$\ce{Mn}$ in $\ce{KMnO4}$ is in $+7$ oxidation state, has no d electrons so d-d transition is not possible.
Then, how is $\ce{KMnO4}$ coloured?
$\ce{Mn}$ in $\ce{KMnO4}$ is in $+7$ oxidation state, has no d electrons so d-d transition is not possible.
Then, how is $\ce{KMnO4}$ coloured?