21.6 Color

Next, we briefly talk about why many transition metal complexes are very brightly colored. Now it turns out that this crystal field splitting energy here, tends to correspond to the energy of colored light. And so, all of a sudden electronic transitions become possible when we have these two energy levels that weren't possible when we had five degenerate D orbitals. And so this electron here, if it absorbs just the right energy of light- light corresponding to the energy of the crystal field splitting energy- this electron would get promoted to this upper set. So and when it falls back down it emits a photon. But the idea is that it can absorb certain photons and those photons happen to correspond to colored light. So, and typically the bigger this crystal field splitting energy is, the higher the energy of light- the bluer the color if you will, the more violet the color- will be absorbed. Now the color absorbed is not actually equal to the color that you perceive it to be with your eye. So typically if it absorbed one color, we observe the complementary color on the other side of the color wheel. So this ultimately results in two places where we don't expect transition metal complexes necessarily to be colored. And they might be colored for other reasons but there won't be colored for this reason. So we might just say we don't expect them to be colored. And that's if you have no D electrons. If you have no D electrons, then there's no electrons down in the lower set here that could get promoted to the upper set. And so we won't absorb light corresponding to that crystal field splitting energy in this case. So- or if you have ten D electrons, then you definitely still have electrons in the lower set, but there's none of these four spots is available in the upper set. So no transition can happen here either. And we won't absorb light corresponding to the crystal field splitting energy here. So let's take a look at the following question. So this question says "Decide whether complexes with the following ions would be colored." So  and essentially, we just need to take a look at their electron configurations and make sure that they aren't D-0 or D-10 essentially. So if we look at chromium 3 plus. Might remind you that that chromium is [argon]4s1 3d5. So therefore chromium 3 plus is [argon]3d3. And so since he's not d-0 or d-10 he has d electrons, but not a complete set. We do expect him to probably be colored. Let's look at copper plus. So again, chromium and copper here are exceptions. Don't forget that. So [argon] 4s1 3d10. And so copper plus 1, will lose the 4s electron and would just simply be [argon] 3d10. And having 10 D electrons, no we do not anticipate complexes containing copper plus 1 to be colored. So scandium 3 plus. We look at scandium. We got [argon] 4s2 3d1. So scandium 3 plus will be isoelectronic with argon. So notice he does not have any d electrons. He's d-0. And so again no we do not expect complexes containing scandium 3 plus to be colored. And now things are going to get a little bit fun here. When you've got 4 to 7 d electrons you've got to deal with low spin and high spin. So if we look at Fe 2 plus for a second. So Fe, we might recall, is [argon] 4s2 3d6. And so Fe 2 plus, take away 2 electrons, is going to be argon 3d6. And so here because we're between 4 & 7 d electrons it really matters on whether or not we're low spin or high spin here. And so in the first example we're going to do low spin and then in part E we'll look at high spin. So with low spin here, filling in 60 electrons, the lower set each gets 1 and then we start pairing them up down low for low spin. And with 6 of them we see that in this case, we've got empty spots up above. We're not d-4 or d-10. So, and we will say "Yes this guy expects to be colored." Notice, did we really need to fill in the electrons? No. Because they weren't d-0 or d-10, but we'll find out that this is going to play a role in magnetism in a little bit. So same thing with a high spin. The electrons fill in differently. So- but even though we fill them in differently, we're not d-0 or d-10, and so we still have some electrons down low. We still have a couple empty spots up above. And so yes, we expect them to still be colored. Nickel 2 plus. So nickel is [argon] 4s2 3d8. And so nickel 2 plus, remove the two 4s electrons, is just [argon] 3d8. And again we're not d-0 or d-10, we're somewhere in between. And so yes we expect nickel 2 plus complexes to be colored. All right finally. This last example we're dealing with iron. So- but now in the context of giving you the actual complex ion. And so you've got to figure out "Okay, what charge does he have?" And things of a sort. So if FeF6 is a -3 charge overall, and each of the fluoride ions is minus 1, that means iron has got to be Fe3+. So we've got Fe right up here still, so I'll reference that. So if we move three electrons, we'll take away the 2 4s's and then one of the 3d's, and we'll just be left with [argon] 3d5. And it says weak field ligand here. Weak field means it's going to be high spin. And 1, 2, 3, 4, 5. And yet again we're going to expect this thing to be colored. Now this one's kind of a funky exception potentially, with all the spins lining up we might actually have a tough time doing an excitation. The electron that gets excited would actually have to flip it spin. Which is somewhat of a rare event. So technically, some professors might go so far to say that no this one's not colored. But most aren't expected to be colored anyways. But most of the time, we're just looking for d-0 or d-10 in that regard. But technically, this one may not be colored either.