The pentahydrate salt of copper (II) sulfate exists as a blue solid while the anhydrous and monohydrate salts appear as white solids. The structures of these salts are discussed here. The blue colour of the copper (II) sulfate pentahydrate salt arises due to a similar reason why $\ce {Cu^2+}$ in aqueous solution is blue in colour. Based on Crystal Field Theory, this is due to the phenomena of d orbital splitting in the octahedral complex. As discussed in the post linked, in the pentahydrate salt, the $\ce {Cu^2+}$ ion is octahedrally coordinated, with 4 $\ce {H2O}$ molecules forming dative covalent bonds with it while the other two sites are occupied by the oxygen atoms of the sulfate ion. This leads me to wonder if the sulfate ions contribute to any significant degree of splitting of the d orbitals of the $\ce {Cu^2+}$ ion. As the colour of the anhydrous $\ce {CuSO4}$ salt is white, we can infer that the sulfate ions do not contribute to the d orbital splitting. Consequently, we can infer that the sulfate ions do not really form dative covalent bonds with the $\ce {Cu^2+}$ ion.

But why would this be the case? In the sulfate ion, there is rather poor delocalisation of the lone pair due to long $\ce {S-O}$ bond lengths and thus, ineffective p orbital overlap between the p orbitals of the sulfur and oxygen atoms. Hence, the negative charges, as well as the lone pairs of electrons, are rather localised on the oxygen atoms. This analysis all points to the conclusion that the sulfate ion should be able to act as a ligand.

Similarly, the fact that anhydrous, neutral $\ce {CuCO3}$ exists as a gray solid, suggests that the $\ce {Cu^2+}$ ion does not have dative covalent bonds with the carbonate ion , i.e. $\ce {CO3^2-}$ does not act as a ligand. Although for the carbonate ion, due to the stronger delocalisation of electrons within the ion, it may be more convincing to say that it may not be able to act as a ligand effectively.

  • $\begingroup$ No, you can't infer that if anhydrous "salts" are colorless then there's no dative bonds, on contrary, every single transition metal compound has more or less covalent bonds, often more covalent then ionic. $\endgroup$
    – Mithoron
    Commented Oct 16, 2019 at 17:25

1 Answer 1


First and hopefully obvious things first: sulphate ions do form (weak) coordinate bonds to metal centres and thus by definition cause some d orbital splitting. In a simplified crystal field model, you might imagine them as weaker negative charges but nonetheless the do interact or we would not see the structures we see.

Comparing the structures of anhydrous $\ce{CuSO4}$ and its pentahydrate is very much non-trivial. For starters, they don’t even share the same surrounding. While the copper in $\ce{CuSO4 . 5 H2O}$ is approximately in a $D_\mathrm{4h}$ environment with $\ce{Cu-OH2}$ bonds of $\pu{197pm}$ and approximately $\pu{195}$ for the two different copper atoms in the unit cell, the environment of copper in anhydrous sulphate is better described as $D_\mathrm{2h}$ because the $\ce{Cu-O}$ distances average out to $\pu{187pm}$, $\pu{215pm}$ and $\pu{236pm}$, respectively—around the same copper atom. Two symmetry-identical oxygens with a $\ce{O-Cu-O}$ angle of approximately $180^\circ$ will be equidistant, all others will not.

That already means the corresponding molecular orbital schemes will look very different, even if we treat the oxygen atoms identically. Group theory tables tell us, that while in the case of the pentahydrate at least some degenerate orbitals remain, all orbitals will be independent in the case of the anhydrous sulphate. This in turn means that we are dealing with a potentially much more complicated MO scheme and all excitation discussions will get progressively more complicated. I would not be willing to base an argument on low coordinative ability of sulphate only on the difference between these two crystal structures.

Finally note that the carbonate $\ce{CuCO3}$ itself opens another can of worms: as Seidel et al. show, the copper is in a pentacoordinated environment so we cannot even use our usual octahedral arguments there.


  • Structure of $\ce{CuSO4 . 5 H2O}$: J. N. Varghese, E. N. Maslen, Acta Cryst. B 1985, 41, 184–190. DOI: 10.1107/S0108768185001914.

  • Structure of $\ce{CuSO4}$: P. A. Kokkoros, P. J. Rentzeperis, Acta Cryst. 1958, 11, 361–364. DOI: 10.1107/S0365110X58000955.

  • Structure of $\ce{CuCO3}$: H. Seidel, H. Ehrhardt, K. Viswanathan, W. Johannes, Z. Anorg. Allgem. Chem. 1974, 410, 138–148. DOI: 10.1002/zaac.19744100207.


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