 Okay, let's go ahead and get started. So today we're going to see an application of the eigenvalue problem, and it's going to require a lot of introduction, but first what I wanted to do was to correct a mistake that I'm pretty sure I made yesterday when telling you about something on the board. So yesterday we were talking about the matrix eigenvalue problem, which wrote it like this, a matrix multiplied by a vector is equal to a number times the vector, that is the definition of the eigenvalue problem, where we seek to find vectors and eigenvalues, eigenvectors and eigenvalues that solve such equations. Now yesterday what I said was that if A is n by n, square matrix, then we will have n eigenvectors and n eigenvalues, okay? And I said we could collect them all together and write down a single equation, which I wrote like this, capital X is equal to lambda capital X. This is actually not correct. I wrote it backwards, and I said something stupid about what you can do when you multiply by X on the left. So this is not right. It should be X times lambda, where lambda is the diagonal matrix containing the eigenvalues. Okay, and then we can multiply by X minus 1 on both sides, and so we get the final result, which tells us that solving the eigenvalue problem is equivalent to finding a matrix X, which contains the eigenvectors that diagonalizes the matrix A where the diagonal elements are the eigenvalues. Okay, so that's a little detail that maybe you don't care about, but in the interest of being correct, I thought I'd clarify that for you. It's done properly in the notes. Okay, now today what we're going to do is talk about an eigenvalue problem that arises in quantum mechanics and the discussion of, among other things, the electronic structure of atoms and molecules. Okay, so those of you who are in Chem 131A are familiar with the Schrodinger equation, which says H psi is equal to E psi. H is the Hamiltonian operator, which contains the kinetic energy, the Laplacian, and potential energy. Psi is the wave function, so this says you operate on the wave function by the Hamiltonian operator, and you get back the wave function multiplied by a number, which is the energy. And the energies, well, this is an eigenvalue equation, something times wave function or H operating on wave function gives value times wave function. Okay, so notice the similarity. All right, so when you want to solve the Schrodinger equation, you are essentially solving an eigenvalue problem. Now, in Chem 131A, unless they've changed the way they're doing things these days, H is always going to be the Hamiltonian operator. Okay, so that's going to give you a differential equation for psi, and then you see how to solve that for various model problems in Chem 131A. Now, in Chem 131B, you'll learn that you can write down a matrix equation that has the same form as our matrix eigenvalue problem that's an alternate way of expressing, well, formulating quantum mechanics. Okay? Now, the way that's going to work is, let's see here, well, I have to first introduce you to where we're headed with this. Okay? So let's, what we're going to talk about now is a model, a simple model, a simple but powerful and very informative model for discussing pi bonding in organic molecules. Okay? Now, let's try to remember what do we mean by pi bonding? What's a pi bond? Pardon? Yes, a double bond or a triple bond, and how are pi bonds formed? What, yes, and in particular, what kind of overlap? So we have different kinds of overlaps. We can have an overlap like this. That gives what kind of bond? Sigma bond, which has cylindrical symmetry around the bond axis. And when we talk about pi bonds, we're talking about if this is a pi bonding combination, these might be 2p orbitals on carbon atoms, right? So the sigma bond would form the first bond with cylindrical symmetry, and then the pi bond forms this thing that's sort of perpendicular to the bond axis. Okay? Now, do you remember the molecular orbital picture of pi bonding? So what's a molecular orbital? A molecular orbital is in the usual way that we think about them is that it's some kind of combination of atomic orbitals that's distributed over all or part of the molecule. Okay? And a convenient, this is a convenient way of discussing bonding in molecules is in terms of molecular orbitals formed by appropriate combinations of atomic orbitals. Okay? So do you remember what happens then if I take 2p orbitals on carbon? All right? So I have, and now I want to combine them together to make a pi bond in addition to this combination which we just talked about. So this would be an orbital where the positive and negative amplitudes overlap constructively, and so we might draw that something like that. And it goes down in energy relative to the atomic orbitals. And then what else do I get when I combine these 2p orbitals? I get the so-called anti-bonding combination. All right? So that would be something that looks like, and there would be a node down the middle. Okay? Now suppose I wanted to occupy, populate those orbitals with electrons. Where