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 incorrect and statement-II is correct

Both statements I and II are incorrect

Statement-I is correct and statement-II is incorrect

answer is D.

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Detailed Solution

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 t_2g and e_g. The unpaired electron undergoes d-d transitions due to which the complex absorbs red colour and emits blue-green colour.

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