You are here
CHEM 125a: Freshman Organic Chemistry I
Lecture 14
- Checking Hybridization Theory with XH3
Overview
This lecture brings experiment to bear on the previous theoretical discussion of bonding by focusing on hybridization of the central atom in three XH_{3} molecules. Because independent electron pairs must not overlap, hybridization can be related to molecular structure by a simple equation. The “Umbrella Vibration” and the associated rehybridization of the central atom is used to illustrate how a competition between strong bonds and stable atoms works to create differences in molecular structure that discriminate between bonding models. Infrared and electron spin resonance experiments confirm our understanding of the determinants of molecular structure.
Professor McBride’s website resource for CHEM 125 (Fall 2008)
http://webspace.yale.edu/chem125_oyc/#L14
This website may include third-party materials pertaining to relevant topics, provided for the user’s convenience. Yale does not control or take responsibility for the content of any off-site pages or linked sites.
Transcript | Audio | Low Bandwidth Video | High Bandwidth Video |
---|---|---|---|
htmlFreshman Organic Chemistry ICHEM 125a - Lecture 14 - Checking Hybridization Theory with XH3Chapter 1. A Relationship between Hybridization and Molecular Structure [00:00:00]Professor Michael McBride: Okay, so we’re in the payoff period of having spent all this time on quantum mechanics, and we want to see first whether what we’ve done is realistic or not. So last time we looked at H-H and H-F, in terms of what the bond strengths were, and we saw that indeed one of them was more stable for breaking into ions and the other more stable with respect to breaking into atoms. But here’s a more extensive reality check today, which involves a lot more experimental evidence. And it has to do with XH_{3}; that is, the molecules BH_{3}, CH_{3} and NH_{3}. And we’re going to look at their structure, and how they move, and see how that checks out with what we expect theoretically. So first, what do we expect theoretically? Well BH_{3}, CH_{3, }NH_{3 }are the same, except boron has three valence electrons, carbon has four, and nitrogen has five. Now each of them uses three of their valence electrons to share with hydrogens to make the three XH_{3} bonds. But boron then has no leftover electrons, carbon has one and nitrogen has two. So we want to see what the implications of this are going to be for the structure and dynamic behavior of these molecules, and then we’ll check it against reality. First we need to look at something that shows that the hybridization that you use to make bonds should be related to structure. Now suppose there is one of these atoms that uses an sp hybrid orbital, that looks like that. You see the little dot where the nucleus is. And suppose it happens also to use some other hybrid orbital to make another bond. Now, you want to have, because of the Pauli Principle — and I’m sorry, we don’t have time to go into this now — the orbitals that you’re using must not overlap one another. Because if there’s net overlap between two orbitals then — and you have electrons in both of them — those electrons are, to a certain extent, the same thing, if the orbitals overlap. If they don’t overlap, then they’re independent. So to have only two electrons per orbital, the orbitals must not overlap. So we can use that as a criterion to say something about what other orbitals could be used at certain angles. For example, suppose the other hybrid orbital that this atom was using was sp^{3}. Now you’ll notice that — remember — let’s suppose the green and the blue are positive, and the red and the yellow are negative. Now how much overlap is there? Well there’ll be positive overlap where green overlaps blue, and there’ll be positive overlap where red overlaps yellow. But there’ll be negative overlap where green overlaps red, or where yellow overlaps — no, where blue overlaps red, or where green overlaps yellow. Okay? Is that right? Yeah, where blue overlaps red, or green overlaps yellow; that’ll be negative. And this, there’ll be some angle at which those things exactly cancel out and you get no net overlap, no overlap integral. And we’re interested in knowing what that angle would be; that is, how should hybridization be related to angles, be related to structure? Well, so the question is, what’s that angle? And there’s a nice formula, that isn’t too hard to work out, why it should be that way, that relates sp^{m} and sp^{n}, the values m and n, to the angle that should be between them, in order to have no overlap. So let’s use that formula and look at several mn’s, and what the angles is. Let’s look at sp with sp; sp^{2} with sp^{2}; sp^{3}, sp^{3}; and sp-anything with sp^{infinity}. What’s sp^{infinity}? That says the ratio of p to s is infinite. So it’s a pure p. So what angle would be pure p orbital make with any hybrid? Okay, so those are the things we’re going to talk about. So first let’s try the first one. You can do the math for me. Okay, so it’s sp^{1}, sp^{1}; m and n are one. So the denominator’s the √1, that’s one. The cosine of the angle is minus one. What’s