital; and you'll see why very shortly. So the LUMO and the HOMO are very important. In this particular case it turns out there are two HOMOs that have the same energy. Not surprisingly, because it's such a symmetrical molecule. Okay now, what do you think the very lowest orbital looks like? Katelyn, you're our expert. What?



Student: 1s.



Professor Michael McBride: It's the 1s orbital of boron. So there it is. No surprise there. Not to diminish your contribution, but it's not surprising. Okay, so there's 1s. That's the boron core. Now what's going to come next do you think? After we've got the 1s molecular orbital, what'll be next?



Students: 2s.



Professor Michael McBride: 2s. But it's a molecular orbital, it'll be made up of everything. So here it is, right? And it's all the hydrogens and also the boron 2s are contributing to that. Again there's a dimple in the middle, more of the hydrogens than of the boron. Why? Because the hydrogen 1s is lower than the boron 2s. The boron 1s was lower; that was the first one, right? Okay. And then, so there's that radial node, the spherical sort of node down the middle, showing it's a 2s. Right? And then there's a 2px molecular orbital, and a 2py molecular orbital. And that's four, and that's all eight electrons. So that's what we care about. They're the occupied molecular orbitals of BH₃. Now how about the vacant orbitals? The lowest unoccupied orbital is the last 2p, the 2pz, that's pointing in and out of the plane of the molecule. And then we're going to have 3s, which you see has that node in the middle; so two spherical sort of nodes. And then a dx^2-y^2, and finally dxy. Right? So that's the molecular orbitals that the computer generates for this thing, using what's called a "minimal basis set"; that is, the lowest number of orbitals you can use to get -- that is, through the valence level; you don't put the higher ones in.



Okay, now that's how the computer does it. Right? And you can get good numbers out of the computer, like the energy of the molecule, or the shape of the molecule, where the nuclei would like to be to minimize the energy. That's fine, but for understanding it's not so good. What we want to do, as chemists to understand it, is to partition the total electron density, not into these molecular orbitals, but into atom-pair bonds, and anti-bonds, as you'll see, and lone pairs; the same stuff Lewis talked about. Okay. That is, we usually do this, but sometimes it doesn't work. And when doesn't it work? When it doesn't work, we must use more sophisticated orbitals, where we say there is a certain phenomenon going on. You know what that phenomenon is? When we can't use localized bonds?



[Students speak over one another]



Professor Michael McBride: That's what resonance is, when this approach doesn't work. Okay, remember that's where Lewis Theory had to get sort of complicated, Baroque. Okay, so there's that 2pz orbital, a vacant orbital. Right? Here's the boron core. But now we're doing it. We're happy to use a vacant orbital. That's part of Lewis's idea. So 2pz we'll share with the computer; both of us see that as a lowest vacant orbital. The boron core we understand quite well. Okay? But how are we going to treat those other three pairs of electrons, in the picture Lewis would draw? Where would Lewis put those three pairs of dots? Angela? Can't hear.



Student: He'd make them the bonds between --



Professor Michael McBride: Those would be the bonds, between B and H. Okay? So what will those look like in our crummy picture? Right? We're not as good, at least I'm not as good as the computer at drawing these things. But what I'd say, it's going to be an orbital that's big on hydrogen, smaller on boron; some hybrid sp² hybrid of boron is going to overlap with the 1s orbital of hydrogen. It's going to be bigger on hydrogen than on boron. Why? Why should the lowest, the bonding orbital, be bigger on hydrogen than on boron? Lucas?



Student: The 1s is more shielding.



Professor Michael McBride: Yes. That's another -- that that orbital, the hydrogen orbital, is lower in energy than the boron it's mixing with. Right? So it's uneven mixing. Right? Poor energy match. So the one that went down looks mostly like the hydrogen, and the one that went up looks mostly like the boron. So we see one -- in fact, we see three of these, one pointing toward each hydrogen. Right? From our point of view they all have the same energy, they're all B-H bonds, shown as their energy in red there. Okay? But notice, you get the same total energy, as the computer would. The computer divides the pie completely differently. Right? But you get the same answer -- right? -- and the same geometry that the computer does, from the point of view of what direction do the hybrids point and so on. So we're getting the same structure, we're getting the same energy. You get the same total electron density too, if you add all the electrons together. Three bonds add together to give the -- the same way, remember, three p orbitals added up together to give a sphere. Right? Our three bonds add up to give the same energy as the computer. So who needs a computer? Right? Our bonds work just as well, if we view them right. Okay? And then there would be the anti-bond, the one that's -- what's left over, that's big on boron, small on H, and a node between hydrogen and boron; it's antibonding, the one that went up in energy when the atoms came together. And again there'd be three of these σ*'s, big on boron, small on H. They would all have the same energy. But again their average energy is the same as the average energy of unoccupied molecular orbitals that the computer gets.



So we're not doing so bad, following Lewis in this respect, looking at the electron density in terms of bonds rather than in terms of molecular orbitals. Okay? So for many purposes, localized bond orbitals are not bad, and they're certainly a heck of a lot easier to think about. But, beware of resonance. Right? Because sometimes this isn't going to work. So we'll have to understand how that works. So the localized bond orbital picture, pairwise molecular orbitals; that is, two atoms coming together to overlap. And also there'll be isolated atomic orbitals, like unshared pairs, or isolated vacant orbitals, like the pz orbital of boron that we just saw, that will be -- this picture will be our intermediate between hydrogen-like atomic orbitals, which we really understand -- you just look up in the table and you've got them, right? -- and computer molecular orbitals, which are too complicated really to understand in any kind of detail.



It's handy to sometimes -- for some purposes it's very handy to be able to calculate them and look at them, right? But you don't do it very often in your head. So when must we think more deeply? When you have these localized orbitals, but they mix, to cause reactivity; that is, a molecular orbital from this molecule mixes with a molecular orbital from this molecule, and one goes down and one goes up. That's what a reaction is. Or when you have two pairwise orbitals in the same molecule, that we thought of as bonds or vacant orbitals or something like that, and those interact with one another, within the same molecule. We were too simplistic in thinking of them as independent. They interact with one another. That's what resonance -- that's when resonance occurs is when you have important interactions within a molecule, between orbitals, or between what you thought were orbitals; but were thinking too simply.



So where are we? We've gone through atoms, we've gone through molecules, and looked at how we can understand the electron density in molecules. And now we're to the final payoff, which is reactivity. How can we understand reactivity in terms of SOMOs, Singly Occupied Molecular Orbitals, high HOMOs, and low LUMOs? And then how can we recognize functional groups? So even if we never saw them before, we know how reactive they're going to be, and what with; that's the payoff. Okay, so now we know that if you have two molecules and you're interested in their reacting, there'll be molecular orbitals of the two that come together, interact. And if the molecule gets -- if the energy goes down because of that, that'll be a favorable reaction. If the energy goes up, they'll bounce off one another. Okay? So, but how many molecular orbitals are there on a molecule?



