# Does iron(VI) fluoride exist?

Why does $$\ce{FeF6}$$ not exist? There are hexavalent iron compounds, so that is not the problem. There is the $$\ce{[Fe(CN)6]^3-}$$ ion, so there doesn't seem to be a steric problem.

• Maybe reading about bonding characteristics of Potassium Ferrate may help. Apr 17, 2021 at 4:43
• No, it doesn't, as far as it is possible to prove a negative. Nor do CrF6, MnF7 or CoF5. And note your steric argument is not correct, Fe3+ is appreciably bigger than Fe6+, as far as those terms have any meaning in chemical environments. Apr 17, 2021 at 8:51
• Not an answer. But $\ce{FeF3}$ as a crystal already is something like $\ce{FeF6}$, in the sense that each iron is bound to six fluorine atoms; the lattice is linked via $\ce{-F-}$ bridges. The existence of $\ce{[Fe(CN)6]^3-}$ is more of an argument for $\ce{[FeF6]^3-}$ than $\ce{FeF6}$. Apr 17, 2021 at 12:50
• Food for thought: The dimer of iron(III) fluoride doesn't exist(unlike iron(III) chloride), but the anion, $\ce{[Fe2F6]^-}$ does exist consisting of $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions and it form salts like $\ce{[NH4]Fe2F6}$. Apr 22, 2021 at 15:04

Besides steric factors related to the small size of the transition metal core, we could be seeing an electronic effect described in this answer. Iron is fairly early in the transition metal series, so when pushed beyond the $$+3$$ oxidation state it has few $$d$$ electrons in the central core. As explained in the referenced answer, this makes the iron strongly pi-electron accepting, and therefore more favorable for combining with a stronger pi-donor ligand such as oxide instead of fluoride.
• @ThomasForster Not exactly FeO2F2 but there are similar types of compounds: $\ce{Fe3(OF)2}$, $\ce{Li2Fe4OF8}$, $\ce{FeSbO2F2}$. Apr 22, 2021 at 14:57