# Crystal field splitting in complexes other than octahedral or tetrahedral [closed]

Is there a methodology to developing the crystal field splitting diagram for various crystal fields? I know the octahedral and tetrahedral diagrams, but if I'm asked to draw something like square pyramidal, trigonal prismatic, etc. that I haven't memorized, then is there a process I can go through? I searched Google, but am having little luck finding anything that shows the process.

Thank you!

• If you know the process to create the octahedral and tetrahedral crystal fields, why can’t you apply that same process to other fields? – Jan Nov 11 '17 at 4:21
• I don't know the process, I just have them memorized. Memorization without the process is quite honestly worthless at this point. – MatthewSpire Nov 11 '17 at 4:45
• Yes, memorisation without knowing the process is always worthless; especially in this case where memorising requires a lot of brain capacity to do while deriving would be a very simple job. – Jan Nov 11 '17 at 4:49

Crystal field theory assumes that the ligands will approach the central metal in a certain manner and that these ligands will be point-shaped negative charges. Before the ligands approach, all orbitals of the metal’s same subshell will be degenerate, i.e. have the same energy. Because these orbitals have an orientation in space (e.g. the $\mathrm d_{z^2}$ orbital can be seen as ‘pointing’ its lobes in the direction of all the coordinate axes), the approach of the ligands will cause interactions between them and the negatively charged electrons that make up the orbitals. Therefore, any orbital that is pointed towards the approaching ligand will be destabilised (and all others stabilised to a certain degree).
In the octahedral case, the ligands approach from $\pm x,\pm y$ and $\pm z$. I have already pointed out that the $\mathrm d_{z^2}$ orbital would be pointing towards them; the $\mathrm d_{x^2-y^2}$ orbital does, too. Therefore, these two are destabilised while the other three orbitals $\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$ are stabilised.