MHTCET 22 Apr 2025 Shift 1 Paper

Concept:Strong-field ligand $\mathrm{CN}^-$ forces pairing of $3d$ electrons, leading to $\mathrm{dsp}^2$ hybridisation in a square planar complex.Explanation:In $\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}$, let the oxidation state of Ni be $x$.$x + 4(-1) = -2 \Rightarrow x = +2$.So, $\mathrm{Ni}^{2+}$ has the configuration $[\mathrm{Ar}]\,3d^8$.Since $\mathrm{CN}^-$ is a strong field ligand, it causes pairing of $3d$ electrons.The configuration becomes $3d^8$ with one vacant $3d$ orbital.One $3d$, one $4s$, and two $4p$ orbitals hybridise to form four equivalent $\mathrm{dsp}^2$ hybrid orbitals.Four $\mathrm{CN}^-$ ligands donate lone pairs to these vacant hybrid orbitals, forming four $\mathrm{Ni}-\mathrm{CN}$ coordinate bonds.These bonds are directed towards the corners of a square, giving square planar geometry.Answer:$\mathrm{dsp}^2$ hybridisation is present.Correct option: B. $\mathrm{dsp}^{2}$