The colour of $K M n O_{4}$ is due to

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$M \to L$ charge transfer transition

$d \to d$ transition

$L \to M$ charge transfer transition

$σ \to σ$ transition

answer is C.

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Detailed Solution

The oxidation state of Mn in KMnO4 is +7, which means that Mn does not have any d electrons left in its orbital. Actually, Mn7+ complexes should not be colored because electronic transitions are not possible. However, KMnO4 has a deep purple color because of charge transfer from the ligand or we can say O2– to the metal center. This is called a ligand-to-metal charge transfer, which leads to absorption of energy and color

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