# How do you tell whether fluorine goes to axial or equatorial position in iodine dioxide difluoride?

I need to find the $\ce{F-I-F}$ bond angle of $\ce{[IO2F2]^-}$. But the thing is I can put 2 fluorines in the equatorial or axial positions so I get 3 different answers. I'm guessing they're all three different molecules with slightly different properties and I have to pick out the most stable one. Looking it up on the internet I find the two in axial position which I guess is because the two fluorines pull charge towards themselves and repel the other fluorine so they want to stay far away from each other, am it right?

As we are deaing with a single 4e3c bond, iodine will need to be hybridised as $\mathrm{sp^2}$ or less. This means we have one direction (axial) with a $\mathrm p$ type bonding orbital and one (equatorial) with something like $\mathrm{sp^2}$ or similar. We can now use Bent’s rule to deduce that fluorine should occupy the axial positions as it is the more electronegative substituent.
We could also have arrived there by realising that oxygen hardly ever takes part in 4e3c bonds. These bonds are basically the incorporation of a central atom into an existing $\ce{X-X}$ bond on a formal basis. Oxygen, however, needs two electrons per atom to reach an octet meaning that there needs to be another additional bond. This cannot be the case in simple compounds such as $\ce{IO2F2-}$.