# IBrF2 - van der Waals repulsions - and number of unique bond angles

I understand that the molecule $\ce{IBrF2}$ is T-shaped, with three ligand atoms and two non-bonding pairs.

However, how many unique bond angles can this molecule exhibit?

My professor argues for the existence of only 2 unique bond angles in this molecule, with the somewhat $< 180^\circ$ axial $\ce{F-I-F}$ bond angle and the two equivalent and somewhat $< 90^\circ$ $\ce{F-I-Br}$ bond angles.

However, can't we also arrange this molecule as to have this axial arrangement of atoms: $\ce{Br-I-F}$?

My professor argues that this arrangement is impossible due to van der Waals repulsions. I suppose this makes sense; the two lone-pairs $\ce{<->}$ large bromine atom repulsions would make this shape unfavorable.

Impossible? I'm not sure about that. Dammit, nothing in chemistry is impossible.

What do you think? I'm arguing for the existence of either two or three bond angles because we can either have a $\ce{F-I-F}$ axial arrangement or a $\ce{Br-I-F}$ axial arrangement. Plus, axial bonds are longer than equatorial bonds. So that would lower the energy of the $\ce{Br-I-F}$ axial conformer. Side note: why are axial bonds longer than equatorial bonds?

In sum:

• Certainly not impossible. I bet they're relatively close in energy. With the bromine axial you replace a 90 degree F-lone pair interaction with a 90 degree Br(longer bond)-lone pair interaction - that might be energetically favorable. Maybe Martin can do a calculation? – ron Jul 26 '14 at 23:00
• Right, I was saying that the axial bonds are generally longer than the equatorial bonds. This length also minimizes repulsions. Side question: why are axial bonds longer than equatorial bonds? – Dissenter Jul 26 '14 at 23:06
• I could very well be mistaken here, and literature on ternary interhalogens is virtually nonexistent, but my suspicion would be that there's a non-trivial energy difference between the two conformers, with the equatorial bromine favorable chiefly for steric reasons. – Greg E. Jul 26 '14 at 23:08
• @Dissenter Equatorial bonds have fewer destabilizing 90 degree interactions than axial bonds. Therefor, they are at lower potential energy, e.g. more stable. That translates into shorter bond length. – ron Jul 26 '14 at 23:22
• @Dissenter Yes. Jahn-Teller distortion in octahedral coordination compound usually, but not invariably, works in the direction of a lengthening of axial bonds (here defined as the 2-out-of-6 ligands, as opposed to the equatorial 4 remaining). Exceptions include $\ce{KCu3(OH)2(H[AsO4]2)}$ (H. Effenberger, Z. Kristallogr., 1989, 188, 43); see Figure 5 here: pubs.rsc.org/en/content/articlehtml/2003/dt/b309242a – Silvio Levy Jul 27 '14 at 17:43

I suspect the molecule $\ce{IBrF2}$ does not exist, and is purely conjectural (theoretical) at this point. If it ever had been synthesized or observed, it would have made the cover of Science or Nature. "First neutral ternary interhalogen compound found!" (A handful of anionic ternary interhalogen species are known.)
This is true in binary compounds like $\ce{PCl5}$, and apparently a sufficient explanation is that each axial chlorine wants to avoid 3 other chlorines at 90$^\circ$ angles, while the equatorial chlorines have 2 at 90$^\circ$ but the other 2 are further away anyway, at 120$^\circ$. In other words, if you just want to maximize the average Cl-Cl distance, subject to the constraint that the P-Cl distance is reasonable, you're better off with a slightly elongated bipyramid instead of one inscribed in a sphere.
In the hypothetical $\ce{IBrF2}$ of course the axial I-F bond would be shorter than the "equatorial" I-Br bond, since Br is a bigger atom.