Question

Hello. I have a question about ligands.

How do you figure out how many ligands a central metal cation can have?

For example, Cu^2+ and NH3. When they combine, it forms Cu(NH3)4^2+

Why does it form 4 ligands oppose to 6?

A general rule, which isn't always followed, is to double the charge to find the number of ligands. One reason, for Cu^+2, is the s, p, and d orbitals available.
29Cu is 1s2 2s2 2p6 3s2 3p6 3d9 4s2.
The Cu^+2 loses those two 4s electrons, promotes the 1 unpaired 3d electron so it has available a 3d, a 4s, and 4p orbitals and it forms dsp2 which is a square planar complex. A 6-coordinate compound would be d2sp3 and there is only 1d available in the 3d shell. So the d2sp3 is not a good option.

Ah thank you for the speedy response!

So... let me apply what I learned from you.

My next problem just so happens to be Zn^2+

e- config for Zn: [Ar] 3d10 4s2
therefore, Zn^2+: [Ar] 3d10

now i'm stuck. What gets promoted?

I could apply that unofficial rule of yours and say that because it's Zn^2+, it should have 4 ligands.

There is nothing to promote. All of the 3d orbitals are filled. My first educated guess is to use the sp3 and a coordination number of 4. You will find this in Zn(NH3)4^+2, Zn(OH)4^-2, etc. I would guess sp3 and tetrahedral. By the way, it has been found in studies that Cu^+2, does in fact, often add two more ligands to make a coordination number of 6 and the complex is octahedral. To do this it uses the 4s, 4p, 5d for an "outer" complex of d2sp3, or more accurately, sp3d2.