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Question
\[\ce{CuSO4 . 5H2O}\] is blue in colour while \[\ce{CuSO4}\] is colourless. Why?
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Solution
In \[\ce{CuSO4 . 5H2O}\], water acts as ligand as a result it causes crystal field splitting. Hence d–d transition is possible in \[\ce{CuSO4 . 5H2O}\] and shows colour. In the anhydrous \[\ce{CuSO4}\] due to the absence of water (ligand), crystal field splitting is not possible and hence no colour.
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