Understanding Crystal Field Theory

Crystal field theory is going to 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. 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. There will be one more chapter after this one, which I'll finish up next week. But if you'd 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 headlong into crystal field theory here, I've just got to 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, 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 going to 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 want to do one of these cations, start with just a plain old neutral element first, and we've got scandium right here, atomic number 21, and scandium's gonna be argon, and then you might recall 4s2 3d1. So plain old scandium would be argon 4s2 3d1. So that's scandium, but we've got scandium 3-plus here, and so we've got to remove three electrons, and so you just remove the last three, and so... You gotta remember though that you remove 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 these, gone. And so scandium 3 plus is just isoelectronic with argon here. But the big thing I want to point out here is that scandium 3 plus would be a D0 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 Fe3+. So go into iron here, and iron again, argon, and then 4s2, and then 3d1, 2, 3, 4, 5, 6. And so plain old iron would be argon, 4s2, 3d6. And we've got to remove three electrons. And again, the thing you need to remember is that you pull out the 4s's before you pull out the 3d's. And so if you come up with argon, 4s2, 3d3, you've done it wrong. So you've got to pull out that... those 4s's first, so that's two electrons gone, and then you've got to pull out one of the d's, so 3d5. And we'll just write this all over again as argon 3d5. And the big thing to take away here that we needed to figure out is that it's got five d electrons in Fe3+, and so the different complex ions and coordination compounds that Fe3 is involved in, there are five d electrons, and that's going to be relevant here. Moving on to manganese 2+, and so manganese, we take a look, is argon 4s2 3d12345. Just more of the same here. So 4s2, 3d5. We just need to lose two electrons and that's going to be the 4s's. And so we're just going to be left with argon 3d5 and I'll take the time to write that over again. And once again, 5d 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 going to 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, so we'd have argon in the case of chromium, and then 4s2, 3d1, 2, 3, 4, that's what we think it would be if it wasn't an exception. So, but it turns out that with chromium's column, the idea is that half-filled subshells 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 going to be 4s1 3d5. And chromium and then molybdenum right below him does the same thing a row down. Alright, so that's where we got to start. Now we've got to remove three electrons. And you got to remember, remove the 4s before the 3d. So the first one we take out is that 4s, gone. Got to remove two more, and so we're going to end up with 3d3. And once again, I'm going to 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 3d electrons. All right, moving on to copper plus here, and again we'll start off with argon. And again if we didn't know that copper was an exception we'd say argon and then 4s2 and then 3d1, 2, 3, 4, 5, 6, 7, 8, 9. So argon, 4s2, 3d9, and again copper knows something that maybe we didn't initially and that completely full or half-filled subshells 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, would be 4s1, 3d10, 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 one though, so we've got to remove one electron, and you've got to remember definitely that you remove the 4s before the 3d. So it's just going to be argon 3d10. For copper plus one, 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've got to remind ourselves what the d orbitals look like, and we're going to take a look at the five different 3d 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. 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 in dxy. So it's four lobes in that four-leaf clover, and they lie in the xy plane. But important here is that they don't lie on the x and y axes, but lie in between the x and y axes. Similarly, dyz, four leaf, the four leafs of that four-leaf clover if you will, four lobes that are orbital. lie in the yz plane, but again in between the y and z axes, not on the y and z axes. dxz lies in the xz plane, but again in between the x and z axes, not on the axes. And so these three, the big important point here, is that the lobes of electron density for these d orbitals lie in between the axes. Super important we understand that. Now dx squared minus y squared, four lobes again, in the xy plane but right on the x and y axes. And then dz squared most of the electron density is on the z-axis. And so we get really a differentiation here between two types of orbitals. These three that lie in between the axes, and then these two where the electron density is mostly on the axes. 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 axes. So, 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, 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. So we're going to split into two levels here and again x, y, y, z, x, z are going to be the lower set and then x squared minus y squared and z squared the higher set again because these two are on the axes and the ligands are coming in on the axes so the ones that get in the way shift to higher energy. Alright, so this difference in energy now is called the crystal field splitting energy sometimes abbreviated either delta or delta q so and again it's the crystal field splitting energy 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's CFSC for short, or more commonly either delta or delta Q, which is what I'm going to 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, it might actually be relatively large instead. And depending on if it's relatively small or relatively large, it's going to be a profound difference here. Now, 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 Fe3+, A little bit ago, electron configuration was argon 3d5. And so we've got to fill in 5d electrons for say Fe3+. Well in our original where we had just the degenerate, which recall means all equal energy orbitals, we would have just filled in five electrons and been like one, two, three, four, five, and life would have been good. So, but now we have a problem with two different energy levels. And so we learned 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, it turns out it's not going to 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 before we fill in the higher energy ones. So, 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 it's going to cost energy. 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, 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 going to be a little bit different. Now we are most likely going to 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. 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 them in this way is called low spin. 