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 $O^{2-}$ to empty $d$-orbitals of metal ion $\left((Mn{)}^{7+}\right)$
The colour in $KMn{O}_4$ arises from an electronic transition. In the permanganate ion $\left(Mn{O}_4^{-}\right)$, 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 $M$.