should I put electrons? First of all, how many do I have? How many electrons are there in each p orbital? Remember the electron configuration of carbon? Okay? So each of these guys comes in with an electron, and then in the molecular orbital we would put them both into the lowest energy or bonding combination. Okay? So you remember that from general chemistry and probably also organic chemistry. All right? Now, so this is pretty simple to work this out, you know, sort of qualitatively without any numbers involved on the board. But what now if I want to come up with a corresponding picture for something that has many pi orbitals? So suppose I wanted to do the molecule butadiene, where now I have 2pi bonds and 4p orbitals. Can you write those, all the orbital combinations down for butadiene? Probably if I forced you to by making it worth a lot of points and gave you an arbitrary amount of time to sit down, you could probably either reconstruct it from memory if you already learned it or kind of reason your way through it. But it gets harder and harder to write down these simple pictures as you get more and more pi orbitals. All right? Has anybody ever seen all of the pi orbitals combinations for benzene, for example? We're all familiar with the doughnut-shaped lowest energy molecule orbital, which has these constructively interfering p orbitals that are above and below the plane of the ring. All right? That makes sense, the resonance around the ring. All right, so what we're going to talk about today is a simple model for pi bonding in organic molecules called the Huckel model. And it will enable us to do actually quantitative calculations of the energies and also determine the way the orbitals overlap when we put them all together. All right? And what we're going to have to solve in order to get those orbitals is an eigenvalue problem. Okay, so now what I want to do is walk you through how we turn the problem of molecular orbitals into a simple eigenvalue problem that we can solve using techniques that we've already learned here. All right? So the first thing we're going to do is we're going to write down an expression for what's known as the linear combination of atomic orbital, molecular orbital model. Is that ring a bell from general chemistry? Hopefully. All right, so with this we're going to now write down an equation that expresses a molecular orbital which we'll call psi sub mu in terms of a linear combination of atomic orbitals which we're going to call phi sub i. Okay? So in our Huckel description of pi bonding in organic molecules, we're going to suppose that these phi's are 2p orbitals, all right? Now what I mean by a linear combination is that we have to add up a bunch of these orbitals. So we're going to sum from i equals 1 to n where n is the number of p orbitals we're going to combine. And then each of those is going to have a weight in the molecular orbital. And we'll call that weight c mu sub i. Okay? So I'm going to call these orbital coefficients. So these are the coefficients of the atomic orbitals that determine how they combine in order to produce the molecular orbital. And you may recall from general chemistry in this LCAOMO picture when you combine n atomic orbitals, you get n molecular orbitals. So mu goes from 1 to n. Okay? Now when you, essentially when you, you can think of it as like plugging these orbitals into the Schrodinger equation, what you end up getting out is a matrix equation that looks like follows. Okay? So that's essentially the Schrodinger equation applied to the linear combination of atomic orbital equation. All right? Now what are these matrices? So these are all n by n matrices. Okay? So h is what's called the Hamiltonian matrix and its elements, h ij, are equal to the integral of phi i of the Hamiltonian operator, phi j integrated over all space called dv. Okay? And this matrix S is what's called the overlap matrix. It measures how much each of the atomic orbitals overlap and that's given by that integral. Okay? So if you actually wanted to fill up these matrices, you'd have to evaluate these things. E is a diagonal matrix with the orbital energies. Okay? And C, this is a matrix of all the orbital coefficients. Okay? So it's going to have, it's basically going to be a column vector containing the orbital coefficients of molecular orbital 1, 2, dot, dot, dot, n. Okay? So this is sort of a general formulation of Schrodinger equation for the case where we express the wave functions in this form. All right? Now, this is essentially what you're solving when you do a Spartan calculation. Spartan uses a very similar approach to determine the orbitals and their energies in the ground state of a molecule that you sketch in on the screen. Okay? I mean, there's a little more to it than just solving this problem directly, but you're essentially doing this. Okay? Now, that's a bit too complicated for expressing as in a simple eigenvalue equation, which is our goal here today. So we're going to bring in the so-called Huckel model, which works surprisingly well for how simple it is. Okay? So