[Students speak over one another]



Professor Michael McBride: Suppose methane, how many molecular orbitals does it have? It has an infinite number, right? The 13z orbital and so on. Okay? So we have to pick out which orbitals are going to be important in interacting with one another, or else we're just going to be at a loss, because there'll be so many. So you have these orbitals on molecule B, say, and those are occupied, and the ones above are vacant, unoccupied. And on molecule A you have a bunch of occupied and a bunch of unoccupied molecular orbitals. And now we're going to look at orbitals of B mixing with orbitals of A. Right? And which ones? We might look at those two interacting, for example, mix those two, between A and B. Or we could mix those two, or those two, or those two, or those two, or those two, or those two, or those two, or those two -- and it goes up infinitely. So this is a problem. Even if we understand how orbitals interact with one another, we're going to have to narrow our focus, if we want to deal with this. Okay? So there are myriad possible pairwise mixings.



Now there's one that's going to be obviously important. Suppose that molecule B has an odd number of electrons, and therefore a singly occupied molecular orbital, and the same is true of molecule A; just suppose that happens. Right? Then when they come together and mix, those two electrons will both go down in energy. So that's bound to be favorable, substantially favorable, and you'll get a new bond. That's like two hydrogen atoms coming together. So if there are SOMOs, on the two molecules -- and notice, suppose B had a SOMO but A didn't. But you're not just talking about two molecules, you're talking about two flasks. If you mix -- you had to have two B's that both had SOMOs. So A could be another B. So if you have any SOMOs, you can pair them; B with B, if A doesn't have any. Right? So if they exist, if there's a SOMO, then -- and that's true of many atoms, or things called free radicals, like the hydrogen atom, the chlorine atom, the methyl radical -- then they're going to be very reactive. Okay? And for that very reason they're not very common. You can't buy a bottle of methyl radicals because they will all have reacted with one another. The same for hydrogen atoms, the same for chlorine atoms. So that's an easy case to understand, when you have SOMOs. But it's very rare to encounter it, unless you're talking about flames, where lots of things are atoms and free radicals, as they're called. We'll talk about that a lot more at the end of the semester. Okay, now how about -- there are two that could mix. It's not bad energy match, right? And suppose they have good overlap. Do you care about that one? Why not? Lucas?



Student: It doesn't exist.



Professor Michael McBride: Oh it exists, it's an orbital.



Student: Oh, there are no electrons.



Professor Michael McBride: There are no electrons to have that energy. It's an available energy, but nobody availed themselves of it. So it doesn't contribute anything. Right? So there's nothing there. So forget that one. And you can forget most of them for this reason, of the infinite number. Okay, now how about this pair? Now they have -- let's suppose they have good overlap. They're not bad energy match. Right? So two electrons will go down and two will go up. Do you care? Steven?



Student: No, because since there's two that are higher energy, they're antibonding.



Professor Michael McBride: Yes. So they'd more or less cancel. Do they exactly cancel?



Students: No.



Professor Michael McBride: In which direction do they not cancel? Corey?



Student: The lower one is bigger, so it would be even worse.



Professor Michael McBride: The lower one is bonding, but bigger -- by bigger, you mean greater than 1/√2 kinds of thing; or no, pardon me, less than 1/√2, it's smaller, is what you should have said. Okay, it's smaller. So it doesn't go down as much as the upper one goes up. So pretty much they cancel. But to the extent they don't cancel, it's repulsive, it's bad to come together, which means the molecules will bounce off. Right? So when you have filled orbitals hitting one another, the molecules repel one another. Okay? So that's weak net repulsion. That's not going to explain a reaction. Now how about here? Now we're mixing a filled orbital with a vacant orbital. So two electrons go down. So you don't have this canceling. Is this going to be important, this one? Josh?



Student: No, bad energy match.



Professor Michael McBride: Ah, the energy match is so bad that you don't get very much going down; not enough to make up for the fact that if they'll come together, there are going to be a lot of other orbitals, molecular orbitals, overlapping, and they're going to be repulsive, the kind we just looked at. So this weak attraction is not going to be worth much. So it's negligible mixing because of bad energy match. Now how about here? Now that's not bad energy match; I mean, it's not great energy match, but it's certainly better than any other pair of occupied with vacant. Okay? So that means that pair will go down, right? And you'll get bonding, if the energy match is good enough that the amount that they go down is significant. Okay? So remember, when we tried the same thing the other direction, from the highest one on the right to the lowest one on the left, the energy match was not good. So that didn't help. But this one, where it's high on the left and low on the right, that one will help. Okay, so what it requires, to have a reaction, is that you have a high HOMO, an unusually high HOMO, on one molecule, and an unusually low LUMO on the other, so that you have good energy match, or at least as good as you can get. Right? Then you could get a reaction. And it turns out that there are simple names for things that have unusually high HOMOs and things that have unusually low LUMOs. And you know what those names are?



[Students speak over one another]



Professor Michael McBride: Something that has an unusually high HOMO is called a base, and something that has an unusually low LUMO is called --



Students: Acid.



Professor Michael McBride: An acid. Right? So that's a generalization of the idea of acid and base. Most mixing of MOs affects neither the overall energy, nor the overall electron distribution, for one or more of these reasons. Okay? First the electron occupancy can be four; that is, both were occupied, which means two go up and two go down, and the ones that go up, go up a little more. So that's not a bond. Or zero, and you don't care about the energy. Or the energy match is poor, so there's not much mixing all together. Or there's poor overlap. So even if the energy match were good, you wouldn't get anything out of it. But, if you have high HOMO and low LUMO mixing, then you get reactivity; unusually high HOMO and unusually low LUMO. And that's the secret we've been going for all semester.



So here's increasing generality in the concept of acid and base. The names were introduced by Lavoisier, in 1789, and we'll talk about that. And from his point of view, a base was an element, a substance that could be oxidized, and an acid was something that had been oxidized. Like sulfur was a base; oxidized sulfur, sulfuric acid, was an acid. Phosphorus was a base; oxidized phosphorus was phosphoric acid. Carbon was a base; oxidized carbon was carbonic acid. Okay? Then 100 years later, Arrhenius, in Sweden, said that an acid was something that was -- had the idea that there were ions, things that were charged, and that a thing that would give H+ was an acid, and a thing that would give OH- was a base. That's Arrhenius's Theory. And that was further generalized, in 1923, by Brønsted and Lowry -- Brønsted was a Dane and Lowry an Englishman -- who said that an acid was an H+ donor -- that's the same as Arrhenius -- but that a base was something that could accept H+; not only OH-, but ammonia, for example, could accept a base [correction: proton]. And Lewis then, in the same year, had the idea that an acid was something that could accept an electron pair, like BH3, and a base was something that could donate an electron pair, like ammonia. Right?



And other names for those acceptors and donors are electrophile, something that loves electrons, will accept them; and a nucleophile, something that loves a nucleus, is something that will give electrons to it. But in the 1960s this was generalized finally to where it is now, that an unusually low vacant orbital is something that can accept electrons; electrons like to be in a low energy orbital. So that is the ultimate definition of what an acid is. And by the same token, an unusually high HOMO has electrons that are eager to avail themselves of some vacant orbital, to lower their energy; and that is a base. Okay? So as so often in science, as the science matures, you get more and more generality to the idea. And HOMO and LUMO, unusually high HOMO, unusually low LUMO, are the ultimate form of acid-base; and that's what we'll be talking about. But they have to be unusual. Why do they have to be unusually high and unusually low?