the angle? What angle has a cosine of minus one? Student: 180. Professor Michael McBride: 180. Okay, good. So that would be linear. And you knew that, I bet, already. Didn’t they teach you that in AP Chemistry, that if you have two sp orbitals, they’re linear, or if you have bonds that point opposite direction you use sp hybrids for them. If they didn’t, that’s true anyhow. Okay? Now how about sp^{2}, sp^{2}? So we’re going to have m and n are both two; √4 is two. So it’s the angle whose cosine is minus one half. Anybody know what that is? I don’t — pardon me? Sixty? That’s +½. It’s 120, to be -½; it has to be greater than ninety to be negative. So that’s going to be 120º, and that’s trigonal. So sp^{2} hybrids make an angle of 120º. Now how about sp^{3} hybrids? That’s √9; so minus a third. That one probably isn’t stored in your head, but it’s an interesting angle. It’s 109.5º, which is the angle in a tetrahedron. Right? So that methane, which makes sp^{3} bonds at the carbon, has tetrahedral angles, 109.5º. Now, how about sp-anything, with n=infinity? What’s the denominator? Students: Infinity. Professor Michael McBride: It’s infinity. So what’s the angle whose cosine is… Student: 90º Professor Michael McBride: Okay, 90º. Right? So that means — that is what makes things σ and π. Right? A p orbital, with any other hybrid, has to be perpendicular to it. So you won’t — that’s what we talked about, σ and π, last time. So the p orbital is π, and it’s perpendicular to anything else, 90º, no matter what other hybrid. Okay, so now how about this particular one we have here where it’s sp^{3} and sp^{1}? That turns out to 125.3º. So that’s what the angle should be, if the carbon were making one bond with sp and one with sp^{3}. Right? But the important thing here is that’s there’s a relationship between hybridization of two orbitals on the same atom and what the angle should be between them. Yes, Angela? Student: When would you have an sp^{3} and an sp hybrid; like be at angles? Professor Michael McBride: Well we’re going to have to decide what determines the hybridization. Right? And it could be different for two bonds from the same atom, and we’ll see examples of that. So that will become answered soon. It’s if they’re bonding to different kinds of things, they may have different hybridizations; and we’ll see an example of that, several examples. So we want to optimize the hybridization, that is to get — yes, Lucas? Student: Does different weightings of the hybridized orbitals change the angle, degrees? Professor Michael McBride: Yes, there are two ways. You could change the angle by moving the nuclei around, for some other reason, whatever it is, and that would then change the hybridization of the carbon that’s making the bonds. So it works both ways. If you change the hybridization, you change the angle; if you change the angle, you change the hybridization. Chapter 2. Optimizing Hybridization in XH_{3} Molecules [00:08:31]Okay, now we want to optimize hybridization. What do mean, optimize? We mean we want to get the lowest possible energy. We want to choose the hybridization that’ll give the lowest hybridization [correction: energy], and therefore a particular geometry, for XH^{3}, certain angles. Okay, now let’s think about that. We’re going to have one of the valence electrons of the X atom in each of three bonding atomic orbitals, bonding to the hydrogen, and whatever’s left over goes into the fourth atomic orbital, and we want to minimize the energy. Okay? Now let’s first look at it from the point of view of the bonds. We want to make the strongest possible bonds. We want to maximize the overlap in the bonds. What hybrid gives the best overlap? sp. But we have to make three bonds; sp you can only make two bonds. So what’s the best you can use if you want to make three bonds? Student: sp^{2}. Professor Michael McBride: sp^{2}. So no matter whether it’s BH_{3}, CH_{3 }or NH_{3}, all of them will overlap best in making three bonds, if you use sp^{2}_{. }So the bonds cry out, “Use sp^{2}”; that’ll maximize the overlap and minimize the energy of the electrons in the bonds. Okay? But that’s not the only game in town, because there’s also the other electron that’s on the atom, if there is another electron. Okay? Because carbon has one and nitrogen has two other electrons. So they should get a say in it too. So the X atom, which has to accommodate whatever’s left over, says, “Okay, go ahead, make three bonds, but get the lowest energy for the other electron, that I’m the only one responsible for.” Right? So what’s the lowest energy you could make for a pair of electrons, or for a single electron; what’s the lowest energy valence orbital? Students: s. Professor Michael McBride: s. So the atom says, “Okay, make your three bonds, but make sure you don’t let s orbital go begging.” Right? Use the maximum you can of s orbital, because otherwise you’re using high-energy atomic orbitals when you might not need to. Now sometimes these would agree and sometimes they would fight with one another, the bonds and the atom. Okay? So from the point of view of the atom, boron, with only three valence electrons, all being used in bonds, doesn’t