Question

Can anybody help me with this please??????

Answer 1

Why are the transition metals of Mn^+2 compounds almost colourless?

Answer 2

@TuringTest any idea regarding this??

Answer 3

I'm not sure, but I'm in the middle of taking inorganic chemistry right now, so I have a fairly good explanation that I'm sure explains it.

So if you look at Mn2+ you'll notice it will have 5 electrons in the d-orbitals, since in oxidizing it to the +2 state, the electrons will come from the s-orbital. Now most commonly when transition metals bind to ligands, it usually will have 6 partners, which is called an octahedral shape because if you connect all 6 of them, you will get a shape that has 8 faces. So if you imagine that octahedral shape, the manganese is in the center and the other 6 ligands are on the corners, all equal distances from each other and the manganese.

Now, the ligands have their own orbitals that will interact with these d-orbitals on the manganese. This causes the energy of the d-orbitals to split into higher and lower energy depending on if they're where we bond or not. The d-orbitals are really just 5 places to put 10 electrons, a pair in each, and they occur such that if you put the d-orbitals on a three dimensional graph, you could then bind those ligands on the x axis, y-axis, and z-axis. There are three axes and we can approach from the positive or negative direction for each. This lets us bind all 6 of our ligands conveniently, meaning we have 3 shared orbitals of pretty low energy. The other 2 d-orbitals are sort of inbetween these, and are of higher energy because of this. It's pretty much the same reason s-orbitals are lower energy than p-orbitals. They're just more stable! Remember, all 5 of these are bonding orbitals. So we have orbitals that look like this, and if you're curious, I wrote their names next to them.



So these are the orbitals we have to fill with 5 electrons. Now we have to see if it's high or low spin. Since Mn is in the first row of the d-block, and is in the +2 oxidation state, this tends to mean it's a high spin configuration, which means that the orbitals will fill the eg orbitals before placing another electron in the t2g orbitals with an opposite spin. So basically, you'll have something like this:



Now if you don't understand any of that garbage I said before, it doesn't really matter. What matters is this is what the highest energy orbitals look like in this picture. So when light comes in to excite an electron from t2g to eg, it can't since putting an "up" spin electron into another orbital with another "up" spin electron is forbidden, since you can't have two "up" electrons there! This means it won't absorb light and if it doesn't absorb it, then it won't emit light either. Thanks Pauli exclusion principle!

Answer 4

Also, you might think, well how come we don't use the HOMO instead since the LUMO will surely not have any electrons to forbid it. However, I'm pretty sure that transition doesn't generally occur in the visible region.

Answer 5

Also, I realize I just put a lot of words up there, and more than likely you're not going to understand any of it. But if you ask me any questions and want to understand, I will do my best to help fill in the gaps for you! :)

Answer 6

thank you very much... though the explanation was long but very effective.. Thank you very much.

Answer 7

@Kainui... In Mn atom for CFT both the eg and t2g is filled with the same spin, Can you explain me this?