what the Huckel model is going to say is that there's two types of interactions going on between the orbitals. Okay? Or two energies involved. Okay? The first energy is the energy of an electron in an atomic orbital on an atom. Okay? And that's given by a number, which we'll call alpha. It's just a number. It's an energy unit. Okay? And the value we're going to use today, is going to be minus 11 electron volts. Okay? And then the other energy involved is that between orbitals that are on adjacent atoms, neighboring atoms, we're going to call that beta. And this is going to be equal to beta only when J is equal to I plus or minus 1. Meaning I and J are neighbors. The two P orbitals are next to one another. And this is going to have the value today of minus 0.7 electron volts. Okay? And then if J is not equal to I or I plus or minus 1, this will be zero. Okay? Now the next simplification that comes in for the Huckel model is we're basically going to say that the overlap matrix is equal to the identity matrix. Okay? So diagonal elements are 1 and off diagonal elements are 0. Okay? So now this approximation here actually makes life nice because what that means is we can just get rid of S here. All right? And now we have something that looks like the problem that we know how to solve. Remember that was AX was equal to X lambda. All right? So that's the way we wrote the general eigenvalue problem. And now we see that within the Huckel model, the equation that we need to solve in order to get the coefficients of the orbitals and then the energies of the molecular orbitals is identical to that problem. HC equals CE. All right? And so essentially what we need to do is find the matrix C that diagonalizes the Hamiltonian matrix. There's the similarity transform. And we're done. Make sense? Sound like fun? No? Well, the reason I'm, well, one of the reasons I'm teaching of this in addition to the fact that I think it's a very nice illustration of application of what we're learning to a problem, useful problems in chemistry is also that you're going to be asked to do this in Chem 131B. Okay? So you'll have nice examples of how to do it. Okay. So now let's, there's in one other little item of business we have to deal with before we start cranking away. And that is let's, via a couple of simple examples, see how it is that we actually construct H. All right? Because the whole starting point for doing a Huckel calculation and getting orbitals and energies is we have to be able to properly construct the Hamiltonian matrix. And once we've done that, then we just plug and chug in Mathematica and you'll see it's a piece of cake. Okay? So let's see if we can figure out how to construct the Hamiltonian matrix for a couple of simple molecules. All right? Actually I want to check one little thing here. Okay, good. All right. So let's start simple. Let's do ethylene. Okay? And I'm going to suppose I have a 2P orbital on the first carbon which will label 1. And the second P orbital which I'll label 2. All right? So first of all, what's the dimension of our Hamiltonian matrix? How many atomic orbitals are we combining? Two. So we're going to get two molecular orbitals and our matrix is going to be 2 by 2. Okay? So it's going to look like this. H11, H12, H21, H22. All right? Now, within the Huckel model, what should I put in as the elements of this matrix? What's H11? Alpha. What's H12? Two is the nearest neighbor to one. Or if you like, 2 equals 1 plus 1. So that should be beta. How about 2, 1? Also beta. And 2, 2? Alpha. Okay? All right? So if we diagonalize that matrix, then we get the energies of the two molecular orbitals which we already wrote down sort of using what we learned in GCAM, like to get values for those energies in terms of these parameters, alpha and beta. And then we'll get orbital coefficients for the p orbitals and we can guess that one of them is going to be, they're going to be both positive in the case of the bonding combination and one of them will be negative in the case of the anti-bonding combination. And we'll see this in just a minute because we're actually going to solve that problem. Okay, let's do one more. Now I want to look at the conjugated pi system in butadiene. So I'll label this carbon one or this 2p orbital one, this one 2, this one 3, and this one 4. So what's the dimensionality of my Hamiltonian matrix now? Four by four. Okay? All right? And what's the 1, 1 element? Alpha. All the diagonal elements are always alpha. So we can write alpha. Alpha, alpha, and alpha. How about 1, 2? Beta. 1, 3. Zero. Because 1 and 3 are not nearest neighbors. And 1, 4? Zero. All right? How about 2, 1? Beta, alpha, and then 2, 3, they're neighbors. So beta and then zero. And then 1, 3, or I guess it's 3, 1, 0, 3, 2, beta, 3, 4, beta. And then we have 0, 0, beta, alpha. Okay? And if we were to diagonalize this matrix, we would get coefficients of the four molecular orbitals and their energies. All right? So here's what we're going to do next. We're going to do the simple ethylene calculation. And the reason I want to do that is because later, I want to compare, we're going to do benzene, and I want to compare how stable is the ground state of benzene compared to three isolated ethylenes. All right? So that gives us a way to quantify the aromatic stabilization that you learned about in organic chemistry. All right? So let's see if we can construct because we're going to need it in just a minute, the Hamiltonian matrix for benzene. All right? So we're going to suppose there's a 2p orbital on each of the carbons. There's a total of 6. So we have 6 by 6. Matrix, 1, 1, alpha, 1, 2, beta, 1, 3, 0, 1, 4, 0, 1, 5, 0. 