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

Explanation

Electronic configuration of `Cu^+` ion is [Ar] `4s^0 3d^9`
In `CuCl, Cu^(+)` ion has no unpaired d- electron . Thus d-d transition is no possible in CuCl and hence it is colourless.
In `CuSO_4. 5H_2O` the `Cu^(2+)` ion has one unpaired d-electron in 3d -subshell . Thus, d-d transition is possible and hence it exhibits colour.