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# CHEM 125a: Freshman Organic Chemistry I

## Lecture 11

## - Orbital Correction and Plum-Pudding Molecules

### Overview

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 CH_{4} and NH_{3} 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.

Professor McBride’s website resource for CHEM 125 (Fall 2008)

http://webspace.yale.edu/chem125_oyc/#L11

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html## Freshman Organic Chemistry I## CHEM 125a - Lecture 11 - Orbital Correction and Plum-Pudding Molecules## Chapter 1. Introduction [00:00:00]
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 XH ## 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 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, 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?
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?
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 Z that you should use for _{eff}2s is a little bit bigger than the one you should use for 2p. Kevin?
a, 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 _{o}Zfor the _{eff }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]
Z 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?_{eff}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 Z. 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 _{eff}Z, or whatever type of guess. What do you do next?_{eff}
[Students speak over one another]
[Students speak over one another]
## 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?
[Laughter]
[Students speak over one another]
1s electrons of carbon. Right? Bingo! And that gives us 2*10^{4} 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? 10 Z to guess how the _{eff}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 n as well. So you could scale the energy and find that it’s an order of magnitude less. Okay.^{2}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 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 10 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?
[Laughter]
[Students speak over one another]
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?changesSo 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 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.
p and the one that’s in the _{y}p 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 _{z}x times ^{2}+y^{2}+z^{2}e; after we’ve squared. Right? And how can you simplify that? What’s ^{-ρ }x? It’s ^{2}+y^{2}+z^{2}r. Okay? So how does it depend on ^{2}θ? How does it depend on φ? Elizabeth?
x,which is ^{2}+y^{2}r, in two dimensions. Right? So it’s cylindrical, doesn’t depend on the angle around the bond axis.^{2}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?
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.
## 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,
Notice that the difference between the CH 2p, but you have to rotate it to see it. If we rotate it 90 degrees here, you can see that it’s a _{y}2p orbital. And again, the CH_{y}_{4} 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]
The same thing is true of the NH 3d. 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?_{x}^{2}-y^{2}
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