Why does $ KMn{O_4} $ show a $d-d$ transition?

Hint: To understand the reason whether $ KMn{O_4} $ undergoes d-d transition, we first need to understand what is d-d transitions and what are the various types of transitions and what are the criteria for these types of transitions. Also, we need to understand what is the filling order of electrons of the central metal atom in the given compound, i.e. $ M{n^{ + 7}} $ in $ KMn{O_4} $ .

Complete answer:
As the name of the transition itself suggests, the d-d transition is the type of transition in which the molecule would undergo transition between two d orbitals. It means that an electron that is present in the d orbital of the molecule, when a photon is given to it, gets excited and moves to a different d orbital of the molecule that is of a higher energy.
In the molecule of $ KMn{O_4} $ , we see that the molecule undergoes ligand to metal charge transfer. It means that the ligand $ {O^{2 - }} $ gives its lone pair of electron the central metal atom, i.e. $ M{n^{ + 7}} $ . This happens since there is a nonbonding $ 2p $ orbital of oxygen atom and an unoccupied molecular orbital of the compound. Hence, we can clearly see that there is no such transition in the $ KMn{O_4} $ . This can also be suggested from the fact that in the $ KMn{O_4} $ molecule, the central atom $ M{n^{ + 7}} $ has no electron in the outermost shell to perform d-d transition.

Note:
Clearly, although the question suggests that $ KMn{O_4} $ shows d-d transitions, however, we can see from the above deductions that no such d-d transitions take place. The reason being unavailability of d orbital electrons in the $ M{n^{ + 7}} $ form. However, the compound shows purple color, which is due to ligand to metal charge transfer.