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 going to 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 axes and the two orbitals that are on the axes get shifted to higher energy. Well, the more tightly the ligands bind, the bigger this splitting energy is going to be. And so as a result, for every metal, it's a little bit different 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 in this low spin fashion, we will call that a strong field ligand. 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. So it turns out there's what we call a spectrochemical series that kind of shows how a ligand is going to 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-forms a very, 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, whether they're weak field or strong field. Well, because that also depends on the metal ion as well, because 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 going to be weak field versus strong field really depends on the metal as well. Where's the cutoff between what's weak field and strong field will be different for every metal ion. Alright, so I can't really tell you, you know, if a ligand is going to be weak-filled or strong-filled. We just have tendencies. Cyanide's more tenden-you know, has a stronger tendency to be strong-filled. Something like SCN-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, you know how to fill in the electrons. So, in the high spin, you're going to 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 going to fill up everything down low before you put any electrons up high. So you got to know how to fill in these electrons. And so if you notice, like in this case for Fe3+, in a low spin complex, there's only one unpaired electron. Whereas in a high spin complex, there are five unpaired electrons. And it turns out the number of unpaired electrons is going to have an effect on some of the actual physical properties of this and stuff like that. We'll find out in the next lesson that Fe3 plus high spin complexes are more paramagnetic, as the word will use, than Fe3 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, 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. Assuming we start off with a complex that only had one d-electron and work our way up. So, in this case, let's say we have zero d-electrons. Can I tell the difference? No, because 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 going to get one unpaired electron and I'm largely not going to be able to tell the difference. So, fill in two electrons, and now I've got two unpaired electrons in either case, and I still can't tell the difference. Fill in three electrons. 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, 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 them up and it's easier to pair them up down low, the lower energy set here. So we'll pair those up before we pair the upper ones up and then one here as well. And again I've got no unpaired electrons, 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 unpaired 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 unpaired electrons i can't tell the difference go to nine and i still can't tell the difference go to ten and i still can't tell the difference and so what i want 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 ten 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. 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 in tetrahedral shape, the ligands come in and bind in between the axes. And so now dxy, dyz, and dxz, which are in between the axes, Those are the ones that now shift to higher energy. And then x squared minus y squared and z squared, which are on the axes, so are going to be lower in energy. And so it turns out for a tetrahedral complex, this delta Q here, this crystal field splitting energy is always going to be small, smaller than the pairing energy. So it turns out that all tetrahedral complexes, every last one of them, they are all high spin. 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 going to go up high before ever pairing anything in. And so it turns out, you know, you'd be pairing them in, or filling them in similar to how you would if they were just degenerate, since that, again, crystal field splitting energy is relatively small. So it turns out if you took 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 the splitting pattern for square planar is just unique and it's not the most important thing in the world. These are actually not the most common things in the world. It turns out that square planar complexes only exist for metal ions that have 8d electrons like nickel 2 plus or platinum 2 plus and and again even then sometimes those end up being tetrahedral instead of square planar. It's not an easy way for you to tell. So you've kind of got to be told for something with a coordination number four, whether it's tetrahedral or square planar. However, so if you're D8, there's no guarantee that you're square planar. But if you're square planar, there is a guarantee that you're going to 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 going to 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 going to put its electrons. it can put it in one of the d orbitals. Well, where is it going to put the other? 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 going to use a d, and then we're going to 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 and 9 in terms of hybridization. With hybrid orbitals, we get better bond angles when we start using these hybrid orbitals. And the same thing is going to happen with these coordination complexes. So we're going to use the s and p orbitals We're not actually going to use just the d orbital and the s orbital and the two p orbitals. We're going to mix them all together. And if you mix 1d, 1s, and 2p 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. 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 got to 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. Because if you recall there's two different hybridizations for octahedral. Okay so here we've got the splitting patterns for both high spin and low spin octahedral complexes. And let's say we're dealing with complexes with Fe2 plus involved. So an Fe2 plus is argon. 3d6. So Fe is argon 4s2 3d6. Well, we lose the two 4s's to get Fe2+. So it's 3d6. And so in this case, if we're high spin, it's 1, 2, 3, 4, 5, 6. And if it's low spin, it's 1, 2, 3, 4, 5, 6. Alright, now we need some empty orbitals for the ligands. Well, in the case of the low spin, the first, the lowest energy empty orbitals we have, well, the first two are going to 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, we're going to 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... 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 octahedral. It's D2sp3. 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 going to take a look and see how crystal field theory allows us to explain a couple of different phenomenon we see with coordination compounds. Most of them 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 2 here studying for your final exams. And if that's the case, then you might take a look at my Gen Chem Master Course. In addition to the normal part of the course, there are final exam rapid reviews where I review all of Gen Chem 1 in about six hours and all of Gen Chem 2 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.

