 OK. What I've been recruited to do is to do some Marcus theory and to do that basically toward electrode to get a garusher model of electron transfer at an electrode. And since I'm basically a solution chemist, we'll do more Marcus theory and solution prior than we'll do garusher model. But that's just the way to do it. So what I'm going to try to do is I'm first going to try to introduce Marcus theory in a way a little more standard than Andy introduced it and to try to motivate it a little bit physically. So we're going to start off thinking about Marcus theory. And the first thing to say is that Marcus theory was actually not uniquely developed by Marcus, but Noel Hush in Australia also developed a model very similar at very much the same time. But because his model didn't talk in terms of these parabolic surfaces that we're going to talk in, it was much harder to visualize. And so as a result, everybody has used Marcus theory. And even Hush now uses the Marcus parameters rather than his parameters. So it's sort of a little bit. OK. So let us start and think about the simplest electron transfer reaction. And the simplest electron transfer reaction we're going to think about is the ferrous ferric, AQUO, going to ultimately Fe H2O 6, 3, 3 plus plus Fe H2O 6, 2 plus. And these are for homogeneous ions in solution that are floating around. Now clearly when they're far apart, they can electron transfer. They can electron transfer until they come together into some sort of solvent cage and form what I'm going to call precursor complex. I need more space. My board isn't big enough. I have to erase that. And so now I have the precursor complex. I'm going to elect the electron transfer. And I'm going to form where I'll write it. Maybe I'll just write it simply. This will become the iron 3. And this will become the iron 2. And then they'll come out of the cage into a fluid solution. OK. And that's the reaction I have to formulate. So in order to formulate that reaction, what I want to do is I want to consider this basically an equilibrium step. And I'm going to say it as a rate of k1 and k minus 1. And this as the electron transfer step with a rate of ket. And so what I'm going to do is then set up a steady state for the reaction. I'll calculate the steady state of the iron 2, 3 plus. So if I calculate d, d, dfe, 2, fe, 3, dt, it's going to be k1 times fe, h2o, 6, 2 plus fe, h2o, 6, 3 plus minus k minus 1 times my fe, 2, fe, 3 minus ket times the fe, 2, fe, 3. OK. And I can set that equal to 0. So I set this equal to 0. I solve for the precursor. And I get that my fe, 2, fe, 3 will equal k1 over k minus 1 plus ket times fe, h2o, 2 plus fe, h2o, 6, 3 plus. And in most cases, the electron transfer rate is slower than k minus 1. And so I can reformulate this as k1 over k minus 1 times my two concentrations. And the rate of the overall reaction is going to be just the rate of this step going there. So the rate of the overall reaction is going to be ket times the fe, 2, fe, 3, which is going to be k1 over k minus 1 ket times fe, 2, fe, 3. OK. So this is now, my k observed, my rate constant, is going to be k1 over k minus 1 times ket. Or I can just call it k1. It's an association constant. It's the equilibrium constant to form the precursor complex times ket. OK. Now, that big equilibrium constant, the association constant, I don't want to really talk about. That's not really of great interest to us. Well, I'm really interested in the ket. So primarily, I'm going to talk about the ket. But if you deal with fluid solution, you're going to have to worry about that ket. Now, we want to ask when these reactions were originally thought about in the late 40s and 50s, the question that was asked was, why aren't these reactions instantaneous? Why don't the reactions happen every time the reagents come together? Why aren't they diffusion controlled? And this reaction is known, it has a rate constant of about 1. Molar minus 1, second to the minus 1. So we want to try to first understand why these reactions, the physics of these reactions, that make them so they aren't instantaneous. And the first thing we can ask is what is the difference between the ferrous ion and the ferric ion? And the first thing we know, where this is 2, the bond length is about 2.17 angstroms. Not 1.7, 2.1, that's good enough for it. The ferric has a bond length of 2.0 angstroms. So the ferrous and the ferric look different. If we transfer an electron from the ferrous to the ferric, we would create a ferrous that had the wrong bond length, and we create a ferric that also had the wrong bond length. So there's an energy question when you transfer an electron. So what we want to do is we want to talk about that. OK? So let's ask, so what we have now is let me draw this. Let me, this is in our precursor complex. We have this like this. And what I want to do is I want to consider this as a super molecule. OK? And I want to draw a potential energy surface. So what I want to do is I'm going to ask, if I take this as a super molecule, the electron transfer now is no longer moving an electron from one center, one molecule to another. It's moving an electron within the molecule itself. And that's basically where we're talking about an electronic transition. And we can apply what we know about electronic transition. So the first thing I want to do is draw a potential energy surface for that. And the, so I'm going to draw a potential energy surface. And all potential energy surfaces in chemistry are drawn with what kind of curve? With a parabola. That's the best shape that God gave us. And the question is, what am I drawing? OK. What I'm drawing is the first question is what is this coordinate that I wanted to do? And what I want to draw is I want to consider the coordinate that consists of the anti-symmetric stretch for the bond. That we have the long ferrous and the short ferric. And we want the ferric to expand at the same time that the ferrous is shortening. And this will give us this sort of potential energy surface. And this is going to be the position at which one nucleus, which I'll call the a nucleus, is the ion 2. And the other nucleus is the ion 3, which I'll call the b nucleus. OK. On the other hand, this ultimately goes to the ion 3, where this one's short. And the ion 2, where this one's long. OK. And that's going to have a different set of coordinates. And if we're talking about the asymmetric stretch, it's going to have a different position. So this is going to be a, where the ion 3. And we'll have another set of coordinates. So what we've drawn is a potential energy surface where this is what we would call the reactants. And this is the products. And we want to talk now about moving along that potential energy surface. OK. Now, this coordinate is a nuclear coordinate. It's the motion of nuclei. And when I move along this coordinate, what is the time scale that I need to move along that coordinate? OK. So that's our next question. What are the time scales involved? And how can I move on this surface? OK. So let me ask, what are the time scales of motion? OK. So let's say we'll talk about frequencies. And we'll talk about the frequencies of electronic motion. And the frequencies of electronic motion, the electron moves very fast. It's very light. And that has a frequency of something like 10 to the 16th seconds. That's already, yeah, that's it's frequent. No, that's it's, I want it per second. 