The color of KMnO_4 is due to: | vedclass

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(A) $KMnO_4 	o K^{+} + MnO_4^-$. In $MnO_4^-$,$Mn$ is in the $+7$ oxidation state,which means it has a $d^0$ electronic configuration (no electrons i

Vedclass · Accepted Answer

$L 	o M$ charge transfer transition

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(A) $KMnO_4 	o K^{+} + MnO_4^-$. In $MnO_4^-$,$Mn$ is in the $+7$ oxidation state,which means it has a $d^0$ electronic configuration (no electrons in $d$-orbitals). The intense purple color of $KMnO_4$ is not due to $d-d$ transitions because there are no $d$-electrons. Instead,it arises from $L 	o M$ (Ligand to Metal) charge transfer,where an electron from the oxygen $2p$ orbital is excited to the vacant $Mn$ $3d$ orbital. Since this charge transfer is both Laporte-allowed and spin-allowed,it results in an intense color.

The color of $KMnO_4$ is due to:

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