The colour of $$KMnO_4$$ is due to: [where M $$\rightarrow$$ metal, L $$\rightarrow$$ ligand]

The colour of potassium permanganate ($$KMnO_4$$) is due to electronic transitions in the permanganate ion ($$MnO_4^-$$). In this ion, manganese is in the +7 oxidation state. The electron configuration of manganese in this state is $$[Ar] 3d^0$$, meaning there are no electrons in the d orbitals. Since d-d transitions require electrons to move between d orbitals, and there are no d electrons present, option C (d-d transition) is incorrect.

Now, consider charge transfer transitions. Charge transfer involves the movement of an electron from one part of the complex to another. In $$MnO_4^-$$, manganese (+7) is highly electron-deficient due to its high positive charge, while the oxygen ligands (L) are electron-rich because oxygen is more electronegative and has lone pairs. This creates a situation where an electron can be transferred from the ligand (oxygen) to the metal (manganese). This is known as a Ligand-to-Metal Charge Transfer (LMCT) transition.

Option B describes an M → L charge transfer (metal-to-ligand), which would require the metal to donate an electron. However, manganese in +7 state has no electrons to donate (empty d orbitals), so this is not possible. Option A ($$\sigma - \sigma^*$$ transition) is typically high-energy and occurs in saturated compounds without lone pairs, but $$MnO_4^-$$ has oxygen ligands with lone pairs, making $$\sigma - \sigma^*$$ transitions unlikely to cause visible colour.

Therefore, the intense purple colour of $$KMnO_4$$ arises from an L → M charge transfer transition, where an electron moves from the oxygen ligand to the manganese metal. This transition absorbs light in the visible region, giving the compound its characteristic colour.

Hence, the correct answer is Option D.