1, 6, they're neighbors. So it should be beta. All right? So when you go around the ring, see you populate this guy with a beta. All right? What about 2, 1, beta, 2, 2, alpha, 2, 3, beta, 2, 4, 0, 2, 5, 0, 2, 6, 0. See how it works? It's not so bad, right? You don't like it. All right? Where are we? 3, 3, 1. Zero, 3, 2, beta, 3, 3, alpha, 3, 4, 3, 5, 3, 6. You see this little cluster moving across the matrix, right? Okay? So then we should be able to just say zero, zero, beta, alpha, beta, zero. And then we have to be careful here. This is going to be 6, 1, which is beta. It's their neighbors. Zero, zero, zero, 6, 5, beta, 6, 6, alpha. Got it? Look right? Did I skip the 5? Oh, yeah, we missed one in there, didn't we? All right. Left to move this one down. Beta, zero, zero, zero, beta, alpha. Right. It should be 6 by 6, huh? Okay. So then 5, 1, 0, 5, 2, 0, 5, 3, 0, 5, 4, beta. 5, 5, alpha, and 5, 6, beta. Okay, now we're good. Okay, so here, that's what we're going to do then. We are going to solve the eigenvalue problems for that one and that one and do some comparisons. And for benzene we'll even draw the molecular orbitals very schematically. And you maybe have seen them before, but if not, hopefully it will be interesting to you. Okay, so now let's go ahead and get going on that. Any questions on the Huckel model? Okay, so what I'm going to do is first we'll start with ethylene and we're just going to input the Hamiltonian matrix which I'll call H. So H equals, curly, curly, and we can use the pallet to put in alphas and betas. All right, so what did we say? We need an alpha, comma, beta for the first row. And then for the second row, beta, comma, alpha. Okay, semi-colon and we can just have a look, make sure it looks like what we wrote on the board. So there you have it. All right, next thing we're going to do is we're going to find the energies. Energies which I'll call E, those are going to be equal to the eigenvalues of H. So there you have it. Which one of those is the lowest energy? If we use the values that are written down over there on the board, alphas minus 11, betas minus 0.7, they're both negative. So actually this alpha plus beta is the lowest energy. All right, so let's draw, I'm sorry to go back and forth so much, but let's draw the energy diagram for this thing. So here's what we just determined. All right, so this is just like what we did at the start, right? We said the two p orbitals come together. They combine in bonding and anti-bonding combinations and then we have a lower energy orbital and a higher energy orbital, the bonding combination which we'll see in a minute, exactly how that works in the anti-bonding. Okay, but now we actually have energies. We can plug in numbers and make specific predictions. The other thing that we can do is we can populate it with electrons. So we already decided earlier that we have two pi electrons and so they go there. And so if I was to write down the energy of the ground state, what would that be? Alpha plus beta is the energy of the orbital and I have two electrons. So it's two. Everybody got that? Because it's going to be a little more complicated when we do benzene so I want to make sure you got it. I guess so. All right, so there's the energies and the next thing we can do is look at the orbital coefficients. What do we have to do to get the orbital coefficients? What do we have to find? Eigenvectors maybe? Okay. All right, so I'm going to put those into a variable I call C. All right? So we'll just say C equals eigenvectors of H. Okay? Now, we're going to have to do a little massaging here to make them into proper orbital coefficients in just a second. But what we can already see, so what are these things? These are the numbers by which we multiply two p orbitals and combine them into the pi orbitals. So what this eigenvector tells us is that in the LCAO-MO formulation, we take one times the first p orbital and we add it to one times the second p orbital. So this is a positive combination of the p orbitals or in other words, a bonding combination, right? And that bonding combination corresponds with the lowest energy eigenvalue. And the minus 1, 1 says we're actually combining them in the destructive interfering way, right? We're flipping, if you like, or changing the amplitudes by minus 1, a factor of minus 1 in one of the p orbitals. So this is the anti-bonding combination. Now, there's something wrong with these coefficients. And you aficionados of quantum mechanics in