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0:01
- Crystal field theory gonna be the topic of this lesson. My name is Chad and welcome to "Chad's Prep", where my goal is to take the stress out of learning science. Now, in addition to high school and college science prep, we also do MCAT, DAT, and OAT prep. I'll be sure to leave a link in the description below for where you can find those courses. Now, this lesson is part of my new general chemistry playlist, which is almost complete. Will be one more chapter after this
0:21
when we shall finish up next week. But if you like to be notified every time I post a new lesson, or when I get started on my next playlist, then subscribe to the channel, click the bell notification. All right, before we can get going head long into crystal field theory here, we've just gotta do a quick refresher on the electron configurations of the transition metal cations. And so a couple things you need to remember about some exceptions,
0:42
and then when you remove electrons to make the cations. So super important for this chapter, because how many d electrons these transition metals have is gonna be kind of relevant both in this lesson, as well as the next one. So let's take a look here. We'll start with scandium here. And if you wanna do one of these cations, start with just the plain old neutral element first, and we've got scandium right here, atomic number 21. And scandium is gonna be argon. And then you might recall 4s2 3d1.
1:08
So plain old scandium
1:11
would be argon 4s2 3d1.
1:15
So that's scandium but we've got scandium 3 plus here. And so we've gotta remove three electrons. And so you just move the last three. And so you gotta remember though that you move the s's before the d's, but in this case, that wouldn't really make a difference. So we'll move those s's before the d. And in this case, then that's all three of theses gone. And so scandium 3 plus is just isoelectronic with argon here. But the big thing I wanna point out here is that scandium 3 plus
1:40
would be a d zero transition metal here, has zero d electrons, and that'll be important for some properties we look at in the next lesson. All right, moving on to Fe 3 plus. So going to iron here and iron again, argon, and then 4s2 and then 3d one, two, three, four, five, six.
1:59
And so plain old iron would be argon 4s2 3d6,
2:06
and we've gotta remove three electrons. And again, the thing you need to remember is that you pull out the four s's before you pull out the three d's. And so if you come up with argon 4s2 3d3, you've done it wrong. So you've gotta pull out those four s's first, so that's two electrons gone, and then you've gotta pull out one of the d's, so 3d5. And we'll just write this all over again as argon 3d5.
2:28
And the big thing to take away here that we needed to figure out is that it's got five d electrons in Fe 3 plus. And so the different complex ions and coordination compounds that Fe 3 plus is involved in, there are five d electrons, and that's gonna be relevant here. Moving on to manganese 2 plus. And so manganese would take a look, is argon 4s2 3d one, two, three, four, five.
2:51
Just more of the same here, so 4s2 3d5. We just need to lose two electrons, and that's gonna be the four s's. And so we're just gonna be left with argon 3d5. And I'll take the time to write that over again.
3:03
And once again, five d electrons is the key. So next, we'll take a look at chromium and copper here and their ions. And you gotta remember that chromium and copper are exceptions on the periodic table and as well as some of the elements below them. So turns out chromium and molybdenum are gonna be exceptions here, but not tungsten. And then copper, silver, and gold will be exceptions as well. So if we take a look and just ignore the fact that they're exceptions real quick,
3:24
so we'd have argon in the case of chromium and then 4s2 3d one, two, three, four. That's what we think it would be if it wasn't an exception.
3:35
But it turns out that with chromium's column, the idea is that half-filled sub-shells are more stable. And so if he steals one from the s and puts it up into the d, these will both be half filled, and that's exactly what happens. And so instead of 4s2 3d4, it's gonna be 4s1 3d5.
3:52
And chromium and then molybdenum right below him does the same thing a row down. All right, so that's where we gotta start. Now, we've gotta remove three electrons. And you gotta remember remove the 4s before the 3d. So the first one we take out is that 4s, gone. Gotta remove two more and so we're gonna end up with 3d3.
4:08
And once again, I'm gonna take the time to write this out nice and neat. So argon 3d3, and we find out that chromium 3 plus, and the complexes that it contains have three d electrons. All right, moving on to copper plus here. And again, we'll start off with argon.
