$\mathrm{Ti}^{2+}$ is purple while $\mathrm{Ti}^{4+}$ is colourless because

• Correct answer
  A

  $\mathrm{Ti}^{2+}$ has $3 d^2$ configuration
• B

  $\mathrm{Ti}^{4+}$ has $3 d^2$ configuration
• C

  $\mathrm{Ti}^{2+}$ is very small cation when compared to $\mathrm{Ti}^{2+}$ and hence, doesn't absorb any radiation
• D

  There is no crystal field effect in $\mathrm{Ti}^{4+}$

Solution

$\begin{aligned} \mathrm{Ti}^{2+} & =[\mathrm{Ar}] 3 d^2 \\ \mathrm{Ti}^{4+} & =[\mathrm{Ar}] 3 d^0 \end{aligned}$ $\mathrm{Ti}^{2+}$ has two unpaired electrons in $3 d$ orbital and hence $d-d$ transition is possible due to absorption of light in visible region whereas $\mathrm{Ti}^{4+}$ is diamagnetic hence colourless.

Hence, the correct answer is option: 'A'.