have anything else. So it would certainly want to use three sp^{2} and leave the p orbital vacant. That gets the maximum utility out of the s; puts the s where the electrons are. So that’s what boron says, use three sp^{2}’s, for the bonds. What does carbon say — no, pardon me, what does nitrogen say? Nitrogen, if it wanted to be the best atom, would put two electrons, the unshared pair, into the s orbital, and one into each of the remaining three p orbitals. Right? That gets the maximum use of out of s. You don’t want to waste s by putting it where there’s only one electron on the nitrogen atom. You want to use it where there are two electrons. Okay, so nitrogen says, “Well, I’d prefer to use three p orbitals to make my bonds, and put two electrons into the s orbital.” Okay? Now, what does carbon say? Carbon says, “Look, I got four electrons; three of them are going to be in some kind of hybrid, that’s going to make the bonds, and one of them’s going to be something else. But if I take the s out of one place, I’m going to put it into the other one. Over all the four, I’m going to use up all the s character. There’s one electron in each of them. Who cares?” Right? So carbon really doesn’t care, from the point of view of the carbon atom, about hybridization. It’ll care for making — for the bonds, but it doesn’t care for the unshared electron. Okay, but now we have to reconcile these two considerations. So the boron will make the best bonds if it uses sp^{2} and it’ll make the best atom if it uses sp^{2}. So great, they agree. So boron should really want to use sp^{2} hybrids for its bonds. Remember, those are the ones that form 120º. So it’ll be the pattern of a Mercedes star. Yes, Yoonjoo? Student: Can you explain again why we use three p instead of like, the hybrid orbitals for nitrogen? Professor Michael McBride: If you’re just looking at a nitrogen atom, it’s going to have two electrons in one of its orbitals, and one in each of three other orbitals. Right? So those two electrons, that orbital that holds two electrons, should be as low as possible in energy; and that’s s. So put two in s and one in each of the three p’s, and that’ll give the lowest energy atom, with one electron in each of the four orbitals — pardon me, with two in one and three in the other three; one in each of the other three. Okay? Okay, so but anyhow, with BH_{3 }they agree, from both points of view: the strongest bonds and the best atom. So it’ll certainly want to be flat, and 120º bond angles. Now how about CH_{3}? CH_{3}, from the point of view of the bonds, wants to be sp^{2}, and the atom doesn’t care. So fine, be sp^{2}, be flat, be planar, with 120º bond angles. But it won’t care quite as much as boron does. So carbon says it’ll prefer to be planar, but not as strongly; so it’ll be easier to bend it, because it doesn’t have this consideration of leaving a p orbital vacant. Okay? And now about nitrogen? Now you got a fight on your hands, right? Because the bonds want to be flat and 120º bond angles, but the atom wants to use p orbitals, 90º angles for the bonds; so like the corner of a cube would be the nitrogen, the bonds would be going out at 90º angles. So here there’s going to have to be some compromise. Okay? So NH_{3}, since it must compromise, won’t be either perfectly flat or 90º angles, but something in between. It’ll be pyramidal. Okay? Now are the predictions clear; and the reason for them? Yoonjoo, do you see now, why it’s the way it is? Don’t let me bully you into it, if you don’t really understand. Dana? Student: Can you explain the difference in what the atoms and the bonds want for nitrogen again? Professor Michael McBride: Yes, what we want to do is get the lowest possible energy. The lowest energy for the bonds will be the best overlap. You’re making three bonds. So sp^{2} will give the lowest energy, as far as the bonds go. But there also might be — and that’s the only game in town for boron. Right? But in carbon you have another electron you have to accommodate, and there’s a question of what energy it will have, even though it’s not being shared. And in nitrogen there are two of those other electrons. Now, the best orbital of an atom, for an electron to be in, the lowest energy, is s. And in nitrogen there are going to be two electrons that want to be in an s orbital. If they do that, they use up the s orbital, and the only thing you have left over to make the bonds is the p orbital. So now you’ve got a fight on your hands. Are you going to have sharp angles and have the unshared pair in a low-energy, s orbital but weak bonds, or are you going to have strong bonds, sp^{2}, but put the unpaired electrons in a p orbital? Okay, Kevin, did you have a question or did that answer it? Okay. Yes? I can’t hear very well. Student: Is there such a thing as an sp^{3} orbital? Professor Michael McBride: Yes. What does sp^{3 }mean? It means a hybrid orbital, a mixture of s and p, with three times as much p as s. Okay? You could have that. You could have three sp^{3} bonds, for nitrogen, for example, and then an sp^{3} hybrid, in which the unshared pair lives. Okay? That would be a possibility, sure. And that’s the kind of thing