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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?

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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.

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