Question

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

• A. 0.0
• B. 6
• C. 8
• D. 4

Answer 1

The correct option is A 0.0


In $Br{F}_5$, there are 5 bond pairs and 1 lone pair on $Br$ atom and hence the hybridization is $s{p}^3{d}^2$, 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 ${90}^{\circ }$.