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5.111 Principles of Chemical Science, Fall 2005 Transcript – Lecture 28
All right. We have a few more PowerPoint things before I am going to attempt using the board. Again, we are in the transition metal unit. And today we are going to introduce something called crystal field theory.
And this is in Chapter 16 in your book. Let me just tell you about two different theories. Again, chemistry is an experimental science. We collect data and then we try to come up with theories that explain the data, so the theories are not sort of 100%.
And some are more simple and some are more complicated to try to explain what we observe. And some of these theories, although they are pretty simple approximations of what is really going on, do a pretty good job of explaining the things that we are observing.
All right. There are two different kinds of theories that you will hear about with transition metals. You will hear about crystal field theory and ligand field theory. And so these were, again, developed to explain certain types of properties of coordination complexes of transition metals.
There is observed data. And then people tried to think about how you could rationalize what was observed.
The basic idea behind these theories is that if you have your metal ion with its given oxidation number and it is in the center of a coordination sphere, the energy levels of the d-orbitals for that particular metal are going to be altered when ligands are bound compared to sort of a free system.
The binding of ligands around that metal, where there are those d-orbitals associated with that metal, is going to have an effect and create special properties.
Crystal field theory, which is what we are going to be talking about, is based on an ionic description.
Basically you are going to be considering your ligands as negative point charges, a negative charge. And you are going to ask the question is that negative charge pointed toward a d-orbital? If so, that will be repulsive.
And that is going to have an effect. Ligand field theory includes covalent, as well as ionic description. It is a more powerful theory and it can describe things a little bit better, but we don't have enough time to cover it here.
If you are interested in transition metals, you will probably go on and take another chemistry course called 5.03 which is inorganic chemistry. At this point, we are just going to be doing the more simple theory which just considers the ionic descriptions.
But it does pretty well. You can explain some things with it, but you should be aware that you can do a little bit better with another theory. You should just be aware of that, that this is really the most simplistic kind of representation or theory to explain some of the properties of coordination complexes.
All right. All of these theories require you to think in 3-dimensions about d-orbitals. Because they are all about the interactions with ligands, either covalent or ionic interactions, but they are all about the interactions of ligands with those d-orbitals.
I mentioned last time that you need to know the shapes of the d-orbitals and you need to be able to draw them to some limited ability. We went through the names of the d-orbitals last time. Let's just review it now because we are going to be talking about this more today.
What is this guy up here? That is dz2. And here is our reference frame. z is up here, y along here, and x is coming out toward you. What about this one? Dx2 minus y2. Down here, what is that one? dxy.
The amplitudes are 45 degrees off the axes in the xy plane. Over here we have dyz, so now we are along z. And the last one is? xz over here. All right. Those are the five d-orbitals that we will be talking about a lot in the next few lectures.
All right. Let me just show you. This is really 3-dimensional, so let's take a look at the 3-dimensional image of these d-orbitals.
This is dz2. This isn't in your handout, obviously. It is a movie, but you can sort of get a little better sense of the 3-dimensions here.
All right. The movie first rotates the axes. I am not sure why it does that. And then it is going to rotate the d-orbitals. You can see the doughnut in the middle with the hole and that the maximum amplitudes are along the z axis here.
This is dz2. And the orbitals are right along the z axis, which is important. Now let's take a look at dx2 minus y2. The important point here is that the orbitals are directly on axis, and the axes are the x and the y. You can see how they are directly along those axes. All right. Here is the next one. Here we have dxy. And now the orbitals are not along the axes. They are 45 degrees in between, so they are not along, they are in between, which is the key point for this d-orbital in terms of what we are going to be talking about.
Now we have some amplitude along the z axis. We have z and x. Again, the orbitals are 45 degrees off the axes. And we have one more. Now we have dyz, again 45 degrees off the axes. All right. Now you should have these sort of orbitals floating around inside your head.
And, as we talk about bringing ligands in toward the orbitals, you can think about how those ligands are coming in and what orbitals they are going to be near. Again, crystal field theory is just considering the ionic interactions.
