# Why can't O form OF6?

At first I thought that the reason is what I was taught, the absence of vacant d-orbitals but then I did some research. Most of the answers stated what I initially thought but one particular answer told that it's a common misconception (d-orbital theory). According to it, there can be two possible reasons-

1. Not much electronegativity difference between O and F.

2. The atomic size of Oxygen is not sufficient for it to hold 6 atoms.

So which one is likely to be more appropriate, the d-orbital theory or the electronegativity and atomic size theory?

• See this question and the links therein. – Nilay Ghosh Sep 11 '19 at 7:27
• Specifically this answer: chemistry.stackexchange.com/a/100938 – Karsten Theis Sep 11 '19 at 10:11
• While "OF6" looks pretty absurd, there's possibility of existence of OF3+ cation which would have IV+ ox. state for oxygen! – Mithoron Sep 11 '19 at 14:02
• @mithoron wow! Inquiring minds want to know your references! – Oscar Lanzi Sep 11 '19 at 22:34
• – Nilay Ghosh Sep 12 '19 at 1:55

This reference published by the University of British Columbia gives the molecular orbitals for sulfur hexafluoride. A key feature of these molecular orbitals is a doubly degenerate pair, labeled $$\mathrm{2e_g}$$ in the above source, which effectively contains the "extra" valence electrons in fluorine-based orbitals.
The picture below, taken from a different source due to limited downloading options from the free one, includes one of these combinations, shows thus orbital (here labeled $$\mathrm{3e_g}$$, as different nomenclature is used) among others.