JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Chemical Bonding and Molecular Structure

In $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$ the nickel is $\mathrm{N}{\mathrm{i}}^{2+}$ ($3d^8$), and the strong-field cyanide ligands force the electrons to pair so the ion is square planar with $ds{p}^2$ hybridisation, which draws on just one $d$ orbital. Each of the other species uses $s{p}^3{d}^2$ hybridisation, involving two $d$ orbitals: $\mathrm{Xe}{\mathrm{F}}_4$, $[\mathrm{Cr}{\mathrm{F}}_6{]}^{3-}$ and $\mathrm{Br}{\mathrm{F}}_5$. Hence only the nickel complex uses a single $d$ orbital.