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. There will be one more chapter after this one, which I'll finish up next week. But if you'd 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 headlong into crystal field theory here, I've just got to 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, 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 going to 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 want to do one of these cations, start with just a plain old neutral element first, and we've got scandium right here, atomic number 21, and scandium's gonna be argon, and then you might recall 4s2 3d1. So plain old scandium would be argon 4s2 3d1. So that's scandium, but we've got scandium 3-plus here, and so we've got to remove three electrons, and so you just remove the last three, and so...

You gotta remember though that you remove 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 these, gone. And so scandium 3 plus is just isoelectronic with argon here. But the big thing I want to point out here is that scandium 3 plus would be a D0 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 Fe3+.

So go into iron here, and iron again, argon, and then 4s2, and then 3d1, 2, 3, 4, 5, 6. And so plain old iron would be argon, 4s2, 3d6. And we've got to remove three electrons. And again, the thing you need to remember is that you pull out the 4s's before you pull out the 3d's.

And so if you come up with argon, 4s2, 3d3, you've done it wrong. So you've got to pull out that... those 4s's first, so that's two electrons gone, and then you've got to pull out one of the d's, so 3d5.

And we'll just write this all over again as argon 3d5. And the big thing to take away here that we needed to figure out is that it's got five d electrons in Fe3+, and so the different complex ions and coordination compounds that Fe3 is involved in, there are five d electrons, and that's going to be relevant here. Moving on to manganese 2+, and so manganese, we take a look, is argon 4s2 3d12345. Just more of the same here. So 4s2, 3d5.

We just need to lose two electrons and that's going to be the 4s's. And so we're just going to be left with argon 3d5 and I'll take the time to write that over again. And once again, 5d 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 going to 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, so we'd have argon in the case of chromium, and then 4s2, 3d1, 2, 3, 4, that's what we think it would be if it wasn't an exception.

So, but it turns out that with chromium's column, the idea is that half-filled subshells 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 going to be 4s1 3d5. And chromium and then molybdenum right below him does the same thing a row down. Alright, so that's where we got to start.

