The complexes and are both diamagnetic. However, their hybridizations differ. What are their respect — Coordination Compounds Chemistry Question

💡 Solution & Explanation

In $[Ni(CO)_4]$, $Ni$ is in the $0$ oxidation state ($3d^8 4s^2$). The strong field $CO$ ligands force the $4s$ electrons into the $3d$ orbitals, making it $3d^{10}$. The $4s$ and three $4p$ orbitals hybridize to form an $sp^3$ tetrahedral complex. In $[Ni(CN)_4]^{2-}$, $Ni$ is in the $+2$ oxidation state ($3d^8$). The strong $CN^-$ ligands force the pairing of the two unpaired $3d$ electrons, leaving one $3d$ orbital empty. This results in $dsp^2$ hybridization (square planar).