Chemistry - Coordination Chemistry Question with Solution | TestHub

In ${\mathrm{K}}_3\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]$, Fe is in the +3 oxidation state.

Electronic configuration: $\mathrm{Fe}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^{6}4{\mathrm{s}}^2$, so ${\mathrm{Fe}}^{3+}=\left[\mathrm{Ar}\right]3{\mathrm{d}}^5$.

${\mathrm{CN}}^{-}$ is a strong-field ligand, so it pairs the 3d electrons, leaving two 3d orbitals vacant.

These two 3d, one 4s and three 4p orbitals hybridise to six ${\mathrm{d}}^2{\mathrm{sp}}^3$ orbitals, which form six coordinate bonds with the six ${\mathrm{CN}}^{-}$ ligands.

Thus $\mathrm{Fe}$ shows ${\mathrm{d}}^2{\mathrm{sp}}^3$ hybridisation in ${\mathrm{K}}_3\left[\mathrm{Fe}\right(\mathrm{CN}{)}_6]$ - option (D).