Chem 131 have learned that orbitals should be, wave functions should be orthonormal, meaning their magnitude, the magnitude of the coefficient vector should be 1. And these two guys should be perpendicular or orthogonal to one another. So if I dot them together, I should get 0. So these guys dotted with themselves should give 1. And then if you dot them with each other, you should get 0. Is that the case here? What's the length of this vector? It's not 1. It's 1 squared plus 1 squared is 2 square root, 2 squared is 2. So we should normalize those if we want to have proper orbitals. So if I call psi 1 the lowest energy orbital and I want it to be normalized, I could say psi 1 equals normalize bracket, bracket, or sorry, bracket C. And I want the second element of C, bracket, bracket 2, all right? And so now if I do that, you see I get the orbital coefficients have been normalized such that now the magnitude of that vector is equal to 1. And similarly, I can normalize the second wave function to higher energy 1, which I'll call psi 2, by choosing the first vector in the list, all right? So that's just minus 1 over the square root of 2, 1 over the square root of 2. Now let's check to see if they're orthonormal. So I can say let's look at psi 1 dot psi 1 to 1. Psi 2 dot psi 2, you have to spell it right? 1. And then finally psi 1 dot psi 2, 0. Okay, so these are what we call orthonormal vectors. These collections of orbital coefficients. Okay, so now we've basically completely solved the problem of ethylene. The last thing that I want to do, we're going to use this later. So I'm going to store in a variable called e ground ethylene. This is going to be the ground state energy of ethylene, which we said should be 2 times the lowest energy orbital, which is, its energy is e bracket, bracket 2 bracket, bracket. Okay, so that's 2 times alpha plus beta as we already noted. All right, now we're going to do benzene. So we have to type in the matrix over there. Okay, so we type in H equals bracket, bracket, alpha, beta, and then we have three zeros and then beta. So that's the first row. Second row is beta, alpha, alpha, beta, zero, zero, zero. Third row is zero, beta, alpha, beta, zero, zero. And then the fourth row is zero, zero, beta, alpha, beta, zero, zero. Actually, that's too many zeros, one zero, all right? And then the fifth is going to be zero, zero, zero, and then beta, alpha, beta. And then the sixth will be beta, zero, zero, oops, zero, zero, beta, alpha. Okay, a lot of work, but it'll be worth it. All right, now let's just have a look at it in matrix form to make sure it looks right. Does that look like what we wrote on the board? Looks good to me. Everybody got it in right? Because if not, you won't be studying benzene. Okay, now I want the energies. So what do I need to do? Eigenvalues, so let's do it. E equals eigenvalues of H. It's that simple. And as expected, we get six values. Which one is the lowest energy? The last one. Which one is the highest energy? The first one. They actually decrease like this. There's something else interesting that we can observe here. Notice these and these. What do we call that when two orbitals have the same energy? Degenerate, okay? So let's go ahead and write down an energy level diagram for this. And if you're really lucky, you've got to see it when you were taking GKM, but probably not. Okay, so let's go over here and draw our energy level diagram for benzene. Okay, so we have the lowest is alpha plus 2 beta. All right, and we have one of those. And then the next highest is alpha plus beta, one beta. So it's higher in energy by one beta, or minus one beta. And now we have two of those, okay? And then the next one is alpha minus beta. So that's going to be higher by 2 beta, minus 2 beta I guess I should say, since beta is negative. And then finally, excuse me, alpha minus 2 beta. And we have one of those. Has anybody seen that diagram before for the molecular orbitals in benzene? No? Okay, so you learned something new. All right, now, how do I populate that with electrons? How many electrons do I have to put in the pi orbitals? Six, okay? And we follow the usual rules. One, two, three, four, five, six. So if we want to calculate the energy of the ground state of benzene, what do we need to do? We need to take 2 times alpha plus 2 beta plus 4 times alpha plus beta, that would be the ground state energy of benzene, the electrons in the pi orbitals of benzene in the Huckel model, okay? All right, so we'll do that in a minute. But the next thing I want to do is find the orbital coefficients, in other words, the eigenvectors. And then we'll actually make a little very crude but hopefully informative diagram of what those orbitals look like. And then we'll calculate the ground state energy of benzene, compare it to 3 times ethylene, and get an estimate of the energy stabilization due to the resonance or aromatization. Is that ring a bell from organic chemistry? Yeah, okay. All right, so that's the plan. Okay, so orbital coefficients, C equals eigenvectors of H, okay? So we get six of