[Students speak over one another]



Professor Michael McBride: Because otherwise you don't have good energy match, to change the electron energies when they come together. Okay, now here's what you should have asked. Unusual: compared to what? Okay? So here is what -- we're organic chemists. If we were some other kind of chemists in some other universe, dealing with different elements, we'd have different points of comparison. But what we deal with is things that have a lot of carbon and hydrogen in them. And by far our most common bonds are among -- between carbons, or between carbon and hydrogen. So it turns out that the valence orbitals of carbon, the sp hybrid orbitals of carbon, and the 1s orbital of hydrogen, are roughly the same in energy. For our purposes we'll consider them to be the same. Okay? So you bring two of these together, they overlap, and you get σ and σ*. And we've talked about that, the bonding and the antibonding orbital. So most of the occupied orbitals we're dealing with are σCC or σ CH, and most of the anti-bonding orbitals are σ*CH or σ*CC. So those are our point of comparison. Those are the plain vanilla electrons and holes. Okay?



Now let's look at other things and see whether they're unusual with respect to this. Now there will be several classes that they fall into. Right? So those are the usual LUMO and the usual HOMO, or the CC analogs. Now what could make things weird? One thing could make things weird is that if you had valence-shell atomic orbitals, that didn't get mixed with anything. So they're not σ or σ*, bonding or antibonding; they're the original orbital, in between these. So that would make things unusual. A second one would be a little bit in between these two. If there's no overlap, then you have unmixed orbitals. If there's a lot of overlap, you get bonds. But there could be poor overlap, in which case you don't go down as far with your bonding orbital, or up as far with your anti-bonding orbitals. So a second case would be poor overlap. A third case would be that the actual atomic energy, the atomic orbital energy that you're using, the atom you're using, is not like carbon and hydrogen. So it starts at a completely different place from where the standard orbitals start, before they make the molecular, the bonding orbitals. And fourth can be if there's charge around. Because obviously if you have a negative charge, it's not such a good place to put electrons; that means the orbitals are high in energy. If you have a net positive charge, it's a good place to put electrons; it's low in energy. So charge can make it. So we're going to show examples of these four contributors that will allow you to identify when a molecule has an unusual orbital energy. Right? Unusually high HOMO, unusually low LUMO. And then you know that it'll be a functional group, and you also know what it'll react with. Because if you have an unusually high HOMO, it'll react with an unusually low LUMO; and vice versa.



Okay, so first let's look at the first one, unmixed valence-shell atomic orbitals. So H+, Arrhenius's acid. Right? That's a vacant orbital. Right? No electron in it. So that's obviously unusually low. It's where this started, the 1s of hydrogen, before it went up or down. So if you had H+, it would be an unusually low vacant orbital. How about NH₃? Right? It has an unshared pair, only on the nitrogen. Now, it likes being on nitrogen, compared to carbon or hydrogen. Why? What makes it better to be on nitrogen, in a valence-shell orbital than on -- the unshared pair -- than to be on an unshared pair on carbon or on hydrogen? Angela?



Student: There's more protons in nitrogen.



Professor Michael McBride: There are more protons in the nucleus. It's a lower energy orbital, to start with. Right? But it didn't mix with anything. There are N-H bonds, in ammonia. Right? And they mixed and went way down in energy, and a vacant orbital went up in energy. But the unshared pair didn't mix with anything. So it still has that original energy. So it's unusually high. Compared to what? Compared to what?



[Students speak over one another]



Professor Michael McBride: σCH or σCC. That's our point of comparison. Okay, or the flip side of that is BH₃, which has a vacant orbital. It's high, to start with, compared with a hydrogen or a carbon, because boron doesn't have a very big nuclear charge. Right? And its nucleus is screened. Right? So it's unusually high for an atom. But it's still atomic, it didn't mix with anything and go up further. Right? So it's unusually low for a vacant orbital. So BH₃ is an unusually low LUMO. So is it an acid or a base?



[Students speak over one another]



Student: An acid.



Professor Michael McBride: Which?



Student: Both.



Professor Michael McBride: I can't hear you.



Student: Both.



Professor Michael McBride: You've read through the lecture. You don't get to vote [laughs]. It's like H+, right? An unusually low LUMO. So it's an acid. How about NH₃? It's a base, unusually high HOMO. Okay, how about water? Right? It has unshared pairs of electrons. Yes Lucas?



Student: Sort of a side question. Didn't we just say that they both had these properties because they had low nuclear charge?



Professor Michael McBride: No, nitrogen has high -- which has low nuclear?



Student: Nitrogen is, low nuclear charge compared to carbon.



Professor Michael McBride: Yes. [apparently did not understand the student's erroneous claim about nitrogen and thought he was speaking of hydrogen vs. boron]



Student: And BH₃ is even lower, compared to carbon.



Professor Michael McBride: Yes.



Student: But the energies are completely different because it's -- is it just because of the unshared pair?



Professor Michael McBride: But remember, you have to bear in mind which row of the periodic table you're in. Right? H, true, it only has one proton, but you're talking a 1s orbital, low in energy. For carbon, nitrogen, oxygen, boron, you're talking 2s. Right? So you're changing the game.



Student: Between BH₃ and NH₃?



Professor Michael McBride: No, no, not between BH₃ and NH₃. NH₃ is unusually high energy for an occupied orbital. BH₃ is unusually low; although it's high, compared to nitrogen, it's low for a vacant orbital. It's an unusually low LUMO. Nitrogen's an unusually high HOMO. Now oxygen is lower than nitrogen. So water is a base. But it's not as good a base as ammonia, because its electrons aren't as high, because it has a bigger nuclear charge, and the same n; it's in the first complete row in the periodic table. Okay, OH- is even higher. It's a stronger base than water. Why? Why are its electrons higher in energy than water?



Student: Less protons.



Professor Michael McBride: Got less protons. Right? It's got a net negative charge, which makes all the electrons high in energy; and specifically the pair we care about, the highest one, unusually high, because it's got a negative charge. [Technical adjustment] Okay, CH₃- is even more of a base, because carbon doesn't have as many -- as big a nuclear charge as oxygen. So when it has a negative charge it's a really high HOMO. Okay, I got to quit there. We'll go on to two, three and four next time.



[end of transcript]

This lecture continues the discussion of the HOMO/LUMO view of chemical reactivity by focusing on ways of recognizing whether a particular HOMO should be unusually high in energy (basic), or a particular LUMO should be unusually low (acidic). The approach is illustrated with BH3, which is both acidic and basic and thus dimerizes by forming unusual "Y" bonds. The low LUMOs that make both HF and CH3F acidic are analyzed and compared underlining the distinction between MO nodes that derive from atomic orbitals nodes (AON) and those that are antibonding (ABN). Reaction of HF as an acid with OH- is shown to involve simultaneous bond-making and bond-breaking.

