Lecture 21. Bigger Atoms, Hund's Rules and the Aufbau Principle
Shaka, Athan J.
 This course provides an introduction to quantum mechanics and principles of quantum chemistry with applications to nuclear motions and the electronic structure of the hydrogen atom. It also examines the Schrödinger equation and study how it describes the behavior of very light particles, the quantum description of rotating and vibrating molecules is compared to the classical description, and the quantum description of the electronic structure of atoms is studied. 
 Today's lecture in Chem 131A is going to be about bigger atoms, not just two electrons. Hun's rules, which are some rules to figure out the most stable state of an atom when it can have different kinds of electron configurations. And the Aufbau, or building up principle, which is the basis for putting atoms on the periodic table. The periodic table, of course, is one of the great intellectual triumphs of chemistry, is to take all these things that we find in nature and first figure out that they're all made up of these atoms, and then figure out that the atoms follow patterns, and then to figure out what the patterns mean, and then to figure out everything else that has followed on. And that's the kind of inexorable nature that science has to it. Once you find out something, then that something else follows from it, and a lot of other stuff is discarded as completely wrong, even though it seemed to be plausible before you found out whatever it is that you found out. A closed shell atom, which we said could be written as one slater determinant in the last lecture, can, in which you've got n orbitals, and they're occupied by 2n electrons, half of which are spin up, and the other half are spin down, otherwise it wouldn't satisfy the Pauli principle. We can write down the Hamiltonian for a system like this. What do we have? Well, we have minus 1 half times all these del squared. All that is is the kinetic energy. We know how to calculate that. There's a formula for it. It's got r, theta, and phi in it, and unfortunately for bigger atoms other than helium and hydride and those guys that we've been doing, we're going to have to get into theta and phi as well, and you can imagine that the equations get more complicated quickly. And we've got 2n of those because we've got 2n electrons in n orbitals, and therefore they've all got kinetic energy buzzing around. We have to keep track of that energy. So those terms are in the Hamiltonian. Then we have the potential energy of this big positive charge, z, which depends on the atomic number, divided by the distance of each electron from the positive charge. When we write it like this, we're kind of implicitly assuming that the nucleus is a point charge, and that's not quite true because the nucleus has a finite volume. And if you look very closely at spectral lines, there can be slight shifts and various things that have to do with the nuclear volume. But for now, those are very, very, very tiny things compared to, in fact, the accuracy with which we would be able to calculate things with the orbital approximation anyway. So we'll ignore those. And then we have this giant sum of all the repulsive parts, which I've written here as a double sum, from j equals 1 to 2n, and then on i equals j plus 1 to 2n, just so we don't count things twice. We could count them twice and divide by 2. Sometimes if you're writing a computer program, that might be easier as long as you aren't actually calculating some terrible integral twice for no reason, in which case, in that case, you're wasting a lot of time to calculate it this way and that way, which is the same, in fact, that the computer is not going to correct your bad algorithm and then divide it by 2. That's pretty easy to understand. If I've got five electrons, I've got the repulsion between 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3. I don't want to count 2 and 1 again. I don't want to count 2 and 2 because that doesn't count. 2 and 4, 2 and 5, and so on. And that's all that double sum means, is count them all up, put them in there, and now we've got to figure out how we're going to solve this thing with the kinetic energy as the first sum. The potential energy is the second sum and then the sum of all the mutual repulsion terms between the electrons as the third term, the double sum. We can write this a little bit neater if we introduce, instead of the capital H for the Hamiltonian, we introduce small H and we index it with the electron J and we say, look, that's just 1 minus 1 half, del squared J, take the derivative with respect to the coordinates of electron J, XJ, YJ, ZJ, or RJ theta J, phi J, and then minus E over RJ, which is the potential. And if we do that, then we configure the energy of the atom as a sum of three distinct terms. Now, I've written out this giant integral, which has a lot of integrals over R1, R2, depending how, out to 2n, and then I've also put in this funny notation, d sigma 1, d sigma 2, out to d sigma 2n. What does that mean? Well, the d sigma part has to do with the spin. We need to take into account the spin as well, because all these 2n electrons have spin, and we have to integrate over the spin part, make sure that the spin functions are normalized, and keep track of that as well as the spatial part. And then we've got the wave function, psi star, which depends on all the coordinates of all the electrons and principle, the Hamiltonian, and then psi. And that breaks up into 2 times a sum of what I'm going to call i sub j plus a double sum from i equals 1 to n, j equals 1 to n of 2 times ij, j, capital J, ij, minus capital K, ij. Let's see what that, what this means. So let's look at these three terms. Well, the i part is just the energy of each single electron. Ignore the others. The j part is the, what we get when we take the integral of psi i r1, psi j r2, then the repulsive term, psi i r1, psi j r2, that we've seen before. And then K is an integral that's a new one. It's the integral of psi i star r1, psi j star r2, psi i r2, not r1, psi j r1 so that the indices are swapped in the second part. So the i is the single electron kinetic and potential energy, jij is the Coulomb integral that we've encountered before. And Kij is called the exchange integral. Kij is a new player in the game. It has no analog and classical mechanics. And here's why I've written it out here on the bottom of slide 523. It has, if we wanted to interpret things as charged clouds interacting, we can't do it because we've got psi i star of r1, which is the coordinate of the first electron. And that's going on to psi i of r2, which is the second electron. And likewise, we've got psi star j of r2, then the repult one over r12, and then psi j of r1. So I think you can see why it's called the exchange integral because you swap the indices on the other side of the function. With larger atoms, the self-consistent field method still works, but there's this exchange contribution which Vladimir Falk saw, and that's why it's called the Hartree-Falk method. And keep in mind, there's kind of a shorthand here by using 2ji, and that is that when k or j, if both the indices are the same, then it doesn't matter that they were swapped. So k11 is j11, and k22 is j22, and so on. And that lets you write it in a tricky way with 2 minus the other one because when they're the same, you get the right answer. And therefore, your self-consistent field equation becomes this new equation rather than just the Hamiltonian like we had before with the effective potential. We've got this new player in the game, f, f hat, which is a function of r1 on some trial function psi of r1 is equal to epsilon j psi j of r1. And this operator, f, has our little Hamiltonian, h, which has the kinetic and potential energy of the electron, and then has this sum of these terms, which are the Coulomb part, 2j, 2 capital J hat sub j of r1, and minus k hat. And the reason we want to write these as operators because we want to operate on the wave function with them. We don't want to just leave things as integrals. Then we've kind of already operated and done the integral. So let's look at what these operators are. The Coulomb operator, when the Coulomb operator with subscript j operates on some wave function psi k, it returns psi k times the integral of psi j integrated out over the other coordinate. And the exchange operator is similar. When the exchange operator operates on a function, it returns the function and it integrates over its own index. So if I have k sub j on psi k, I return psi k times the integral of psi j with the repulsive term on, with the exchange term rather, on psi k. And that then lets us set up all of our equations. And in fact, it's nice to set them up like these because you can set them up as matrices. And then you can use some very, very powerful linear algebra techniques to solve for the energies and to minimize the energy. As before, however, and it just gets worse, in fact, the little energies epsilon j are not, if we sum them up, they're not the energy of the atom. So we have to be careful when we do this kind of thing that we keep track of the algebra appropriately. We can't just blindly sum the things up and expect to get the energy of the atom. We have to look at what it actually comes to and what the energy of the atom actually comes to and then make adjustments to get the thing right. It turns out, so I don't want to solve those, but I do want to point out that optimized orbitals are available online and you can go look. Here's an example I've taken here on slide 526 of a screenshot from this website at ccl.net and which talks about what they call Routhon-Hartree-Fock Ground-State Atomic Wave Functions. And what I've given here is what they have for helium, which is written. The total energy is minus 2.8, 616, et cetera. And I think you know now that that's in Hartree and they give some coefficients here. They give 1s and they give two columns. They give 1.4595 and then 1.3479. Then they give 3s and then they give 2s and then they give another 2s. The question is what does this mean? So these are the accepted Hartree-Fock orbitals. Question is then who was Routhon? And Clemens Routhon was at the University of Chicago in the 1950s and that was around the time that full-fledged but huge and slow computers were being developed. There were a number of just absolute visionaries like Alan Turing and John von Neumann and others who were so good with math that they worked out that you could make a machine that only used the integers and that such a machine could simulate any other kind of machine which is a very interesting idea. Of course now we recognize that as a computer but before you invent a computer it's not at all clear that such a thing could exist even. And here's a quote then from Popular Mechanics in March 1949 and this is a cautionary tale on how hard it is to predict the future. Here's what they said. While a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs 30 tons, computers in the future may have only 1,000 tubes and weigh only 1 1⁄2 tons. And just imagine what happened in the next 65 years. Routhon championed the use of Slater orbitals. Remember you have to maybe go back a couple of lectures but I said that we wanted orbitals that were easy to integrate, right? So we want tuna that tastes good and we're going to be doing a lot of orbitals and therefore we want to have a canned formula for the antiderivative. And it just turns out for the radial parts it's much easier to use the Slater orbitals even though they aren't orthogonal because they're just much easier to integrate. And so in this database then when they refer to 1S and so forth they are not referring to the hydrogen type orbitals. They're referring to Slater orbitals SNLM that I gave you which have different radial parts than the hydrogen atoms. And using these and computers they could systematically improve the energy of multi-electron atoms. It's very, very hard once you get things in the exponent to minimize if you just have a fixed, if you have a linear combination of things with fixed exponents that's okay. That is long as you aren't monkeying with the exponents. But once you put Zeta up there and you want to minimize and find the minimum of these, how much of this and how much of that plus these guys can shift around like shifting sand dunes as well. Then it gets much trickier and you need some numerical methods to actually find the minimum. So you may be able to calculate the integrals analytically but you still then need to find the minimum of the energy which is as you saw when we just had a couple of Zetas there, it was a mess. You can imagine what it gets like. And so you don't try to do that thing by writing out a giant formula and then staring at it. You do it numerically. It's much quicker. You just calculate it and then you minimize it by moving to a lower value. The orbitals then that you extract if you look at those columns are the Slater orbitals. And therefore the atomic orbital in helium then, psi of r, explicitly I've written it out here. It's 1.347900 times the Slater 1s type orbital which coincidentally is exactly the same as the hydrogen 1s with a value of Zeta in the exponent of 1.4595. And then there's a small correction minus, so you can see these numbers tell you about how much of this other stuff is in there. That's the major one, 1.3 minus.001613 times S3s, that has a different value of r, with a big value of Zeta of 5.32. And then about a tenth of an S2s with a 2.6 and then minus about.27 of a Zeta with a 1.75. And this then becomes the approximation for our helium atom and you can look up some of the other atoms on the website and it's very interesting reading. Wouldn't be interesting to most people of course because if you don't know what anything means, of course nothing's interesting. A book isn't interesting, nothing is. But if you know what these things mean and you've tried to calculate them yourself or you've