MHTCET 13 Apr 2026 Shift 2 Paper

Concept:For a coordination complex, the type of hybridisation and geometry depend on the oxidation state, d-electron count, and the nature of the ligand (strong or weak field).Explanation:Let the oxidation state of manganese in $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$ be $x$.Each cyanide ligand has a charge of $-1$, so:$$x+6(-1)=-3$$$$x-6=-3$$$$x=+3$$Thus, manganese is in the $+3$ oxidation state.The electronic configuration of Mn is $[\mathrm{Ar}]\,3d^{5}4s^{2}$.For $\mathrm{Mn}^{3+}$, three electrons are removed, giving $[\mathrm{Ar}]\,3d^{4}$.Since $\mathrm{CN}^{-}$ is a strong field ligand, it causes pairing of the $3d$ electrons.This pairing leaves two inner $3d$ orbitals empty, available for hybridisation.With six ligands, the complex is octahedral.For an inner orbital octahedral complex, the hybridisation is $d^{2}sp^{3}$.Therefore, $\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]^{3-}$ is $d^{2}sp^{3}$ hybridised and octahedral.Answer:The correct statement is: It is $d^{2}sp^{3}$ hybridised and octahedral.Thus, the correct option is Option B.