I’m talking about here, by saying it would be intermediate, between being 90º bond angles and 120º bond angles, flat. Any more questions? Yes? Student: Why are we concerned about the atomic orbitals rather than molecular orbitals? Professor Michael McBride: Ah, I’m going to get to that just shortly here. So, because we’re talking about bonds. Bonds are local. Bonds don’t go over the whole molecule. Molecular orbitals go over the whole molecule. And we’ve been talking thus far, mostly, in terms of molecular orbitals, but now we’re shifting to talk about bonds, because what we wanted to do was talk about bonds. And I’ll do a little about the philosophy of that shortly. Okay? So here are the predictions. BH_{3} should strongly want to be planar and be hard to bend. CH_{3} should want to be planar but be easy to bend. And NH_{3 }should be easy to distort out of plane, so that it’s not flat; in fact, not only should it be easy to distort, it should prefer to distort in its lowest energy form. Okay, so now let’s see if that’s true. Is it true that that unshared pair will compete for s character, and therefore make nitrogen pyramidal? Now, how do we know? There’s our prediction: distortion from the plane will weaken the bond and deprive the electrons of s character; so it doesn’t want to bend. Here, distortion from the plane weakens the bonds, but it shifts s character to the lone electron, where at least it’s not wasted, because there is at least an electron there. And in NH_{3}, it weakens the bonds but it shifts s character from single electrons to a pair of electrons; which is something good. Chapter 3. Infrared Spectroscopy on the Structures of the XH_{3} Molecules [00:19:18]Okay, so are these predictions true? Well, we need experiment. How about X-ray for seeing the structure of BH_{3}, CH_{3}, NH_{3}? How about NH_{3}? Have you ever seen NH_{3}? Have you ever smelled NH_{3}? It’s ammonia. So how about doing X-ray on it? Student: No. Professor Michael McBride: Why not? Shai? Student: Because you can only use X-ray to see electrons — Professor Michael McBride: Can’t hear very well. Student: You can only use X-ray to see electrons; those are the ones that are going to move when you — - Professor Michael McBride: No, that’s not the main problem. Yes? Student: You can’t make it a crystal. Professor Michael McBride: You have to have a crystal to do X-ray crystallography, and NH_{3} is a gas. Now you might say, okay, we’ll cool it way down to liquid nitrogen temperature and then maybe it’ll crystallize. It’ll be difficult to grow a crystal of ammonia, but it’s conceivable we might be able to grow a crystal of ammonia. Okay? Can we use the same trick for CH_{3} and BH_{3}? That one’s a gas. Okay? CH_{3}, unfortunately, if you get two of them together they react, to get C_{2}H_{6}. So you’re never going to be able to cool them down and get a crystal. And the same thing is true of BH_{3}. It forms B_{2}H_{6}. Right? So forget crystals for this. We can’t use X-ray crystallography, the technique we’ve used so far to do this. But there are other techniques for finding structure, and the one we’re going to talk about — actually we’re going to talk about two — but we’re going to talk about, not X-ray, we’re going to talk about IR — we’re going to talk about spectroscopy, infrared spectroscopy, and electron spin resonance spectroscopy. Okay? We’re going to talk more about infrared spectroscopy next semester, but I’ll show you just enough to get through the question that we’re interested in now. Does that remind you of anything that moves like this in a high wind? Student: A weathervane. Professor Michael McBride: It’s like an umbrella, right? In a high wind it blows inside out. Okay, so this is sometimes called the “umbrella” vibration of XH_{3}. And you can treat it with Erwin Meets Goldilocks, as if it’s one-dimension; just how much motion of all these atoms, how much motion. You know what the masses of the atoms are. You know the kind of motion you’re interested in. It’s all coordinated. So you’re changing really just one-dimension, in moving all these things in a coordinated way back and forth. Right? So you can treat it as an Erwin problem, with a fictitious mass that reflects the amount of motion of each of the four atoms, while that’s happening. Okay, so here’s BH_{3} doing that out-of-plane-bending vibration, the umbrella vibration, and CH_{3} doing the same thing with its electron. Now, it turns out that BH_{3} vibrates 34-trillion times a second. Okay? And that’s called 34.2 terahertz; that’s how often it vibrates. CH_{3} vibrates 18.7 terahertz; so only half as fast as BH_{3}. Does this make sense? Suppose the springs that you’re bending when you do this vibration — suppose it’s springs — which set of springs is stiffer? The BH_{3} or the — does a stiff spring make a vibration faster or slower? Student: Faster. Professor Michael McBride: Faster. So it says the springs are stiffer in BH_{3}. It prefers to be planar more strongly than CH_{3} does. Right? Now, you might ask — what might you ask at this stage, when I tell you that it vibrates that fast? [Students speak over one another] Professor Michael McBride: No, that’s