Question

The colour of $KMnO_{4}$ is due to: [where $\text{M} \rightarrow $ metal, $\text{L} \rightarrow $ ligand]

• A. $\sigma - \sigma ^{\text{*}}$ transition
• B. $M \rightarrow L$ charge transfer transition
• C. $\text{d} - \text{d}$ transition
• D. $L \rightarrow M$ charge transfer transition

Answer 1

Answer: (D)

Charge transfer from $O ^{2-}$ to empty $d$-orbitals of metal ion $\left(( Mn )^{7+}\right)$
The colour in $KMnO _{4}$ arises from an electronic transition. In the permanganate ion $\left( MnO _{4}^{-}\right)$, manganese is in the $+7$ oxidation state, so it has no $d$ electrons. Photons promote an electron from the highest energy molecular orbital to an empty $d$-orbital on the manganese. When a photon of light is absorbed, this charge transfer takes place from $L$ to $M$.