Chemistry - Complete d-block Question with Solution | TestHub

The hydrated copper sulphate 

${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ 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 ${\mathrm{CuSO}}_4$
 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.

${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ 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.

${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$

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

unpaired electron present so it show colour due to $\mathrm{d}-\mathrm{d}$ transition.