puzzled about what these orbitals actually look like in these complicated atoms, then it is interesting because now you've got these functions and with Mathematica or some other program, you can plot them. You can look at them, you can square them, you can make contours of them and you can learn an awful lot. The energy that we get out of this function for helium with these four terms is pretty good. It's limited more by the orbital approximation, the fact that we are not allowing electron correlation in this approach than it is by the number and choice of the functions. So we can't necessarily improve the energy by just adding more and more slater functions because we've made a fatal flaw in the beginning in that we've averaged out all the other electrons. Well, that's because we're chemists and we like to think in terms of orbitals. I'll just note that the simple two symmetric sum that I initially suggested earlier which had one electron bound and the other one loose plus one electron, the other electron bound and one loose, that that did better, even for helium, than this much more complicated looking thing. But that's not a product of one electron orbitals and so that's a different kind of thing. And I would not necessarily be able to write down something so clever for a barium atom, for example, whereas this will be a systematic way to get something reasonable for that. Now, if every time we wanted to figure out what kind of configuration of an atom out of all the possible ones there might be is the most stable. If we had to actually go through and solve all these equations, it would be pretty slow going, that's for sure. But luckily we do not because there are some rules and like all rules there may be broken in a way. But most of the time they're quite good especially for ground state atoms. They're quite good and I don't even know of any ground state atom that doesn't follow these rules which are called Hund's rules. And they tell us of which of the possibilities where the electrons could be in which orbitals, which one's the lowest energy. That's very important to know what the ground state of the atom is going to be. If we're going to talk in orbitals, we want to know the configuration of the atom. For example, Hund's rules predict that if we have a nitrogen atom and a vacuum by itself that the 3P electrons are going to be distributed one for each of the 2P orbitals. One's going to be in PX, one's going to be in PY, and one's going to be in PZ. Or we could think of one is going to be in M sub L plus 1, 0 and minus 1. Same thing in the end. But they're going to be in these 3 orbitals. They are not going to be jammed together with 2 and 1 orbital and then one in the other or some other combination. And furthermore, Hund's rules is going to predict the following. That if these 3 electrons are in these 3P orbitals for nitrogen, that there are bar magnets are all going to be aligned. They're all going to be either spin up or spin down. If we don't have an external magnetic field, there's no difference. But there is a difference between having them all aligned and then having one different than the others. That could be different. And Hund's rules predict that the electron bar magnets will all align each other when there are unpaired electrons. Well, let's go through these rules. So here are Hund's rules, the 3 rules. And they're meant to be applied in order. And only in the case that the rule doesn't adjudicate what happens, do you go to the next rule. The first rule is the state with the largest value of big S, which is the total spin multiplicity, 2S plus 1. Remember we've got S as 1 half and we add them up. And we can add them up, couple them together to get big S, the overall spin. And the multiplicity is 2S plus 1. Those are the number of substates that you can have for a certain value of S, just like you could have 3 substates for L equals 1 and only 1 substrate for L equals 0. Then the state with the highest multiplicity or the biggest value of S is the most stable. And stability decreases with decreasing S. If I have 4 unpaired electrons in 4D orbitals, this will be all of them up, will be the most stable, and 2 up and 2 down will be worse than 3 up and 1 down, for example. If you have two possibilities with the same multiplicity, the same value of big S, then the state with the largest value of the orbital angular momentum, big L. So both S, the spin first, align the bar magnets, then L, get them going around together is going to be the most stable. And finally, if you have states that have the same value of L and of S, then if the sub-shell is less than half full, for example, if you have one or two electrons in the P-shell, then the state with the smallest value of J is the most stable. And if the sub-shell is more than half filled, then the state with the largest value of J is the most stable. At first, these rules seem to be completely arbitrary, but we can give some rationale for them that makes some physical sense. Of course, you don't invent the rules before you know the answer, where you get the rules from looking at how things turn out, right? So Hundt knew the answer before proposing the rules. The rules apply to excited state atoms too, but not quite as well, but they do, for example, allow us to predict the ordering of terms in atomic spectroscopy. If we have energy levels there, the best way to keep track of them is with the term. Why? Because it's got 2S plus 1. We want to know that to know where the energy is. It's got L, we want to know that to know what the energy is. And it's got J, and that's the final arbiter. And so if we've got the term symbol for the state, we know what we've got. And it, we can use Hundt's rules to predict that 2P3 halves, for example, is higher in energy than 2P1 half, at which we saw with sodium giving the 589 doublet. But what's the physical basis if there is any for these rules? And we have to be a little bit careful because you might be rationalizing something that seems reasonable, but unless you've done a calculation of some type to show that it is true, you have to be a bit careful. Well, we can sort of understand why the total spin S should be maximized by, let's look at the case of 2 spins in a, something like helium. Now, the 2 spins cannot be parallel if they're in the same orbital in helium, so it'll have to be an excited state. But let's put one electron in an excited orbital so that it can have the same spin state. And then let's compare this one to that one. And let's see if there's any argument that we could make. So let's look at a 1S2S state, for example. And what's the reason why parallel spins could be lower in this state? Well, you might argue, well, maybe it's a magnetic energy. So the bar magnets are lining up because of the, they're bar magnets. That might be an argument. That's not the argument that's usually given. The argument that's usually given is that if the spin part is symmetric, that means because of the polyprinciple that the space part is anti-symmetric because if these two, when I swap them, nothing changed sign, then that means the space part has to be anti-symmetric and vice versa. Therefore, we could have two combinations here. We could have psi plus and psi minus. And one of them is 1SR1, 2SR2 plus 2SR1, 1SR2. And the other one is minus. The question is, do these spatial parts have any difference? And the answer is, suppose the spatial part's anti-symmetric. So it's 1SR1, 2SR2 minus 2SR1, 1SR2. What happens to that function as R1 and R2 become the same? What does it mean when R1 and R2 become the same? It means the electrons are getting very close together. Well, sort of like magic, when R1 equals R2, psi vanishes. That means that this anti-symmetric spatial wave function automatically causes the electrons to avoid each other in some sense because the way, because of its very structure, it doesn't allow the electrons to be at the same point in space without vanishing. That means the probability of that is very low. And that means that the electron-electron repulsion that could occur if they got close like that is gone. If they're anti-symmetric, there's no such restriction. And they could, in principle, get quite close. And we would expect then to have a big electrostatic repulsion. That's the conventional argument. That may or may not be true. However, there have been some more detailed calculations which indicate that there may be other effects like penetration and shielding that might be players in the game here. But for now, we'll accept that at least this is kind of the stock argument is this anti-symmetric spatial part. And therefore, the electrons pair only if they have to. If they have to be in the same orbital, they pair. If they do not have to be in the same orbital, then they're either both up or both down. What about rule two? Well, this is even more wooly in my opinion and quite qualitative. And the idea is this. If the value of big L is high, that means the value of little L for the electrons is all adding up. And we can in some sense imagine then that the electrons are moving the same way around. And so if they're moving the same way around the atom, then they're avoiding each other. If they're moving opposite ways so that L is zero or a lower value of L, some are going one way, some are going the other way so that the total is not the maximum, then they can crash. They can crash into each other and get close. And that means that the energy could be higher on average because the crashes push up the potential energy. And therefore, we want a high value of L to avoid these crashes and push up the energy. Now, if you, I don't think you want to believe that too literally, but anyway, it's a qualitative argument that gives you a feeling for why the repulsive forces may be lowered. The justification for rule three is even more subtle. And that harks back to remember what we talked about with sodium. Why was there the splitting? Because we argued that we have an unpaired electron in the orbital up there. And it's the gut nonzero L if it's in a P state. That means that in some sense is going around something. Imagine the nucleus from the perspective of the electron going around the other way. And that looked like a current loop and that created a magnetic field that acted on the magnetic moment of the electron. And the electron could either be aligned with or against the magnetic field. And we called that the spin orbit coupling interaction that we introduced when we talked about atomic spectroscopy. If we look at the state with the lowest value of J, it's lowered by the spin orbit coupling. And therefore, if everything else is equal, then we look for this small magnetic effect, the spin orbit coupling. If the shell is less than half full, it's going to be lowered. If the shell is more than half full, remember when we were working out the terms, I said, if you're going to do fluorine, you don't put 5P electrons and then drive yourself mad, you put in one hole and you get the configurations for one hole. But since you're putting in a hole, that reverses the algebra on J. And so, when the shell is more than half filled, you work with holes and then the rolls are reversed and therefore you should go with the highest value of J, rather than the lowest value of J. And the way to do it always is look at the term symbol. A configuration of electrons gives rise to certain terms. The terms are what you want to know because the term symbol is going to tell you what ground state of the atom you've actually got. Let's then predict stability here on practice problem 27. Let's consider an excited state, beryllium atom, 1S2, 2S1, 3S1, you see how I go back to the orbitals right away. Don't even consider actually the real thing. And the ground state carbon atom, 1S2, 2S2, 2P2. And let's predict the most stable state for each of these atoms. Well, with 2S electrons, L must be 0. And that's called an S state, but that's not the S of the spin. That's the S of the S state of L is equal to 0. So it's big S. And there are two possibilities for that state. There's the triplet state, triplet S1, because if 2S plus 1 is 3, then J has to be 1. And then there's the singlet state, 1S0 or singlet S0. The term with the highest multiplicity is the triplet term, triplet S1, and so that one is the most stable. That's what we predict with beryllium. And that physically means that the electrons are aligned. With carbon, we have to figure out the terms that can arise with a P squared configuration. And the terms that can arise are 1, singlet S0, triplet P0, triplet P1, triplet P2, and singlet D2. Now we look at all these. Well, according to Hunn's rule number one, the triplet states are lower in energy. So it has to be one of the triplet P states. The value of L is the same in all of them, because they're all P. We don't have to worry about different kinds. We don't have a triplet D or a triplet S to worry about. And so therefore we have to go all the way to the third rule and the shells less than half filled because there's two P electrons and there can be six P electrons in the shell. And therefore the most stable configuration, the most stable term is triplet P0 because that has the lowest value of J. Now Hunn knew the answers before he proposed the rules. And the rules are generalizations. They work best to predict the ground state and not for, necessarily for excited states. We can build up then the periodic table by adding electrons to orbitals, which we very loosely associate with the solutions for the hydrogen atom. We sort of think of them as having similar shapes, but really what we ought to do is go to the website and look at the shapes at least for the self-consistent field orbitals because then we'd have a much better feel. And then we'd see that they change as the atom goes. So 1S becomes different as, well 1S doesn't become much different, but some of them do become different as you add more electrons and get more repulsion. The real situation, of