$\underset{\text{ excess }}{{\mathrm{CuSO}}_4+{\mathrm{NH}}_4\mathrm{OH}}⟶$$\underset{\mathrm{complex} \mathrm{salt}}{\mathrm{Deep} \mathrm{blue} \mathrm{soluble}}$

The correct statement regarding the complex is/are:

Since $\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{SO}}_4$ has one unpaired electron, square planar and$ds{p}^2$ hybridization.

$\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right]{\mathrm{SO}}_4\stackrel{{\mathrm{H}}_2\mathrm{S}}{⟶}\mathrm{CuS}\downarrow \text{ (Black) }$

Hence all the given options are correct.