MHTCET 19 Apr 2026 Shift 1 Paper

Concept:Coordination complexes with weak-field ligands and $d^{8}$ metal ions often show outer orbital hybridization.Explanation:In $\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]^{2-}$, let the oxidation state of nickel be $x$.Since each chloride ion has a charge of $-1$, we get $x + 4(-1) = -2$.So, $x = +2$.The electronic configuration of nickel is $\mathrm{Ni}: [\mathrm{Ar}]\, 3d^{8}\, 4s^{2}$.For $\mathrm{Ni}^{2+}$, the configuration becomes $\mathrm{Ni}^{2+}: [\mathrm{Ar}]\, 3d^{8}$.Chloride ion $\mathrm{Cl}^{-}$ is a weak-field ligand, so it does not force the pairing of electrons in the $3d$ orbitals.Therefore, nickel uses one $4s$ orbital and three $4p$ orbitals for bonding.The hybridization involves $4s + 4p_x + 4p_y + 4p_z$, which gives $sp^{3}$ hybridization.This results in a tetrahedral geometry for the complex.Hence, the correct option is A, which is $sp^{3}$.Answer:$\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]^{2-}$ shows $sp^{3}$ hybridization, so the correct option is A.