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I faced a question like this:
The number of $90$ degree $\ce{F-Br-F}$ angle in $\ce{BrF5}$ according to VSEPR theory is:
The answer is given $0$ or $8$. I know $\ce{BrF5}$ has $\pu{sp3d2}$ hybridization. But there is no $90$ degree $\ce{F-Br-F}$ angle in this geometry due to the lone pair electron in $\ce{Br}$. But I find no other alternative structure to present $8$ $\ce{F-Br-F}$ angle of $90$ degree.
Any hints?
The starting structure would be an octahedron with 5 of the 6 positions occupied by an F atom and the last being the lone pair. Since the F is quite small, and VSEPR is "all about" lone pair repulsion, you can assume the lp would push the four equatorial F atoms up away from it (assuming it is in the south polar position) and so their angles would be less than 90° (both their angles between two adjacent equatorial Fs and their angles to the (north) polar F). According to various internet sources (of questionable reliability) the angles are nearly 90° anyways: 86° and 85° respectively. The key thing would be to realize this is square pyramidal rather than a triangular bipyramid. I think it's arguable whether or not VSEPR would or would not "predict" the bond angles to within a couple of degrees, and that's the difference we're talking about here (between 85° and 90°), so I'd characterize this as a very poor homework "first principles" question. I think it would be in the "too close to call" category.
mhchem. You've written over 38 posts, so by now you should be aware how to use proper formatting. Please respect other people's time by taking the trouble to make the post as readable as it can be. $\endgroup$