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# CHEM 125a: Freshman Organic Chemistry I

## Lecture 13

## - Overlap and Energy-Match

### Overview

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.

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

http://webspace.yale.edu/chem125_oyc/#L13

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html## Freshman Organic Chemistry I## CHEM 125a - Lecture 13 - Overlap and Energy-Match## Chapter 1. Distance and Hybridization in the Overlap Integral [00:00:00]
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
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]
[Laughter]
[Students speak over one another]
[Students speak over one another]
p? 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?_{σ}
p. Can you think of any other way we should talk about it? We’ve talked about _{σ}s with s. We saw that s with pwas orthogonal; _{π }s with pwe’ve got there; _{σ }p with _{σ}p. Any others we need to think about?_{σ}
p?_{π}
p? Can anybody guess that one? _{π}Pis the same sign, top and bottom; _{σ }pis opposite signs, top and bottom._{π }
p. Okay, so there’s _{π}p with _{π}p. 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?_{π}
[Students speak over one another]
p,or _{σ}p with _{π}p? Any guesses? Well, I’ll give you the answer. Right? _{π}p with _{π}p is approaching one. At big distance, it’s smaller than _{π}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 2pbetter than _{σ }either 2s with 2s or 2pwith _{σ }2p. 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 _{σ}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 sp, 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 ^{3}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 sp, how about if we use ^{3}sp? Remember what happens as we approach ^{2}sp? The orbitals extend more. So what do you expect?
sp, a little better. How about ^{2}sp with sp? Better still. Could we get any better? How about s with ^{2}ps, or ^{2}psp with ^{½}sp? Right? It’s a little better at short distance, and a little not so good at long distance. So it optimizes around ^{½}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?
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 spisn’t much worse than ^{3 }sp. 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 ^{2}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-p_{σ} is so good in its overlap. Now, so sp gives the best overlap, but only allows two orbitals, with 50% in each; sp gives four, and nearly as much overlap. Okay?^{3}## Chapter 2. Influence of Overlap on Molecular Orbital Energy [00:18:49]Now we’re going to look at the influence of overlap on molecular orbital energy. But we’re going to use Now, suppose we increase the overlap. What will happen if we shorten the distance? You did this on a problem set with 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 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
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 ## Chapter 3. “InferiorÂ Orbitals and Energy-Matching [00:29:45]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 So we already actually saw this in the case of the 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, bparts of the B density, and also this overlap term, ^{2 }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, A^{2 }is higher energy than B^{2}. 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, A^{2}. 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 A a is tiny. But the amount of ab you put in is a times b, not a. 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, ^{2}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 a you get is only 0.001. The amount of ^{2}b is 0.96. And the amount of ^{2}2ab is 0.06. So you’re getting sixty times as much of ab as you get of a. 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.^{2}So let’s look at
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 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
## Chapter 4. Experimental Evidence and Conclusion [00:46:59]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 But there’s also an antibonding orbital, That’s why you call it hydrofluoric _{2} hydrogenic acid or something like that. Right? But if you break it into two , 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 XHatoms_{3}, which will have lots of experimental data that will show that this stuff is sensible.[end of transcript] Back to Top |
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