Transcript



October 10, 2008



Professor Michael McBride: So this is where we were at the end last time, trying to decide when a high HOMO, or a low LUMO, is unusual in its energy; because, as the world is, things like to be at low energy. We'll talk about that later, why that's so. But that means that electrons tend to be in the occupied, the low -- occupy the low-energy orbitals and not the high-energy orbitals. So that empty orbitals are almost always higher than any occupied orbitals. So if you want to get good energy match, and therefore mixing, and therefore bonding, the occupied orbital that you're interested in must be unusually high, and the vacant orbital must be unusually low. So the task we have is to recognize what orbitals should be unusually high and what ones should be unusually low, and if we can find the unusual ones, then we'll know what's reactive, be able to identify functional groups. Okay, so unusual, compared to what? Compared to normal occupied and vacant orbitals. That is, in organic chemistry, or in most kinds of chemistry actually, carbon-carbon, carbon-hydrogen bonds are occupied orbitals; not unusually high, they're our standard of usual. And the antibonds, corresponding antibonds are unusual -- are the usual vacant orbitals.



So what can make things unusually low or unusually high? When the valence-shell orbitals that, if they bonded, would go down and up, don't bond with anything, don't mix with anything, then they'll be unusual. Second, if the orbitals don't overlap, then they don't go down and go up. Third, if you have an unusual atomic orbital. These are made of carbon and hydrogen, going down. But if you start with something that's unusual, unusually high or unusually low, then the things that come from it will be unusually high or unusually low. And finally, electrical charge. So right at the end last time we were looking at the first example: unmixed valence-shell atomic orbitals. So we saw that the simplest of all acids, H+, of course is unusually -- if it were occupied, it would be unusually high for an occupied orbital. Right? Because it hasn't gone down. But it's vacant, and it's unusually low, for a vacant orbital, because it hasn't gone up. And so on with these others: an unshared pair on nitrogen; or a little less so on oxygen. A vacant orbital on boron that's not bonding with anything is an unusually low LUMO; even though orbitals on boron are not unusually low, because the nuclear charge isn't very big for something that's using the second, n=2, quantum level orbitals, valence orbitals. It's an unusually low nuclear charge for that. So the orbitals are unusually high in energy, not being attracted by more protons. But it's low for a vacant orbital, if it hasn't mixed with anything.



Okay, and finally notice that some of these are charged: H+; plus means that's a good place to put an electron, so unusually low; OH-, unusually high because of the negative charge; CH₃-, especially high. So this slide also shows the fourth point, that electrical charge makes a difference on how normal occupied and vacant orbitals are. Okay, so these unusually low vacant orbitals are acids, like H+. Unusually high occupied orbitals are bases, like OH-, as we just showed on the previous slide. That means that you can mix the high HOMO of OH-, with the low LUMO of H+, and two electrons, the ones that were on the OH-, go down in energy, and that makes a bond. And our notation for showing that is to draw a curved arrow, and it gives water.



Now think carefully about what a curved arrow means; it's not the same as a straight arrow. A curved arrow designates a shift of electron pairs. It doesn't show that this atom moves from here to here. It shows that a pair of electrons shifts, and that pair of electrons goes from being on the oxygen, to being between oxygen and hydrogen, to form the bond. So you start the curved arrow where the electron pair is at the beginning, and you end the arrow, put its point, where it's going to be in the product. Right? So it's not showing that an atom moves from here to there. It's showing that the electrons that were formerly on oxygen are now between oxygen and hydrogen. The effect of that is, of course, to pull all the hydrogen to the oxygen. But that's not what's being shown by the arrow. What's being shown by the arrow is how the electrons are moving in our picture. Okay, so there's a case.



Now here's another base with the same acid, and we can draw a curved arrow there and show making an ammonium ion out of ammonia. Or we could use the same base with a different acid, the low vacant orbital of BH₃, and draw the same curved arrow, and make this anion, make the boron-oxygen bond. Or we could start with completely different ones. We could start with ammonia, not OH-, not Arrhenius's base, but ammonia; not Arrhenius acid, H+, but BH₃. But it's exactly the same reaction. The high HOMO mixes with the low LUMO to form a new bond. Okay? So that's what curved arrows show, and don't confuse it by trying to draw molecules or atoms moving, and showing a curved arrow. If you want to show them moving, draw some other kind of arrow; draw dotted arrows or a straight arrow or a wiggly arrow, or something. But the curved arrows have a very specific meaning.



Now the second reason that things could be unusually high or low is because they don't have much overlap. Notice in the first case, it was just an extreme case of that. There was no overlap at all, the orbitals were just there. But even though the orbitals are mixed with something else, if the overlap is poor, they haven't changed very much, they still look very much like they did originally. Right? So a good example of not having very good overlap is the side-to-side overlap of p orbitals. The π overlap -- you remember from looking at those curves of amount of overlap for different orbitals versus distance -- the π overlap isn't very big. So even though the p orbitals start a little higher on carbon than they do on hydrogen, or normal bonds of carbon that have s character in them, they start a little higher, but they don't mix very much. Right? So the π orbital is unusually high, and the π* antibonding orbital is unusually low, because they didn't go down or up very much. Okay? Now, which of those do you think is more remarkable, in its highness or lowness? So for this reason, the carbon-carbon double bond could behave either as an acid or as a base. Right? High occupied orbital makes it a base; low vacant orbital makes it an acid. And indeed, it does have both kinds of reactivity. But which do you think is more pronounced? Which would be more familiar, how high the occupied one is, or how low the vacant one is? Lucas, what do you say?



Student: How low the vacant one is.



Professor Michael McBride: And why do you say so?



Student: Because electrons have to sort of -- they would likely move into a position of an energy as close as possible to it.



Professor Michael McBride: Yes, but on the other hand, the electrons of the π want to move into something else.



Student: There are lots and lots and lots of electrons going up.



Professor Michael McBride: Well there are lots -- there are even more vacant orbitals than there are electrons. Right?



Student: Because you also have starring electrons that will -- like in a normal carbon atom.



Professor Michael McBride: Let's just look at the picture. Yes Catherine?



Student: I think the HOMO is more unusual because it's farther away from the usual --



Professor Michael McBride: Ah, the HOMO is further above the normal bonds, than the LUMO is below the normal antibonds. Why?



Student: Because the p orbital started --



Professor Michael McBride: Can't hear very well.



Student: Because the p orbital started a little higher.



Professor Michael McBride: Ah, because they started a little higher. Right? So most reactions that we'll study, or at least many reactions that we'll study of the carbon-carbon double bond, it's reactive because of the high HOMO, not because of the low LUMO; although there are cases where the low LUMO makes it reactive. But it's this π orbital, here, that makes it particularly remarkable, because they started a little high. Okay, so the high HOMO makes it unusually reactive. Now how about the C=O double bond? Right? So it starts with the same p orbital on the carbon, which has π overlap and therefore not much shifting up and down, with a p orbital of oxygen, which of course is lower than that of carbon, because of the higher nuclear charge. Which of these will be most unusual? Somebody got a suggestion? So what would you think the characteristic reactivity of a C=O double bond -- would it be reactive mostly as a high HOMO -- that is, as a base -- or mostly as a low LUMO? Which is more unusual, compared with what we compare things with? Andrew?



Student: I think it's a low LUMO.



Professor Michael McBride: Why?



Student: First of all it can't -- it's further away from its origin.



