JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Coordination Compounds

$\mathrm{ds}{\mathrm{p}}^2$ hybridisation corresponds to a square planar complex, which a $d^8$ ion adopts only when a strong field ligand pairs its electrons and so frees an inner $d$ orbital. In ${\mathrm{K}}_2[\mathrm{Ni}(\mathrm{CN}{)}_4]$ nickel is $\mathrm{N}{\mathrm{i}}^{2+}$, $3d^8$, and the strong field cyanide ions force the two unpaired electrons to pair, leaving one $3d$ orbital empty to mix with $4s$ and two $4p$ orbitals. In $\mathrm{N}{\mathrm{a}}_2[\mathrm{NiC}{\mathrm{l}}_4]$ the weak chloride ligands leave the $3d^8$ set as it is, so the ion is tetrahedral and $\mathrm{s}{\mathrm{p}}^3$; $\mathrm{NiC}{\mathrm{l}}_2\cdot 6{\mathrm{H}}_2\mathrm{O}$ contains the octahedral $\mathrm{s}{\mathrm{p}}^3{\mathrm{d}}^2$ hexaaqua ion, and in $[\mathrm{Ni}(\mathrm{CO}{)}_4]$ nickel is zerovalent $3d^{10}$ and again $\mathrm{s}{\mathrm{p}}^3$. The $\mathrm{ds}{\mathrm{p}}^2$ compound is ${\mathrm{K}}_2[\mathrm{Ni}(\mathrm{CN}{)}_4]$.