Question

The correct increasing bond angle among $B{F}_3,P{F}_3$ and $Cl{F}_3$ follow the order

• A. $B{F}_3<P{F}_3<Cl{F}_3$
• B. $P{F}_3<B{F}_3<Cl{F}_3$
• C. $Cl{F}_3<P{F}_3<B{F}_3$
• D. All have equal bond angle

Answer 1

Answer: (C)

In $B{F}_3$, $B$ is $s{p}^2$ hybridised with bond angle $120^{\circ }C$. In $P{F}_3$, $P$ is $s{p}^3$ hybridised but bond angle is less than $109^{\circ }28$ due to lone pair bond pair repulsion. In $Cl{F}_3$, $Cl$ is $s{p}^{3}d$ hybridised having $T$-shape with bond angle $90^{\circ }C$.