4:26
And again, if we didn't know that copper was an exception, we'd say argon, and then 4s2 and then 3d one, two, three, four, five six seven eight nine. So argon 4s2 3d9. And again copper knows something that maybe we didn't initially, and that completely full or half filled sub-shells are more stable. And so while the 4s is full, the 3d is not half full or completely full, but if it steals one from that 4s puts it up into that 3d, we'd be 4s1 3d10,
4:55
and now the 4s it's not full, but it is half full. And the 3d is full as well and this is more stable lower energy. And this is the electronic configuration for copper. Now, we've got copper plus 1 though, so we've gotta remove one electron. You've gotta remember definitely that you remove the 4s before the 3d. So it's just gonna be argon 3d10
5:15
for copper plus 1. And again, big thing to note is that it's got 10 d electrons. And now that we've done this review, we are ready to start diving into crystal field theory. So for crystal field theory, we gotta remind ourselves what the d orbitals look like. And we're gonna take a look at the five different three d orbitals. And in shell number three here, the five d orbitals are named dxy, dyz, dxz, dx squared minus y squared, and dz squared. Now, you might recall what these look like.
5:41
The first four here look like a four leaf clover, and I'll put the image up on the other side of the board here so like a four leaf clover and dxy. So there's four lobes in that four leaf clover and they lie in the x-y plane. But important here is that they don't lie on the x and y-axis, but lie in between the x and y-axis. Similarly, dyz, the four leafs of that four leaf clover, if you will, four lobes that are orbital lie in the y-z plane, but again,
6:06
in between the y and z-axis, not on the y and z-axis. Dxz lies in the x-z plane, but again, in between the x and z-axis, not on the axis. And so these three, the big important point here, is that the lobes of electron density for these d orbitals lie in between the axis. Super important we understand that. Now, dx squared minus y squared, four lobes again in the x-y plane but right on the x and y-axis. And then dz squared, most of the electron density is on the z-axis.
6:37
And so we get really a differentiation here between two types of orbitals, these three that lie in between the axis and then these two where the electron density is mostly on the axis. And that's important because when the ligands come in to bind, like in an octahedral complex, the ligands come in and bind right on the x-axis, right on the y-axis, and right on the z-axis from all six sides. Whereas in a tetrahedral complex, they come in and bind right in between the axis,
7:03
and there's gonna be a profound difference. And so it turns out when these areas of electron density get in the way of where the ligands are coming in, they get shifted to higher energy. And so in an octahedral complex, these two are on the axes and the ligands are coming in on the axes, and so these two are gonna get shifted to higher energy. Whereas in a tetrahedral complex, again, in a tetrahedral complex, the four ligands don't bind on the axes,
7:26
they bind in between the axes 109.5 degrees apart. And so since these are the ones that are in between the axes, it's these three that shift to higher energy instead. And so we're gonna get, instead of having five degenerate d orbitals, so once we bind ligands, we're gonna get some splitting into at least two different levels or energy levels here. And so let's take a look at the octahedral first.
8:00
And so we're gonna split into two levels here. And again, xy, yz, xz are gonna be the lower set and then x squared minus y squared and z squared the higher set, again, because these two are on the axis and the ligands are coming in on the axis, so the ones that get in the way shift to higher energy. All right, so this difference in energy now is called the crystal field splitting energy, sometimes abbreviated either delta or delta q. And again, it's the crystal field splitting energy.
8:29
So with this splitting between the once degenerate d orbitals into two levels now. And some people call it the crystal field stabilization energy as well. Just note, you'll see both used pretty commonly here. So that CFSE for short or more commonly, either delta or delta q, which is what I'm gonna use here for the symbol. All right, so we've got that difference in energy now. And that difference in energy might be relatively small or in an octahedral complex,
8:57
it might actually be relatively large instead.
9:03
And depending on if it's relatively small or relatively large, it's gonna be a profound difference here. So we call this the octahedral splitting when you see this pattern. Again, we'll see the tetrahedral splitting and the square planar splitting will be a little bit different when we get there. But the key is how we fill in electrons in the d orbitals might be a little bit different now. So if you recall, when we had Fe 3 plus a little bit ago, electron configuration was argon 3d5.
