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

Explanation

Electronic configuration of `Cu^(2+)` ion is `[Ar]4s^(0)3d^(10)`
Electronic configuration 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 `CuSO_(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.