The $Cu$ atom is in form of $Cu ^{+2}$ in the compound.

In $Cu ^{+2}$, the electronic configuration would be $1 s^2 \,2 s^2 \,2 p^6\, 3 s^2\, 3 p^6 \,3 d^9 \,4 s^0$

So, there would be a rearrangement of electrons in $Cu ^{2+}$ because of the $NH _3$ 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.