course, for any of these multi-electron atoms is much more complex than this, but at least this is a good start. The periodic table itself can help us decide which orbital we're going to fill next because the problem is the ordering gets jumbled by all the electron-electron repulsion. We can't solve for the hydrogen atom where there's no electron-electron repulsion and then expect that all our solutions are going to be the same and the same relative even ordering of energies when we've got 18 electrons orbiting around and all these repulsive terms and all these effective charge distributions. What we have to do is go through and see what happens. And the periodic table is the best way to do that because the periodic table, by looking at the atomic number and where it comes out, you know exactly which orbital got filled because things that have the same outer electron configuration have very similar properties and that's why we put them into columns and that's why we recognize that there were patterns in the first place. The most dominant effect as you go up the periodic table is that s-electrons are lower energy than p or d. p is lower than d or f and so on. The s-electrons as we saw when we did those radial distribution functions can be found very near the nucleus. If you have a big atom like xenon or cesium or something like that then those electrons way out there, those 5s and 6s electrons can spend some time quite near the nucleus and then they see that it is not at all like a hydrogen atom where you have the charge and all the other electrons canceling it out so to speak and just one charge. It's much different than that. The s can sample that. The p has a node there so it cannot sample it very much but the d has a crisscross at least there and so the sampling is very small and f is even more nodes at the nucleus and therefore it hardly ever sees the nucleus. Why should it? It's an f electron and so it has a high angular momentum and it's out there orbiting around. It doesn't know much about the nucleus. It's far away like Pluto. For example then the ground state of the potassium atom is closed shell argon. You put the closed shell of an atom of a rare gas in square brackets to indicate well that's those 18 electrons taken care of and then 4s1 even though argon is 3p6 instead of going 3d1 you go 4s1. We know that because we know that potassium is under sodium on the periodic table which is under lithium and all those are s1 and they have very similar chemistry, very similar properties, soft, silvery, ultra reactive metals and that doesn't have anything to do with something like boron or which would be 3p1 but scandium has totally different chemistry which has 3d1. The order of the occupation of the orders of the orbitals then in these multi-electron systems as we go up the periodic table and keep in mind this is as we fill them. If we take a light atom and we look high up in its energy and say g is 5s above 4d or not for a lighter atom where those energy levels aren't occupied yet, the ordering may be different but this is the ordering if we're going to fill them as we go. So 1s, obvious, 2s, 2p, 3s, 3p, first jumble, 4s, 3d, then 4p, then 5s, 4d, 5p finally, 6s and so we get to 6s, we get to cesium before we get to 4f. They're that far apart and that really has to do with the penetration, shielding and the electron-electron or the propulsion in these systems that have a lot of atoms. They're just different and we have to look at them differently. In hydrogen where you don't have any of these issues, 4f has very, very, very similar energy to 4f. There's always small effects but basically they're degenerate and they're both well below 6s. There's no confusing that but by the time you get to cesium where you actually want to fill 6s, it's below 4f. Now there's a cautionary note. The order of the orbitals is not always predictable without a detailed calculation and sometimes even the idea that you know which orbitals are occupied is a little bit flawed. Remember this is quantum mechanics so if there's a choice, the atoms pick both choices at once and you have to always be prepared for that. If you have nearby energy levels with the same term symbol as one you're looking at then these can interact. These are like neighbors that both play loud music. They don't like each other and they tend to move apart and the same thing happens. You have to have the same kind of term symbol though. If you have a different kind of term symbol then it's like you're on a shift work and I'm playing loud music at night but you're at work and so you don't care and then I'm working in the day when you're doing it and so we don't interact at all. But there are some systems for example, this was worked out in the 30s even, that in magnesium, the excited state 3s3d, the triplet d is higher in energy than the singlet d and that violates Hund's rule number one. Now that's for an excited state atom and I said that it works best for ground state atoms but nevertheless, it's kind of interesting to say well why did that fall apart for magnesium, it's not like it's a really heavy atom or something where you could argue that there's some other effect maybe coming in. And the reason why it happens is that there's another state nearby that's 2 3p electrons rather than 3s3d, it's 2 3p's and that also gives rise to a singlet d state and it's up here in the singlet d states here, the triplet d was supposed to be lower but these guys hate each other. So they pull apart and this ends up being pushed lower. So as usual, there's some subtle effects that go on and sometimes it can be a little bit hard to predict every little detail. That's what makes the periodic table sort of interesting is that there are all these countervailing trends sort of like predicting which politician is going to vote on which thing, are they going to get elected again and so on and so forth, same thing here. There's repulsion, there's this, there's that and then there are these fine effects. Okay, this slide I've called fission, no, not atomic fission as in nuclear power but suppose we begin now and we do a mental experiment, we do a thought experiment. We spent a lot of time on a helium atom. Let's take a helium atom now and let's break the nucleus apart slowly and we can just imagine. So we've got our two electrons there and now we're going to break this alpha particle, this helium nucleus apart into two deuterons, one proton and one neutron each. The neutron we don't care about changes the reduced mass but the question is what would happen if we did that experiment? And what the resulting thing that has two positive charges and two electrons, we know helium stable but suppose we pull it apart, would it be energetically stable compared to two deuteron deuterium atoms which is just a heavy hydrogen atom? It's just like a hydrogen atom with an extra neutron in the nucleus which as far as we are concerned just changes the reduced mass and nothing else. In other words, if we imagine mentally pulling this helium nucleus apart, would it hold together even though the things aren't held together by the strong force? In other words, would the electrons hold it together or would there be a chemical bond between these two positively charged particles? Well clearly when the two nuclei are close but they're not in contact, once they're in contact they're held together by the strong force which we don't write a potential for and we don't ever treat excitation in the nucleus explicitly. But let's just say when they're close enough they're held like super glue so even though they're two positively charged particles they still hold together and they're very stable and they're not going to come apart. But once we mentally pull them apart so they don't touch, the strong force is like a big wrestler but he's got a limited reach. If he can grab a hold of you, look out, you're never going to get away but if he can't reach you then you're safe and once the particles get far enough apart there is no strong force but there certainly is electrostatic repulsion and they hate each other because they both have positive charges and so they push apart like crazy. And if we mentally pull them way apart then they have nothing to do with each other then they're two atoms. The wave functions of the electrons have died off, they don't even know the other ones there. So that will just have the same energy as two atoms, that's not going to be more stable. But so initially it's very bad if they're too close and then something happens and then it's zero if they're way apart. The question is could there be a sweet spot where the nuclei are far enough apart that they aren't repelling each other and destroying just moving apart but the electrons can overlap and share an orbital with each other and then create a stable configuration and if so what does that? Well, look, H2 is stable. Hydrogen is probably going to be the economy of the future for auto transportation one day. So we know for sure H2 does exist and D2, heavy hydrogen exists and even H2 plus exists. It's very important in astrophysics for example to look out and see that there is H2 plus way out there. So we know they exist. What we're going to find out in the next series of lectures is why they exist.
 Today's lecture in Chem 131A is going to be about bigger atoms,  not just two electrons.  Hun's rules, which are some rules to figure  out the most stable state of an atom when it can have different kinds  of electron configurations.  And the Aufbau, or building up principle,  which is the basis for putting atoms on the periodic table.  The periodic table, of course, is one  of the great intellectual triumphs of chemistry,  is to take all these things that we find in nature and first figure  out that they're all made up of these atoms, and then figure  out that the atoms follow patterns, and then to figure  out what the patterns mean, and then to figure  out everything else that has followed on.  And that's the kind of inexorable nature that science has to it.  Once you find out something, then that something else follows  from it, and a lot of other stuff is discarded as completely wrong,  even though it seemed to be plausible before you found  out whatever it is that you found out.  A closed shell atom, which we said could be written  as one slater determinant in the last lecture,  can, in which you've got n orbitals, and they're occupied  by 2n electrons, half of which are spin up,  and the other half are spin down,  otherwise it wouldn't satisfy the Pauli principle.  We can write down the Hamiltonian for a system like this.  What do we have?  Well, we have minus 1 half times all these del squared.  All that is is the kinetic energy.  We know how to calculate that.  There's a formula for it.  It's got r, theta, and phi in it, and unfortunately  for bigger atoms other than helium and hydride  and those guys that we've been doing, we're going to have  to get into theta and phi as well, and you can imagine  that the equations get more complicated quickly.  And we've got 2n of those because we've got 2n electrons  in n orbitals, and therefore they've all got kinetic energy  buzzing around.  We have to keep track of that energy.  So those terms are in the Hamiltonian.  Then we have the potential energy  of this big positive charge, z, which depends  on the atomic number, divided by the distance  of each electron from the positive charge.  When we write it like this, we're kind of implicitly assuming  that the nucleus is a point charge, and that's not quite true  because the nucleus has a finite volume.  And if you look very closely at spectral lines,  there can be slight shifts and various things that have  to do with the nuclear volume.  But for now, those are very, very, very tiny things compared  to, in fact, the accuracy with which we would be able  to calculate things with the orbital approximation anyway.  So we'll ignore those.  And then we have this giant sum of all the repulsive parts,  which I've written here as a double sum, from j equals 1  to 2n, and then on i equals j plus 1 to 2n,  just so we don't count things twice.  We could count them twice and divide by 2.  Sometimes if you're writing a computer program,  that might be easier as long as you aren't actually calculating  some terrible integral twice for no reason,  in which case, in that case, you're wasting a lot of time  to calculate it this way and that way, which is the same,  in fact, that the computer is not going  to correct your bad algorithm and then divide it by 2.  That's pretty easy to understand.  If I've got five electrons, I've got the repulsion  between 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3.  I don't want to count 2 and 1 again.  I don't want to count 2 and 2 because that doesn't count.  2 and 4, 2 and 5, and so on.  And that's all that double sum means, is count them all up,  put them in there, and now we've got to figure  out how we're going to solve this thing  with the kinetic energy as the first sum.  The potential energy is the second sum and then the sum  of all the mutual repulsion terms between the electrons  as the third term, the double sum.  We can write this a little bit neater if we introduce,  instead of the capital H for the Hamiltonian,  we introduce small H and we index it with the electron J  and we say, look, that's just 1 minus 1 half,  del squared J, take the derivative with respect  to the coordinates of electron J, XJ, YJ, ZJ, or RJ theta J,  phi J, and then minus E over RJ, which is the potential.  And if we do that, then we configure the energy of the  atom as a sum of three distinct terms.  Now, I've written out this giant integral,  which has a lot of integrals over R1, R2, depending how,  out to 2n, and then I've also put  in this funny notation, d sigma 1, d sigma 2,  out to d sigma 2n.  What does that mean?  Well, the d sigma part has to do with the spin.  