Question

Now let's consider BrF₆^- as VSEPR would predict it's structure. There are two different choices for the structure. What are these two structures and what symmetry point groups do they belong to?

Answer 1

          Evidence that this is indeed the case is provided by SeF₆²⁻, IF₆⁻, and XeF₆ where, because the central atom is larger than the Br atom the fluorine ligands are not quite close-packed, allowing room for some of the nonbonding electron density to move into the valence shell to form what has been called a partial or weak lone pair, resulting in
a small C_3v distortion of the octahedral structure observed for BrF₆⁻. Because of the close, or near close, packing of the ligands none of these molecules has a structure based on a pentagonal bipyramidal valence shell of seven electrons with a fully active lone pair, such as that found for the IF₇ molecule.
            It is interesting to point out that on the basis of the VSEPR model I predicted in 1962 that the XeF₆ molecule would not be found to be octahedral contradicting the claims by theoreticians based on qualitative MO theory, that it would have a regular octahedral geometry. When my prediction was later verified by the determination of the structure by Bartell and Gavin who found it to have a C3v distorted octahedral structure it significantly increased interest in the VSEPR model

1) R.J. Gillespie, in: H. Hymman (Ed.), Noble-Gas Compounds, University of Chicago Press, 1963, p. 333.
2) L.S. Bartell, R.M. Gavin, J. Chem. Phys. 48 (1968) 2466.