Assertion (A): Anhydrous copper sulphate is white, but CuSO4·5H2O is blue in colour.
Reason (R): In the absence of ligands, crystal field splitting of the d orbitals of the metal ion does not occur.
In the light of the above statements, choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both (A) and (R) are true and (R) is the correct explanation of (A).
Crystal field theory explains the colour of a complex by a d–d transition: the ligands split the d orbitals, and an electron absorbs visible light to jump from the lower set to the higher set. The colour seen is the complementary colour of the light absorbed.
Assertion (A): Anhydrous copper sulphate is white, but CuSO4·5H2O is blue in colour. — True. The hydrated salt is blue; without its water it is white.
Reason (R): In the absence of ligands, crystal field splitting of the d orbitals of the metal ion does not occur. — True. The splitting is caused by the field of the ligands; with no ligand, the d orbitals stay degenerate.
Why R explains A?
In CuSO4·5H2O the water ligands split the d orbitals of Cu2+, and a d–d transition gives the blue colour.
Without the water there is no splitting and no d–d transition, so the anhydrous salt is white.
NCERT (2026-27 edition):
5.5.5 Colour in Coordination Compounds – “…The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.”
“…in the absence of ligand, crystal field splitting does not occur and hence the substance is colourless. For example, removal of water from [Ti(H2O)6]Cl3 on heating renders it colourless. Similarly, anhydrous CuSO4 is white, but CuSO4.5H2O is blue in colour.”