The compound $KMnO_4$ (potassium permanganate) is known for its intense purple color. The distinctive color arises due to an electronic transition known as a charge transfer transition. Specifically, in $KMnO_4$, the color is attributed to a $L \rightarrow M$ charge transfer transition.
In the $KMnO_4$ molecule, manganese is in a high oxidation state (+7), and the oxygen atoms are part of the ligands bound to manganese. This setup allows the transition of electron density from oxygen (ligand, L) to the empty d-orbitals of manganese (metal, M).
During this $L \rightarrow M$ charge transfer, an electron is excited from the highest occupied molecular orbital associated with the oxygen (ligand) to the lowest unoccupied molecular orbital associated with the manganese (metal).
This process absorbs light in the visible region, and the complementary color, which is purple, is seen.
Let's consider why the other options are incorrect:
$M \rightarrow L$ charge transfer transition: This would require electron transfer from metal to ligand, which is not the mechanism in $KMnO_4$.
$\sigma \rightarrow \sigma$ transition: This is irrelevant for $KMnO_4$, as the involved transitions are not between sigma orbitals.
$\sigma \rightarrow \sigma ^*$ transition: Similar to above, this type of transition is not responsible for the color observed in $KMnO_4$.
Therefore, the correct and typically responsible transition for the purple color in $KMnO_4$ is the $L \rightarrow M$ charge transfer transition.