10 to the minus, so I guess I want this is, which do I want it as? I want it to go as, I guess, 10 to the minus 16. Well, then it's a velocity. OK, so we'll take it as 10 to the 16th one over seconds. OK. So that's for the electron. The frequency of vibration of metal ligand bonds is something of the order of, say, 300 centimeters to minus 1. Well, it's something of that sort. That's for a metal ligand bond. That's for the ion oxygen. And that gives us a timescale of about 10 to the minus 13. Well, OK, 10 to the 13th per second. OK. And this is for a metal ligand bond. The vibration of a something like a CC bond is something of the order of, say, let us say about 1,000 centimeters to minus 1. And this will be something like 10 to the 14th per second. And the last one I want to ask about is the frequency of the solvent. OK. The solvent is the water round. And the solvent can rotate its dipolar. Let's think of water. And it can rotate. And it has a frequency that's lower energy still. And it has a frequency of the order of about 3 centimeters to minus 1. And so this is about 10 to the 11th per second. OK. So what we have is we have a large timescale from a 10 to the 16th frequency to a 10 to the 11. And what that means when you have a large timescale is you basically can't couple those sorts of modes. That the electron is going to move much, much faster than the nuclei are going to move. And so they're going to move independently. So what that says is that when you move on this surface, OK, when the electron is going to move from one surface to the other from the reactants to the products, it's going to move very, very rapidly. And if you start out here and you transfer the electron, you would end up at this point up here. And there would be a net energy gain. And you'd have more energy up here than you had down there. And the problem is, how does that energy get in? Well, in a thermal reaction, the only way you can input energy into the reaction is by collisions of nuclei. And so there's no way in the timescale for the electron to jump that you could go there. So you can't do it that way. So the way you have to do it is you have to move along the ground state surface with collisions. And you get here where it's the generate for the electron to be on the reactants or the products. Then the electron can jump. And then it can relax down to products. So that if we are doing a thermal reaction, we have to go over this barrier. This is the barrier to the thermal reaction. However, there's another way you can actually transfer an electron. And that other way allows you to actually go vertically up. And that way is with a photon. If you use a photon, you put the energy in on a timescale of the electron transfer. And so you can excite it with optically. So optically, you can see this transition. And thermally, you see that transition. And clearly, they are related. OK. So when we move vertically, we have fixed nuclear positions. The nuclei don't change. And it happens essentially instantaneously. When we move horizontally, we're moving at the timescale of nuclear motions. And which nuclear motions that we're going to be moving at is a question we need to discover. OK. Now, if I've drawn the system like this, one of the postulates that Rudy made was that these surfaces would be parabolic, that they would be harmonic. And that was sort of a guess. But it turns out that actually that's a reasonably good guess. And from the very fact that they're parabolic, we can develop a lot of information about the system. So let us now just consider these curves. OK. And this is energy. And I'm going to define this as a parameter. I'm going to call lambda, the reorganization parameter. And that parameter is the energy it takes to, well, it's a thermonutrile reaction because I'm just considering self-exchange reactions. So there's no driving force. And it's the energy needed to reorganize the reactants till they have the equilibrium configuration of the products or the products till they have the equilibrium configuration of the reactants. And arbitrarily said this is my axis as 0 and 1. And I'll call this my energy of my reactants. And this is the energy of my products. And the energy of my reactants is going to be equal to lambda x squared. And the energy of the products is going to be equal to lambda x minus 1 squared. OK. And you can see that when x is 0, the energy of the product is exactly lambda as I say it should be. OK. The next thing we want to do is I want to say if this is the thermal barrier, we don't call delta G star for the reaction. And you see, as long as I've assumed that they both have the same force constant of the two parabolas, that this is going to occur at 1 half. And so delta G star is going to be 1 fourth lambda. So that there's inherently a relationship between the barrier you'll see for the thermal electron transfer and the optical transition. OK. Now how do we get the rate? So what I now want to do is I want to get the rate of the electron transfer. What I need is the KET. And to do that, I'm going to go and use transition state theory much like Andy did. And what we'll do is, I guess, in transition state theory, we have say we have a transition state there. We say the transition state is an equilibrium with the equilibrium configuration. And so if we say this is C, this is just the concentration of my reactant C. Can I make a C? And this will be C star. I'm going to say there's a equilibrium constant that equals C dagger over C. And then once I'm up here, the question of whether I go over is going to be defined by some frequency at the top of the barrier. That at the top of the barrier, I have the transition state