Which of the following compounds show colour due to d - d...

The hydrated copper sulphate 

${CuSO}_{4} . 5{H}_{2}O$ is blue in color. The ligand (water molecules) cause splitting of d – orbitals. We know that, due to this, the process of d – d transition and color transition is facilitated.

Whereas, on the other hand the anhydrous copper sulphate ${CuSO}_{4}$
 is colourless. In the absence of ligand (water molecules), splitting of d orbitals is not possible. Hence, d – d transition is not possible.
We can understand the above from the following statements.

${CuSO}_{4} . 5{H}_{2}O$ is blue in colour because its oxidation state $+ 2$, thus it has unpaired electrons in it. So, d –d transition is possible so it exhibits colour where on the other hand the anhydrous copper sulphate complex, d – d transition is not possible that’s why it is colourless.

${CuSO}_{4} . 5{H}_{2}O$

${Cu}^{2 +}: 3{d}^{9}4{s}^{0}$

unpaired electron present so it show colour due to $d - d$ transition.