We need to take into account the spin as well,  because all these 2n electrons have spin, and we have  to integrate over the spin part, make sure  that the spin functions are normalized, and keep track  of that as well as the spatial part.  And then we've got the wave function, psi star,  which depends on all the coordinates of all the electrons  and principle, the Hamiltonian, and then psi.  And that breaks up into 2 times a sum of what I'm going  to call i sub j plus a double sum from i equals 1 to n,  j equals 1 to n of 2 times ij, j, capital J, ij,  minus capital K, ij.  Let's see what that, what this means.  So let's look at these three terms.  Well, the i part is just the energy  of each single electron.  Ignore the others.  The j part is the, what we get when we take the integral  of psi i r1, psi j r2, then the repulsive term,  psi i r1, psi j r2, that we've seen before.  And then K is an integral that's a new one.  It's the integral of psi i star r1, psi j star r2,  psi i r2, not r1, psi j r1 so that the indices are swapped  in the second part.  So the i is the single electron kinetic  and potential energy, jij is the Coulomb integral  that we've encountered before.  And Kij is called the exchange integral.  Kij is a new player in the game.  It has no analog and classical mechanics.  And here's why I've written it out here on the bottom  of slide 523.  It has, if we wanted to interpret things as charged clouds  interacting, we can't do it because we've got psi i star  of r1, which is the coordinate of the first electron.  And that's going on to psi i of r2,  which is the second electron.  And likewise, we've got psi star j of r2,  then the repult one over r12, and then psi j of r1.  So I think you can see why it's called the exchange integral  because you swap the indices on the other side of the function.  With larger atoms, the self-consistent field method  still works, but there's this exchange contribution  which Vladimir Falk saw, and that's why it's called the  Hartree-Falk method.  And keep in mind, there's kind of a shorthand here  by using 2ji, and that is that when k or j,  if both the indices are the same, then it doesn't matter  that they were swapped.  So k11 is j11, and k22 is j22, and so on.  And that lets you write it in a tricky way with 2 minus the other  one because when they're the same, you get the right answer.  And therefore, your self-consistent field equation becomes  this new equation rather than just the Hamiltonian  like we had before with the effective potential.  We've got this new player in the game, f, f hat,  which is a function of r1 on some trial function psi of r1 is  equal to epsilon j psi j of r1.  And this operator, f, has our little Hamiltonian, h,  which has the kinetic and potential energy of the electron,  and then has this sum of these terms,  which are the Coulomb part, 2j, 2 capital J hat sub j of r1,  and minus k hat.  And the reason we want to write these as operators  because we want to operate on the wave function with them.  We don't want to just leave things as integrals.  Then we've kind of already operated and done the integral.  So let's look at what these operators are.  The Coulomb operator, when the Coulomb operator  with subscript j operates on some wave function psi k,  it returns psi k times the integral of psi j integrated  out over the other coordinate.  And the exchange operator is similar.  When the exchange operator operates on a function,  it returns the function and it integrates over its own index.  So if I have k sub j on psi k, I return psi k times the integral  of psi j with the repulsive term on,  with the exchange term rather, on psi k.  And that then lets us set up all of our equations.  And in fact, it's nice to set them up like these  because you can set them up as matrices.  And then you can use some very, very powerful linear algebra  techniques to solve for the energies  and to minimize the energy.  As before, however, and it just gets worse, in fact,  the little energies epsilon j are not, if we sum them up,  they're not the energy of the atom.  So we have to be careful when we do this kind of thing  that we keep track of the algebra appropriately.  We can't just blindly sum the things up and expect  to get the energy of the atom.  We have to look at what it actually comes to and what the  energy of the atom actually comes to  and then make adjustments to get the thing right.  It turns out, so I don't want to solve those,  but I do want to point out that optimized orbitals are  available online and you can go look.  Here's an example I've taken here on slide 526 of a screenshot  from this website at ccl.net and which talks  about what they call Routhon-Hartree-Fock  Ground-State Atomic Wave Functions.  And what I've given here is what they have  for helium, which is written.  The total energy is minus 2.8, 616, et cetera.  And I think you know now that that's in Hartree  and they give some coefficients here.  They give 1s and they give two columns.  They give 1.4595 and then 1.3479.  Then they give 3s and then they give 2s  and then they give another 2s.  The question is what does this mean?  So these are the accepted Hartree-Fock orbitals.  Question is then who was Routhon?  And Clemens Routhon was at the University of Chicago  in the 1950s and that was around the time that full-fledged  but huge and slow computers were being developed.  There were a number of just absolute visionaries  like Alan Turing and John von Neumann and others  who were so good with math that they worked out  that you could make a machine that only used the integers  and that such a machine could simulate any other kind  of machine which is a very interesting idea.  Of course now we recognize that as a computer  but before you invent a computer it's not at all clear  that such a thing could exist even.  And here's a quote then from Popular Mechanics in March 1949  and this is a cautionary tale on how hard it is  to predict the future.  Here's what they said.  While a calculator on the ENIAC is equipped  with 18,000 vacuum tubes and weighs 30 tons,  computers in the future may have only 1,000 tubes  and weigh only 1 1⁄2 tons.  And just imagine what happened in the next 65 years.  Routhon championed the use of Slater orbitals.  Remember you have to maybe go back a couple of lectures  but I said that we wanted orbitals  that were easy to integrate, right?  So we want tuna that tastes good and we're going  to be doing a lot of orbitals and therefore we want  to have a canned formula for the antiderivative.  And it just turns out for the radial parts it's much easier  to use the Slater orbitals even though they aren't  orthogonal because they're just much easier to integrate.  And so in this database then when they refer to 1S  and so forth they are not referring  to the hydrogen type orbitals.  They're referring to Slater orbitals SNLM that I gave you  which have different radial parts than the hydrogen atoms.  And using these and computers they could systematically improve  the energy of multi-electron atoms.  It's very, very hard once you get things in the exponent  to minimize if you just have a fixed,  if you have a linear combination of things  with fixed exponents that's okay.  That is long as you aren't monkeying with the exponents.  But once you put Zeta up there and you want to minimize  and find the minimum of these, how much of this  and how much of that plus these guys can shift  around like shifting sand dunes as well.  Then it gets much trickier  and you need some numerical methods  to actually find the minimum.  So you may be able to calculate the integrals analytically  but you still then need to find the minimum of the energy  which is as you saw when we just had a couple of Zetas there,  it was a mess.  You can imagine what it gets like.  And so you don't try to do that thing by writing  out a giant formula and then staring at it.  You do it numerically.  It's much quicker.  You just calculate it and then you minimize it  by moving to a lower value.  The orbitals then that you extract if you look  at those columns are the Slater orbitals.  And therefore the atomic orbital in helium then,  psi of r, explicitly I've written it out here.  It's 1.347900 times the Slater 1s type orbital  which coincidentally is exactly the same  as the hydrogen 1s with a value of Zeta in the exponent  of 1.4595.  And then there's a small correction minus,  so you can see these numbers tell you about how much  of this other stuff is in there.  That's the major one, 1.3 minus.001613 times S3s,  that has a different value of r, with a big value  of Zeta of 5.32.  And then about a tenth of an S2s with a 2.6 and then minus  about.27 of a Zeta with a 1.75.  And this then becomes the approximation for our helium  atom and you can look up some of the other atoms  on the website and it's very interesting reading.  Wouldn't be interesting to most people of course  because if you don't know what anything means,  of course nothing's interesting.  A book isn't interesting, nothing is.  But if you know what these things mean and you've tried  to calculate them yourself or you've puzzled  about what these orbitals actually look  like in these complicated atoms, then it is interesting  because now you've got these functions  and with Mathematica or some other program, you can plot them.  You can look at them, you can square them,  you can make contours of them and you can learn an awful lot.  The energy that we get out of this function for helium  with these four terms is pretty good.  It's limited more by the orbital approximation,  the fact that we are not allowing electron correlation  in this approach than it is by the number and choice  of the functions.  So we can't necessarily improve the energy by just adding more  and more slater functions because we've made a fatal flaw  in the beginning in that we've averaged  out all the other electrons.  Well, that's because we're chemists and we like to think  in terms of orbitals.  I'll just note that the simple two symmetric sum  that I initially suggested earlier which had one electron bound  and the other one loose plus one electron,  the other electron bound and one loose, that that did better,  even for helium, than this much more complicated looking thing.  But that's not a product of one electron orbitals  and so that's a different kind of thing.  And I would not necessarily be able to write  down something so clever for a barium atom, for example,  whereas this will be a systematic way  to get something reasonable for that.  Now, if every time we wanted to figure  out what kind of configuration of an atom out  of all the possible ones there might be is the most stable.  If we had to actually go through and solve all these equations,  it would be pretty slow going, that's for sure.  But luckily we do not because there are some rules  and like all rules there may be broken in a way.  But most of the time they're quite good especially  for ground state atoms.  They're quite good and I don't even know  of any ground state atom that doesn't follow these rules  which are called Hund's rules.  And they tell us of which of the possibilities  where the electrons could be in which orbitals,  which one's the lowest energy.  That's very important to know what the ground state  of the atom is going to be.  If we're going to talk in orbitals, we want  to know the configuration of the atom.  For example, Hund's rules predict that if we have a  nitrogen atom and a vacuum by itself  that the 3P electrons are going to be distributed one  for each of the 2P orbitals.  One's going to be in PX, one's going to be in PY,  and one's going to be in PZ. Or we could think  of one is going to be in M sub L plus 1, 0 and minus 1.  Same thing in the end.  But they're going to be in these 3 orbitals.  They are not going to be jammed together with 2  and 1 orbital and then one in the other  or some other combination.  And furthermore, Hund's rules is going  to predict the following.  That if these 3 electrons are in these 3P orbitals  for nitrogen, that there are bar magnets are all going  to be aligned.  They're all going to be either spin up or spin down.  If we don't have an external magnetic field,  there's no difference.  But there is a difference between having them all aligned  and then having one different than the others.  That could be different.  And Hund's rules predict that the electron bar magnets will all  align each other when there are unpaired electrons.  Well, let's go through these rules.  So here are Hund's rules, the 3 rules.  And they're meant to be applied in order.  And only in the case that the rule doesn't adjudicate what  happens, do you go to the next rule.  The first rule is the state with the largest value of big S,  which is the total spin multiplicity, 2S plus 1.  Remember we've got S as 1 half and we add them up.  And we can add them up, couple them together to get big S,  the overall spin.  And the multiplicity is 2S plus 1.  Those are the number of substates that you can have  for a certain value of S, just like you could have 3 substates  for L equals 1 and only 1 substrate for L equals 0.  Then the state with the highest multiplicity  or the biggest value of S is the most stable.  And stability decreases with decreasing S. If I have 4 unpaired  electrons in 4D orbitals, this will be all of them up,  will be the most stable, and 2 up and 2 down will be worse  than 3 up and 1 down, for example.  If you have two possibilities with the same multiplicity,  the same value of big S, then the state with the largest value  of the orbital angular momentum, big L. So both S, the spin first,  align the bar magnets, then L, get them going  around together is going to be the most stable.  And finally, if you have states that have the same value of L  and of S, then if the sub-shell is less than half full,  for example, if you have one or two electrons in the P-shell,  then the state with the smallest value of J is the most stable.  And if the sub-shell is more than half filled,  then the state with the largest value of J is the most stable.  