CHEM 125a: Freshman Organic Chemistry I
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Freshman Organic Chemistry I
CHEM 125a - Lecture 13 - Overlap and Energy-Match
Chapter 1. Distance and Hybridization in the Overlap Integral [00:00:00]
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?
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.’
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?
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?
Professor Michael McBride: Zero. Okay? Now, as you bring it together, as they start coming together, will you get positive or negative overlap?
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?
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?
Professor Michael McBride: Zero, because it’ll be canceling right to left. Okay, so here’s a plot of that, for s-p σoverlap. 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?
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?
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 sp½? 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?
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 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?
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?
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 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?
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?
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 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.
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