them, each of them six in length, right? Because each one of these numbers is for a given molecular orbital the weight of the 2p orbital on each atom, okay? So one of them looks familiar, I hope, and that's the lowest energy orbital. That's the one where we make the fully bonding combination of 2p orbitals. And that's the one that gives those donut shaped orbitals around the benzene ring. You see that? Because all of these are plus one. So that means all the p orbitals are parallel to one another and they're all adding up in phase, okay? Is this normalized? Do they have a length of one? No, has a length of square root of six. All right, so now what we're going to do is we're going to label these from lowest to highest, one, two, three, four, five, six. We'll normalize them. Another little issue, I don't know if you guys in Kim 131A talked about this. It turns out that the degenerate combinations corresponding to these two energies and these two energies as they come straight out of the eigenvectors they are not orthogonal to one another. So we're going to add and subtract them to make combinations that are orthogonal, all right? So that's a little detail. Sorry about that. Okay, so let's go ahead and normalize them and make orthonormal combinations for the degenerate ones. All right, so the highest energy orbital, I'm going to call psi six and it's already, I mean it's non-degenerate so to make a proper normalized orbital, we'll say normalize C bracket, bracket one, okay? So that's this one corresponding to the highest energy, all right? And we need another bracket. Actually, that's another thing that I meant to point out to you and we'll see this when we draw the diagrams. Notice that the highest energy, so the lowest energy orbital was when all the two p orbitals were adding up in phase, right? All their amplitudes are plus one or they're going to turn out to be one over square root of six, but whatever. Notice here these alternate in sign, so these correspond to one p orbital with positive amplitude in that direction then the next one that direction, next one that, so this is the fully anti-bonding orbital and it happens to be the highest energy, which we expect. Okay, now when we go to the next one we get to a degenerate orbital and so what I'm going to do is I'm going to define psi five as a combination of the two degenerate orbitals corresponding to the same energy here. I'm going to say normalize C bracket, bracket two, bracket, bracket plus C bracket, bracket three, bracket, bracket. All right, so that's what we get and then I'm going to call psi four the same thing except instead of adding we're going to subtract them and now we can do, whoops, I meant to call that psi four, so let me re-enter that and call this psi four, okay? And now let's look at psi four dot psi five. We see zero, all right? So we've created by these adding and subtracting, we've taken non-orthonormal eigenvectors and created two orthonormal orbitals with the same energy. All right, the next one is psi three. This is another degenerate orbital so we're going to do the same trick, say normalize, bracket and this time we're going to add C bracket, bracket four, bracket, bracket plus C bracket, bracket five, bracket, bracket and then we'll prepare psi two by taking the same two eigenvectors but this time we'll subtract them and psi two and psi three now will be orthonormal, okay? And then finally psi one is non-degenerate so we don't have to worry about doing anything special and we'll normalize C bracket, bracket six, bracket, bracket. So now we have the full set of energies and orbital coefficients properly normalized and made to be orthogonal for the six molecular orbitals in benzene constructed from six 2P orbitals on each of the carbons within the Huckel model, all right? Now what I want to do is draw the orbitals, okay? So what we're basically going to do here and what I'm going to do is type in the first one quickly and then we'll just copy paste the rest of them and I'll explain to you exactly what I'm typing in but the strategy here is going to be that we're going to suppose that we're looking down on the ring, okay? Now if we're looking down on the ring we're going to see the end of a P orbital which will be sticking out of the screen at us, all right? Now depending on the amplitude that's facing us, in other words whether there's a minus or plus coefficient, we'll either draw a gray disk to represent that low of the P orbital or we'll draw a black disk, all right? So we'll be able to see when the amplitude changes by when the disk changes from gray to black or vice versa, all right? And you may notice that when we created these orthonormal combinations we actually got molecular orbitals that have no contributions from some of the atomic orbitals. They have zero coefficients. So we're going to indicate that by just drawing a little dot, okay? So we're going to organize these disks around a hexagon and draw and color them according to the coefficients. Their diameters will be proportional to the size of the amplitude here, all right? So let's do one of them and then I'll try to explain to you what we're actually doing and then once we have one in it will be a simple matter to draw the rest of them just by copy pasting, all right? So the command we're going to use to draw disks is called graphics, all right? And then within graphics we're going to, well, we're going to be using the graphics command to display a table of disks that we're going to define, okay? So I'm going to say table, all right? And then I'm going to set the color using gray level and the value is going to be given as follows. 