Professor Michael McBride: And why is it further down, compared to what we usually compare? Why does the π*, C double bond O, why is that orbital so very low? Pardon me, one reason it's low is because the overlap isn't very much. So it didn't go up so much. Right? That's the same, true with a C-C double bond. But what's special about the C-O double bond Andrew?



Student: I thought the atoms are different, there's a --



Professor Michael McBride: Because of the bad energy match, it didn't go up very much from the carbon, or down very much from the oxygen. But the average is low, that it starts from. So what's special here is the π is not so unusual, not so far from here. But this one, π*, is very far from here. So what should make a carbonyl especially reactive is its low vacant orbital, because of poor overlap; and that it started with an oxygen, an unusually low orbital, atomic orbital. Okay? We saw another example here. The three-membered carbon ring is unusually reactive, for a carbon-carbon bond. That is, the electrons in that carbon-carbon bond are not shifted down so much, and the antibond is not shifted up so much, as in normal carbon-carbon bonds. Why? What's special about the bonds in the three-membered ring? Kevin?



Student: Well two of them are bent.



Professor Michael McBride: Ah, they're bent. And what does that mean?



Student: Greater than 90°.



Professor Michael McBride: Yes, but they're bent, so what?



Student: It overlaps, there are better overlaps.



Professor Michael McBride: Does it overlap as well? So it doesn't go up and down as much. Right? So another example of poor overlap. Okay, or you could have an unusual atomic energy orbital in the molecular orbital. For example, how about a C-F single bond? Right? So the overlap can be perfectly good. It's a regular old σ bond, straight on. Right? But the fluorine starts very low. There's bad energy match. So the π* isn't so very different from a vacant orbital on carbon; unusually low as a vacant orbital. So it should act as an acid. Things that have high HOMOs should come up and react with that σ* of C-F. It's an unusually low LUMO. Can you think of what the flip side of that -- how could you have something like this, a carbon bonding to something, that for the same or analogous reason would have an unusually high occupied orbital? What would you want the carbon to bond with? Choose an element.



Student: Lithium.



Professor Michael McBride: Lithium. Why lithium?



Student: Because --



Professor Michael McBride: Because you looked at the PowerPoint?



Student: No.



Professor Michael McBride: [Laughs]



Student: I didn't look at it.



Professor Michael McBride: Why lithium?



Student: Because its valence electron -- it's easy for the valence electron to be given away.



Professor Michael McBride: Why?



Student: Because it would be as stable as a lithium plus one ion.



Professor Michael McBride: Why? That's restating the same thing, right? What's different about lithium? Why are orbitals on lithium, as an atom, unusually high in energy, compared to carbon-hydrogen? Catherine?



Student: Because it has a low nuclear charge.



Professor Michael McBride: Because it has a low nuclear charge. Right? It's right at the left end of the table. Okay? So its orbitals aren't very stable because it doesn't have very high nuclear charge. So it's a high energy. And therefore it has bad energy match. Or it could've been lithium that I drew here. I happened to draw boron, which is, for the same reason, a lower nuclear charge than carbon; therefore higher energy as an atomic orbital. When they mix, what's special here is the occupied orbital, σ, which is unusually high, because the average started high. Okay. So boron. You may remember that in the first slide we showed, boron had a vacant orbital; that p orbital on boron is vacant. Unusually low. Therefore what, acid or base? Unusually low vacant orbitals, which does that make it?



Student: Acid.



Professor Michael McBride: Acid. So BH₃ should be an acid. Right? Well what I just showed you is what Shai? On the previous slide; we'll go back there.



Student: That it had a big B-H; yes the B-H.



Professor Michael McBride: It has an unusually high HOMO, BH₃. So what does that make it Shai?



Student: It would make it a base but --



Professor Michael McBride: So is it an acid or a base?



Student: Both.



Professor Michael McBride: Wilson?



Student: Both.



Professor Michael McBride: It's both. And what does that suggest, if the same molecule is both an acid and a base? What do acids react with?



Students: Bases.



Professor Michael McBride: So what if the same molecule is an acid and a base?



[Students speak over one another]



Professor Michael McBride: It's going to react with itself. Two of the molecules will react with one another. Okay, and that's why you can't do a crystal structure of BH₃, because BH₃ becomes B₂H₆. Let's look at how it happens. So there's the low LUMO that makes BH3 an acid. What's the high HOMO? What's the high HOMO of BH3? In localized terms. I don't mean these big things that go over the whole molecule. It's the B-H bond, right? Which was poorly matched in energy and so on. So there. Notice that the B-H bonding electrons are big on hydrogen, small on boron. We saw that last lecture. Okay? So they should overlap like this, right? And the vacant orbital on the B should stabilize these high-energy electrons in the B-H bond. But now, you could imagine many different orientations of the top molecule that would allow overlap. Do you know why I chose this particular orientation? Which one acted as the acid and which one is the base? The one on the bottom, the vacant orbital, acted as an acid. The one on the top acted as a base. So what other possibility is there? Yoonjoo?



Student: So then there also -- you could reverse the roles of them.



Professor Michael McBride: Ah ha! You could have -- so that would make -- notice incidentally, what I forgot to mention here, is that there are three nuclei being held together by that pair of electrons now. Originally the top molecule, that pair of electrons, mostly on hydrogen but partly on boron, held the hydrogen and the boron together. Now that same pair of electrons is attracted -- is helping form a bond with a boron, the bottom boron. So in fact that's an unusual kind of bond, because it's bonding three nuclei together, not just two. It's doing double, or perhaps triple duty. So we need a new symbol to talk about such bonds, and a reasonable one would be a bond that looks like that. It's a Y bond. We could make it blue. Okay? But now we have, as Yoonjoo said, the vacant orbital on the top and the high occupied orbital on the bottom, and they can do exactly the same thing. So we have two of these three-center, two-electron bonds. Right? Two electrons holding three atoms together. Right? Or we can draw it that way. Right? And that's the structure of B2H6. Two of the pairs of electrons are each bonding three atoms together instead of two. Any questions about this? Yes Alison?



Student: Is that what the dotted line means for --



Professor Michael McBride: Yes, the dotted just means it's a little different. You could draw the solid Y if you want to. There's no really standard notation for that. But it's clear what it means. It's just that a pair of electrons is being shared among three nuclei, rather than two, and their electron density is -- correspondingly, if you did a difference density map, if you could do an X-ray of this, you'd expect electron density to build up in the middle of all three. Yes Claire?



Student: This may seem like a stupid question but a high HOMO, we've talked about as a base, and a low LUMO we've talked about as an acid.



Professor Michael McBride: Yes.



Student: And the low LUMO is unoccupied. If you think about it, it's sort of receiving electrons.



Professor Michael McBride: Right.



Student: But generally, bases are supposed to receive electrons, instead of acids, and --



Professor Michael McBride: No, you got it backwards.



Student: Do I?



Professor Michael McBride: Yes.



Student: Oh okay.



Professor Michael McBride: Because H+ is an acid. Right?



Student: Right.



Professor Michael McBride: So it obviously can't give up electrons. There aren't any electrons. An acid accepts electrons. Okay? Think about it a little bit in the privacy of your room, and you'll see that. Okay?



