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    I have read somewhere that CuSO4 (white solid), when water is added, forms the complex ion [Cu(H2O)6]2+ and this is why the solution is blue.

    But then I read elsewhere that what actually happens is CuSO4.5H2O is formed, which is blue, and so the solution (if it's a solution is blue), and the crystals themselves are blue as well.

    What actually happens, then? Perhaps my confusion comes because I don't really understand the relationship (even though I know it must exist) between water of crystallisation and complexation with H2O.
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    I think the second one is correct.
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    (Original post by Big-Daddy)
    I have read somewhere that CuSO4 (white solid), when water is added, forms the complex ion [Cu(H2O)6]2+ and this is why the solution is blue.

    But then I read elsewhere that what actually happens is CuSO4.5H2O is formed, which is blue, and so the solution (if it's a solution is blue), and the crystals themselves are blue as well.

    What actually happens, then? Perhaps my confusion comes because I don't really understand the relationship (even though I know it must exist) between water of crystallisation and complexation with H2O.
    When copper(II) ions are surrounded by water ligands the degeneracy of the 'd' orbitals is broken, allowing d-d transitions to absorb energy in the visible region of the spectrum. The colour that is seen is the complementary colour that isn't absorbed.

    This is known as crystal field (or ligand field) splitting.

    The hydrated copper(II) sulhpate crystals have water molecules surrounding the copper ions as does the aqueous complex.
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    Thanks, I got it now.
 
 
 
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