$C u S O_{4} .5 H_{2} O$ is blue in colour while $C u S O_{4}$ is colourless. Why?

Open in App

Solution

In $C u S O_{4} .5 H_{2} O,$ water acts as a ligand and causes crystal field splitting. Hence, d-d transition is possible thus $C u S O_{4} .5 H_{2} O$ is coloured. In the anhydrous $C u S O_{4}$ due to the absence of water (ligand), crystal field splitting is not possible and hence, it is colourless.

Suggest Corrections

Similar questions

C u S O_{4} \cdot 5 H_{2} O

C u S O_{4}

C u {S O}_{4} \cdot 5 H_{2} O

C u {S O}_{4}

C u S O_{4} .5 H_{2} O

C u S O_{4}

(C u S O_{4} \cdot 5 H_{2} O)

Join BYJU'S Learning Program