not the question I want. Student: How can you tell? Professor Michael McBride: How do you know? How do I know that it vibrates at that frequency? Well, so it would be a stiffer spring, but how do I know? Well, when it vibrates, the electric charge is moving back and forth. Right? So the direction of the — there’s a dipole there, +/-, and it’s moving back and forth. And what does that create, when you have charges moving back and forth? Electromagnetic radiation. Right? So electromagnetic radiation is going to come out, as this thing vibrates, like that. And it comes out with a shorter wavelength for BH_{3} than for CH_{3}. Right? And it turns out that the wavelength is such that for BH_{3} there are 1100 vibrations, waves, in a centimeter; short wavelengths. And it’s longer wavelengths, only 600 waves in a centimeter for CH_{3}. Now, we can also — so different light would come out, infrared light, of these two; or different colors of light, infrared light, get absorbed. That’s how you actually measure it. And remember, the frequency of light has to do with the energies of the photon. So in, if you take the quantum mechanical view that you go from one quantized state to another quantized state, when you absorb, or emit, light, then we know, from this frequency, that in one case the two energy levels you’re talking about are separated by 3.26 kcal/mole, and in the other case by 1.73 kcal/mole; only about half as much. Is everybody on board with this so far? So it’s from the absorption of infrared light that we know how fast they vibrate, and therefore how stiff the springs are. Okay? Now just how stiff are they? Well we can use Erwin for this point of view. We have the amount of deformation, which is our one-dimension, as it goes back and forth, and we can assume that it’s something like Hooke’s Law — that’s an assumption — the distortion, these springs. And then we can adjust the potential, how sharp the parabola is, of potential energy, until we get two energies that are spaced like that. Does everybody see what we’re doing? So here we do it. On the left, we get these lowest two energy levels and we — there’s the parabola that does the trick. So that’s a strong, a stiff spring, strong planar preference. And in the case of CH_{3}, to get that lower energy separation, you need a broader parabola. It’s easier to bend; the springs are softer. Okay? Everybody see how that worked? Okay, so infrared spectroscopy then tells us that first, that it prefers to be planar, and second, that it’s stiffer for boron than it is for carbon; just as we predicted. Okay? Now how about for NH_{3}? For NH_{3}, you see two absorptions in the infrared, and they’re very near one another; 968 and 932 wavenumbers; that is, waves per centimeter. If we translate that, it means that there are two energy levels — there’s a low energy level — and then there are two that are very close to one another, in order to have these two different colors of infrared light being absorbed. How can we get this? How can we get — this is not Hooke’s Law. Remember what happens with Hooke’s Law in the energy spacing? What happens? They’re even. Remember? So this isn’t anything like Hooke’s Law. Where do you get two that are very similar to one another? Do you remember? Students: Double minimum. Professor Michael McBride: What? Students: Double minimum. Professor Michael McBride: A double minimum gives that. So NH_{3 }must be a double minimum, like this. Right? And those must — the blue and the red one, must be the one with three nodes and the one with two nodes. There’s also one with zero nodes, and one with one node. But those are both down at the bottom. Right? They have the same energy, because there’s not enough overlap between the two to make them split. Right? So the IR spectroscopy of ammonia tells you it’s a double minimum, and also, in order to get that spacing, tells you how big the barrier is. It costs three kcal/mole to make NH_{3} planar. It prefers to be pyramidal. Okay? And that splitting, that “tunnel” splitting, that we talked about before, for those two levels, is thirty-seven wavenumbers. Remember, there’s also the two that are down at the bottom, with zero and one node, and they’re separated by only one wavenumber; they’re very close to one another. But we can translate that into the tunneling frequency, when you’re down in the lowest energy levels. How fast does it go back and forth? And you remember how to do that. If one wavenumber is two small calories/mole, not kilocalories, and we use that thing that we talked about in lecture nine, about how to go from splitting to tunneling rate; and it turns out that in the ground state, in the lowest energy state, ammonia is flipping back and forth 10^{11} times a second. And in fact for a while that was used as a fundamental clock; here was something that you really knew the frequency of. Hooke, when he was trying to devise a clock that could tell where you were in longitude, could’ve used ammonia, if he’d had the technology to do it, which would be an absolute frequency. Okay, so IR shows — completely confirms what we said before, that BH_{3 }is stiff and flat; CH_{3} is easily bent and flat; and NH_{3} is