$CuSO_4 \cdot 5 H_2O$ is blue in colour while $CuSO_4$ is colourless. Why?

(N/A) In $CuSO_4 \cdot 5 H_2O$,the $H_2O$ molecules act as ligands that cause the splitting of $d$-orbitals in the $Cu^{2+}$ ion.
This splitting allows for $d-d$ transitions of electrons when light is absorbed,which results in the blue colour of the hydrated salt.
In anhydrous $CuSO_4$,there are no ligands present to cause $d$-orbital splitting.
Consequently,no $d-d$ transition can occur,making $CuSO_4$ colourless.