9

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.

andselisk
  • 37,604
  • 14
  • 131
  • 217
Thomas Forster
  • 463
  • 2
  • 3
  • 1
    Maybe reading about bonding characteristics of Potassium Ferrate may help. – Rishi Apr 17 '21 at 04:43
  • 2
    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. – Ian Bush Apr 17 '21 at 08:51
  • 2
    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 $\ce{FeF6}$. – Linear Christmas Apr 17 '21 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 $\ce{[NH4]Fe2F6}$. – Nilay Ghosh Apr 22 '21 at 15:04

1 Answers1

5

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.

Oscar Lanzi
  • 56,895
  • 4
  • 89
  • 175