not bent, but can tunnel back and forth, by the umbrella motion. Chapter 4. Electron Spin Resonance on Hybridization and Electron Density [00:29:44]Okay, so now the other technique I want to tell you about is electron spin resonance spectroscopy. And it turns out — and we don’t have time to go into this — well you know already that electrons are magnetic, and some nuclei are magnetic, and you can get a magnetic interaction between the nucleus and the electron, which influences how hard it is to change the electron from being pointing one way to being pointing another way. That’s a kind of spectroscopy; see what color of light you need to make an electron flip from being pointed one way to the other way. But it’s influenced by the magnet, which is the nucleus. But there’s a neat thing about it, which is that influence can occur only when the electron is inside the nucleus. Where, in a 1s orbital, is the highest electron density? Remember the formula for a 1s electron? What’s the atomic orbital? It’s a constant times what? [Students speak over one another] Professor Michael McBride: e^{-ρ}. Where’s it biggest? Student: At zero. Professor Michael McBride: Inside the nucleus is the most dense part of the electron. Right? So when the electron is inside the nucleus it feels this effect. When it’s not inside the nucleus it does not feel the effect, for the purpose of this spectroscopy. Okay? So we can measure how much s character there is in a hybrid orbital that holds an electron by seeing how strongly the nucleus interacts with the electron; because it happens only to the extent that it’s s. Right? So we could measure the percent of s character by measuring the light that’s absorbed by the electron. Christopher? Student: How could the electron ever actually be inside the nucleus? Wouldn’t that have two particles occupying the same space? Professor Michael McBride: But they’re different particles. And in fact it’s a little more complicated than that, because what’s the kinetic energy, if the distance is zero? Coulombic energy is 1/r. Suppose r is zero. What’s the kinetic energy? Student: It’s infinite. Professor Michael McBride: Infinite. Right? Now, that means the electron is really zinging, which means it becomes relativistic and its mass changes and all sorts of wild things occur. So it’s actually that which causes this interaction. So we’ll leave that across the way here in Sloane Physics Lab. Okay? And take it from me, that the amount of this splitting that you see in the ESR spectrum depends on how much s character there is on the orbital. Okay? And you can worry about whether it’s precisely inside the nucleus, or just so heavy and fast that it’s behaving weirdly. Okay? But anyhow, that’s the trick. Okay, so the singly-occupied molecular orbital, the one where there’s only one electron, in CH_{3}, will give rise to electron spin resonance, and the magnitude of this effect will tell you how much s orbital there is. Okay, so let’s look at it. And this is what I just said. There’s a line separation due to magnetic interaction, and it occurs only if the electron spends time on the nucleus, which happens only for an s orbital; so we can measure how much s orbital. Okay, so here’s a picture of planar CH_{3}, and the orbital, the singly occupied molecular orbital, SOMO, that contains the electron. How much time is it spending on the nucleus? Zero; there’s a node at the nucleus, it’s in a p orbital. Now suppose we bend it, so it looks like that; or we could look inside and it looks like that. Now it’s got some s character, in that orbital. Right? And if we make little boxes and blow them up to see what it looks like down near the nucleus, one of them has no density at the nucleus and the other has electron density at the nucleus. So you’ll see magnetic interaction, in this case. If it’s bent you’ll see magnetic interaction, and the more bent, the smaller that angle from 120 — remember, the smaller the angle, so it’ll go down toward ninety; the more s character, there’ll be more of it. Okay, so here’s what you see for CH_{3}. Bear in mind it has to be magnetic C, it has to be C-13H_{3}, not C-12, in order to see the spitting. But at any rate, the amount of this effect is thirty-eight gauss, which are the units it can be measured in. And that implies that, on average, it’s 2% s. Now, remember we already said that its lowest energy form is flat, with no s. But it’s constantly undergoing this umbrella vibration. So it’s spending time with different amounts of s character, not zero, as it vibrates. And the average is 2%. Now suppose I made the hydrogens heavier, by making them deuterium. So I make the atoms heavier. Do you remember how that changes the shape of the wavefunction, if I make things heavier? Here it is for H is moving. What if it were deuteriums that are moving, how will the thing change? Student: [inaudible] Professor Michael McBride: It’ll get narrower. So as it gets narrower, that means it’s not bending as much, on average. And what should happen to the thirty-eight gauss, if it doesn’t bend as much? Student: [inaudible] Professor Michael McBride: I’ll let you think about it a second. It’s only when it’s bent that