Student: Does the bond between three nuclei only happen --



Professor Michael McBride: I can't hear very well.



Student: Does the bond between the three nuclei only happen here because of the geometry of the molecule?



Student: Yes. And to get that, you have to have overlap. So if you didn't have -- if the top BH3 were oriented so that the B-H bond were vertical and the boron was way up at the top, it wouldn't overlap the other one. So you have to -- always to get a bond, you have to have overlap, because otherwise the orbitals don't mix and you just have the original orbitals, as we've seen.



Student: Is this kind of bond very common?



Professor Michael McBride: Pardon me?



Student: Is this kind of bond very common?



Professor Michael McBride: No, it's not very common, because there are not many really low energy vacant orbitals running around. Right? Boron is a very special case. But a lithium can do the same thing. But lithium doesn't have energy, orbital energies as low as those of Boron, because it doesn't have as big a nuclear charge. So boron is particularly good at getting this kind of thing. And this answers the puzzle about Lewis structures that we raised in Lecture two, about how can BH3 react with BH3? Right? That's how it does it, by making three-center bonds.



Now here's a True and False quiz. On the basis of what you know, is it true that low energy molecular orbitals result in bonding? True or false? I don't think you trust me anymore. [Laughter] We've been saying that when you get those low orbitals -- things come together, you get a low orbital -- that results in a bond. Lots of overlap. [Laughter] I can't talk you into it? Good. That's false. What makes a bond is lowered energy orbitals. It's when things come together and the electrons get more stable. It's not how low they are, it's how much they're lowered by the coming together, because then pulling apart they have to go back up again. So it's not whether they're high or low, it's whether they get lowered. Okay? Now, compared to what? What do they have to get lower compared to?



Student: By the size.



Professor Michael McBride: Yoonjoo?



Student: So it's kind of like how in Erwin and Goldilocks you showed the antibonding and bonding. So wouldn't it be the reactants?



Professor Michael McBride: You can say something in fewer words than that. What do you compare to? When you say energy is lowered, and that makes a bond, what's it lowered, compared to? Christopher?



Student: The atoms in their standard states.



Professor Michael McBride: Well yes. It wouldn't necessarily be between atoms, it could be between two molecules, like BH3 with BH3. But you're right. What it's lowered compared to is the things it was before they came together. Right? Now these things, before they came together, one of them had electrons, one didn't. Those might've been very high. Right? So they came together. The electrons went substantially down, but still aren't so very low. They could've been very low-energy orbitals to begin with, but not gone down very much, because there was bad overlap say. So these, even though they're lower than the ones we talked about first, would not be so bonding. It's these that were bonding, because they came down a lot, when the mixing happened. So it's lowering. Compared to what? Compared to the separated components, before you made this bond. Right? So when things come together, and that results in the electrons going way down in energy, that's a strong bond. Yes, Chenyu? Pardon me?



Student: How far does it have to go down?



Professor Michael McBride: Well that's what we're going to have to learn. That's a question of lore. Right? And it has to be at least enough to overcome the fact that when they come together other electrons, other orbitals, filled orbitals are overlapping, which is net repulsive. So it may not be that there's an absolute criterion. Right? It may be that if there are a lot of other things opposing the coming together, filled orbital with filled orbital, then you have to have really enormous going down. So you can't make a simple answer to that. But we'll learn as we see examples. Okay. So now, HOMO/LUMO mixing, for reactivity and resonance. So reactivity means between molecules. So far we've been talking mostly about atoms coming together and forming a bond. But molecules have high orbitals and low orbitals, as in the case of BH3. The B-H was a molecular orbital, or not an atomic orbital. Right? So when things come together, orbitals are orbitals. If you have an unusually high energy and an unusually low energy, and they overlap and go down, that makes a bond. So that's between molecules. But it turns out that what resonance is, is HOMO/LUMO mixing, within a molecule.



Now you might say the molecules have certain molecular orbitals. How can you mix them? Right? The idea is that we made our first analysis on the basis of localized orbitals: σ,σ* here; σ,σ* here; not these big Chladni things that go over the whole thing. But it may be that this σ,σ*, and this σ,σ*, are near one another and overlap. So that when we made our initial analysis, and looked only at this and only with this, we didn't take into account that this one might interact with this one, and give still lower energy. When that kind of thing is important, that's when you have to draw other resonance structures. And I'll show you an example. But first I'm going to show you reactivity, and then we'll go on to resonance. Okay, now let's look at the frontier orbitals for H-F. Okay? So it has four valence electron pairs and five valence atomic orbitals; 2s, 2pxyz on fluorine, and a 1s on hydrogen, and four pairs of electrons. So there are going to be four occupied orbitals. And this is what the lowest orbital looks like. What does it look like?



[Students speak over one another]



Professor Michael McBride: Well it's a 2s orbital, the 1s being the core on fluorine. But it's made up of atomic orbitals. Is it exactly a 2s of fluorine? Does it look like sphere on fluorine? Alison, you're shaking your head.



Student: It's a little bit distorted.



Professor Michael McBride: And how did it get a little distorted? What did we mix with the F orbital?



Student: The H.



Professor Michael McBride: A little bit of the 1s on hydrogen. It's mostly the F on fluorine. Why? Because the fluorine's way down in energy. So the best combination is mostly this. But you can see that it's a little bit egg shaped, a little bit drawn out toward the hydrogen. Okay, so it's mostly a 2s of fluorine, but a little bit of 1s on hydrogen. Right? Now here's the next one. What's that mostly? Can you see?



[Students speak over one another]



Professor Michael McBride: Tyler, what do you say?



Student: I would say a 2p.



Professor Michael McBride: It looks like a 2p on fluorine. Does it look exactly like a 2p on fluorine, or is it just hard for you to see that it's not? It's very similar.



Student: I don't know, it looks pretty close, but the blue one might be a bit bigger.



Professor Michael McBride: Yes, the blue one is a little bit bigger, because it's got a little bit of the 1s of hydrogen lowering the energy of the 2p of fluorine. Okay? And now the next two orbitals are these. What are those? Steve, what do you say?



Student: The 2py and 2pz.



Professor Michael McBride: Yes, the 2py and 2pz of fluorine. Is there hydrogen in those too?



Student: Not very much.



Professor Michael McBride: Why not?



Student: Because they don't overlap with the one --



Professor Michael McBride: Ah, they're orthogonal. It's a π versus a σ. There's no overlap, therefore no mixing. So those are the occupied orbitals. And the 2p's have the same energy. Okay, and then remember there are going to be five molecular orbitals. And now you make the last one, which is going to have another node, with the leftovers. What's left over, after we made the occupied orbitals? What's it mostly left over? We used up the 2p's of fluorine to make those HOMOs. Those were pure, right? But what was left over from the bottom?



Students: 1s.



Professor Michael McBride: We used very little of 1s on hydrogen, and there's a little bit of 2p fluorine and 2s fluorine that we didn't use in the bottom ones. So they're back in the top. So what we have is mostly a 1s on hydrogen, but a little bit of some kind of sp hybrid on fluorine. Is that clear to everyone? So that's the vacant orbital. Is it unusual energy, that vacant orbital?



