Question

The colour of $ KMn{O_4} $ is due to
(A) M $ \to $ L charge transfer transition
(B) d-d transition
(C) L $ \to $ M charge transfer transition
(D) $ \sigma - {\sigma ^*} $ transition

Answer 1

Hint :The colour in most of the ionic compound is due to the presence of d orbital. As d orbital is not involved in bonding and also they are partially filled. The colour of the ion depends upon the complement of the colour absorbed.

Complete Step By Step Answer:
This question belongs to the concept of d and f block elements. Most of the d and f block elements that are transition elements are coloured whether they are in solid or liquid state. As they have incompletely filled d orbitals. In order to understand why $ KMn{O_4} $ is coloured we will first get familiar with d-d transition.
So, when a transition metal compound is formed the d generated orbitals of the metal splits into two sets. One set has dxy, dyz and dxz orbitals which are known as $ {t_2}g $ orbitals and the other set is $ {d_{{x^2} - {y^2}}} $ and $ {d_{{z^2}}} $ orbitals which are known as eg orbital. These orbitals are slightly higher in energy level in the presence of the octahedral field. This phenomenon is called crystal field splitting. When white light falls on these compounds, some wavelength is absorbed by the electrons from one set of lower energy orbitals to another set of slightly higher energy within the same d-subshell. This is called d-d transition. Also, the remaining light is reflected back.
Now let us see the case of $ KMn{O_4} $ .
In the case of $ KMn{O_4} $ the $ Mn $ atom is responsible for the colour as it undergoes d-d transition. The $ Mn $ atom has $ + 7 $ oxidation state and the configuration is $ [Ar]3{d^0}4{s^0} $ thus we can see that no unpaired electrons are present. So we can conclude that the molecule should be colourless.
The colour of the $ KMn{O_4} $ is due to ligands to metal transfer not due to d-d transition as there are no unpaired electrons which can result in d-d transition.
The charge transfers from $ 2p(L) $ of oxygen to $ 3d(M) $ of $ Mn $ .
Hence we can conclude that it is due to L to M charge transfer.
So option C is the correct answer.

Note :
d-d transition has no role to play in the colour of $ KMn{O_4} $ compound. Manganese gets oxidised from $ + 7 $ to $ + 6 $ . Though $ KMn{O_4} $ is non-combustible it catalyses burning of other combustible compounds.