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: