Question

$CuSO_{4} \cdot 5H_{2}O$ is blue in colour while $CuSO_{4}$ is colourless due to

• A. presence of strong field ligand in $CuSO_{4} \cdot 5H_{2}O$
• B. absence of water (ligand), $d - d$ transitions are not possible in $CuSO_{4}$
• C. anhydrous $CuSO_{4}$ undergoes $d -d$ transitions due to crystal field splitting
• D. colour is lost due to loss of impaired electrons

Answer 1

Answer: (B)

absence of water (ligand), $d - d$ transitions are not possible in $CuSO_{4}$

Answer 2

In $CuSO_{4} \cdot 5H_{2}O$, water acts as ligand and as a result it causes crystal field splitting making $d-d$ transitions possible in $CuSO_{4} \cdot 5H_{2}O$ Hence, it is coloured. In anhydrous $CuSO_{4}$, due to absence of ligand crystal field splitting is not possible hence no colour is observed.