sitting there. And it can either vibrate and go back through reactants, or it can vibrate and go down to products. And that's some sort of nuclear frequency that's destroying the transition state. How fast it destroys. So I'll say that's a nuclear frequency new end. And so my rate is going to be new end times C star. And that's going to be new end KC. This is my rate. This is my rate constant. This is my total rate. Excuse me. That's my total rate. And so my rate constant K, this is my KET, is going to be new end K. And K is going to be equal to e to the minus delta G dagger over RT. So that's all very simple. And in my case, what I'm going to say is that I'm going to say this is the star, and this is new end e to the minus delta G star over RT, or new end e to the minus lambda over 4 RT. So that's all very simple. Now there's a little bit of flim flam here. Transition state theory is really developed for bimolecular reactions. And my KET is really a unimolecular reaction. I've preformed everything. And that's basically taken into account why transition state theory uses this dagger. And Rudy used the star because he was dealing with a unimolecular reaction. But it's not much different. And so we'll just leave it at that. OK. There is, though, one thing that at this transition state, the electron is hopping back and forth. It can transfer. It's either on the reactant side or the product side, but that's degenerate. And so it can go back and forth very easily. So what it's actually doing, it's transferring between states. I mean really what I have is I have, at the transition state, I'm either on the reactant's curve or the product's curve and it transfers back and forth. And that's really one of the questions I've got to ask about. And it's sort of a delocalized transition. OK. So what am I going to ask? So I have this, I guess, my next consideration is what do I want to use for my, I can see Chalk has certain advantages. What I want to use for the new N. And that's the frequency that I break apart the activated complex. And the frequency I'm going to use for new N is a nuclear frequency because it's basically, what I've done is I've had this anti-symmetric stretch that goes like that. OK. And when it happens to get to the situation where they are the same distances, then you have a degenerate state where the electron can transfer from one side to the other. And so what's going to destroy that is basically this frequency of the symmetric stretch when that then starts stretching so that you have the right-hand one as the iron two, I guess it is, and the left-hand one as the iron three, then it will destroy it. And so this is basically, I've written it new N because it's a nuclear frequency. And so that's going to have a frequency of somewhere between 10 to the 11th to 10 to the 13th, seconds to the minus one. OK. OK. Now to go back to the point that I mentioned before, that you also have that the electron can hop from one surface to the other. If this is all there is, what I'm assuming then is that every time they get to the transition state, the electron can hop back and forth. But you know that when you have transitions between electronic states, sometimes they're allowed and sometimes they're not allowed. So there is the possibility that the probability that the electron could actually transfer would be very small. OK. And so there are two types of situations. There's an adiabatic electron transfer. And that has a probability of electron transfer as one that it can hop across from one curve to the other. And then there's non-adiabatic electron transfer. And there the probability is much, much less than one. OK. And to ask about what you have to do is you need to somehow put this probability in here. And so you can then go back and look at how do you calculate the probability for curve crossing. And so we go back and we're going to take results that were developed by Landau and Zainer. And they developed a theory to calculate how likely you are to transfer between electronic states. And then rather than write the rate constant like that, they write the rate constant for KET as new electronic times e to the minus delta G star over RT. OK. And now new electronic will write as rather than a nuclear frequency, it's going to involve the electron motions. OK. And it's the transfer probability. It's going to be 4 pi squared HAB squared over H times 1 over square root of 4 pi lambda RT. OK. HAB is the interaction term between these two states. This is one state and these are another state. And as we're going to see with quantum mechanical considerations, they're going to interact. And their interaction is going to be this HAB. And this is going to tell you. Now the frequency clearly can go from something very small when this is very, very small, this frequency is low. So if HAB is of the order of 100 centimeters to minus 1, I calculated it, you get a frequency, new EL is about 10 to the 12th seconds to the minus 1. OK. And if HAB goes down much lower, you get a much lower frequency. On the other hand, what happens when HAB gets much bigger? What happens when HAB becomes of the order of 1,000 centimeters to minus 1? What happens is now the probability has gone up to 1 and you can always cross the curve. So you don't get extra. You can never go faster than being able to cross the curve. It may be willing to jump it back and forth a million times. So then you go over to the nuclear frequency. So if we plot the frequency, if we just say our rate constant is equal to some sort of times meaning the minus delta G star over RT, and we look at this and we plot it versus HAB and this will be that frequency, it's going to be quadratic in HAB initially and then it's going to level off at new N. So the largest that pre-factor can be is the nuclear frequency. But the electronic probability of transferring between curves can be much less. Now the next question we want to do is ask what happens when we don't have a self-exchange reaction? Oh, now I get to use colors. Here is our self-exchange reaction and now I'm going to add a driving force. It will be delta G zero, a driving force for the reaction. And so since this offset is delta G zero, despite my drawing, this offset will be delta G zero. Also this distance is lambda. And now what we want to know is delta G star. Now since everything's parabolic, we're assuming the same force constant, we can work that all out. We again say this is x, this will be zero, this will be one, this will be energy. We can work that all out. And so how do we do that? We're