Why are some anhydrous transition metal salts colored?
Here are some examples I found.
- Anhydrous cobalt(II) chloride ($\ce{CoCl2}$), with sky-blue color.
- Anhydrous chromium(III) chloride ($\ce{CrCl3}$), with purple color.
- Anhydrous copper(II) chloride ($\ce{CuCl2}$), with yellow-brown color.
I already know that complex ion could undergo d-d transition from its unfilled d orbitals, and that's what makes them pretty-colored. (The wavelength of absorbed light corresponds to the gap in energies between d orbitals.)
But in anhydrous salt, which I think isn't a coordination compound, the d orbitals are degenerate. Thus, no energy gap and excitation could occur, no light is absorbed, and the compound would be colorless / white-colored as it reflects all wavelength.
I summarize my understanding in this problem as below.
- Solution of complex ion could give color (with exception of $\ce{[Zn(H2O)6]^2+}$ and some other complex ions). e.g. $\ce{CuSO4 (aq)}$, which form $\ce{[Cu(H2O)6]^2+}$ complex ion with blue color.
- Hydrated salt is indeed a coordination compound (found related question here), so it could give color. e.g. $\ce{CuSO4.5H2O (s)}$ with blue color.
- Anhydrous salt, because it doesn't have any ligands coordinating with the metal ion, it's not a coordination compound. It's a mere ionic salt instead. Shouldn't it be colorless?
Is there any explanation to this case?
Note that in all compounds mentioned above, the metal ions are transition metal. We don't consider s-block metal ions such as $\ce{NaCl}$ and $\ce{MgSO4}$. (Additional question: They should be colorless, am I right?)