it has s character and you see this effect. If it were rigidly flat, the splitting would be zero; thirty-eight would be zero. What would happen if I change H to D, so that it doesn’t bend as much, on average? Sophie, what would you say? Student: [inaudible] Professor Michael McBride: Can’t hear. Student: If it was more rigid it would — Professor Michael McBride: If it were more rigid it would — how big would the effect be? Bigger or smaller, the gauss? Student: Smaller. Professor Michael McBride: Pardon me? Student: Smaller. Professor Michael McBride: Thirty-six. Right? So again it confirms what we’re saying, that it prefers to be flat. It’s only when it bends that you get this effect, and it bends less with heavier atoms than with lighter atoms. Wonderful. So again it confirms what we said. There’s what CD_{3 }looks like. It’s a little bit narrower and therefore a little bit less of this magnetic interaction. So this is a structural isotope effect. CH_{3} spends more time more bent than CD_{3}, and thus uses more s character, and thus gives a bigger magnetic interaction. Okay, so again we’ve confirmed what we predicted. Now here’s one that we didn’t work on in predicting, is CF_{3}. Now what’s changing H to F going to do? Well, it’s going to be heavier. So the same deal that would happen with deuterium should happen here. But there might be more to it than that, because now we have more protons and a bigger nuclear charge of the atoms that were H before. So they’re going to attract the electrons in the bond more. So, but there’s another thing too. Because the fluorines have a partial negative charge, and are bigger, they have more electrons altogether, they’ll repel one another. Now what would — if the things attached to the central carbon get bigger and repel one another, what effect should that have on it being bent? Should it make it easier or harder to bend? Student: Harder. Professor Michael McBride: If these things are big here? Easier to bend from flat, or harder, if these things are — Students: Harder. Professor Michael McBride: Harder, right? So valence electron pair repulsion, and all that, should say that if CH_{3} is flat, what about CF_{3}? Student: It’s flatter. Professor Michael McBride: It should really be flat, harder to bend, if these things are bigger. Okay. But there’s another point of view that you can take, which is that fluorine has the lion’s share of the bonding electron pair. The bonding electron pair, in each of the C-F bonds, is spending more time near fluorine than near carbon. So carbon, being a cagy sort of character, says, “Why should I waste my valuable s character trying to stabilize these electrons that are spending their time on fluorine anyway?” Right? “Why don’t I use it to stabilize that other electron, that’s only on me and not on fluorine?” Got the idea? So if carbon uses its s character to go to the odd electron, forgetting the ones that are mostly on fluorine anyway, then that would make it more bent. That would favor bending, if you put the s character in the odd electron rather than in the bonds. That’ll favor using more p character in the bonds and therefore more bending. So this is exactly the kind of thing that Bacon would have liked. Right? This is a crucial experiment. There are two opposite predictions, and you can do an experiment and find out which one is true. Is it that fluorine is bigger and negatively charged, so it makes it more flat? Or is it that the carbon says, “I want to put my electron, the only electron I’m really solely responsible for, in the orbital with the most s?” Right? So this is a real horserace and you place your bets. Okay? Now how many people think it’s — we’ll guess ahead of time — how many people think that fluorines are bigger and negatively charged, so it will be harder to bend away from planarity? And how many people think that it’s the odd electron, that you want to put s character in? Well this is very close. Okay? So we’ll have to look at experiment. Okay, so this is the singly-occupied molecular orbital of CF_{3}. It’s mostly on the central carbon, but a little bit on the fluorines. And if you look in the middle, you see that there’s the s character on the carbon. And if we measure the splitting, it’s 271 gauss. What was it for methane? Students: Thirty-eight. Professor Michael McBride: Thirty-eight. This is ten times almost, or whatever number of times, maybe like eight times more s character than there was in methane. What’s the conclusion? It’s much more bent, much more of the time, than CH_{3 }is. Right? Okay, so that means it’s 20% s. That means that it’s sp^{4} in that, almost sp^{3}. Right? So it’s almost perfectly pyramidal, like methane would be. Right? So it’s the character of that odd electron, wanting to be s, that’s controlling the geometry. Okay, now we’re going to get to what Christopher wanted, which is why the heck are we talking about bonds when we’ve been talking about molecular orbitals all the time? Well one reason is we wanted to talk about bonds. That’s how we got into this whole thing, asking what bonds are. Okay? But there are three different operators here, and they have different motivation and methods. First