At first, these rules seem to be completely arbitrary,  but we can give some rationale for them  that makes some physical sense.  Of course, you don't invent the rules before you know the answer,  where you get the rules from looking at how things turn out, right?  So Hundt knew the answer before proposing the rules.  The rules apply to excited state atoms too, but not quite as well,  but they do, for example, allow us to predict the ordering  of terms in atomic spectroscopy.  If we have energy levels there, the best way  to keep track of them is with the term.  Why? Because it's got 2S plus 1.  We want to know that to know where the energy is.  It's got L, we want to know that to know what the energy is.  And it's got J, and that's the final arbiter.  And so if we've got the term symbol for the state,  we know what we've got.  And it, we can use Hundt's rules to predict that 2P3 halves,  for example, is higher in energy than 2P1 half,  at which we saw with sodium giving the 589 doublet.  But what's the physical basis if there is any for these rules?  And we have to be a little bit careful  because you might be rationalizing something  that seems reasonable, but unless you've done a calculation  of some type to show that it is true,  you have to be a bit careful.  Well, we can sort of understand why the total spin S should be  maximized by, let's look at the case of 2 spins in a, something  like helium.  Now, the 2 spins cannot be parallel if they're  in the same orbital in helium, so it'll have  to be an excited state.  But let's put one electron in an excited orbital  so that it can have the same spin state.  And then let's compare this one to that one.  And let's see if there's any argument that we could make.  So let's look at a 1S2S state, for example.  And what's the reason why parallel spins could be lower  in this state?  Well, you might argue, well, maybe it's a magnetic energy.  So the bar magnets are lining up because of the,  they're bar magnets.  That might be an argument.  That's not the argument that's usually given.  The argument that's usually given is  that if the spin part is symmetric, that means  because of the polyprinciple that the space part is  anti-symmetric because if these two, when I swap them,  nothing changed sign, then that means the space part has  to be anti-symmetric and vice versa.  Therefore, we could have two combinations here.  We could have psi plus and psi minus.  And one of them is 1SR1, 2SR2 plus 2SR1, 1SR2.  And the other one is minus.  The question is, do these spatial parts have any difference?  And the answer is, suppose the spatial part's anti-symmetric.  So it's 1SR1, 2SR2 minus 2SR1, 1SR2.  What happens to that function as R1 and R2 become the same?  What does it mean when R1 and R2 become the same?  It means the electrons are getting very close together.  Well, sort of like magic, when R1 equals R2, psi vanishes.  That means that this anti-symmetric spatial wave function  automatically causes the electrons to avoid each other  in some sense because the way, because of its very structure,  it doesn't allow the electrons to be at the same point  in space without vanishing.  That means the probability of that is very low.  And that means that the electron-electron repulsion  that could occur if they got close like that is gone.  If they're anti-symmetric, there's no such restriction.  And they could, in principle, get quite close.  And we would expect then to have a big electrostatic repulsion.  That's the conventional argument.  That may or may not be true.  However, there have been some more detailed calculations  which indicate that there may be other effects like penetration  and shielding that might be players in the game here.  But for now, we'll accept that at least this is kind of the stock  argument is this anti-symmetric spatial part.  And therefore, the electrons pair only if they have to.  If they have to be in the same orbital, they pair.  If they do not have to be in the same orbital,  then they're either both up or both down.  What about rule two?  Well, this is even more wooly in my opinion and quite qualitative.  And the idea is this.  If the value of big L is high, that means the value  of little L for the electrons is all adding up.  And we can in some sense imagine then  that the electrons are moving the same way around.  And so if they're moving the same way around the atom,  then they're avoiding each other.  If they're moving opposite ways so that L is zero  or a lower value of L, some are going one way,  some are going the other way so that the total is not the maximum,  then they can crash.  They can crash into each other and get close.  And that means that the energy could be higher on average  because the crashes push up the potential energy.  And therefore, we want a high value of L to avoid these crashes  and push up the energy.  Now, if you, I don't think you want to believe that too literally,  but anyway, it's a qualitative argument that gives you a feeling  for why the repulsive forces may be lowered.  The justification for rule three is even more subtle.  And that harks back to remember what we talked about with sodium.  Why was there the splitting?  Because we argued that we have an unpaired electron  in the orbital up there.  And it's the gut nonzero L if it's in a P state.  That means that in some sense is going around something.  Imagine the nucleus from the perspective  of the electron going around the other way.  And that looked like a current loop  and that created a magnetic field that acted  on the magnetic moment of the electron.  And the electron could either be aligned with  or against the magnetic field.  And we called that the spin orbit coupling interaction  that we introduced when we talked about atomic spectroscopy.  If we look at the state with the lowest value of J,  it's lowered by the spin orbit coupling.  And therefore, if everything else is equal, then we look  for this small magnetic effect, the spin orbit coupling.  If the shell is less than half full, it's going to be lowered.  If the shell is more than half full,  remember when we were working out the terms, I said,  if you're going to do fluorine, you don't put 5P electrons  and then drive yourself mad, you put in one hole  and you get the configurations for one hole.  But since you're putting in a hole,  that reverses the algebra on J. And so,  when the shell is more than half filled, you work with holes  and then the rolls are reversed  and therefore you should go with the highest value of J,  rather than the lowest value of J. And the way to do it always is  look at the term symbol.  A configuration of electrons gives rise to certain terms.  The terms are what you want to know  because the term symbol is going to tell you what ground state  of the atom you've actually got.  Let's then predict stability here on practice problem 27.  Let's consider an excited state, beryllium atom, 1S2, 2S1,  3S1, you see how I go back to the orbitals right away.  Don't even consider actually the real thing.  And the ground state carbon atom, 1S2, 2S2, 2P2.  And let's predict the most stable state for each of these atoms.  Well, with 2S electrons, L must be 0.  And that's called an S state, but that's not the S of the spin.  That's the S of the S state of L is equal to 0.  So it's big S. And there are two possibilities for that state.  There's the triplet state, triplet S1,  because if 2S plus 1 is 3, then J has to be 1.  And then there's the singlet state, 1S0 or singlet S0.  The term with the highest multiplicity is the triplet term,  triplet S1, and so that one is the most stable.  That's what we predict with beryllium.  And that physically means that the electrons are aligned.  With carbon, we have to figure out the terms that can arise  with a P squared configuration.  And the terms that can arise are 1, singlet S0, triplet P0,  triplet P1, triplet P2, and singlet D2.  Now we look at all these.  Well, according to Hunn's rule number one,  the triplet states are lower in energy.  So it has to be one of the triplet P states.  The value of L is the same in all of them,  because they're all P. We don't have to worry  about different kinds.  We don't have a triplet D or a triplet S to worry about.  And so therefore we have to go all the way to the third rule  and the shells less than half filled  because there's two P electrons and there can be six P electrons  in the shell.  And therefore the most stable configuration,  the most stable term is triplet P0  because that has the lowest value of J. Now Hunn knew the answers  before he proposed the rules.  And the rules are generalizations.  They work best to predict the ground state and not for,  necessarily for excited states.  We can build up then the periodic table  by adding electrons to orbitals, which we very loosely associate  with the solutions for the hydrogen atom.  We sort of think of them as having similar shapes,  but really what we ought to do is go to the website  and look at the shapes at least  for the self-consistent field orbitals  because then we'd have a much better feel.  And then we'd see that they change as the atom goes.  So 1S becomes different as, well 1S doesn't become much different,  but some of them do become different as you add more electrons  and get more repulsion.  The real situation, of course,  for any of these multi-electron atoms is much more complex  than this, but at least this is a good start.  The periodic table itself can help us decide which orbital we're going  to fill next because the problem is the ordering gets jumbled  by all the electron-electron repulsion.  We can't solve for the hydrogen atom  where there's no electron-electron repulsion and then expect  that all our solutions are going to be the same  and the same relative even ordering of energies  when we've got 18 electrons orbiting around  and all these repulsive terms  and all these effective charge distributions.  What we have to do is go through and see what happens.  And the periodic table is the best way to do that  because the periodic table, by looking at the atomic number  and where it comes out, you know exactly which orbital got filled  because things that have the same outer electron configuration  have very similar properties and that's why we put them  into columns and that's why we recognize  that there were patterns in the first place.  The most dominant effect as you go up the periodic table is  that s-electrons are lower energy than p or d. p is lower  than d or f and so on.  The s-electrons as we saw when we did those radial distribution  functions can be found very near the nucleus.  If you have a big atom like xenon or cesium  or something like that then those electrons way out there,  those 5s and 6s electrons can spend some time quite near  the nucleus and then they see that it is not at all  like a hydrogen atom where you have the charge  and all the other electrons canceling it out so to speak  and just one charge.  It's much different than that.  The s can sample that.  The p has a node there so it cannot sample it very much  but the d has a crisscross at least there  and so the sampling is very small  and f is even more nodes at the nucleus  and therefore it hardly ever sees the nucleus.  Why should it?  It's an f electron and so it has a high angular momentum  and it's out there orbiting around.  It doesn't know much about the nucleus.  It's far away like Pluto.  For example then the ground state  of the potassium atom is closed shell argon.  You put the closed shell of an atom of a rare gas  in square brackets to indicate well  that's those 18 electrons taken care of and then 4s1  even though argon is 3p6 instead of going 3d1 you go 4s1.  We know that because we know that potassium is under sodium  on the periodic table which is under lithium  and all those are s1 and they have very similar chemistry,  very similar properties, soft, silvery,  ultra reactive metals and that doesn't have anything to do  with something like boron or which would be 3p1  but scandium has totally different chemistry which has 3d1.  The order of the occupation of the orders  of the orbitals then in these multi-electron systems  as we go up the periodic table and keep in mind this is  as we fill them.  If we take a light atom and we look high up in its energy  and say g is 5s above 4d or not for a lighter atom  where those energy levels aren't occupied yet,  the ordering may be different but this is the ordering  if we're going to fill them as we go.  So 1s, obvious, 2s, 2p, 3s, 3p, first jumble, 4s, 3d, then 4p,  then 5s, 4d, 5p finally, 6s and so we get to 6s,  we get to cesium before we get to 4f.  They're that far apart and that really has to do  with the penetration, shielding and the electron-electron  or the propulsion in these systems that have a lot of atoms.  They're just different and we have to look at them differently.  In hydrogen where you don't have any of these issues,  4f has very, very, very similar energy to 4f.  There's always small effects but basically they're degenerate  and they're both well below 6s.  There's no confusing that but by the time you get to cesium  where you actually want to fill 6s, it's below 4f.  Now there's a cautionary note.  The order of the orbitals is not always predictable  without a detailed calculation  and sometimes even the idea that you know which orbitals are  occupied is a little bit flawed.  Remember this is quantum mechanics so if there's a choice,  the atoms pick both choices at once  and you have to always be prepared for that.  If you have nearby energy levels with the same term symbol  as one you're looking at then these can interact.  These are like neighbors that both play loud music.  They don't like each other and they tend to move apart  and the same thing happens.  You have to have the same kind of term symbol though.  If you have a different kind of term symbol then it's  like you're on a shift work and I'm playing loud music at night  but you're at work and so you don't care  and then I'm working in the day when you're doing it  and so we don't interact at all.  