Question

Why (cu(en)2)so4 is more stable than(cu(nh3)4)so4 complex .

Answer 1

Just want to make an amendment to my original answer as I got it wrong. If the electron hops into d(z2) and pairs with the electron already in there, that leads to additional shielding of the positive metal from the attraction of the negative l (in this theory we assume the metal to be positively charged and ligands to be negative). That shielding, relative to the shielding provided by the single electron in d(x2-y2), leads to the z ligands being pushed further from the metal. That means the z ligands are further from the z electrons and repel them less. That stabilises the z electron, hence Jahn Teller effect. Same arg for d(x2-y2).

The important thing is that Cu(II) is d9, Cu(I) is d10. Since there is no Jahn Teller effect in Cu(I), no stabilisation occurs, so Cu(II) is more stable.