My book says that the structure of blue vitriol is the following:

Blue Vitriol Structure

From the structure, I figured that since there are 4 water molecules coordinated to the cuprate ion, the hybridization should be $sp^3$. But when I googled, I saw that people were saying its hybridization is $sp^3d^2$. Why would that be true?

  • $\begingroup$ Can the downvoter explain what mistake I made? $\endgroup$ Sep 17 at 13:14

enter image description here

The picture you provided shows half the story.

In the diagram above you can see that copper has a charge of +2 i.e. Cu$^{2+}$ which leads to a 3d$^9$ configuration. Since H$_2$O will here act as a weak field ligand, no pairing of electrons will be done in d-orbitals which will lead to sp$^3$d$^2$ hybridization.

enter image description here

  • 1
    $\begingroup$ Not a believer in hybridization of transition metals, but the structure is worth a +1. $\endgroup$ Sep 17 at 10:47
  • $\begingroup$ I think sulphate ion doesn't form any coordinate covalent bond with cupper. But there exists an ionic bond between [Cu(H20)4]+2 and (SO4)-2. Because coordination number of Cu is 4 not 6. $\endgroup$
    – Infinite
    Sep 17 at 15:27

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