But there are some systems for example,  this was worked out in the 30s even, that in magnesium,  the excited state 3s3d, the triplet d is higher in energy  than the singlet d and that violates Hund's rule number one.  Now that's for an excited state atom and I said  that it works best for ground state atoms but nevertheless,  it's kind of interesting to say well why did that fall apart  for magnesium, it's not like it's a really heavy atom  or something where you could argue  that there's some other effect maybe coming in.  And the reason why it happens is  that there's another state nearby that's 2 3p electrons rather  than 3s3d, it's 2 3p's and that also gives rise  to a singlet d state and it's up here  in the singlet d states here, the triplet d was supposed  to be lower but these guys hate each other.  So they pull apart and this ends up being pushed lower.  So as usual, there's some subtle effects that go on  and sometimes it can be a little bit hard  to predict every little detail.  That's what makes the periodic table sort of interesting is  that there are all these countervailing trends sort  of like predicting which politician is going  to vote on which thing, are they going to get elected again  and so on and so forth, same thing here.  There's repulsion, there's this, there's that  and then there are these fine effects.  Okay, this slide I've called fission, no,  not atomic fission as in nuclear power but suppose we begin now  and we do a mental experiment, we do a thought experiment.  We spent a lot of time on a helium atom.  Let's take a helium atom now and let's break the nucleus  apart slowly and we can just imagine.  So we've got our two electrons there  and now we're going to break this alpha particle,  this helium nucleus apart into two deuterons,  one proton and one neutron each.  The neutron we don't care about changes the reduced mass  but the question is what would happen  if we did that experiment?  And what the resulting thing that has two positive charges  and two electrons, we know helium stable  but suppose we pull it apart, would it be energetically stable  compared to two deuteron deuterium atoms  which is just a heavy hydrogen atom?  It's just like a hydrogen atom with an extra neutron  in the nucleus which as far as we are concerned just changes  the reduced mass and nothing else.  In other words, if we imagine mentally pulling this helium  nucleus apart, would it hold together even though the things  aren't held together by the strong force?  In other words, would the electrons hold it together  or would there be a chemical bond  between these two positively charged particles?  Well clearly when the two nuclei are close  but they're not in contact, once they're in contact they're held  together by the strong force which we don't write a potential  for and we don't ever treat excitation  in the nucleus explicitly.  But let's just say when they're close enough they're held  like super glue so even though they're two positively charged  particles they still hold together and they're very stable  and they're not going to come apart.  But once we mentally pull them apart so they don't touch,  the strong force is like a big wrestler  but he's got a limited reach.  If he can grab a hold of you, look out,  you're never going to get away but if he can't reach you  then you're safe and once the particles get far enough apart  there is no strong force but there certainly is electrostatic  repulsion and they hate each other  because they both have positive charges  and so they push apart like crazy.  And if we mentally pull them way apart then they have nothing  to do with each other then they're two atoms.  The wave functions of the electrons have died off,  they don't even know the other ones there.  So that will just have the same energy as two atoms,  that's not going to be more stable.  But so initially it's very bad if they're too close  and then something happens and then it's zero  if they're way apart.  The question is could there be a sweet spot  where the nuclei are far enough apart  that they aren't repelling each other  and destroying just moving apart  but the electrons can overlap and share an orbital  with each other and then create a stable configuration  and if so what does that?  Well, look, H2 is stable.  Hydrogen is probably going to be the economy of the future  for auto transportation one day.  So we know for sure H2 does exist and D2,  heavy hydrogen exists and even H2 plus exists.  It's very important in astrophysics for example to look  out and see that there is H2 plus way out there.  So we know they exist.  What we're going to find out in the next series  of lectures is why they exist.
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Today's lecture in Chem 131A is going to be about bigger atoms, not just two electrons. Hun's rules, which are some rules to figure out the most stable state of an atom when it can have different kinds of electron configurations. And the Aufbau, or building up principle, which is the basis for putting atoms on the periodic table. The periodic table, of course, is one of the great intellectual triumphs of chemistry, is to take all these things that we find in nature and first figure out that they're all made up of these atoms, and then figure out that the atoms follow patterns, and then to figure out what the patterns mean, and then to figure out everything else that has followed on. And that's the kind of inexorable nature that science has to it. Once you find out something, then that something else follows from it, and a lot of other stuff is discarded as completely wrong, even though it seemed to be plausible before you found out whatever it is that you found out. A closed shell atom, which we said could be written as one slater determinant in the last lecture, can, in which you've got n orbitals, and they're occupied by 2n electrons, half of which are spin up, and the other half are spin down, otherwise it wouldn't satisfy the Pauli principle. We can write down the Hamiltonian for a system like this. What do we have? Well, we have minus 1 half times all these del squared. All that is is the kinetic energy. We know how to calculate that. There's a formula for it. It's got r, theta, and phi in it, and unfortunately for bigger atoms other than helium and hydride and those guys that we've been doing, we're going to have to get into theta and phi as well, and you can imagine that the equations get more complicated quickly. And we've got 2n of those because we've got 2n electrons in n orbitals, and therefore they've all got kinetic energy buzzing around. We have to keep track of that energy. So those terms are in the Hamiltonian. Then we have the potential energy of this big positive charge, z, which depends on the atomic number, divided by the distance of each electron from the positive charge. When we write it like this, we're kind of implicitly assuming that the nucleus is a point charge, and that's not quite true because the nucleus has a finite volume. And if you look very closely at spectral lines, there can be slight shifts and various things that have to do with the nuclear volume. But for now, those are very, very, very tiny things compared to, in fact, the accuracy with which we would be able to calculate things with the orbital approximation anyway. So we'll ignore those. And then we have this giant sum of all the repulsive parts, which I've written here as a double sum, from j equals 1 to 2n, and then on i equals j plus 1 to 2n, just so we don't count things twice. We could count them twice and divide by 2. Sometimes if you're writing a computer program, that might be easier as long as you aren't actually calculating some terrible integral twice for no reason, in which case, in that case, you're wasting a lot of time to calculate it this way and that way, which is the same, in fact, that the computer is not going to correct your bad algorithm and then divide it by 2. That's pretty easy to understand. If I've got five electrons, I've got the repulsion between 1 and 2, 1 and 3, 1 and 4, 1 and 5, 2 and 3. I don't want to count 2 and 1 again. I don't want to count 2 and 2 because that doesn't count. 2 and 4, 2 and 5, and so on. And that's all that double sum means, is count them all up, put them in there, and now we've got to figure out how we're going to solve this thing with the kinetic energy as the first sum. The potential energy is the second sum and then the sum of all the mutual repulsion terms between the electrons as the third term, the double sum. We can write this a little bit neater if we introduce, instead of the capital H for the Hamiltonian, we introduce small H and we index it with the electron J and we say, look, that's just 1 minus 1 half, del squared J, take the derivative with respect to the coordinates of electron J, XJ, YJ, ZJ, or RJ theta J, phi J, and then minus E over RJ, which is the potential. And if we do that, then we configure the energy of the atom as a sum of three distinct terms. Now, I've written out this giant integral, which has a lot of integrals over R1, R2, depending how, out to 2n, and then I've also put in this funny notation, d sigma 1, d sigma 2, out to d sigma 2n. What does that mean? Well, the d sigma part has to do with the spin. We need to take into account the spin as well, because all these 2n electrons have spin, and we have to integrate over the spin part, make sure that the spin functions are normalized, and keep track of that as well as the spatial part. And then we've got the wave function, psi star, which depends on all the coordinates of all the electrons and principle, the Hamiltonian, and then psi. And that breaks up into 2 times a sum of what I'm going to call i sub j plus a double sum from i equals 1 to n, j equals 1 to n of 2 times ij, j, capital J, ij, minus capital K, ij. Let's see what that, what this means. So let's look at these three terms. Well, the i part is just the energy of each single electron. Ignore the others. The j part is the, what we get when we take the integral of psi i r1, psi j r2, then the repulsive term, psi i r1, psi j r2, that we've seen before. And then K is an integral that's a new one. It's the integral of psi i star r1, psi j star r2, psi i r2, not r1, psi j r1 so that the indices are swapped in the second part. So the i is the single electron kinetic and potential energy, jij is the Coulomb integral that we've encountered before. And Kij is called the exchange integral. Kij is a new player in the game. It has no analog and classical mechanics. And here's why I've written it out here on the bottom of slide 523. It has, if we wanted to interpret things as charged clouds interacting, we can't do it because we've got psi i star of r1, which is the coordinate of the first electron. And that's going on to psi i of r2, which is the second electron. And likewise, we've got psi star j of r2, then the repult one over r12, and then psi j of r1. So I think you can see why it's called the exchange integral because you swap the indices on the other side of the function. With larger atoms, the self-consistent field method still works, but there's this exchange contribution which Vladimir Falk saw, and that's why it's called the Hartree-Falk method. And keep in mind, there's kind of a shorthand here by using 2ji, and that is that when k or j, if both the indices are the same, then it doesn't matter that they were swapped. So k11 is j11, and k22 is j22, and so on. And that lets you write it in a tricky way with 2 minus the other one because when they're the same, you get the right answer. And therefore, your self-consistent field equation becomes this new equation rather than just the Hamiltonian like we had before with the effective potential. We've got this new player in the game, f, f hat, which is a function of r1 on some trial function psi of r1 is equal to epsilon j psi j of r1. And this operator, f, has our little Hamiltonian, h, which has the kinetic and potential energy of the electron, and then has this sum of these terms, which are the Coulomb part, 2j, 2 capital J hat sub j of r1, and minus k hat. And the reason we want to write these as operators because we want to operate on the wave function with them. We don't want to just leave things as integrals. Then we've kind of already operated and done the integral. So let's look at what these operators are. The Coulomb operator, when the Coulomb operator with subscript j operates on some wave function psi k, it returns psi k times the integral of psi j integrated out over the other coordinate. And the exchange operator is similar. When the exchange operator operates on a function, it returns the function and it integrates over its own index. So if I have k sub j on psi k, I return psi k times the integral of psi j with the repulsive term on, with the exchange term rather, on psi k. And that then lets us set up all of our equations. And in fact, it's nice to set them up like these because you can set them up as matrices. And then you can use some very, very powerful linear algebra techniques to solve for the energies and to minimize the energy. As before, however, and it just gets worse, in fact, the little energies epsilon j are not, if we sum them up, they're not the energy of the atom. So we have to be careful when we do this kind of thing that we keep track of the algebra appropriately. We can't just blindly sum the things up and expect to get the energy of the atom. We have to look at what it actually comes to and what the energy of the atom actually comes to and then make adjustments to get the thing right. It turns out, so I don't want to solve those, but I do want to point out that optimized orbitals are available online and you can go look. Here's an example I've taken here on slide 526 of a screenshot from this website at ccl.net and which talks about what they call Routhon-Hartree-Fock Ground-State Atomic Wave