$$Statement-I$$ : Anhydrous $$CuSO4$$ is blue $$colouredStatement-II$$ : $$CuSO4.5H2O$$ is colourless

Correct option is 4Both statements I and II are incorrect

Questions

Statement-I : Anhydrous CuSO₄is blue coloured
Statement-II : CuSO₄.5H₂O is colourless

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Both statements I and II are correct

Statement-I is correct and statement-II is incorrect

Statement-I is incorrect and statement-II is correct

Both statements I and II are incorrect

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detailed solution

Correct option is D

Cu+2(3d9) have one unpaired electron. Presence of unpaired electron is not the only condition for a substance to exhibit colour. The ligands (H2O molecules) are absent in anhydrous CuSO4. So the degenracy of 3d orbitals is unaffect. So there is no possibility of d-d transitions. No energy changes occur.∴anhydrous CuSO4 is colourless.In CuSO4 .5H2O, the ligands (H2O molecules) split the 3d orbitals into two sets of orbitals t2g and eg. The unpaired electron undergoes d-d transitions due to which the complex absorbs red colour and emits blue-green colour.