going to say E of the reactants now is going to be lambda x squared. That's what we had before. And E of the products is going to be lambda times x minus one squared plus delta G zero. Because that's driven it down. We want to know the crossing point where these are equal. So we can solve for that. We can say lambda x squared at the crossing point is going to be lambda times x minus one squared plus delta G zero. And we'll solve for that. And we find out that x equals delta G zero plus lambda over two lambda. I think there's a minus sign in there maybe, too. But that doesn't go any way. Bother us too much. And so now we can ask what delta G star is. And that's going to be the energy of my reactants at this crossing point. So that's going to be, so this is lambda at the, we'll call it what, lambda at the, probably we can call it x at the transition state. So this is going to be lambda x transition state squared. OK. And that's going to be equal to delta G plus lambda squared over four lambda. OK. So now just by, I mean, it's sort of magic that you have assumed parabolic curves and you can get out all these relationships. Now what is going to happen as you increase the driving force more? Initially as we increase the driving force, what happens is the barrier goes down. Initially it's here and I increase the driving force and it goes down to there. But it clearly, if I keep increasing the driving force that ultimately I get a place where there's no barrier. And then if I increase the driving force even more, I start to re-establish a barrier. OK. That prediction is the prediction of the inverted region. That the rate constant as you increase the driving force should get faster and faster until it reaches a limit and then it should go down. OK. What's an interesting fact was that both Marcus and Hush predicted that. Hush didn't believe it. He said it was just an artifact of the assumptions. And so he didn't really believe it. Marcus, on the other hand, believed it and 25 years later they actually showed that it was true. Actually, there's another comment to me. So if we plot KET times the minus delta G0, where that's the driving force, the more negative the free energy is, the more driving force you have, you would expect that it's going to follow this sort of relationship and it's going to go something like that. It's going to be Gaussian. OK. And the Gaussian's going to have a peak at lambda. OK. The more common way people tend to plot it is they plot the log of KET versus minus delta G0. And then you have a parabolic relationship, again, with a peak at lambda. It's very hard to reach this region, the inverted region, in solution experiments because you need such strong reactants that they tend to chew up themselves before they actually react. So there aren't many places that in fluid solution you can see that. And so it was very hard. The first place it was seen was in a frozen solution where you have everything fixed. And they could see that as they changed the driving force for the couples that they ultimately went down. But, well, there's another place you can see it too. And they're wondering, OK. Well, let me just say one more thing about that. Since we're dealing with the transfer from one electronic surface to another, you don't have to be really dealing with electron transfer reactions for the formalism. You can almost be dealing with any sort of reaction that you transfer from one surface to the other. And one common place you're doing that is in excited states. And so when you go from an excited state to a ground state, you're transferring from one electronic surface to another. But those surfaces in general are highly exothermic. You're up there. You're in very nested parabolas. And you're talking about a transition from down here to the ground state. And in those sorts of excited states, you can ask, what's the lifetime of this excited state? And that gives you a measure of the rate of going from the excited state. And there's a dependence on the energy gap between the two surfaces and the lifetime. The larger the energy gap, the longer the lifetime. And that's just a result of being in the inverted region for the Marcus relationship. So excited state lifetimes that they have a length is basically due to that they have a big energy gap. And so there are all sorts of rules and photochemistry that come out about that. But that's another story. So now we want to talk about this reorganization energy, the lambda, and say what it's about. So we're going to talk about the reorganization energy, lambda. And generally, I said there were different time scales in the molecule. There's the sort of time scale of the electron, which we've dealt with in a sense. There's the time scale of the metal ligand vibrations. And there's the time scale of the solvent motion. And so we're going to say they basically are decoupled. And we're going to say that there's a lambda for the solvent, which I'm going to call lambda O for outer sphere, meaning it's the solvent outside our complex. Plus, then there's going to be a reorganization energy for the vibrations within the metal complex. Primarily metal ligand vibrations, but they can be other things. And in organic molecules, they may be something entirely different. Although in organic molecules, the forces holding your bonds are much, much stronger than the metal ligand bonds. So they don't tend to change much. So they usually aren't too important. For metal complexes, precisely because the metal ligand bonds are weak, they tend to change a lot in these reactions. And so they're important. So first I'll talk about lambda I, the inner sphere. And what I want to do is I'm going to think about it totally in terms of metal ligand bonds. And so I'm going to say that lambda I is going to equal to the reorganization energy for the, say for the 2 plus species, where my iron is going, but say I'm dealing with my iron 2 plus and iron 3 plus, for the 2 plus species. And that has to go from its equilibrium configuration, the metal 2 to the metal 3. Plus I'm going to have a reorganization piece for the ferric, which is going to go from the m30 to dm20. And if I basically say they're harmonic oscillators, like I say everything is parabolic, I'll say that for this is going to be 1 half times the summation of the force constants times the change in each of those bonds. Plus 1 half, this is for the 2 plus and this is for the 3 plus. These are squared. And if I deal with an octahedral complex that has six bonds, there'll be six of these. And that's the only