And it is really considering the ligands as negative point charges. And so negative point charges are going to be repulsive with the d-orbitals. dorbitals don't like it when you have these negatively charged ligand coming in and being near. It will be repulsive near the d-orbital.
This is just another way to sort of draw that. And we have our octahedral frame. Here we are talking about a octahedral geometry.
And we can think about how ligands are going to be affecting the d-orbital in these various places.
If you had ligands on all of these positions in an octahedral arrangement around the metal. And you can think of these ligands as these negative point charges. And so this is just another depiction.
Here are the d-orbitals again. It is a slightly different xyz frame, but not too much different. And so here are our orbitals. And these little green dots represent the point charges. What we will be thinking about today is the effect of ligands being close to the d-orbitals.
You see here there is a ligand that is quite close up here and one down here. In this picture, these negative point charged ligands are also really close to the orbitals. Here they are not really as close.
They are sort of off axis. And that becomes really important. All right. Let's now think about the different dorbitals and the ligands when the ligands are arranged in an octahedral geometry. In this case, we have these ligands, these negative point charges coming in on all the different axes. And so, if we think about it, these are close. And I mentioned that in the last slide. A negative point charge along z is going to be very close to dz2 and it will be repulsive. For dx2 minus y2 the negative point charges along y and along x are also going to be very close to these d-orbitals.
And that will, again, be repulsive. We are going to have a large repulsive interaction for these two orbitals with an octahedral geometry.
These two d-orbitals are destabilized, and they are destabilized by the same amount. When they are destabilized by the same amount, what is that called? The energy of the orbitals are the same, what is that called? Yeah, they are degenerate. And so these two orbitals then will be more destabilized by those ligand negative point charges than the other sets of orbitals.
Because they are off axis and the ligands are on axis. The negative point charges will be farther away from these guys. Let's take a look at that as well. Here we have our three other d-orbitals. And if ligands are coming in as negative point charges on axis, they are not going to be directly pointing at any of these orbitals.
Because they are 45 degrees off axis. Ligands are on axis. It is not directly pointing toward any of them. These three orbitals then are stabilized relative to the other sets of orbitals we talked about and they are stabilized by the same amount.
Those three are also degenerate. All right. To summarize then, you will have a large repulsion for dx2 minus y2 and dz2. And they will be more destabilized compared to the other orbitals. There will be a smaller repulsion for these three.
And they will be stabilized compared to these sets of orbitals. And so here you are really thinking about comparing the d-orbitals to a case where the ligands are sort of all around.
We will talk about that in a minute. If you localize your ligands on axis, how is this affected? And this just reminds you and shows you the pictures again. Over here they are pointing much more directly toward and here they are away.
If it helps you, depending on how you like to think about it, I had these orbitals in your handout on the first page, but just to show them again.
You can think about how close they are. See, that ligand is very close, not nearly as close here.
It is all about the geometry around the metal, the geometry of the d-orbitals, the geometry of the ligands. Now I am going to try writing on the board. Here we are going to think about how you would draw a diagram to represent the energy level when you consider crystal field theory.
You would have what's called a crystal field splitting diagram. Here energy would be going up. And we have our case. We have a hypothetical case where we have five d-orbitals. And so this is the average energy of the d-orbitals. Average energy of d-orbitals with the ligands bound in what is known as a spherical crystal field.
This does not exist. It is hypothetical. If the ligands were everywhere in a sphere around your metal, this is our hypothetical spherical crystal field, and you imagine the metal is the inside. And on the outside you have all the ligands.
The ligands are everywhere. They are equally distributed everywhere. And if they are equally distributed everywhere, all the d-orbitals would have the same energy. It is pointing toward everything equally, so it is spherical.
It is completely surrounding the metal. This is a hypothetical case because that does not happen. Ligands are bound in discrete positions, but if it did these would be of the same energy. Let me try to get his a little more flat.
All right. They are all the same energy. Well, what happens then when you bring your ligands into their discrete positions, when you bring your ligands in, in this case into octahedral geometry then things change.
Then you have a destabilization. You have a destabilization of your dx2 minus y2 and dz2 orbitals. They are going to go up in energy. And you will have a stabilization of your dxy, dyz and dxz orbitals.