9:31
And so we've gotta fill in five d electrons for say Fe 3 plus. Well, in our original where we had just the degenerate, which recall means all equal energy orbitals, we would've just filled in five electrons and been like one, two, three, four, five, and life would've been good.
9:49
But now we have a problem with two different energy levels. And so we learn that when you've got different energy levels, you fill up the lower energy ones first and then move on to the higher energy. Well, turns out it's not gonna be so simple. Now, if this difference in energy is large, then that's exactly what we'll do. And so we'd fill in one, two, three, and these aren't full yet, so four, five. And so we'd fill in all the lower orbitals here in the d orbitals,
10:15
before we fill in the higher energy ones. But that's assuming that this gap is big. If the gap is small, we see it actually fills in a little bit differently. We'd fill in one, two, three, but then with a small gap, we actually just go four and five. And it's reminiscent of what we did when they were all degenerate. Well, the idea is this, when you've got two electrons in the same orbital, they repel each other. And to put them together in that orbital,
10:42
it's gonna cost energy and we call it pairing energy or electron pairing energy. And so the idea is this, if your crystal field splitting energy is smaller than that pairing energy, well, then it would be easier to just put an electron up here than it would be to pair them up. And that's what we're doing in this top set, when you've got this small crystal field splitting energy. But if that pairing energy is smaller than the crystal field splitting energy,
11:08
well, then it costs less energy to pair them up and that's what we're doing down here. And so all of a sudden now with octahedral, depending on if we have a small or a large delta q, crystal field splitting energy, so how we fill in the electrons is gonna be a little bit different. Now, we are most likely gonna end up with way more unpaired electrons if we fill them in this way than this way. And as a result, this is often referred to as high spin.
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Whereas here, we're not as likely to get as many unpaired electrons so since we start pairing them up sooner here. And so as a result, filling 'em in this way
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is called low spin.
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Now it turns out it's often the ligands and the metal itself has a little role in this, like as well, but how tightly the ligands bind is gonna determine how big this crystal filled splitting energy. And again, this difference is due to the fact that the ligands are coming in on the axis and the two orbitals that are on the axis get shifted to higher energy, while the more tightly the ligands bind, the bigger this splitting energy is going to be. And so as a result, for every metal,
12:12
it's a little bit different in how we identify these ligands and stuff like this. But the way this works is that if you have a ligand that causes a large splitting energy, so large enough to be bigger than the pairing energy so that it fills in this low spin fashion, we will call that a strong field ligand.
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Whereas if the ligand doesn't bind so tightly, so that the difference in energy here, the splitting energy, is smaller than the pairing energy, then we'd call that a weak field ligand.
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And it turns out there's what we call a spectrochemical series that kind of shows how a ligand is gonna affect this difference in energy. And it turns out, you know, like the one I've given you on your handout here on page 92 for the next lesson, shows cyanide binds really tightly and forms a very large splitting, whereas like SCN minus doesn't bind so tightly and gives a relatively small splitting. Now, what it doesn't show on this spectrochemical series though, is, you know,
13:15
whether they're weak field or strong field. Well, 'cause that also depends on the metal ion as well, 'cause with some metal ions you get tighter binding of all ligands and some you get looser binding of all ligands and stuff like this. And so what is actually gonna be weak field versus strong field really depends on the metal as well. Whereas the cutoff between what's weak field and strong field will be different for every metal ion.