0.5 plus 0.5 times psi 1 bracket, bracket I, bracket, bracket where we're going to use I to index around the ring, okay? And then I need one more here, all right? Now what do I need, comma, okay? So now I'm going to specify a disk, okay? And to specify its position I have to give coefficients on the screen, so in other words like an X and a Y coefficient. And in order to generate a hexagon we're going to do that using trigonometric functions. And so we can do that by saying that the X coordinate is cosine of 2 times pi times I divided by 6 and the Y coordinate is going to be sine of 2 times pi times I over 6, all right? So this is going to draw the disks equally distributed around a hexagon, all right? So now we can put in the curly bracket and then we're specifying now the diameter of the disks, all right? So that's going to be the maximum value of either 0.02. So if the coefficient is 0 then we're going to have a really small diameter, 0.02, or and through trial and error I determined that 0.8 times the absolute value of psi 1 bracket, bracket I and then we need bracket, bracket and then another one for the abs, another one for the max, another one for the disk and another one for the table, I have some problem here. This should be 0.8. Let's see, I need one here for the coordinates of the disk. Ah, yeah, you're right, I need one out here, way out here, okay? So that takes care of that one and then I should have one for table and one, wait, oh, I have to put in the counter. I goes from 1 to 6 and then one for the table but I think you're right, I'm missing one more thing here, no? Oh, I'm going to put in an option here, image size, arrow, small, bracket, okay, so that's pretty complicated. By the way, I'm not going to ask you to draw these things. If you want to do it, you have an example now but I will definitely ask you in the next homework to do this kind of calculation, okay? All right, so let's see if this works. Yeah, it does, so what is this? What we're doing is we're looking from one side of the ring, down on the ring, at, if you like, the bottoms of, for the tops of the p orbitals and the fact that they're all gray is meant to indicate that these guys are all going to combine in a constructive way and this is because we've put in psi 1 here, this is the lowest energy orbital. Did everybody get that on the screen? Okay, all right, so that one you've seen before, that's the donut orbital, okay? So now let's go ahead and do the next ones and all we have to do is copy, paste and we just change psi 1 to psi 2 if we want to get the next highest energy orbital so we only have to do that in two places, all right? And if we look up here at psi 2, we can see now that the coefficients are not all the same so we should have discs of different sizes and when we enter that, we see that that is in fact the case. So notice the second highest or second lowest energy orbital has p orbitals with one sign, they're lobes of one sign facing in this direction and the other ones have the opposite sign, that's what the black means, okay? And notice that there's a plane that we can draw through which if you go across the plane, you get this change in the sign of the orbital and what do we call such a plane? A node, okay? And it's also interesting to see that the ones on the end have greater amplitude, greater weight, okay? All right, so you've seen the donut orbital, I don't know if you've seen that one before. If you were going to draw this with resonance, it would be, this would all be connected together and then there would be the node and then this would be connected together, all right? So let's do the next one. So the next one has the same energy as the one we just drew but we'll see that it's going to have a different shape, if you like, so we change psi 2 to psi 3 and we enter, okay? And notice this one's interesting because it's got zero amplitude so the two p orbitals in this particular molecular orbital, the two p orbitals on these two carbons don't contribute to the pi bonding and then we have this bonding combination over here and bonding combination over here but through this plane now we have a node. And so another thing that you can notice is that orbitals with the same energies have the same number of nodes. So each of these two, which would contain two electrons in our energy diagram over there, they each have one node. So how many nodes do you think the next one's going to be? The next one is going to be of higher energy and it's degenerate, all right? So here we're