They are going to go down in energy. In this hypothetical case, the d-orbitals have equal energy. Ligands are everywhere.
They are pointing at all the d-orbitals equally, but in an octahedral case. Then two orbitals are destabilized, go up in energy because those ligands now are pointing right at those orbitals. They are not pointing equally everywhere. They are pointing at these orbitals so they go up in energy.
These orbitals are in between the axes so they are not affected as much and get stabilized. These orbitals have a name, eg.
And these orbitals have a name, t2g for the octahedral case. And the difference in energy between these sets of orbitals is this sign here, sort of this delta zero sign, actually O. And that is for octahedral. The O is for octahedral. And it is the crystal field splitting energy. It is how much the ligands split the field. This theoretical case up here, there is no splitting going on.
You bring in d-ligands and they cause a splitting. They cause some orbitals to go up in energy and some to go down.
And the distance between them, the degree of splitting is this octahedral crystal field splitting energy.
Now, the overall energy of the system is maintained. If these go up and these go down, to get everything to be equal, let's see if I drew that right.
These should go up. I am trying to draw this straight from this line. These are going to go up in energy by plus two-fifths, the octahedral crystal field splitting energy. And these three orbitals are going to go down in energy, will be stabilized by minus two-fifths, the octahedral crystal field splitting energy.
Two orbitals are going up, so it is by three-fifths. Three orbitals are going down, so it is by two-fifths. Because it is going to have to equal out. You are not sort of creating a different system here.
The energy is going to be conserved. The overall energy is maintained. Two orbitals up, two-thirds. Three orbitals down, two-fifths. OK. That is really a simple idea.
Instead of the theoretical where all the ligands are everywhere, when you put them in octahedral geometry it affects some of the d-orbitals differently than others.
Some orbitals go up in energy. Some go down. The overall energy is maintained. Let's look at an example now and figure out how to put electrons into this system. Let's look at a case where we have chromium, (NH3)3 and BR3.
First we need to figure out the oxidation number of chromium. Having everything in the bracket here means that you have six ligands total, three NH3 ligands, three BR ligands or bromide ligands. And those are all in one bracket. There is no number in the corner of the bracket, so that means the overall charge in this system has to be equal to zero.
What will be the oxidation number of bromine? It should be minus one. Three minus ones. What about NH3 as a unit? Zero. Three times zero. And what does that leave you with for the oxidation number of chromium? Plus three.
We have plus three. Now we can figure out the d-electron count. We need to look up the group number. And the group number is what? Can you see it up there? Most people seem to be saying six. That is right. Six minus three is three. So this is a d3 system. That means we are going to put in three electrons. Where would you guess I am going to put those electrons? Down here or up here? Down. What if I do this? Would that be good? No.
We are coming back to some of the roles that you learned earlier in the semester. First you want to put them in the lowest energy orbital. You don't want to ever have one with two parallel spins because that would give you the same four quantum numbers, so that is not good.
And you also by choice, if they are the same energy, put them in singly before you start to pair. Those are all things that you talked about earlier in the semester. We put our three electrons in.
One to each orbital all the spins parallel. We have Aufbau principle, Pauli exclusion and also Hund's rule all at play here with how we put in the electrons. Now we can talk about the dn electron configuration. Maybe I can write over here. That is just a way of expressing how many electrons are down here.
dn electron configuration. That would just be equal to t2g3. You list the name of these sets of orbitals. And that is really why these have labels, because it makes your life easier instead of listing all the orbitals each time.
There are three in this lower energy level. Now we can also talk about crystal field stabilization energy. Crystal field stabilization energy. Don't get this confused with crystal field splitting energy.
This is crystal field stabilization energy, and it is abbreviated CFSE. And that is really just thinking about how much you stabilize by this splitting. If you had this hypothetical spherical crystal field, your electrons would be in orbitals with this energy, but they are not.
After the ligands come in, in the octahedral geometry, the electrons are in the t2g orbitals. They are stabilized.
This is lower in energy than where they would have been before. And so the crystal field stabilization energy is a way of indicating whether you have overall stabilized the system or not by this splitting.