one. So there'll be 1 half times six, and they're all the same. So I'll say this is for my 2, I'll put my 2 down there. And this will be delta d squared v, how do I want to do it? It's going to be delta d0 squared, delta d0 squared plus. And that's a 3, that's 3, that's 3, delta d0. Where this is just the distance to go from the iron 2 to the iron 3 bond length. But this is the reorganization energy. So it's not at the activated complex, it's at the products. So we get this is 3f2 plus f3 times delta d0 squared. OK. Now, if these force constants are not the same, and you wouldn't expect them to really be the same, then they have different contributions to the reorganization energy. So now I want to draw a picture of our surface. I guess I can draw the picture of our surface here. And let's just maybe say what we're going to have. The nucleus I'm going to call A will be the iron 2. And the nucleus I call B will be the iron 3. And that's going to undergo the transition to the opposite. And so if I draw here the radius of the nucleus A, I'll call it the metal ligand. Well, it's the metal ligand bond distance for a nuclei IA. And this will be the metal ligand bond distance for B. OK. So at the reactants, B is an iron 3. And A is an iron 2, right? So iron 3 will tend to have a higher force constant. So what that means is the surface will be more steeply sloped in the direction of the iron 3 than in the iron 2. So it's going to have a surface that looks like this. OK. And its equilibrium configuration is here. So I'm drawing you a three-dimensional surface. Likewise, the product will be FeA3 plus FeB2 plus. And that will look shifted. And that's going to look like this. And the reaction is going to move on that potential energy surface. And the reaction path, if you draw it, the classical reaction path is basically going to be like that. The transition state is going to be here. OK. And you can calculate the motion by dealing with that surface. OK. Now I can ask, what are the positions? What are the radii? What are the metal ligand bond distances at the transition state? And I can attempt to answer that also. OK. And the way I'm going to attempt to answer that is, again, by sort of writing it all out. And rather than writing out for the reorganization energy, which is lambda i, I'm going to write it out for the delta G. OK. And again, I'm going to have 1 half times 3 since 1 half times 6. I'm going to be using basically the formula phi delta D squared. So there are 1 half times 6. There's six bonds. There's going to be the force constant for the 2 times the D 2 0 minus D transition state squared plus 6 iron 3 plus force constant D 3 0 minus D transition state. OK. And to find, now, one thing I have assumed is I assume that the iron 2 and the iron 3 rearrange so that they're at the common distance. Because I want the energy to be degenerate. If they aren't at a common distance, you won't have the same energy when the electron transfers. So I've assumed that in there. And so now what I want to do is I want to minimize this with respect to the position of the transition state. So I'm going to take D delta G in star D, D dagger. And I'm going to set that equal to 0. And we can solve it. And I get D star. I guess it's not dagger, this is star. Oh, I called it dagger. So I guess it's, whether it's dagger or star, I seem to have no consistency. OK. So it's going to be equal to F 2 D 2 0 plus F 3 D 3 0 all over D upon D. OK. Now, again, we get that there's sort of contributions from the two sides. Normally, the difference in the force constants is of the order of 10 or 20%. So it doesn't make a lot of difference. But it does provide a interesting result that we can now go back first. And we can calculate delta G star in. And that's 3 F 2 F 3 over F 2 plus F 3 times delta D 0 squared, where I plug this back in to these and work it through. And I end up with no more D dagger, but only D 2 0, the equilibrium. And I end up with the difference. OK. Now, an interesting problem is you see this result. And you see this result. And what do you notice? There was supposed to be a relationship between this, the delta G star and the lambda. That delta G star was supposed to equal lambda over 4. But in fact, it doesn't. OK. What's going wrong? It isn't our path is different. But we assumed when we started out that these are parabolic surfaces. And we inherently assumed that everything had the same force constant. Here, we suddenly started to let the force constants be different on the two surfaces. And that messes everything up. So that you don't quite get this to be right. It's not off by a lot, because the force constants aren't all that much different. But they're off. And you don't get this relationship doing it this way. So what you do is it was so nice having these simple parabolic surfaces that Rudy went back and redefined the force constants. And he defined an average force constant. And he defined it as 2f2f3 over f2 plus f3. Actually, I don't know if it was Rudy who did this originally, but it's lost in the lure to me. And if you do that and you define them both to have the same force constant, then of course, you will get this relationship and everything. And what the surface looks like is rather than look like the surface over there, the surface now looks like this is RA and this is RB. Now looks like spheres, because the force constants are the same. And the reaction path is just straight across. Now, another comment about lambda n is what is the size of lambda n? Since we're doing this, it's nice to have some feeling of what size quantities we're dealing with. Oh, green doesn't go away. The lambda n has a range from this lambda n from 0 to about 2EVs. That's about its whole range. And most things are around 0.5EVs. You can actually look at the kinetics and try to extract the lambda n. And if you do that for many complexes, well, that wasn't like 0.5EVs. So anyway, it's about 0.5EVs. I don't know, extracting it's hard. For the lambda n we will extract. OK. Oh, there's one more comment. This is going to be directly proportional to the difference in metal, ligand bond distances. So if you look at some complexes like ruthenium bipyridine, the 3 plus 2 plus, the delta r here is about 0. There's no real difference between the, for the iron aqua, we saw it was about 0.1 angstroms, OK? For cobalt amines, H3 taken as 6, 3 plus 2 plus, it's about 0.2 angstroms. This is large. That's really large. You seldom see stuff like that. And as a result, the rate constant for this, if I KET, is somewhere of the order of 