$\ce{Cu^2+} $ has 9 electrons and a d-orbital and is almost completely filled (except 1 electron vacant) and $\ce{NH3}$ donates an electron eventually forming $\mathrm{sp^3}$ hybrid orbitals.

But it has a square planar geometry hence completely contradicting the above reasoning. Please explain what the hybridization is?

$\ce{Cu^2+} $ has 9 electrons and the $\mathrm d$-orbital is almost completely filled (except for 1 electron vacancy), eventually forming $\mathrm{sp^3}$ hybrid orbitals, where each $\ce{NH3}$ donates a lone pair of electrons.

But it has been found to have a square planar geometry, hence completely contradicting the above reasoning. So, please explain what is the hybridization of the $\ce{Cu^2+}$ in the given complex?

$\ce {Cu++} $ has 9 electrons and d orbital and is almost completely filled (except 1 electron vacant) and $NH3$ donates electron eventually forming Sp3 hybrid orbitals
But it has square planar geometry hence completely contradicting the above reasoning.Please explain what is the hybridization.

$\ce{Cu^2+} $ has 9 electrons and a d-orbital and is almost completely filled (except 1 electron vacant) and $\ce{NH3}$ donates an electron eventually forming $\mathrm{sp^3}$ hybrid orbitals.

But it has a square planar geometry hence completely contradicting the above reasoning. Please explain what the hybridization is?

$\ce Cu2+ $ has 9 electrons and d orbital in almost completely filled (except 1 electron vacant) and $NH3$ donates electron eventually forming Sp3 hybrid orbitals
But it has square planar geometry hence completely contradicting the above reasoning.Please explain what is the hybridization.

$\ce {Cu++} $ has 9 electrons and d orbital and is almost completely filled (except 1 electron vacant) and $NH3$ donates electron eventually forming Sp3 hybrid orbitals
But it has square planar geometry hence completely contradicting the above reasoning.Please explain what is the hybridization.