[jkm89]

So on my last chemistry quiz we had to identify the ligand geometry of Ir(NH₃)₂(CN)I^-.

I knew that there were 4 ligands so the structure had to be tetrahedral or square planar, however, square planar only exists for d⁸ transition metals.

Since CN and I both have an oxidation number of -1, and the oxidation number of the whole molecule is also -1, I reasoned that the oxidation number of Ir was +1. Ir is originally a d⁷ metal but subtracting one electron from the d orbital gives you d⁶. So I said that the ligand geometry was tetrahedral, but I was wrong and it was actually square planar. Why is this?

[FeLiXe]

in the complex all the valence electrons move to the d subshell. Ir has 9 valence electrons, 1 is removed. that makes d8

[jkm89]

Why does this happen?

[FeLiXe]

that is a general concept.

in the elctrostatic picture i think it is because the ligands destabilize the outer orbitals

you could also consider it with MO theory. then you can't really say d8 because you moved from AOs to MOs. but you count the total number of electrons. there 16 electrons which means that one of the MOs stays unoccupied. this is favorable for a planar complex because one of the antibonding MOs has a higher energy. 18 electrons is the typical case for a low spin octahedral or tetrahedral complex