CHEM 125a: Freshman Organic Chemistry I

Lecture 12

 - Overlap and Atom-Pair Bonds


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

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Freshman Organic Chemistry I

CHEM 125a - Lecture 12 - Overlap and Atom-Pair Bonds

Chapter 1. The United-Atom “Plum-Pudding” View for Ethane and Methanol [00:00:00]

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

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

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

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

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

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

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

Student: Oxygen has lone pairs.

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

Student: It’s not going to be symmetrical.

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

Student: Towards the electron pair.

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

Student: Away from hydrogen.

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

Student: It brings them.

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

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

Student: Positive and negative signs.

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

Student: Positive.

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

Student: It doesn’t matter.

Professor Michael McBride: Why not?

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

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

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

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

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

Chapter 2. The Orbital Shape of 1-Flouroethanol [00:13:25]

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

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

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

Student: Oxygen.

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

Student: One of the carbons.

Professor Michael McBride: Ah, which carbon?

Student: The left carbon.

Professor Michael McBride: Why?

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

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

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

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

Student: Towards the oxygen.

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

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

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

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

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

Chapter 3. Localized Pairwise Bonding between Atoms and the Idea of Overlap [00:19:38]

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

Student: Because you have multiple electrons.

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

Student: Position.

Professor Michael McBride: Position of what?

Student: One electron.

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

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

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

Student: For probability.

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

Student: You square it.

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

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

Professor Michael McBride: Why not?

Student: It needs to be one or the other.

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

Student: No.

Professor Michael McBride: Where would it be most likely?

Student: Probably nearer to one of the two atoms.

Professor Michael McBride: And which of the two?

[Students speak over one another]

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

Student: Both.

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

Student: Why isn’t there a 2A — or a 2AB?

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

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

Student: [Inaudible].

Student: You can’t just put the square —

Professor Michael McBride: You want that?

Student: Yes.


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

Student: Not very big.

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

Student: Not very big.

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

Student: Very big.

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

Student: No.

Professor Michael McBride: Do we make an error here?

Student: No.

Professor Michael McBride: Do we make an error here?

Student: Yes.

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

Student: Not very big.

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

Student: Difference density.

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

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

Students: One.

Professor Michael McBride: And this one?

Students: One.

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

Students: One.

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

Student: Into it.

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

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

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

Student: No.

Professor Michael McBride: Why not? Speak up.

Student: It’s very small near A.

Professor Michael McBride: What’s very small?

Student: The value of A.

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

Student: The value of B is very small.

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

Student: There’s going to be a change.

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

Chapter 4. Hydrogen at Bonding Distance: A Case for Overlap [00:36:37]

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

Student: Shouldn’t it be H2+ ion?

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

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

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

Student: A sphere.

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

[Students speak over one another]

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

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

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

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

Student: You could hybridize it.

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

Student: It has those lobes.

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

Student: The left one.

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

Student: No change.

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

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

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

[end of transcript]

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