CHEM 125a: Freshman Organic Chemistry I

Lecture 11

 - Orbital Correction and Plum-Pudding Molecules


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.

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Freshman Organic Chemistry I

CHEM 125a - Lecture 11 - Orbital Correction and Plum-Pudding Molecules

Chapter 1. Introduction [00:00:00]

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.

Chapter 2. Correcting for Electron Repulsion When Using Orbitals [00:01:54]

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.

Chapter 3. Correlation Energy and the Limits of Orbital Theory [00:15:26]

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?


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.


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.

Chapter 4. Kinetic Energy’s Effects on the Shapes of Atomic Orbitals [00:30:48]

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 4dorbital. 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; 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.

Chapter 5. Moving Nuclei to Distort “Electric Puddings”: Case Studies with Methane and Ammonia [00:40:33]

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]

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