`CaCl` is colourless, while `CuSO_(4).5H_(2)O` is coloured. Explain?

Explanation

Electronic configuration of `Cu^(+)` ion is `[Ar] 4s^(0) 3a^(10)` and of `Cu^(2+)` ion is `[Ar] 4s^(0) 3d^(9)`
  In `CuCl, Cu^(+)` ion has no unpaired d- electron. Thus d-d transition is not possible in CuCl and hence it is colourless.
In `Cul//8SO_(4).5H_(2)O,` the `Cu^(2+)` ion has one unpaired d- electron in 3d- subshell. Thus, d-d transition is possible and hence it exhibits colour.