The electronic spectrum of an aqueous solution of shows three distinct bands: A(~400nm), B(~690nm) and C(~1070nm).The transitions assigned to A, B and C, respectively are

Correct option is B

In d⁸ octahedral complexes, the d-orbitals split into t_2g (lower energy) and e_g (higher energy) orbitals. Thus, in ground state the electronic configuration is

There is only one electronic arrangement for this electronic configuration

and the ground state term corresponding to this configuration is represented as

In a d⁸ octahedral complex two holes may be considered in e_g orbitals as shown in figure.

When an electron from any of the t_2g orbitals is excited to any one of the e_g orbitals, the electronic configuration becomes

Now there will be two holes one in t_2g and one in e_g orbitals. Corresponding to this electronic configuration, there are two triply degenerate states. The lower energy state will arise when two holes occupy the orbitals as far apart as possible. This state is represented as

The holes may be present either in

orbitals giving rise to lesser repulsions between two holes because two holes occupy more space in x, y and z-directions. Holes are effected in the opposite way as compared to electrons. There is another triply degenerate arrangement of two holes in which the two holes occupy the orbitals which are close together. The holes may be present in either

This is of higher energy and is represented as

When both the electrons are excited, the electronic configuration becomes

Now two holes are present in any of the two t_2g orbitals either in

Therefore, this state is triply degenerate and is represented as

Since double excitation requires higher energy, therefore,

state is of highest energy state. There are four energy states including ground state. Thus, there are three transitions. The transitions of holes is similar to electronic transitions.