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 $\text{O}^{2 -}$ to empty $\text{d}$ -orbitals of metal ion $\left(\text{(Mn}\right)^{7 +} \left.\right) .$

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