Now we've got to remove three electrons. And you got to remember, remove the 4s before the 3d. So the first one we take out is that 4s, gone.

Got to remove two more, and so we're going to end up with 3d3. And once again, I'm going to 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 3d electrons.

All right, moving on to copper plus here, and again we'll start off with argon. And again if we didn't know that copper was an exception we'd say argon and then 4s2 and then 3d1, 2, 3, 4, 5, 6, 7, 8, 9. So argon, 4s2, 3d9, and again copper knows something that maybe we didn't initially and that completely full or half-filled subshells 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, would be 4s1, 3d10, 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 one though, so we've got to remove one electron, and you've got to remember definitely that you remove the 4s before the 3d. So it's just going to be argon 3d10. For copper plus one, 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've got to remind ourselves what the d orbitals look like, and we're going to take a look at the five different 3d 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. 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 in dxy.

So it's four lobes in that four-leaf clover, and they lie in the xy plane. But important here is that they don't lie on the x and y axes, but lie in between the x and y axes. Similarly, dyz, four leaf, the four leafs of that four-leaf clover if you will, four lobes that are orbital.

lie in the yz plane, but again in between the y and z axes, not on the y and z axes. dxz lies in the xz plane, but again in between the x and z axes, not on the axes. And so these three, the big important point here, is that the lobes of electron density for these d orbitals lie in between the axes. Super important we understand that. Now dx squared minus y squared, four lobes again, in the xy plane but right on the x and y axes.

And then dz squared most of the electron density is on the z-axis. And so we get really a differentiation here between two types of orbitals. These three that lie in between the axes, and then these two where the electron density is mostly on the axes.

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 axes. So, 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, 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. So we're going to split into two levels here and again x, y, y, z, x, z are going to be the lower set and then x squared minus y squared and z squared the higher set again because these two are on the axes and the ligands are coming in on the axes so the ones that get in the way shift to higher energy. Alright, so this difference in energy now is called the crystal field splitting energy sometimes abbreviated either delta or delta q so and again it's the crystal field splitting energy 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's CFSC for short, or more commonly either delta or delta Q, which is what I'm going to 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, it might actually be relatively large instead.

And depending on if it's relatively small or relatively large, it's going to be a profound difference here. Now, 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 Fe3+, A little bit ago, electron configuration was argon 3d5.

And so we've got to fill in 5d electrons for say Fe3+. Well in our original where we had just the degenerate, which recall means all equal energy orbitals, we would have just filled in five electrons and been like one, two, three, four, five, and life would have been good. So, but now we have a problem with two different energy levels.

And so we learned 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, it turns out it's not going to 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 before we fill in the higher energy ones.

So, 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 it's going to cost energy. 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, 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 going to be a little bit different. Now we are most likely going to 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. 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 them in this way is called low spin. 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 going to 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 axes and the two orbitals that are on the axes get shifted to higher energy. Well, the more tightly the ligands bind, the bigger this splitting energy is going to be. And so as a result, for every metal, it's a little bit different 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 in this low spin fashion, we will call that a strong field ligand. 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. So it turns out there's what we call a spectrochemical series that kind of shows how a ligand is going to 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-forms a very, 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, whether they're weak field or strong field. Well, because that also depends on the metal ion as well, because 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 going to be weak field versus strong field really depends on the metal as well.

Where's the cutoff between what's weak field and strong field will be different for every metal ion. Alright, so I can't really tell you, you know, if a ligand is going to be weak-filled or strong-filled. We just have tendencies.

Cyanide's more tenden-you know, has a stronger tendency to be strong-filled. Something like SCN-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, you know how to fill in the electrons. So, in the high spin, you're going to 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 going to fill up everything down low before you put any electrons up high. So you got to know how to fill in these electrons.

