Question

The colour of $KMn{O}_4$ is due to: [where $\text{M}\to$ metal, $\text{L}\to$ ligand]

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

Answer 1

Answer: (D)

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

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 $\text{M}\text{.}$