`[Ti (H_2O)_6]^(3+)` is coloured while `[Sc(H_2O)_6]^(3+)` is colourless. Explain.

Explanation

O.S. of Ti in `[Ti(H_2O)_6]^(3+)` is +3, E.C. of Ti(Z = 22) and `Ti^(3+)` are as Ti(z=22): `[Ar]^(18) 3d^2 4s^2`, `Ti^(3+)` : `[Ar]^(18)3d^(1)`
In the presence of six molecules of water which acts as ligand, the five degenerate 3d-orbitals split up into two set of orbitals, one with lower energy `(t_(2g))` and the other with higher energy `(e_g)`.
In the ground state, the single 3d-electrons lower is present in the `t_(2g)` level. The next higher state available for the electrons is `t_(g)` level. If light corresponding to the energy of blue green region is absorbed by the complex, it would excitive electron from `t_(2g)` level to `e_g` level `(t_(2g)^(1) eg^(o) rarr t_(2g)^(o)e_g^(1)`. Hence the complex appear violet in colour (voilet is the complementry colour of blue green).

According to crystal field theory the colour of the complex is due to d-d transition of electron.
In case of `[Sc(H_2O)_6]^(3+)` the `Sc^(3+)` has no electron as in 3d-subshell as explained below.
Sc(Z=21): `[Ar]^(18) 3d^(1) 4s^2, Sc^(3+) : [Ar]^(18) 3d^(0)`
With no eleciron in 3d-subshell, d-d transition of electron is not possible in `[Sc(H_2O)_6]^(3+)` and hence it is colourless.