And so if you notice, like in this case for Fe3+, in a low spin complex, there's only one unpaired electron. Whereas in a high spin complex, there are five unpaired electrons. And it turns out the number of unpaired electrons is going to have an effect on some of the actual physical properties of this and stuff like that. We'll find out in the next lesson that Fe3 plus high spin complexes are more paramagnetic, as the word will use, than Fe3 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, 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. Assuming we start off with a complex that only had one d-electron and work our way up. So, in this case, let's say we have zero d-electrons.

Can I tell the difference? No, because 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 going to get one unpaired electron and I'm largely not going to be able to tell the difference.

So, fill in two electrons, and now I've got two unpaired electrons in either case, and I still can't tell the difference. Fill in three electrons. 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, 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 them up and it's easier to pair them up down low, the lower energy set here. So we'll pair those up before we pair the upper ones up and then one here as well.

And again I've got no unpaired electrons, 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 unpaired 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 unpaired electrons i can't tell the difference go to nine and i still can't tell the difference go to ten and i still can't tell the difference and so what i want 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 ten 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. 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 in tetrahedral shape, the ligands come in and bind in between the axes. And so now dxy, dyz, and dxz, which are in between the axes, Those are the ones that now shift to higher energy.

And then x squared minus y squared and z squared, which are on the axes, so are going to be lower in energy. And so it turns out for a tetrahedral complex, this delta Q here, this crystal field splitting energy is always going to be small, smaller than the pairing energy. So it turns out that all tetrahedral complexes, every last one of them, they are all high spin.

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 going to go up high before ever pairing anything in. And so it turns out, you know, you'd be pairing them in, or filling them in similar to how you would if they were just degenerate, since that, again, crystal field splitting energy is relatively small.

So it turns out if you took 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 the splitting pattern for square planar is just unique and it's not the most important thing in the world.

These are actually not the most common things in the world. It turns out that square planar complexes only exist for metal ions that have 8d electrons like nickel 2 plus or platinum 2 plus and and again even then sometimes those end up being tetrahedral instead of square planar. It's not an easy way for you to tell.

So you've kind of got to be told for something with a coordination number four, whether it's tetrahedral or square planar. However, so if you're D8, there's no guarantee that you're square planar. But if you're square planar, there is a guarantee that you're going to 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 going to 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 going to put its electrons. it can put it in one of the d orbitals.

Well, where is it going to put the other? 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 going to use a d, and then we're going to 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 and 9 in terms of hybridization.

With hybrid orbitals, we get better bond angles when we start using these hybrid orbitals. And the same thing is going to happen with these coordination complexes. So we're going to use the s and p orbitals We're not actually going to use just the d orbital and the s orbital and the two p orbitals.

We're going to mix them all together. And if you mix 1d, 1s, and 2p 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. 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 got to 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. Because if you recall there's two different hybridizations for octahedral. Okay so here we've got the splitting patterns for both high spin and low spin octahedral complexes.

And let's say we're dealing with complexes with Fe2 plus involved. So an Fe2 plus is argon. 3d6.

So Fe is argon 4s2 3d6. Well, we lose the two 4s's to get Fe2+. So it's 3d6. And so in this case, if we're high spin, it's 1, 2, 3, 4, 5, 6. And if it's low spin, it's 1, 2, 3, 4, 5, 6. Alright, now we need some empty orbitals for the ligands. Well, in the case of the low spin, the first, the lowest energy empty orbitals we have, well, the first two are going to 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, we're going to 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... 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 octahedral. It's D2sp3. 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 going to take a look and see how crystal field theory allows us to explain a couple of different phenomenon we see with coordination compounds.

Most of them 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 2 here studying for your final exams. And if that's the case, then you might take a look at my Gen Chem Master Course. In addition to the normal part of the course, there are final exam rapid reviews where I review all of Gen Chem 1 in about six hours and all of Gen Chem 2 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.

Transcript for:Understanding Crystal Field Theory

Transcript for:
Understanding Crystal Field Theory