Question

The number of ${90}^{\circ }$ bond angles in $Br{F}_5$ is :

• A. 0.0
• B. 4
• C. 6
• D. 8

Answer 1

The correct option is A 0

In $Br{F}_5$ there are 5 bond pairs and 1 lone pair on $Br$ atom.
The geometry and shape are octahedral and square pyramidal respectively. Due to lp-bp repulsions, the bond angles are lower than ${90}^{\circ }$ (as in octahedral).
Hence there are no bond angles with ${90}^{\circ }$