10 to the minus, I think 10 seconds to the minus 1. Whereas the iron has a rate constant of about 1, second to the minus 1. And this is about 10 to the 8 seconds to the minus 1. So the rate constants actually span something like 18 orders of magnitude. And they respond very much to the lambda n. And that was one of the first sort of proofs of market theory when that was found. Yeah, these are the self-exchange rate constants. Yes, you're right. My units are wrong. These really are moles to the minus 1, because they're the second order rate constants. So they are mole to the minus 1, seconds to the minus 1. Yes, thanks. OK, now what I want to do is talk about the solvent polarization, since we've talked about the inner sphere and gotten a feeling of the size of the inner sphere. So we're going to talk about the solvent polarization, lambda 0. OK. Now I said that the solvent is sort of slow. It has, if we're talking of water, it has dipoles. And the dipoles can actually reorient. And that's a pretty slow motion, because it's hydrogen bonded. And it's sort of a collective property to reorient its dipole moment. But there are other sorts of solvent polarizations that they're also, they're electronic polarizations. And that the solvent, you can actually, what we do, we can polarize it as you polarize a molecule. In other words, you move the electrons from one side of the solvent molecule to the other. And that we would expect to be very fast. So we would expect that there are sort of two times scales now here. There's a fast time scale electronically, and that can stay in sync with the transfer of the electron. You're transferring the electron quickly, but there's this fast part of the solvent that can stay in sync with that. Then there's a slow part of the solvent, which is this nuclear reorientation motion that can't. And so somehow you've got to separate the fast part from the slow part. And that was one of Rudy's real contributions to develop a cycle to calculate a non-equilibrium type of energy. So we're not going to do that because that would take too long, but we'll sort of just get a feel of it. So we have the fast motions, which are electronic. And we have the slow motions, which are nuclear. OK. And we're going to use the dielectric constants to tell us something about them. There's the static dielectric constant, which tells us something about the total polarization of the solvent. That's a measure we measure at zero frequency. And then if we measure the dielectric constant at high frequency, which I'll call O, if I measure an optical frequency, I can measure the dielectric constant with, say, infrared light, and then I'm measuring it at a frequency of the infrared light. And so that's a high frequency. So this measures sort of the polarization that is fast. And this is the total polarization. And so the formula has to somehow recognize that difference. OK. Now we can ask what the property is. I'm going to try to get the feeling of what the property is of. Here's our, we'll think of our two ions. Here's our iron 2 plus. And here's our iron 3 plus. OK. And what we're talking about is we've polarized the solvent around here. And if we say these are spheres with some radius A, as we increase A, as A goes up, we would think the energy of polarization would go down as the molecules get far away from the center of charge. What's polarizing is that center of charge. You have, say, an iron 3. And it's polarized the solvent around it. If the molecule got real big, it would push the solvent further away. And so we wouldn't have much polarization. So we would expect that the energy, the lambda out, would go down. OK. And so Rudy, as I say, worked out this formula. And the classical formula is that the outer sphere thing is going to be delta E squared times, well, I'm not going to have room there, delta E squared over 4 pi epsilon 0, all times 1 over 2A2 plus 1 over 2A3 minus 1 over R times 1 over d op minus 1 over ds. OK. Where this is the radius of the ferrous. This is the radius of the ferric. This is the distance of electron transfer from one to the other. And these are the dielectric constants, the optical and the static. And by using the inverse of them, you're basically measuring the piece of the dielectric polarization that is slow and so cannot stay in time with the electron transfer. And that's what has to reorganize prior to the electron transfer. OK. Often, traditionally, this is written just when you look at the literature without the 4 pi epsilon 0, because 4 pi epsilon 0 is the constant to go from CGS units to SI units. And traditionally, all the theoreticians use CGS units. And the electrical people were the last to give that up. They hated it. But it's much easier to calculate if you put those in, because otherwise you've got to work in really funny units. OK. Now, to get a feel of this, I also have a little table. I have the dS and the d optical. And we have for water, for acetyl nitrile, for THF, for CH2Cl2, and for hexane. OK. For water, the dS is about 80. And the static is about 1.9. For acetyl nitrile, it's about 37. It's about 1.8. For THF, it's about 7.5. And it's about 2.0. For bichloromethane, it's 8.9, 2.0. And for hexane, it's 1.9, 1.9. OK. So what you notice for hexane is that the static, the low-speed, the low-frequency and the high-frequency dielectric moments are the same. So that will have no reorganization energy for an electron transfer. The chance is that you can't do usually much in hexane because it doesn't dissolve ions, but it would have no barrier, no reorganization barrier to the electric current. On the other hand, water would have a fairly large barrier. And that's why it's important for us to calculate atmosphere reorganization, precisely because water is what we generally use. And now the size of lambda out. By measuring reactions, you can self-exchange rates. You can try to calculate the lambda out if you know the bond length differences in the two oxidation states of the ion because then knowing the bond length differences, you can calculate the lambda in. And from the measuring the reaction, you can get the total lambda. And so you can measure lambda out. And for rubipi, you measure a lambda out of 1.6 EVs. The range that you measure of lambda outs in water is somewhere from about 0.3 to about 1 EV. I don't know if there's anything bigger than 1. And if we calculate it, Tom has done the calculation. And you get somewhere about 0.55 EVs, which is surprisingly good. It's surprisingly good because what we really have is a continuum theory of