$A$ square planar complex is formed by the hybridization of which atomic orbitals?

The shape of $[Cu(NH_3)_4]^{2+}$ is square planar. The $Cu^{2+}$ ion in this complex is:

An octahedral complex of $Co^{3+}$ is diamagnetic. The hybridisation involved in the formation of the complex is

The hybridization observed in the $[PtCl_6]^{2-}$ ion is .......

Nickel $(Z = 28)$ combines with a uninegative monodentate ligand $X^{-}$ to form a paramagnetic complex $[NiX_4]^{-2}.$ The number of unpaired electron$(s)$ in the nickel and geometry of this complex ion are,respectively $:$

Match List $-I$ with List $-II$:

List $-I$	List $-II$
$A$. $[PtCl_4]^{2-}$	$I$. $sp^3d$
$B$. $BrF_5$	$II$. $d^2sp^3$
$C$. $PCl_5$	$III$. $dsp^2$
$D$. $[Co(NH_3)_6]^{3+}$	$IV$. $sp^3d^2$