# 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?

## 1 Answer

First, you would need to find out what the bonding situation is at all; i.e. how many bonds of which type will be formed. You can use a scheme like the one I outlined in this answer. This should give you three bonding electron pairs and 14 lone pairs. Since we have more atom connections to make than we have bonding electron pairs, we cannot consider only 2-electron-2-centre bonds as would be the case e.g. in iodate. We must utilise one 4-electron-3-centre bond to be formed from one of the lone pairs and one of the bonding electron pairs. Read more about this type of bond for example in this well-written answer by Ron.

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 $\ce{IO2F2-}$.

• Im sorry but I didn't quite get what 4ec3 is, I didn't find the Wikipedia link helpful and I'm not able to find anything much on google. I read somewhere that it's 4 electrons shared by 3 atoms which I don't understand how it's applied here – Vrisk Sep 16 '17 at 17:24
• @Vrisk Did you also read ron’s answer linked in mine? I think he explains it rather well. – Jan Sep 17 '17 at 3:05
• okay I read it again - P has 5 valence electrons 2 in s and 3 in p, you form 3 sp2 orbitals and you'll have 2 electrons filling that one p orbital then what? How do two chlorines combine with 1 p orbital? How did he get that MO diagram? – Vrisk Sep 17 '17 at 5:21
• @Vrisk The three atoms (in Ron’s case chlorine, phosphorus and chlorine or in this case fluorine, iodine and fluorine) align in a linear fashion. Each contributes a p-type orbital to the bonding situation. Since we have three (atomic) orbitals we are mixing, we must end up with three (molecular) at the end. It is obvious that the highest and lowest must be fully antibonding and fully bonding, respectively. The central orbital must be something different so one in which the two outer orbitals point the same phase to the central atom. – Jan Sep 17 '17 at 11:37
• Since there is no way a p-type orbital can interact in a symmetry-allowed manner towards both sides, there is no contribution by the central atom to the orbital in the middle. However, due to the (weak) bonding interaction of the two terminal aoms, this is a (weakly) bonding orbital. Therefore, we have more bonding electrons than antibonding ones and thus bond formation is favourable. – Jan Sep 17 '17 at 11:39