Why are compounds of V5+ and Cr6+ colored? [duplicate]

Question

As far as my knowledge goes, transition metal complex ions show a color due to a transfer of electrons between the levels of split $\mathrm d$-orbitals. In the case of $\ce{V^5+}$ and $\ce{Cr^6+}$ however, there are no electrons in the $\mathrm d$-orbitals, and the electronic configuration of both is similar to $\ce{Ar}$, yet $\ce{V^5+}$ and $\ce{Cr^6+}$ show color(yellow/brown and orange respectively).

Answer 1

Firstly, a note about terminology. As Mithoron noted, the naked ions $\ce{V^5+}$ and $\ce{Cr^6+}$ do not exist because they require a prohibitively large amount of energy to make (just take the sum of the first 5 or 6 ionisation energies of V/Cr to see why). As such your question should really be asking about the vanadium(V) and chromium(VI) oxidation states, which can exist in acidic solution as the oxoanions $\ce{VO2+}$ and $\ce{Cr2O7^2-}$ respectively. The same idea can be extended to the manganese(VII) oxidation state, which exists as $\ce{MnO4^-}$.

Now, on to the question itself. As you rightly note, the central atom in these oxidation states have the same electronic configuration as argon; we usually call it a $\mathrm{d^0}$ configuration to indicate that there are no d electrons. The colours of these species therefore cannot arise from d-d transitions, which are the transfer of electrons from one metal orbital to another metal orbital. They are actually associated with electron transfer from the ligand (in this case, $\ce{O^2-}$) to the central metal atom. The name for this is ligand-metal charge transfer (LMCT).