Question

how would be the hybridization of [Cu(NH₃)₄]²⁺ ion?

Answer 1

Cu in its +2 oxidation state has [Ar]3d9 electronic configuration. Ammonia is a strong field ligand hence pairing occurs. Therefore 9 electrons, after pairing, it shows 1 d orbital vacant which hybridized with 4s and 4p orbitals to form dsp2 hybridisation(according to crystal field theory).I think the hybridization should be dsp2.dsp2.

Let's come to the explanation.

The Cu atom is in form of Cu+2 in the compound. In Cu+2, the electronic configuration would be .. 
1s22s22p63s23p63d94s0

So, there would be a rearrangement of electrons in Cu^2+ because of the NH3 ligand (which is a strong one). And the last electron in the d-orbital would be out waiting for the N's electrons to fill up first. 
[because NH3 is a VIP and the electrons donated from it should have got place first--this is only a punctuation]

So, the 4 electron pairs from N would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, the e- from 3d would take place. So, you can say the hybridisation here would be dsp2.

Answer 2

I think the hybridization should be dsp2.

Let's come to the explanation.

The Cu atom is in form of Cu+2 in the compound. In Cu+2, the electronic configuration would be ..  

1s22s22p63s23p63d94s0                                                                                   So, there would be a rearrangement of electrons in Cu^2+ because of the NH3 ligand (which is a strong one). And the last electron in the d-orbital would be out waiting for the N's electrons to fill up first.  

[because NH3 is a VIP and the electrons donated from it should have got place first--this is only a punctuation]

So, the 4 electron pairs from N would be in one 3d, one 4s, & two 4p orbitals and in the third place of 4p orbital, the e- from 3d would take place. So, you can say the hybridisation here would be dsp2.