salvation that is pretty crude, but it does really well. And that's true pretty much across the board. It's surprising how well it does. And people have tried to have much more sophisticated theories of this, and it hasn't really helped much. Now the next thing I want to do is I want to get the cross relationship, which actually Andy sort of did, but we'll do it again sort of from this viewpoint. So now we'll do the cross relationship. We're going to consider, say, the following reaction. An iron aqua, 3 plus, plus ruthenium, say, hexamine, going to iron aqua 2 plus, plus ruthenium hexamine 3 plus. And we're going to call this, the rate constant we measure here is k12. And k12, we're going to say, is equal to the association constant for this reaction times the frequency factor e to the minus delta g0 plus lambda squared over 4 lambda RT. That's the Marcus expression that we got a while ago. And the reorganization energy, this is lambda12, and this is delta g12, this is lambda12 here. And lambda12 is going to be an inner sphere reorganization term for this, an inner sphere reorganization term for this, and then there's going to be the outer sphere. So we're going to have the lambda in of the iron. Well, I'll call it a 1. You'll see plus the lambda in of 2 plus the lambda out. And we're going to compare that now to the self-exchange reactions. The self-exchange reaction is going to be our ferrosperic, fe equal 2 plus plus 3 plus. And we'll call that hazard rate k11, which is equal to the association constant 11, the frequency factor 11 e to the minus lambda11 over 4 RT. And lambda11 is going to be equal to lambda, you're going to have a reorganization term for each of those in inner sphere, so it's going to be 2 lambda1, or lambdaI1, I guess I1 is how I call it, plus an outer sphere term. And now I take the other reaction, the ruthenium. I call this k22, k22A nu22, either minus lambda22 over 4 RT, lambda22 equals 2 lambdaI2 plus lambda0. And the last thing we do is for our cross-reaction, we define the equilibrium constant, k12, which is going to be equal to e to the minus deltaG0 over RT. That's just the equilibrium constant for the overall reaction, it's not the association constant, it's the overall reaction. Now you can see what the game is going to be. You have lambdas and stuff, and so somehow can you consolidate everything into a formula that relates these two reactions to the overall reaction. And you can, and there are numbers of ways of writing it, I'll write it sort of in two ways, but I'll do it in one way so you see it and then I'll write it in a second. So we have our k12, and that's our k12A times nu12, and I'm going to expand the exponent. So this will be deltaG0 squared plus 2 lambda, deltaG plus lambda squared all over 4 lambda RT, and this is lambda12, well, this is lambda12, I can't fit it on the next line. So this is k12A nu12 exp to the minus deltaG0 over 4 lambda12 RT exp of deltaG0 over 2 RT, I guess it's a minus sign. OK, this term is just k12. This term I replace with lambda, well, what is it? This term has lambda1 plus lambda2 plus lambda0, and this term has 2 lambda1 plus lambda0 and 2 lambda2. So I say I can just average those two. So it will be minus lambda11 plus lambda22 divided by 2 over 4 RT. And then I'm going to expand this again over 2 times 4 RT. This isn't k12, who let me do that? You let me get away with a fast one. This is a 2. k12 is just deltaG over RT. So this 2 is a square root. So this is the square root of k12. And I'll take these two twos out there, and I'm going to get the square root of k12 e to the minus lambda11 over 4 RT e to the minus lambda22 over 4 RT. This is a squared. I think I lost that square up there. That was this square. It's here, deltaG0 squared over 4 lambda12 RT. Now, if the driving force isn't too big, you can see that this will be the small term, and these will be the bigger terms. Of course, lambda, by the time you have an inner sphere, now the sphere is about 1 EV. And if the driving force isn't too big, that's going to be the small term. So often you just throw away that term for many reactions because it's not too large. But I've kept it in. So now we rewrite it again as the square root of, this has got to be, I'm going to rewrite it. Well, what I want to say is I'm going to say that new 2, 2, new 1, 1 is approximately equal to new 1, 2. And that k122a, k11a is approximately k12a. That, if they all sort of look the same, you would expect that the frequencies and stuff would all be so. You can actually leave them in and work it all out. But if you make this assumption, you get a k11a, a k22a, a new 1, 1, a new 2, 2, a k12, a new to the minus lambda 1, 1 over 4rt, a new to the minus lambda 2, 2 over 4rt. And there is this term also exp minus delta g0 squared over 4 lambda 1, 2rt. This, this, and this are just the rate constants for k1. So this now becomes k11, k22, k12 times r piece at the end. Now, as I said, if this is small, I could throw away this piece, and I just have a relationship between, a cross relationship between the self-exchange rates and the cross reaction. That's usually how it's used, because when you get to high driving force, the cross relationship usually doesn't work very well anyway. So it's usually used in that. Now, the cross relation is often written today, looks more complicated. So I'll write down how you'll often see it in books, or at least if you read, there's a big Marcus and Sutin review on electron transfers of, they write this all down. And then they write that k12 is equal to k11, k22, k12, f to the 1 half, w12 is 1, 2. And f is the log of equals 1 fourth log of k12 plus w12 minus w21 all over RT. This is squared divided by a log of k11, k22, all over, I'm going to call this 4 pi delta r squared times nu11 nu22 plus w11 plus w22 over RT. And I'll tell you what all of these mean. This is w11 minus w22. OK, you have that complicated expression. And the reason it's much more complicated than what we wrote down is that they worried about the fact of the driving force. We have a reaction and fluid solution that I haven't sort of said too much about. But anyway, if we take, how do I want to do this? Whereas, let's do it in terms of our ion again. Well, let's just do it in A and B. If I have A2 plus plus B3 plus coming together, A2 plus B3 plus going to A plus B4 plus coming apart to A plus plus B4 plus. Now, I haven't written it symmetrically anymore. As long as I write it symmetrically, I save a lot. This has, since these are charged species, there's a work term to bring them