Question

Based on VSEPR theory, the number of 90⁰ F-Br-F angles in $Br{F}_5$ is ___

Answer 1

Answer is 0. If you got 4 or 8, then read on.

Let us use VSEPR to figure this out.

The central atom is clearly Br here. Br has 7 valence electrons and we have 5 F atoms that need to be bonded.

So, electron pairs = $\frac{7+5}{2}=6$

But we have only 5 flourine atoms which need bonding. So, 1 pair is left unbonded which is the lone pair.

So, the geometry will come out to be a square bipyramidal structure with a lone pair at the axial position.

Now it might seem like there are 4 F-Br-F bonds which are at 90⁰. But the catch is that the lone pair at the axial position tends to repel all the bond pairs at the equatorial positions. So, all those bond pairs will be slighlty skewed away from the lone pair reducing the angle. So, none of them will be 90⁰