The chemical formula of copper sulphate pentahydrate, $\ce{CuSO4\cdot 5H2O}$ does not give information about its true structure because it's only an empirical formula.
It forms a network solid. Two units of sulphate ions are bonded with central metal ion $\ce{Cu(II)}$. But each $\ce{SO4^{2-}}$ ions are bonded to two $\ce{Cu^2+}$. Hence, the average number of sulphate ion per cupric ion is one. Similarly, the average number of water molecules per $\ce{CuSO4}$ unit is five. Out of which, one molecule is H-bonded.
Upon heating, first it loses two molecules of water then two more molecules of water and finally the remaining one in following steps:$^1$
$\underset{\underset{\text{(Crystal)}}{\text{Blue}}}{\ce{CuSO4\cdot 5H2O (s)}}\ce{\color{red}{->[63 ^\circ C][-2H2O]}}\underset{\text{Light blue}}{\ce{CuSO4\cdot 3H2O (s)}}\ce{\color{red}{->[109 ^\circ C][-2H2O]}}\underset{\text{Very light blue}}{\ce{CuSO4\cdot H2O (s)}}\ce{\color{red}{->[200 ^\circ C][-H2O]}}\underset{\underset{\text{(amorphous)}}{\text{Grey-white}}}{\ce{CuSO4 (s)}}$
1- Why removal of water molecules following this pattern: 2, 2, 1? Every hydrated salt follows a pattern, e.g., sodium carbonate decahydrate loses 9 molecules of water at once to form monohydrate.
2- In $\ce{CuSO4\cdot H2O}$, the water molecule must be hydrogen bonded but why the coordinated bonded water molecules are breaking bonds prior to H-bonded water? Coordinate bonds are much stronger than H-bonds even if we consider that one water molecule is making multiple H-bonds.
3- What can we say about the structure of $\ce{CuSO4\cdot 3H2O}$?
Reference
- Page 766, Inorganic Chemistry (4E) By Catherine Housecroft, Pearson- 2012, ISBN: 9780273742784