together. There's a work term that takes them apart that's going, if I always think I'm going together. So this is the work term for the products, this is the work term for the reactants. We know the overall free energy change is delta G0. And if this is, I'll call delta G0 prime, that has to be equal to W reactants plus WG0 prime minus WP. That you have to correct the free energy change in the electron transfer step for the fact that you might have an unsymmetric reaction. And the cross reaction tries to do that. Now in high ionic strength solutions, of course, the work terms tend to be small. It's like the double layer and the charging of the electrode and how the potential falls off quickly and high dielectric constant. Here, the energy, the work to bring charged species together in high ionic strength is not much. And so that these work terms aren't very large to begin with when you have high ionic strength. However, in low ionic strength, they're more important. But then the calculation to calculate the work term is usually not very good. The calculation that classically is done is the Debye-Huckel calculation. And that is approximately correct, but only approximately. And actually, they say Debye-Huckel should only be used at very dilute solutions, which is nowhere where we ever were. So it sort of leaves you in somewhat of a quandary of actually how to calculate those with any accuracy. OK. So now we've done the cross-reaction. Now we'll do, I guess, a few more things, and then we'll be done just about done the solution work. And on Thursday, we can actually do electron reactions. OK. All of what I've done is basically classical. The only quantum mechanical is I sort of tacked on to the frequency factor, a Landau-Zehner term that gave me a curve crossing thing. And I want to say a little bit about that. OK. The vibrations of the solvent, we say there of the order of 3 centimeters minus 1, something of that sort within a prior factor of 20 or 30. And the inner sphere vibrations are, say, 1,000 centimeters minus 1. OK. And kT is 200 centimeters minus 1. OK. So kT is in between them. For the solvent, that means that a lot of vibrational levels are filled. And the solvent really looks classical, precisely because kT is big. I mean, kT is really the ruler. If kT is big, you're in a classical world. And if kT is little here, you're in a quantum mechanical. But for the inner sphere vibrations, kT isn't large. And those ought to be treated quantum mechanically. OK. I'm not going to do it. I'll just point out that it can be done. And it's done by attacking the problem as a perturbation problem of transferring between electronic levels. And you use Fermi's golden rule, which is time-dependent perturbation theory. And it's pretty straightforward to do it. But you get formulas that then are more complicated. And in the end, I just don't think for here I'll do it. There is, though, one thing quantum mechanically I feel we should do. And I want to talk a little bit about the surfaces again. We've drawn our surfaces always like this. And they're called diabetic surfaces. They're called diabetic surfaces because they're classical surfaces. And in no way are they real quantum mechanical surfaces. I got them by calculating harmonic oscillators on them. And I'm saying that the surfaces cross and they don't interact. But quantum mechanically, you know that the surfaces are going to repel each other. They're not going to cross. And they're going to interact in some way. And so the actual true surfaces are something like this. I didn't draw them very well. Let's try that again. And these green surfaces are called the adiabatic surface. The adiabatic are the true surfaces. They're the actual surfaces that the real molecule moves whereas the diabetic are the fake surfaces. Now what we have been talking about is the reorganization parameter lambda. What's surprising to many is that when you calculate the difference between the surfaces on the adiabatic surfaces, letting them interact somewhat, you get it is also lambda. That what happens is when the surfaces move, the minimum move together and they split. And the splitting exactly compensates for the moving together. And the difference between the surfaces stays the same. It's sort of magic, but that's what happens. Now I guess now what one wants to do is say something about what HAB is. And the way we do that is I think this is the extent of the quantum mechanics. People who have a quantum phobia will have to sit through this. We're going to say these wave functions are psi A and psi B. And the green ones are going to be psi 1 and psi 2. And psi 1 is going to be, we're going to say, it's a two-state problem. So we only have these two states that we're going to. So we'll have a C A psi A plus C B psi B is one surface, is one wave function. This is the adiabatic. These are the diabetics. And the other adiabatic surface will be minus C B psi A plus C A psi B, where you switch the coefficients. And the coefficients are such that C A squared has to equal 1 and the, no, I'll get it right. Sorry, that doesn't work. C A plus C B squared equals 1. They're normalized. And when I talk about HAB, what I'm talking about is the wave function, how can I say? It's the wave function that is the integral of psi A star H, psi B. And then I integrate over all space. So it's the interaction of the diabatic surfaces. This splitting here is 2 HAB. The splitting of the surfaces, what causes the surfaces to split is that psi A and psi B are not solutions to the Schrodinger equation for the Hamiltonian you have. And so when you get, you can get a better solution by mixing them with arbitrary coefficients that you can then figure out. And this tells you about how the diabatic surfaces interact, OK? And this is going to be proportional to basically the integral of psi A star psi B, so that it's really controlled by the overlap of the wave functions. Now the wave function psi A is the wave function for the reactants, and psi B is the wave functions for the products. And if we're dealing with a one electron problem, what we're doing is we're taking the electron first on the reactant side, and then moving it from the ferrous on the reactant side to the ferric. And so if the nuclei are far apart, they aren't going to overlap. And if they're close together, you'll get much better overlap. And so this is going to have a distance dependence, OK? And so we're going to say that HAB is going to be a function of r, OK? So I think we'll stop there, and on Thursday we'll finally get to electrons.