$T{i^{2 + }}$ is purple while $T{i^{4 + }}$ is colorless, because:
A. There is no crystal field effect in $T{i^{4 + }}$ .
B. $T{i^{2 + }}$has ${d^2}$ configuration.
C. $T{i^{4 + }}$has ${d^2}$ configuration.
D. $T{i^{4 + }}$is a very small cation as compared to $T{i^{2 + }}$ and hence, does not absorb any light.

Hint: Titanium is a chemical element with the symbol \[Ti\] and atomic number 22. It is a lustrous transition metal with a silver color, low density, and high strength. Titanium is resistant to corrosion in sea water, aqua regia, and chlorine.

Complete answer:
Titanium is the second element of the transition series and also the second element of the 3d-series. The atomic number suggests that it has 22 protons and 22 electrons in its ground state and the electronic configuration of the element is:
$Ti = [Ar]4{s^2}3{d^2}$
There are two unpaired electrons in the 3d-subshell. For the formation of $T{i^{2 + }}$ ion, the titanium atom should lose two electrons. Thus, d-d transition is possible.
In the case of formation of the tetravalent titanium ion, $T{i^{4 + }}$ion, the titanium atom loses four electrons. Thus, d-d transition is not possible and it is colorless.
The coloration of ions takes place when there is a transition of electrons in the atom to a higher energy level. Due to the ${d^2}$ electronic configuration, the $T{i^{2 + }}$ ion shows purple coloration.

Thus option B is the correct answer.

Note:
When an electron gains some energy, it jumps to a higher energy level and remains at the excited energy level unless it becomes unstable and starts reacting with the surrounding species and decreases its energy level by releasing the energy. When it does so, it releases photons of a certain wavelength whose wavelength lies in the visible range of the electromagnetic spectrum. This procedure is the coloration effect of the transition metal series.