13:34
All right, so can't really tell you, you know, if a ligand is gonna be weak field or strong field, we just have tendencies. Cyanide is more... You know, has a stronger tendency to be strong field. Something like SCN minus would have a greater tendency to be more on the weak field side. But again, it will depend on which metal ion we're talking about. Cool. However though, when we're talking about octahedral complexes, what you're supposed to take away is if I say high spin or weak field,
13:59
you know how to fill in the electrons. And in the high spin, you're gonna go up high before you start pairing anybody up. Whereas low spin or strong field, if I say you've got this kind of a complex, then you're gonna fill up everything down low before you put any electrons up high. So you gotta know how to fill in these electrons. And so if you notice, like in this case for Fe 3 plus, in a low spin complex, there's only one unpaired electron, whereas in a high spin complex,
14:26
there are five unpaired electrons. And it turns out the number of unpaired electrons is gonna have an effect on some of the actual physical properties of this and stuff like that. And we'll find out that in the next lesson that like Fe 3 plus high spin complexes are more paramagnetic, is the word we'll use, than Fe 3 plus low spin, so just due to the presence of a greater number of unpaired electrons. So I'm alluding to that. I just want you to know that this will have some impact,
14:53
but we'll cover that in the next lesson. Cool, now, one thing to note, in order to tell the difference between these, the way we really tell the difference is by looking at that number of unpaired electrons. So let's just start back from the beginning and start filling in electrons,
15:10
assuming we start off with a complex that only had one d electron and work our way up. So in this case, what's that way of zero d electrons? Can I tell the difference? No, 'cause there's no electrons in the d orbitals in either case. Now, if I fill in one d electron, whether it be strong field, low spin or weak field, high spin, I'm gonna get one unpaired electron and I'm largely not gonna be able to tell the difference. So fill in two electrons,
15:33
and now I've got two unpaired electrons in either case and I still can't tell the difference. Fill in three electrons
15:40
and I still can't tell the difference. They both have three unpaired electrons. It's not until you get to the fourth d electron that you can tell the difference. So here with the weak field high spin, I'll put it up high. And with the low spin, I'll pair it up down low. And now all of a sudden I can tell the difference. And so notice with one, two, or three d electrons, can't tell the difference. But with four d electrons, whether I get two unpaired or four unpaired,
16:02
that's a difference I can tell. All right, let's fill in a fifth electron here. So up high here, we'll put one more and then one here. And here I've got one unpaired electron, here I've got five, I can definitely still tell the difference. Go to six electrons, now I'm ready to pair 'em up and it's easier to pair 'em up down low, the lower energy set here. So we pair those up before we pair the upper ones up, and then one here as well. And again, I've got no unpaired electrons.
16:26
I've got four unpaired electrons, I can tell the difference. Go on to seven electrons. There's number seven, there's number seven. I have one impaired electron here, three up top. I can still tell the difference. And finally go to eight electrons and notice once again, they look exactly the same. They both have two impaired electrons, I can't tell the difference. Go to nine and I still can't tell the difference. Go to 10 and I still can't tell the difference.
16:50
And so what I wanted to indicate here is that from four to seven d electrons, that's when you're gonna have distinct high spin and low spin complexes. If you have zero, one, two, three, eight, nine, or 10 d electrons, there's no distinct high spin and low spin complexes. It only exists for central metal ions that have four to seven d electrons.
17:14
Okay, so that's octahedral complexes. Let's take a look at tetrahedral and square planar. All right, so tetrahedral, we said we got kind of the exact opposite pattern for the splitting of the d orbitals here. And the tetrahedral shape, the ligands come in and bind in between the axis. And so now dxy, dyz, and dxz, which are in between the axis, those are the ones that now shift to higher energy, and then x squared minus y squared and z squared,
17:37
which are on the axis, so are gonna be lower in energy. And so it turns out for tetrahedral complex, this delta q here, this crystal field splitting energy, is always going to be smaller than the pairing energy. So it turns out that all tetrahedral complexes, every last one of 'em, they are all high spin.
17:58
There's no high spin and low spin when it comes to tetrahedral complexes. That's only true for octahedral. So all high spin, which means if you're filling in electrons, there's where your first two go, and then you're gonna go up high before ever pairing anything in. And so it turns out, you know, you'd be pairing 'em in or filling 'em in similar to how you would if they were just degenerate. Since that again, crystal field splitting energy is relatively small.
18:20
And it turns out if you took it an advanced inorganic class, you'd find out that for a tetrahedral complex, the value of the crystal field splitting energy is a little less than half of what it would be in an octahedral complex and all this stuff. So the numbers aren't super important here, but you should know that tetrahedral complexes are always high spin. Okay, so the splitting pattern for square planar is just unique and it's not the most important thing in the world, and these are actually
18:44
not the most common things in the world. But it turns out that square planar complexes only exist for metal ions that have eight d electrons like nickel 2 plus or platinum 2 plus. And again, even then, sometimes those end up being tetrahedral instead of square planar. There's not an easy way for you to tell. So you've kind of gotta be told for something with a coordination number four, whether it's tetrahedral or square planar. However, so if you're d8,
19:06
there's no guarantee that you're square planar. But if you're square planar, there is a guarantee that you're gonna be d8, have eight d electrons. And they fill in just like this. And so the key here is you've got one empty d orbital. And so it turns out when the ligands come into bind, they're gonna be using empty orbitals to put their electrons in when we do the Lewis acid, Lewis base thing. And so for a square planar, one of the ligands is gonna put its electrons.