Functions. And what I've given here is what they have for helium, which is written. The total energy is minus 2.8, 616, et cetera. And I think you know now that that's in Hartree and they give some coefficients here. They give 1s and they give two columns. They give 1.4595 and then 1.3479. Then they give 3s and then they give 2s and then they give another 2s. The question is what does this mean? So these are the accepted Hartree-Fock orbitals. Question is then who was Routhon? And Clemens Routhon was at the University of Chicago in the 1950s and that was around the time that full-fledged but huge and slow computers were being developed. There were a number of just absolute visionaries like Alan Turing and John von Neumann and others who were so good with math that they worked out that you could make a machine that only used the integers and that such a machine could simulate any other kind of machine which is a very interesting idea. Of course now we recognize that as a computer but before you invent a computer it's not at all clear that such a thing could exist even. And here's a quote then from Popular Mechanics in March 1949 and this is a cautionary tale on how hard it is to predict the future. Here's what they said. While a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs 30 tons, computers in the future may have only 1,000 tubes and weigh only 1 1⁄2 tons. And just imagine what happened in the next 65 years. Routhon championed the use of Slater orbitals. Remember you have to maybe go back a couple of lectures but I said that we wanted orbitals that were easy to integrate, right? So we want tuna that tastes good and we're going to be doing a lot of orbitals and therefore we want to have a canned formula for the antiderivative. And it just turns out for the radial parts it's much easier to use the Slater orbitals even though they aren't orthogonal because they're just much easier to integrate. And so in this database then when they refer to 1S and so forth they are not referring to the hydrogen type orbitals. They're referring to Slater orbitals SNLM that I gave you which have different radial parts than the hydrogen atoms. And using these and computers they could systematically improve the energy of multi-electron atoms. It's very, very hard once you get things in the exponent to minimize if you just have a fixed, if you have a linear combination of things with fixed exponents that's okay. That is long as you aren't monkeying with the exponents. But once you put Zeta up there and you want to minimize and find the minimum of these, how much of this and how much of that plus these guys can shift around like shifting sand dunes as well. Then it gets much trickier and you need some numerical methods to actually find the minimum. So you may be able to calculate the integrals analytically but you still then need to find the minimum of the energy which is as you saw when we just had a couple of Zetas there, it was a mess. You can imagine what it gets like. And so you don't try to do that thing by writing out a giant formula and then staring at it. You do it numerically. It's much quicker. You just calculate it and then you minimize it by moving to a lower value. The orbitals then that you extract if you look at those columns are the Slater orbitals. And therefore the atomic orbital in helium then, psi of r, explicitly I've written it out here. It's 1.347900 times the Slater 1s type orbital which coincidentally is exactly the same as the hydrogen 1s with a value of Zeta in the exponent of 1.4595. And then there's a small correction minus, so you can see these numbers tell you about how much of this other stuff is in there. That's the major one, 1.3 minus.001613 times S3s, that has a different value of r, with a big value of Zeta of 5.32. And then about a tenth of an S2s with a 2.6 and then minus about.27 of a Zeta with a 1.75. And this then becomes the approximation for our helium atom and you can look up some of the other atoms on the website and it's very interesting reading. Wouldn't be interesting to most people of course because if you don't know what anything means, of course nothing's interesting. A book isn't interesting, nothing is. But if you know what these things mean and you've tried to calculate them yourself or you've puzzled about what these orbitals actually look like in these complicated atoms, then it is interesting because now you've got these functions and with Mathematica or some other program, you can plot them. You can look at them, you can square them, you can make contours of them and you can learn an awful lot. The energy that we get out of this function for helium with these four terms is pretty good. It's limited more by the orbital approximation, the fact that we are not allowing electron correlation in this approach than it is by the number and choice of the functions. So we can't necessarily improve the energy by just adding more and more slater functions because we've made a fatal flaw in the beginning in that we've averaged out all the other electrons. Well, that's because we're chemists and we like to think in terms of orbitals. I'll just note that the simple two symmetric sum that I initially suggested earlier which had one electron bound and the other one loose plus one electron, the other electron bound and one loose, that that did better, even for helium, than this much more complicated looking thing. But that's not a product of one electron orbitals and so that's a different kind of thing. And I would not necessarily be able to write down something so clever for a barium atom, for example, whereas this will be a systematic way to get something reasonable for that. Now, if every time we wanted to figure out what kind of configuration of an atom out of all the possible ones there might be is the most stable. If we had to actually go through and solve all these equations, it would be pretty slow going, that's for sure. But luckily we do not because there are some rules and like all rules there may be broken in a way. But most of the time they're quite good especially for ground state atoms. They're quite good and I don't even know of any ground state atom that doesn't follow these rules which are called Hund's rules. And they tell us of which of the possibilities where the electrons could be in which orbitals, which one's the lowest energy. That's very important to know what the ground state of the atom is going to be. If we're going to talk in orbitals, we want to know the configuration of the atom. For example, Hund's rules predict that if we have a nitrogen atom and a vacuum by itself that the 3P electrons are going to be distributed one for each of the 2P orbitals. One's going to be in PX, one's going to be in PY, and one's going to be in PZ. Or we could think of one is going to be in M sub L plus 1, 0 and minus 1. Same thing in the end. But they're going to be in these 3 orbitals. They are not going to be jammed together with 2 and 1 orbital and then one in the other or some other combination. And furthermore, Hund's rules is going to predict the following. That if these 3 electrons are in these 3P orbitals for nitrogen, that there are bar magnets are all going to be aligned. They're all going to be either spin up or spin down. If we don't have an external magnetic field, there's no difference. But there is a difference between having them all aligned and then having one different than the others. That could be different. And Hund's rules predict that the electron bar magnets will all align each other when there are unpaired electrons. Well, let's go through these rules. So here are Hund's rules, the 3 rules. And they're meant to be applied in order. And only in the case that the rule doesn't adjudicate what happens, do you go to the next rule. The first rule is the state with the largest value of big S, which is the total spin multiplicity, 2S plus 1. Remember we've got S as 1 half and we add them up. And we can add them up, couple them together to get big S, the overall spin. And the multiplicity is 2S plus 1. Those are the number of substates that you can have for a certain value of S, just like you could have 3 substates for L equals 1 and only 1 substrate for L equals 0. Then the state with the highest multiplicity or the biggest value of S is the most stable. And stability decreases with decreasing S. If I have 4 unpaired electrons in 4D orbitals, this will be all of them up, will be the most stable, and 2 up and 2 down will be worse than 3 up and 1 down, for example. If you have two possibilities with the same multiplicity, the same value of big S, then the state with the largest value of the orbital angular momentum, big L. So both S, the spin first, align the bar magnets, then L, get them going around together is going to be the most stable. And finally, if you have states that have the same value of L and of S, then if the sub-shell is less than half full, for example, if you have one or two electrons in the P-shell, then the state with the smallest value of J is the most stable. And if the sub-shell is more than half filled, then the state with the largest value of J is the most stable. At first, these rules seem to be completely arbitrary, but we can give some rationale for them that makes some physical sense. Of course, you don't invent the rules before you know the answer, where you get the rules from looking at how things turn out, right? So Hundt knew the answer before proposing the rules. The rules apply to excited state atoms too, but not quite as well, but they do, for example, allow us to predict the ordering of terms in atomic spectroscopy. If we have energy levels there, the best way to keep track of them is with the term. Why? Because it's got 2S plus 1. We want to know that to know where the energy is. It's got L, we want to know that to know what the energy is. And it's got J, and that's the final arbiter. And so if we've got the term symbol for the state, we know what we've got. And it, we can use Hundt's rules to predict that 2P3 halves, for example, is higher in energy than 2P1 half, at which we saw with sodium giving the 589 doublet. But what's the physical basis if there is any for these rules? And we have to be a little bit careful because you might be rationalizing something that seems reasonable, but unless you've done a calculation of some type to show that it is true, you have to be a bit careful. Well, we can sort of understand why the total spin S should be maximized by, let's look at the case of 2 spins in a, something like helium. Now, the 2 spins cannot be parallel if they're in the same orbital in helium, so it'll have to be an excited state. But let's put one electron in an excited orbital so that it can have the same spin state. And then let's compare this one to that one. And let's see if there's any argument that we could make. So let's look at a 1S2S state, for example. And what's the reason why parallel spins could be lower in this state? Well, you might argue, well, maybe it's a magnetic energy. So the bar magnets are lining up because of the, they're bar magnets. That might be an argument. That's not the argument that's usually given. The argument that's usually given is that if the spin part is symmetric, that means because of the polyprinciple that the space part is anti-symmetric because if these two, when I swap them, nothing changed sign, then that means the space part has to be anti-symmetric and vice versa. Therefore, we could have two combinations here. We could have psi plus and psi minus. And one of them is 1SR1, 2SR2 plus 2SR1, 1SR2. And the other one is minus. The question is, do these spatial parts have any difference? And the answer is, suppose the spatial part's anti-symmetric. So it's 1SR1, 2SR2 minus 2SR1, 1SR2. What happens to that function as R1 and R2 become the same? What does it mean when R1 and R2 become the same? It means the electrons are getting very close together. Well, sort of like magic, when R1 equals R2, psi vanishes. That means that this anti-symmetric spatial wave function automatically causes the electrons to avoid each other in some sense because the way, because of its very structure, it doesn't allow the electrons to be at the same point in space without vanishing. That means the probability of that is very low. And that means that the electron-electron repulsion that could occur if they got close like that is gone. If they're anti-symmetric, there's no such restriction. And they could, in principle, get quite close. And we would expect then to have a big electrostatic repulsion. That's the conventional argument. That may or may not be true. However, there have been some more detailed calculations which indicate that there may be other effects like penetration and shielding that might be players in the game here. But for now, we'll accept that at least this is kind of the stock argument is this anti-symmetric spatial part. And therefore, the electrons pair only if they have to. If they have to be in the same orbital, they pair. If they do not have to be in the same orbital, then they're either both up or both down. What about rule two? Well, this is even more wooly in my opinion and quite qualitative. And the idea is this. If the value of big L is high, that means the value of little L for the electrons is all adding up. And we can in some sense imagine then that the electrons are moving the same way around. And so if they're moving the same way around the atom, then they're avoiding each other. If they're moving opposite ways so that L is zero or a lower value of L, some are going one way, some are going the other way so that the total is not the maximum, then they can crash. They can crash into each other and get close. And that means that the energy could be higher on average because the crashes push up the potential energy. And therefore, we want a high value of L to avoid these crashes and push up the energy. Now, if you, I don't think you want to believe that too literally, but anyway, it's a qualitative