Pick out the correct statement with respect to $[\mathrm{Mn}(\mathrm{CN}){}_6]{}^{3-}$:

Solution & Explanation

\textbf{Answer:} (C) $[\mathrm{Mn}(\mathrm{CN}){}_6]{}^{3-}$ is $d^{2}s{p}^3$ hybridised and octahedral. \textbf{Solution:} Each $\mathrm{CN}{}^{-}$ carries charge $-1$ and there are 6 of them, so the ligand charge is $-6$. For the overall ion charge $-3$, the metal oxidation state is $+3$: Mn(III). Mn is $[\mathrm{Ar}]3\mathrm{d}{}^{5}4\mathrm{s}{}^2$; Mn(III) loses 3 electrons giving a $3d^4$ configuration. $\mathrm{CN}{}^{-}$ is a strong-field ligand, so it forces the $3d^4$ electrons to pair up, vacating two inner $3d$ orbitals. These two $3d$ orbitals, together with the $4s$ and three $4p$ orbitals, give $d^{2}s{p}^3$ hybridisation (inner-orbital complex). $d^{2}s{p}^3$ hybridisation gives an octahedral geometry. It is a low-spin inner-orbital complex (paramagnetic with 2 unpaired electrons), confirming option (C).