Student: It's low.



Professor Michael McBride: It's low. Is it unusually low? Kate, do you have an idea? What kind of criteria do we have for whether an orbital should be unusually low? What do we look for?



Student: We look for overlap and energy match.



Professor Michael McBride: Overlap. Energy match. How about this case, good overlap? Quite good overlap. It's a hybrid orbital on fluorine pointed right toward a hydrogen. Good overlap. How about the energy match? Kate?



Student: Sure.



Professor Michael McBride: Sure what? Sure it's good or sure it's bad? One of them's hydrogen. What's the other one?



Student: The other one is fluorine.



Professor Michael McBride: Where's fluorine, compared to hydrogen?



Student: Fluorine is going to be lower.



Professor Michael McBride: Why?



Student: It has greater nuclear charge.



Professor Michael McBride: Right. Okay, good energy match or bad energy match?



Student: Not great.



Professor Michael McBride: Not very good. So you don't get much mixing. And you know that already by looking at the picture, because you didn't mix it very much. It's almost all 1s of hydrogen. So it's unusually low for a σ*; it didn't go up very much from hydrogen. So indeed -- now it's got an unusually low vacant orbital. What does that make it, an acid or a base?



Student: An acid.



Professor Michael McBride: It's an acid. Are you surprised that H-F is an acid?



Student: No.



Professor Michael McBride: Why? What's its name?



Student: Hydrofluoric acid.



Professor Michael McBride: Hydrofluoric acid. And that's what makes it an acid. Arrhenius would say it's an acid because it gives up H+. We say it's an acid because it's an unusually low vacant orbital. Right? So notice that those three are made up of three atomic orbitals: the 1s of hydrogen, the 2s of fluorine, and a 2p orbital of fluorine. And they're in different mixtures in three of these. Usually we've looked at just two things, right? There's one atom, atomic orbital, and another one, and they mix. Right? A pair. Here there are three going in, to give three molecular orbitals. But still you can see quite easily why they should be the way they are; why the bottom ones are almost pure fluorine, and the top one is almost pure hydrogen. Okay. And so that top one is σ*; the LUMO, the unusually low LUMO. Lucas?



Student: How can we be sure that the 1s of the low nuclear-charge hydrogen is going to be that much different from the really high nuclear charge 2s of fluorine?



Professor Michael McBride: Yes, this you have to learn. And I'm going to show you very soon how we can tell that kind of thing. Okay, there's a different picture, that was made in 1973, of the same thing. These pictures were drawn just a year or two ago, but this is -- you can see the same thing. It's drawn at a different contour level. It was based on a different calculation. But you can see it's the same thing. And we'll notice that -- how many nodes does this thing have, that are clearly visible?



Students: Two.



Professor Michael McBride: Two nodes, right? Between the H and F. That one is antibonding. Right? When they came together they cancelled in the middle, rather than reinforcing. There's another node there. But notice that that didn't have anything to do with the bonding. That was already there in the atomic orbital you started with. So it didn't have anything to do with lowering. It didn't have anything to do with whether the thing was bonding or antibonding. The pink node had to do with whether it was bonding. That came when they came together. Right? It's unfavorable. So it's better for them to come apart. But if they come apart, you don't do anything with the blue node. It's still there, it's part of the atom. Okay? So you have to recognize that there are two kinds of nodes. There are nodes that were there already, atomic orbital nodes, and there are ones that are associated with the coming together, and that's what makes something bonding, or antibonding; in this particular case it's antibonding. Okay now let's look, instead of H-F, let's look at CH₃-F. Sam?



Student: Why is it called a frontier?



Professor Michael McBride: Because you have occupied orbitals, and then you have vacant orbitals, and the ones you're interested in are the lowest of the vacant and the highest of the occupied, at this border between occupied and vacant orbitals. Okay, so this one has seven valence pairs of electrons. So you're going to occupy seven orbitals. And here's what they look like, the seven that are occupied. What's the very lowest one, mostly?



[Students speak over one another]



Professor Michael McBride: It's the 2s of fluorine. Right? And what's the next one? Well it's C-H bonds, all mixed together. Right? But also a little bit of a p orbital on fluorine. It has the blue, sort of, toward the front, and a little bit of that red behind, as part of the p orbital on fluorine. So it's a mixture of the C-H bonds and of the p orbital on fluorine. And then we have these others, some of them coming in pairs, and those are the HOMOs, because we have seven occupied. And let's look at them a little more closely and compare them, each one, with the one beneath it. And I'll draw another picture too, of that older kind, for making this comparison. And we're interested in what is it that went together, to make these orbitals, and why is one lower in energy, and the other higher? So what do you see on the top orbital, say this one here, the top left; what's that made up of? What is it on the left side? Sophie, what would you say the orbital is on the left side of that?



Student: I think it's 2p over there.



Professor Michael McBride: Right. This part here is a 2p π orbital of fluorine. Now what's this thing on the right here, the dash bit down here? If you just saw that, without any of the rest of it, what would you say that was?



Students: 1s.



Professor Michael McBride: It's more than a 1s on hydrogen; that would be spherical. It's, at a certain contour, the C-H bond. Do you see that? So this is electron density in here, bonding between C and H. And these two on the top are little bits of C-H bonds as well, mixed together. So what this orbital is, is a mixture between some combination of C-H bonds on the right, and the p orbital of fluorine on the left. Now is it a favorable, or an unfavorable combination, of the fluorine orbital with the C-H orbitals? This one up here. Is the interaction between the fluorine orbital and the C-H orbitals bonding or antibonding?



Students: Antibonding.



Professor Michael McBride: Becky, do you have an idea? Are they building up electron density in between, or having a node in between?



Student: A node.



Professor Michael McBride: There's a node in between. So that's antibonding. What's the orbital on the bottom, this one?



Student: Bonding.



Professor Michael McBride: That's the same components, but it's the bonding combination. So this is the favorable combination of those, and this is their unfavorable combination. So this lower-energy one is favorable, and the upper energy is unfavorable combination. How about here and here? Can you see what that is? What's on the left, here, that lump of red and this lump of blue? Eric? Here, this thing, it's very complicated. Okay, will you agree on that?



Student: Okay.



Professor Michael McBride: But part of it, this part on the left, there's a red lump in behind and a blue lump in front. You can see it maybe more clearly here. Red behind, and blue in front, surrounding the green fluorine atom. What is that orbital, that atomic orbital?



Student: Probably the p orbital.



Professor Michael McBride: Can't hear very well.



Student: Probably the p orbital, a different orientation.



Professor Michael McBride: It's the p orbital on fluorine that's pointing more or less toward you. Okay? So this is the p orbital of fluorine, that's mixing with C-H bonds here. Right? In fact, it's the same thing as this, turned on its side. Okay? So this is the bonding combination, built up between the fluorine and the C-H's. This is the antibonding one, with a node between the fluorine and the C-H's. So here we have a node, in this bottom one, but that node came from the atomic orbitals. That's what made it a p orbital here, on fluorine. That one, you don't get rid of, if you pull the fluorine away. That doesn't have anything to do with the bonding, that's just an atomic orbital node. Right? On the top, you again have the same atomic orbital node, because it's the same p orbital that's involved. But what about the top? Sam?



