JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Coordination Compounds

In $[\mathrm{Ni}(\mathrm{CO}{)}_4]$, nickel is in the zero oxidation state ($3d^{8}4s^2$ atomic configuration). CO is a strong-field ligand, forcing pairing of the $4s$ electrons into the $3d$ subshell so that all ten $d$ electrons are paired; the complex is $s{p}^3$ hybridized (tetrahedral) with zero unpaired electrons, hence diamagnetic.

In $[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$, nickel is $\mathrm{N}{\mathrm{i}}^{2+}$ ($3d^8$). $\mathrm{C}{\mathrm{l}}^{-}$ is a weak-field ligand and does not force pairing, so the ion is also $s{p}^3$ hybridized (tetrahedral) but retains two unpaired electrons in the $d$ orbitals, making it paramagnetic.

$[\mathrm{Ni}(\mathrm{CO}{)}_4]$ is diamagnetic and $[\mathrm{NiC}{\mathrm{l}}_4{]}^{2-}$ is paramagnetic.