argument that gives you a feeling for why the repulsive forces may be lowered. The justification for rule three is even more subtle. And that harks back to remember what we talked about with sodium. Why was there the splitting? Because we argued that we have an unpaired electron in the orbital up there. And it's the gut nonzero L if it's in a P state. That means that in some sense is going around something. Imagine the nucleus from the perspective of the electron going around the other way. And that looked like a current loop and that created a magnetic field that acted on the magnetic moment of the electron. And the electron could either be aligned with or against the magnetic field. And we called that the spin orbit coupling interaction that we introduced when we talked about atomic spectroscopy. If we look at the state with the lowest value of J, it's lowered by the spin orbit coupling. And therefore, if everything else is equal, then we look for this small magnetic effect, the spin orbit coupling. If the shell is less than half full, it's going to be lowered. If the shell is more than half full, remember when we were working out the terms, I said, if you're going to do fluorine, you don't put 5P electrons and then drive yourself mad, you put in one hole and you get the configurations for one hole. But since you're putting in a hole, that reverses the algebra on J. And so, when the shell is more than half filled, you work with holes and then the rolls are reversed and therefore you should go with the highest value of J, rather than the lowest value of J. And the way to do it always is look at the term symbol. A configuration of electrons gives rise to certain terms. The terms are what you want to know because the term symbol is going to tell you what ground state of the atom you've actually got. Let's then predict stability here on practice problem 27. Let's consider an excited state, beryllium atom, 1S2, 2S1, 3S1, you see how I go back to the orbitals right away. Don't even consider actually the real thing. And the ground state carbon atom, 1S2, 2S2, 2P2. And let's predict the most stable state for each of these atoms. Well, with 2S electrons, L must be 0. And that's called an S state, but that's not the S of the spin. That's the S of the S state of L is equal to 0. So it's big S. And there are two possibilities for that state. There's the triplet state, triplet S1, because if 2S plus 1 is 3, then J has to be 1. And then there's the singlet state, 1S0 or singlet S0. The term with the highest multiplicity is the triplet term, triplet S1, and so that one is the most stable. That's what we predict with beryllium. And that physically means that the electrons are aligned. With carbon, we have to figure out the terms that can arise with a P squared configuration. And the terms that can arise are 1, singlet S0, triplet P0, triplet P1, triplet P2, and singlet D2. Now we look at all these. Well, according to Hunn's rule number one, the triplet states are lower in energy. So it has to be one of the triplet P states. The value of L is the same in all of them, because they're all P. We don't have to worry about different kinds. We don't have a triplet D or a triplet S to worry about. And so therefore we have to go all the way to the third rule and the shells less than half filled because there's two P electrons and there can be six P electrons in the shell. And therefore the most stable configuration, the most stable term is triplet P0 because that has the lowest value of J. Now Hunn knew the answers before he proposed the rules. And the rules are generalizations. They work best to predict the ground state and not for, necessarily for excited states. We can build up then the periodic table by adding electrons to orbitals, which we very loosely associate with the solutions for the hydrogen atom. We sort of think of them as having similar shapes, but really what we ought to do is go to the website and look at the shapes at least for the self-consistent field orbitals because then we'd have a much better feel. And then we'd see that they change as the atom goes. So 1S becomes different as, well 1S doesn't become much different, but some of them do become different as you add more electrons and get more repulsion. The real situation, of course, for any of these multi-electron atoms is much more complex than this, but at least this is a good start. The periodic table itself can help us decide which orbital we're going to fill next because the problem is the ordering gets jumbled by all the electron-electron repulsion. We can't solve for the hydrogen atom where there's no electron-electron repulsion and then expect that all our solutions are going to be the same and the same relative even ordering of energies when we've got 18 electrons orbiting around and all these repulsive terms and all these effective charge distributions. What we have to do is go through and see what happens. And the periodic table is the best way to do that because the periodic table, by looking at the atomic number and where it comes out, you know exactly which orbital got filled because things that have the same outer electron configuration have very similar properties and that's why we put them into columns and that's why we recognize that there were patterns in the first place. The most dominant effect as you go up the periodic table is that s-electrons are lower energy than p or d. p is lower than d or f and so on. The s-electrons as we saw when we did those radial distribution functions can be found very near the nucleus. If you have a big atom like xenon or cesium or something like that then those electrons way out there, those 5s and 6s electrons can spend some time quite near the nucleus and then they see that it is not at all like a hydrogen atom where you have the charge and all the other electrons canceling it out so to speak and just one charge. It's much different than that. The s can sample that. The p has a node there so it cannot sample it very much but the d has a crisscross at least there and so the sampling is very small and f is even more nodes at the nucleus and therefore it hardly ever sees the nucleus. Why should it? It's an f electron and so it has a high angular momentum and it's out there orbiting around. It doesn't know much about the nucleus. It's far away like Pluto. For example then the ground state of the potassium atom is closed shell argon. You put the closed shell of an atom of a rare gas in square brackets to indicate well that's those 18 electrons taken care of and then 4s1 even though argon is 3p6 instead of going 3d1 you go 4s1. We know that because we know that potassium is under sodium on the periodic table which is under lithium and all those are s1 and they have very similar chemistry, very similar properties, soft, silvery, ultra reactive metals and that doesn't have anything to do with something like boron or which would be 3p1 but scandium has totally different chemistry which has 3d1. The order of the occupation of the orders of the orbitals then in these multi-electron systems as we go up the periodic table and keep in mind this is as we fill them. If we take a light atom and we look high up in its energy and say g is 5s above 4d or not for a lighter atom where those energy levels aren't occupied yet, the ordering may be different but this is the ordering if we're going to fill them as we go. So 1s, obvious, 2s, 2p, 3s, 3p, first jumble, 4s, 3d, then 4p, then 5s, 4d, 5p finally, 6s and so we get to 6s, we get to cesium before we get to 4f. They're that far apart and that really has to do with the penetration, shielding and the electron-electron or the propulsion in these systems that have a lot of atoms. They're just different and we have to look at them differently. In hydrogen where you don't have any of these issues, 4f has very, very, very similar energy to 4f. There's always small effects but basically they're degenerate and they're both well below 6s. There's no confusing that but by the time you get to cesium where you actually want to fill 6s, it's below 4f. Now there's a cautionary note. The order of the orbitals is not always predictable without a detailed calculation and sometimes even the idea that you know which orbitals are occupied is a little bit flawed. Remember this is quantum mechanics so if there's a choice, the atoms pick both choices at once and you have to always be prepared for that. If you have nearby energy levels with the same term symbol as one you're looking at then these can interact. These are like neighbors that both play loud music. They don't like each other and they tend to move apart and the same thing happens. You have to have the same kind of term symbol though. If you have a different kind of term symbol then it's like you're on a shift work and I'm playing loud music at night but you're at work and so you don't care and then I'm working in the day when you're doing it and so we don't interact at all. But there are some systems for example, this was worked out in the 30s even, that in magnesium, the excited state 3s3d, the triplet d is higher in energy than the singlet d and that violates Hund's rule number one. Now that's for an excited state atom and I said that it works best for ground state atoms but nevertheless, it's kind of interesting to say well why did that fall apart for magnesium, it's not like it's a really heavy atom or something where you could argue that there's some other effect maybe coming in. And the reason why it happens is that there's another state nearby that's 2 3p electrons rather than 3s3d, it's 2 3p's and that also gives rise to a singlet d state and it's up here in the singlet d states here, the triplet d was supposed to be lower but these guys hate each other. So they pull apart and this ends up being pushed lower. So as usual, there's some subtle effects that go on and sometimes it can be a little bit hard to predict every little detail. That's what makes the periodic table sort of interesting is that there are all these countervailing trends sort of like predicting which politician is going to vote on which thing, are they going to get elected again and so on and so forth, same thing here. There's repulsion, there's this, there's that and then there are these fine effects. Okay, this slide I've called fission, no, not atomic fission as in nuclear power but suppose we begin now and we do a mental experiment, we do a thought experiment. We spent a lot of time on a helium atom. Let's take a helium atom now and let's break the nucleus apart slowly and we can just imagine. So we've got our two electrons there and now we're going to break this alpha particle, this helium nucleus apart into two deuterons, one proton and one neutron each. The neutron we don't care about changes the reduced mass but the question is what would happen if we did that experiment? And what the resulting thing that has two positive charges and two electrons, we know helium stable but suppose we pull it apart, would it be energetically stable compared to two deuteron deuterium atoms which is just a heavy hydrogen atom? It's just like a hydrogen atom with an extra neutron in the nucleus which as far as we are concerned just changes the reduced mass and nothing else. In other words, if we imagine mentally pulling this helium nucleus apart, would it hold together even though the things aren't held together by the strong force? In other words, would the electrons hold it together or would there be a chemical bond between these two positively charged particles? Well clearly when the two nuclei are close but they're not in contact, once they're in contact they're held together by the strong force which we don't write a potential for and we don't ever treat excitation in the nucleus explicitly. But let's just say when they're close enough they're held like super glue so even though they're two positively charged particles they still hold together and they're very stable and they're not going to come apart. But once we mentally pull them apart so they don't touch, the strong force is like a big wrestler but he's got a limited reach. If he can grab a hold of you, look out, you're never going to get away but if he can't reach you then you're safe and once the particles get far enough apart there is no strong force but there certainly is electrostatic repulsion and they hate each other because they both have positive charges and so they push apart like crazy. And if we mentally pull them way apart then they have nothing to do with each other then they're two atoms. The wave functions of the electrons have died off, they don't even know the other ones there. So that will just have the same energy as two atoms, that's not going to be more stable. But so initially it's very bad if they're too close and then something happens and then it's zero if they're way apart. The question is could there be a sweet spot where the nuclei are far enough apart that they aren't repelling each other and destroying just moving apart but the electrons can overlap and share an orbital with each other and then create a stable configuration and if so what does that? Well, look, H2 is stable. Hydrogen is probably going to be the economy of the future for auto transportation one day. So we know for sure H2 does exist and D2, heavy hydrogen exists and even H2 plus exists. It's very important in astrophysics for example to look out and see that there is H2 plus way out there. So we know they exist. What we're going to find out in the next series of lectures is why they exist.
very, very tiny things compared to, in fact, the game. It has no analog and classical mechanics. to the hydrogen type orbitals. They're referring these, how much of this and how much of that plus its very structure, it doesn't allow the they have very similar chemistry, very similar orders of the orbitals then in these these systems that have a lot of atoms. They're Remember this is quantum mechanics so if there's atom? It's just like a hydrogen atom with an