Student: You have another node but --



Professor Michael McBride: There's another node, that one, and that's an antibonding node. Right? The electrons in that orbital would get lower in energy if you broke the bond, if the fluorine came away.



So the pink nodes are the ones we're interested in; the antibonding nodes, not the ones that are just part of the atomic orbital. Okay. Now, we're getting to Lucas's question. So this thing down here is made up of fluorine and C-H. The one here is made up of fluorine and C-H. Right? This is the favorable combination, and this is the unfavorable combination. So get it right. So here are the two of them. They come together when the F comes up to the methyl. Right? And you get a favorable one and an unfavorable one. But the fluorine and the C-H's may not be at the same energy. How do you know which one's lower? That's your question. Is the fluorine lower, or is the C-H lower, or are they about the same? Now how are we going to tell? If the fluorine is lower, and they come together, what does the lower one look like, mostly?



Student: Fluorine.



Professor Michael McBride: Fluorine. If the fluorine's higher, and they come together, what does the low one look like?



Student: C-H.



Professor Michael McBride: Mostly C-H, right? So by looking at how big these are, we can tell which one is lower. So when we look here, we see that here I would say they're pretty similar, left to right. There's not much difference between a 2p orbital on fluorine and C-H σ bonds; not much difference. But to the extent they're different I would say, looking at this, that the C-H is a little bigger here, and the fluorine is a little bigger here. Would you say that? So which is lower, a C-H σ bond or the p orbital of fluorine?



Students: C-H.



Professor Michael McBride: C-H σ bond would be a little lower than fluorine; not much though, pretty similar. So they're about the same, as we say. But if you have to make a choice, the C-H is a little bit lower in energy; the fluorine's a little higher. Okay. Now there's also a vacant orbital, made up with the leftovers here, some of the leftovers. And what's that? Let's look at a different picture of it. So this has three nodes that are obvious. There's one that's near the fluorine atom, a node, plane, that goes back. The furthest to the left. Everyone see that? Is that an antibonding node, or is that an atomic orbital node?



Student: Atomic orbital.



Professor Michael McBride: That's part of the fluorine atomic orbital of some hybrid on fluorine. Right? And the same is true at the C-H end. There's a node that's an atomic orbital node of the carbon. Right? But what's important? Between those other two is what's between the fluorine and the carbon, and the CH₃. And that's antibonding. So that, the LUMO is a σ* antibond, between C and methyl. And why is it unusually low? The question is whether the LUMO should be unusually low. Why is it unusually low? Is the overlap bad?



Student: It's fine.



Professor Michael McBride: No, the overlap's good; the hybrids are pointing right toward one another. But what makes it low in energy? So I'm not giving you specific problems on this, but look over those things and run your brain as to those four different things that make orbitals unusually high or unusually low. Because that's what you'll be doing next week when you're doing these Wikis, each of you, to decide some functional group, why is it functional? What makes it unusually high or unusually low; or maybe neither? Right? Why is the C-F bond, more properly the C-F σ* orbital, the antibonding orbital, why is it unusually low? Compared to what? What do you compare it to?



Student: C-H.



Professor Michael McBride: A C-H bond. How come C-F is lower, for the σ*? Alex?



Student: Energy mismatch.



Professor Michael McBride: Pardon me?



Student: Energy mismatch.



Professor Michael McBride: Yes, the fluorine is really low, their average is very low. So the vacant orbital is unusually low. It didn't go up much. Right? Where have you seen that before? H-F. The previous slide, H-F, was exactly the same. It's an acid. Remember Kate, you helped us with that. It's an acid because of the bad energy match between fluorine and hydrogen. This is the same thing, but it's the bad energy match between fluorine and carbon. Okay. So CH₃-F is an acid for the same reason that H-F is an acid. Right? There's the low LUMO of H-F. Right? It's got that same antibonding node. Shai?



Student: Why is there no energy mismatch between carbon and hydrogen?



Professor Michael McBride: It just happens that it worked out that way. But it's true. Hydrogen is 1s. That makes it unusually low. But it doesn't have a very big nuclear charge. Carbon has a higher nuclear charge, but you're talking 2s and 2p. And it turns out those just cancel out. If that hadn't been the case, organic chemists -- like if boron happened to match hydrogen, then maybe our organic chemistry would be boron-hydrogen, not -- there would be borohydrates, not carbohydrates. Right? But that's the way things are.



Student: So in this picture we can tell that --



Professor Michael McBride: Actually there are other reasons it wouldn't be boron, because boron has these vacant orbitals and forms B2H6 and so on. But it just happens that that cancellation works that way. Lucas?



Student: Just by looking at this picture we can say that CH3, that there's poor energy match because the orbitals around fluorine are much smaller than those around CH₃.



Professor Michael McBride: Let's look at the next slide here. Is it this one? No. We're going to get to that. If you looked at ones that were very badly mismatched, then the favorable and unfavorable combinations would be very dramatic. Actually, you've already seen that in H-F. Okay, we have one minute to start this. Okay, so we're going to look at how CH₃-F behaves like H-F. Right? Both of them are acids. So first we'll look at H-F, and next time we'll go on to CH₃-F. So here's H-F. So you have to bring -- here's a low vacant orbital. We've talked about that ad nauseam. Okay? Now you want to get good overlap. From what direction will another orbital come, in order to get good overlap, without the nuclei getting too close together? Obviously it'll come from off in the right, where this orbital is big, where you can get a lot of overlap without getting close to the nuclei. So if you had something with a high-energy pair of electrons, it would come and it would overlap, from the right; something like OH-. And you'd draw a curved arrow, to show those electrons; the high HOMO of OH-, being stabilized by the vacant orbital of H-F. And how would you draw the curved arrow? Where would it start, where would it end? To show the electrons of OH- forming a bond with -- OH bond?



[Students speak over one another]



Professor Michael McBride: You'd start from the pair of electrons that you're talking about, and you'd end between H and O. Right? But that means -- this is really important -- that means you're putting electrons, putting electron density, into this orbital. Right? What effect does putting electrons in that orbital have on the H-F bond? Okay, the curved arrows, blah-blah; we did that. Sherwin?



Student: It's like it pushes them out.



Professor Michael McBride: It's what? What do you call that orbital? What's the name of the orbital, that we're putting electrons into?



Student: 1s for --



Professor Michael McBride: It's σ*. It's mostly 1s on Hydrogen, but it's σ*. What does * mean?



Students: Antibonding.



Professor Michael McBride: What does it mean if you put electrons in?



Student: The bond breaks.



Professor Michael McBride: The bond will break. Right? It has that antibonding node. So electrons that go into that will get more stable if the bond breaks. So we're going to draw another curved arrow, like that. So we make a new bond between H and O, but we lose the bond between H and F. So there's our product. That's an acid-base reaction, and it showed H-F acting as an acid; not because it gave H+. H+ never appears in here. Right? What happened is you make a bond and break a bond to the hydrogen. So it's an acid-base reaction. And the same thing, we'll show next time, happens with CH₃-F. Okay.



[end of transcript]