going to change psi 3 to psi 4 in two places. Once again we find that these two carbons don't, their two p orbitals don't contribute to this molecular orbital and now notice we have anti-bonding combination here and here and here and here. How many nodes are in this molecular orbital? If I draw lines through which there are anti-bonding interactions where would they be? Does that look right? All right, so there's two nodes. Okay, so the next one has the same energy as this one so how many nodes do we expect to see in that one? Did I hear somebody say two? Okay, good. All right, so now each of the two p orbitals has some amplitude in this molecular orbital. We have bonding here, bonding here, anti-bonding here, anti-bonding here and here and here. So where are the nodes in this molecular orbital? There should be two, one through there and one through there, okay. And then finally we'll look at the highest energy orbital and we already saw from the coefficients earlier that this should be the totally anti-bonding combination and that means it ought to have three nodes. So now let's do psi 6 and there you have it. So as you can see the sign of the lobe that's facing us changes sign at each carbon, okay. Where are the three nodes in this one? It's like karate chops, right? One there, one there and one there. Okay, now, do those look familiar to anybody? Have you ever seen these in your books somewhere? No? Okay, all right, so you learned something. All right, so the next thing we're going to do and this will be, we'll finish up with this is we're going to see what's the aromatic, we're going to quantify the aromatic stabilization of benzene. Now, if we just draw the normal sort of valence bond of Lewis structure for benzene, we draw three double bonds, right? And then we can draw a resonance structure where those double bonds move around the ring, okay? So if the picture of benzene is having three double bonds was correct, we would expect that it's pi bonding energy should be about the same as three ethylenes, right? If there was nothing special going on in benzene. However, we know that that's not true and there's lots of ways to think about it. Benzene, pi electrons are stabilized by the resonance, if you like, and now we're going to quantify in energy terms how much that is, okay? So to do that, we're going to subtract three times the ground state energy of ethylene from the ground state energy of benzene. So first, let's calculate the ground state energy of benzene. So E ground benzene equals, so we have the lowest, we said we have three electrons, we already drew the diagram over there. The lowest three molecular orbitals are occupied, each with two electrons, so we can get the ground state energy by adding up the energies of those three lowest levels and multiplying it by two, okay? We say two times parentheses and then it's going to be E bracket, bracket six, that's the lowest energy eigenvalue, plus E bracket, bracket five, that's the next lowest energy eigenvalue and then plus E bracket, bracket four, bracket, bracket, the next, all right? And if we enter that, we get a formula. Now we can define E aromaticity equals E ground benzene minus three times E ground ethylene, which we already calculated, okay? And I'm going to simplify it. And so what we find is that according to the Huckel model, the ground state of the pi system in benzene is more stable than that of three times ethylene or three isolated single bonds by an amount two times beta. Why did I say more stable? By more stable, I mean lower in energy and the reason I said that is because I know that beta is negative. In fact, we said a reasonable value for beta is what? Minus 0.7 electron volts, so let's go ahead and put that in. So we'll get a number, minus 0.7. Well, we can kind of do it in our head, can't we? Such minus 1.4 electron volts. Is that a lot of energy? It kind of is. It turns out to be about 32 kilocalories per mole, which is about 130 or so kilojoules per mole. That's pretty substantial. Now, the question is, is it correct? Is it reasonable? Do you remember from organic chemistry how it is that we're different ways in which you can get an energy value from experiment for the aromatic stabilization? If you measure the hydrogenation, the energy of hydrogenation of benzene to cyclohexane, that's one way of doing it. And when you do that, you find that the answer is about 36 kilocalories per mole. All right, so that compares pretty favorably to the Huckel model, which predicts 32. Of course, the Huckel model parameters are chosen to give reasonable results for a bunch of different conjugated systems, so it should be good. All right, but the point is the simple model, first of all, I think provides us with nice insight about the electronic structure of benzene, both in terms of its energy, so we could draw that energy level diagram over there. We could also see what the orbitals look like. And we did that all with very little effort using the eigenvalue solver in Mathematica. All right, so that's going to do it for today. We'll see you on Thursday.