there’s the molecule deciding what to be; and then there’s the computer doing quantum mechanics, deciding what the molecule should do; and then there’s you, trying to understand all this. And these are different. And let’s see the way in which they’re different. The experimental molecule, all it wants to know — it doesn’t care about Schrödinger — all it wants to do is get the lowest possible energy. So it minimized the total energy, kinetic plus Coulomb energies, of the electrons and the nuclei, by just settling down into the most comfortable, lowest energy arrangement. Okay? That’s what the molecule does, it just gets happy. Okay? And once it does that, then it has a certain structure, a certain arrangement of the nuclei and of the electron clouds. It has a total electron density, which you can measure by X-ray. It has certain energies that we showed measuring by IR; although you can measure them other ways as well. It has nuclear electron density, that we just measured by ESR. It has dipole moment, how much there’s a separation of positive and negative, that you can — that if it vibrates gives off IR and so on. So there are these real things that come from the molecule just doing what comes naturally. So it doesn’t care about Schrödinger. Now you have a computer that comes along, armed with Schrödinger’s equation, and it knows it can’t possibly solve the true Schrödinger equation, it’s too complicated. So it’ll do some approximation and try to get an approximate value. So it’ll minimize the total energy, using the Schrödinger equation, with some realistic constraints; for example, using orbitals, or using configuration interaction, or using density functional theory, or some approximation. Okay? So it’ll use like a limited set of atomic orbitals, only say 2s and — it won’t use the 5p orbital or the 5g orbital or something like that. Use Self-Consistent Field, or some correlation, or delocalized molecular orbitals come out. And its goal is to get a useful prediction of the properties; maybe the electron density, maybe the energy, maybe the vibration frequency, maybe absorption of, what colors of light get absorbed and so on. So the goal of the computer is to use any tricks it can get around — it can do, and it doesn’t care how hard it works — it’s not something you have to be able to do on a piece of paper — just to get something useful. And it will be validated, the approximations you made, by experiment. If it agrees with experiment, then you can believe the computer. But that doesn’t mean you understand what the computer did. Right? It’s a really complicated calculation. And that’s different from what you want to do, because your goal is to understand structure and reactivity, with the simplest realistic model; not Schrödinger’s fundamental equation, because you don’t have the brain power to manipulate numbers that way and do all these integrals and so on; leave that to the computer. Your goal is to try to look at things and understand: ah! that should be reactive with this; that bond should be very stiff; that molecular structure should be very stiff because the, quote, “bonds” should be stiff. Okay? That’s what you want to do is to develop an intuition that allows you to understand structure and reactivity. And you want the simplest, obviously, but realistic model. And that, to be both simple and realistic, may not be possible. Right? There’s a certain tension there. Okay? So what kind of things you’ll talk about? You’ll talk about localized bonds. You won’t talk about these big things that are complicated and go over the whole molecule. It’s interesting to see, for example, that those are analogous to atomic orbitals and that kinetic energy plays a part in them, but that doesn’t empower you to predict the properties of things. Okay? So you want to talk about localized bonds, maybe pretend they’re springs or something like that. And what makes a spring stiff or not stiff? And that, Christopher, is why we have been talking about springs and things here, rather than talking about molecular orbitals. So you talk about — these are Lewis’s ideas — bonds, lone-pairs, hybridization. We talk about energy-match and overlap. And we’ll be talking soon about HOMOs and LUMOs — highest occupied molecular orbital, lowest unoccupied molecular orbital — and what a very important thing those are. So we’re going to look at just little bits of things, to try to find out what’s crucial and will allow us to understand. We want them to be realistic. We want them fundamentally to be based on something true, like quantum mechanics, not just on springs. But we have to make it simple enough to understand. So we’re aiming for qualitative insight. And we’ll know when we’re right if we agree with experiment, and also if we agree with what the computer predicts; even though the computer doesn’t know why it predicted it, it just grinds through and gets you a number. But we want to understand why the computer gets that number. Okay. So let me just do one thing quickly here. Pathological Bonding. Or let me let you go to your next class; we’ll do Pathological Bonding next time. [end of transcript] Back to Top |
mp3 | mov [100MB] | mov [500MB] |