19:29
It can put it one of the d orbitals. Well, where is it gonna put the others? Especially if we've got four ligands, I need three more pairs of electrons for a total of four pairs. Well, we'd have to go up to the next shell and use the s and p orbitals. And so we're gonna use a d, and then we're gonna use the s, and then we'd use a couple of the p's. But if you recall, same thing we learned back in chapter 8, 9 in terms of hybridization, with hybrid orbitals, we get better bond angles
19:55
when we start using these hybrid orbitals. And the same thing is gonna happen with these coordination complexes. So we're not actually gonna use just the d orbital, and the s orbital, and the two p orbitals, we're gonna mix them all together. And if you mix one d, one s, and two p orbitals, you mix those four orbitals together, it creates four hybrid orbitals, and you just name them based on what you mixed. And so in this case, it would end up being dsp2.
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And that is why the hybridization of a square planar complex ends up being dsp2. You just have one empty d orbital for every one of these square planar complexes, and then you've gotta use the s and the p orbitals from the next row higher to make those hybrids. Let's go back and take a look at what it would look like for an octahedral in terms of hybridization. 'Cause if you recall, there's two different hybridization for octahedral. Okay, so here we've got the splitting patterns
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for both high spin and low spin octahedral complexes. And let's say we're dealing with complexes with Fe 2 plus involved.
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So an Fe 2 plus is argon 3d6.
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So Fe is argon 4s2 3d6. Well, we lose the two 4s's to get Fe 2 plus. So it's 3d6. And so in this case, if we're high spin, it's one, two, three, four, five, six. And if it's low spin, it's one, two, three, four, five, six.
21:15
All right, now we need some empty orbitals for the ligands. Well, in the case of the low spin, the lowest energy empty orbitals we have, well, the first two are gonna be these two d orbitals, and then we'd have the s orbital of the next shell, and then the p orbitals of the next shell. Well, in an octahedral complex where we're binding six ligands, we have six ligands coming in to donate their pairs of electrons so we need six orbitals. Well, we don't use the original atomic orbitals,
21:39
we're gonna mix them all together and make hybrids again. And you just take the first six available lowest energy orbitals that are empty. And so in this case, the first two are d's, and then an s, and then three p's. And so overall, the hybridization ends up being
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d2sp3 hybridized. Whereas in the case of the high spin, we don't have any empty d orbitals. And so the lowest energy empty orbital you have is the s of the next shell, and then the three p's of the next shell. And you still need more after that so then you have to use the d's from the next shell. And so the lowest energy one you had was an s, and then the three p's, and then the two d's. And that's why there's two different hybridizations for octahedrals, d2sp3,
22:23
if you're using the d orbitals from the current shell, if you will, or it's sp3d2, if it's the d orbitals from the next shell up. In fact, all the orbitals end up being from the next shell higher from where your d electrons are. Cool, so that's why there's two different hybridizations for octahedral. Cool, now, in the next lesson, we're gonna take a look and see how crystal field theory allows us to explain a couple of different phenomenon we see with coordination compounds,
22:46
but that of the most of 'em end up being brightly colored, and they have some interesting magnetic properties that come down to what we call paramagnetism for quite a few complexes. Now, if you found this lesson helpful, a like and a comment let me know, pretty much the best things you can do to support the channel. And odds are you're at the end of gen chem II here studying for your final exams. And if that's the case, then you might take a look at my gen chem master course.
23:09
In addition to the normal part of the course, there are final exam rapid reviews, where I review all of gen chem I in about six hours and all of gen chem II in about five hours, and then include practice final exams to go with it. I'll leave a link in the description for where you can find that course. A free trial is available. Happy studying.
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