Question

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

• A. Presence of strong filed ligand in $CuSO_{4}.5H_{2}O$
• B. Due to absence of water (ligand), d- d transitions are not possible in $CuSO_{4}$
• C. Anhydrous $CuSO_{4}$undergoes d- d transition due to crystal field splitting
• D. Colour is lost due to loss of unpaired electrons.

Answer 1

Answer: (B)

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

Answer 2

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