Question

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

• A. $B{F}_3<P{F}_3<CI{F}_3$
• B. $P{F}_3<B{F}_3<CI{F}_3$
• C. $CI{F}_3<P{F}_3<B{F}_3$
• D. $B{F}_3=P{F}_3=CI{F}_3$

Answer 1

The correct option is C $CI{F}_3<P{F}_3<B{F}_3$

The bond angle in $B{F}_3$ is ${120}^{\circ }$ since B is $s{p}^2$ - hybridized. In $P{F}_3$, P is $s{p}^3$ - hybridized but the bond angle is little less that ${109}^{\circ }{.28}^{{}^{\prime }}$ (but far greater than ${90}^{\circ }$) due to bond pair - lone pair repulsion. In $CI{F}_3,Cl$ is $s{p}^3$ d hybridized. It has T - shape and hence the bond angle is around ${90}^{\circ }$. Therefore the correct order of increasing bond angle is $CI{F}_3<P{F}_3<B{F}_3$