The way that you would represent this is you would just look at where the electrons are and how much they are stabilized, again, three orbitals down minus two-fifths, two orbitals up plus three-fifths. In this case, there are three electrons that are stabilized by minus two-fifths of the octahedral crystal field splitting energy, or you would have a stabilization of minus six-fifths times the octahedral crystal field splitting energy.
That would be the answer to that part. OK. Let's look at another example. I will write this in between here. Here we have manganese with six water ligands in brackets and Cl3 on the outside. What is the charge of just this coordination complex?
It has six ligands to this inside the brackets? What is the charge of just that coordination complex overall? It would be plus three.
And you know that because these are three negatively charged counter ions. When the whole thing is neutral with three Cls it means that had to have been plus three. What is the charge of water?
Zero. What has to be the charge on the metal? Plus three. Now we can figure out the delectron count. What group are we in? Seven minus three is four. We are a d4 system. All right. Now we have four electrons to put in, and we are going to have a choice.
We are going to have a choice as to whether we put the fourth electron up here or keep the fourth electron down here and pair.
And so both of those are possibilities. And it depends on the strength of the crystal field. Let's look at those diagrams. There are two possibilities.
And we have, again, dx2 minus y2 and dz2 up here we have our t2g system, our eg. That is one drawing. Or, we might have a case where we have something more like this where there is a much larger separation between the sets of orbitals.
Between the eg and the t2g. And so here is our crystal field. It is small here. And here is another crystal field splitting energy, and this one is large.
You have a choice. We have to put in four electrons. And depending on whether it is a small splitting or a large splitting you are going to put them in differently.
In this case, we have a crystal field splitting energy that is less than what we call the pairing energy. The pairing energy is how much energy it takes to pair up the electrons.
They prefer to have their own orbital. They would prefer not to be paired. It takes a certain amount of energy to pair them. In this case, because the splitting is small, this set of orbitals up here are really not that much higher in energy than these were. It is not that hard to put one up here. And it takes more energy, in this case, to pair them up than it does to put one in the higher level.
This one would have a drawing like this. You have four electrons you put in. One of them goes over there. In this case, we have a really big difference. It would take a lot of energy to put an electron way up here. In this case, the splitting energy is greater than the pairing energy.
This is really a lot higher up. Here it is small. It makes not much difference. Here the energy is large. In this case, you would want to put your four electrons all down in the lower level. Let's just take a look at some of those rules and say them again.
Here when the splitting is small, this is called a weak field. The ligands here will be weak field ligands. And there will be strong field ligands. And we are going to talk about that next Monday, how you will learn certain ligands and whether they are weak field or strong field.
If they are weak, oh boy, they cannot really split that energy very far. They do not have a lot of strength. They cannot do a huge splitting. They just do a little tiny amount of splitting. They are pretty weak ligands.
And, in this case, you would have a strong field where you would have a large splitting.
Those ligands come in. And, boy, they take those d-orbital energy levels and split them really far apart. Those are strong field ligands. They create a strong field. And this is large. All right.
If we look at some of these rules then, if our pairing energy is greater than that energy difference that means you have a weak field. It is not much different.
It is a weak field not much splitting. The splitting is small. And then what you are going to want to do is put all your electrons in singly to the fullest extent possible. You use both t2g and eg. You put them all singly until you are done before you pair.
On the other hand, if you have a strong field, that is going to have a really large splitting. Big energy difference between t2g and eg sets of orbitals.
Then that splitting energy is going to be greater than the pairing energy. It is more favorable to pair than to make that big jump up to this higher energy level up here. In that case, what you want to do is you want to pair them all up in t2g. And only after this is completely full and you cannot put any more electrons in there do you want to put an electron way up there in the eg sets of orbitals. What this means then is in this case you are going to have the maximum amount of unpaired electrons because you are going to try to keep them unpaired for as long as you possibly can.
And so this is referred to as a high spin system. In this case, you are going to have the minimum number of unpaired. You are trying to pair them up down here before you put one up here.
This is a low spin system, minimum number of unpaired electrons that are possible. These are our two diagrams that you will see. And one question on the problem set has you drawing these diagrams for these different conditions. Now, this particular one that we are talking about, this particular complex is high spin.
