Identify the oxidation state and electronic configuration of manganese in $KMnO_4$

In $KMnO_4$, the anion is $MnO_4^-$. Let the oxidation state of Mn be $x$: $x + 4(-2) = -1$ $x - 8 = -1$ $x = +7$

So manganese is in the $+7$ oxidation state.

Electronic configuration of Mn: $Mn: [Ar]3d^54s^2$

Therefore, $Mn^{7+}: [Ar]3d^0$

Check whether $d$–$d$ transition is possible

A $d$–$d$ transition requires partially filled $d$ orbitals.

But in $Mn^{7+}$, the configuration is $3d^0$, so there are no $d$ electrons.

Hence, option A ($d$–$d$ transition) is not possible.

Consider the origin of colour in $MnO_4^-$

Since $d$–$d$ transition is absent, the intense purple colour must arise from a charge transfer transition.

In $MnO_4^-$, oxygen ligands have filled orbitals, and manganese in the high oxidation state $+7$ can accept electron density.

Thus, an electron is promoted from ligand (O) orbitals to metal (Mn) orbitals: $L \to M$

This is called ligand-to-metal charge transfer (LMCT).

Check remaining options

• B: $L \to M$ charge transfer transition: Correct.
• C: $\sigma$–$\sigma^*$ transition: These transitions generally occur in the far UV region and are not responsible for the visible purple colour.
• D: $M \to L$ charge transfer transition: This would be metal-to-ligand charge transfer, not expected here because Mn is already in a very high oxidation state and electron-poor.

Conclusion

The colour of $KMnO_4$ is due to ligand-to-metal charge transfer transition.

Therefore, the correct option is: $\boxed{B}$

Comparison with stored correct answer

Stored correct answer: B

My derived answer: B

Hence, they agree.