MHTCET 11 May 2024 Shift 1 Paper

Concept:The hybridization depends on the oxidation state of the metal, the ligand field strength, and the coordination number.Explanation:In $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$, the oxidation state of nickel is $+2$.Electronic configuration of $\mathrm{Ni}^{2+}$: $[\mathrm{Ar}]\,3d^{8}$.$\mathrm{CN}^{-}$ is a strong field ligand, so it forces the $3d$ electrons to pair up.Thus, one $3d$ orbital becomes vacant.Four $\mathrm{CN}^{-}$ ligands donate lone pairs into four vacant $dsp^{2}$ hybrid orbitals.This gives a square planar geometry with $dsp^{2}$ hybridization.Answer:The hybridization of $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$ is $\mathbf{dsp^{2}}$.Correct option: B. $\mathrm{dsp}^{2}$