${\mathrm{CuSO}}_4·5{\mathrm{H}}_2\mathrm{O}$ is blue in colour where as anhydrous CuSO₄ is colourless why?

In CuSO₄ .5H₂O water acts as ligand as a result it causes crystal field spliting. Hence d-d transiton is possible thus CuSO₄ .5H₂O is coloured. In the anhydrous CuSO₄ due to the absence of water, crystal field spliting is not possible hence it is colourless.