[jasmine85]

Hi. Now that I understand how the s and p orbitals are responsible for the different molecular shapes in covalent structures, I'm wondering how the d-orbitals are responsible for the shapes of transition metal complexes.

As I understand it, there are these molecules called ligands, which share their lone pair electrons with the metal to fill up it's d-orbitals. And the ligands space themselves out in much the same ways as covalent bonds do in sp3 hybridisation. (tetrahedral/octahedral/etc)

What I don't understand is why a metal ion is sometimes happy to take 4 ligands, and other times it might take 6. How do I work out how many ligands there will be for a given metal cation?

Also, the metal has an ionic bond with a corresponding anion(s). How do these anions affect the geometry of the complex? Do they have an ionic relationship with the metal or with the ligands?

thanks for any help

Jas

[Schrödinger]

Many a time, the strength of the ligand will determine the geometry of the complex, as far as I know.

For example,
1. if there are unpaired electrons in the d orbital of the metal,
2. AND the ligand is a strong one,
3. AND there is a necessity to pair the electrons up so that vacant orbitals can be created to make use of dsp2 or d2sp3 (this is different from the sp3d2 version)

then these electrons will be paired up and the geometry required will be achieved.

On the contrary, if the ligand is a weak one, then no such pairing up will take place. Instead the outer d-orbitals will be used if need be. Else, simple sp3 will be used (i.e, if the number of electrons is less).


The second approach is the crystal field theory...it's easily understandable and interesting. I think you should go through it.