Question

The correct increasing bond angle among $BF_3, PF_3$ and $ClF_3$ follow the order

• A. $BF_3 < PF_3 < ClF_3$
• B. $PF_3 < BF_3 < ClF_3$
• C. $ClF_3 < PF_3 < BF_3$
• D. All have equal bond angle

Answer 1

Answer: (C)

In $BF_3$, $B$ is $sp^2$ hybridised with bond angle $120^{\circ}C$. In $PF_3$, $P$ is $sp^3$ hybridised but bond angle is less than $109^{\circ}28$ due to lone pair bond pair repulsion. In $ClF_3$, $Cl$ is $sp^3d$ hybridised having $T$-shape with bond angle $90^{\circ}C$.