The correct increasing order for bond angles among $\mathrm{BF}_{3}, \mathrm{PF}_{3}$ and $\mathrm{C} \ell \mathrm{F}_{3}$ is :

(C) $\mathrm{C} \ell \mathrm{F}_{3}<\mathrm{PF}_{3}<\mathrm{BF}_{3}$

(A) $\mathrm{PF}_{3}<\mathrm{BF}_{3}<\mathrm{C} \ell \mathrm{F}_{3}$

(B) $\mathrm{BF}_{3}<\mathrm{PF}_{3}<\mathrm{C} \ell \mathrm{F}_{3}$

(D) $\mathrm{BF}_{3}=\mathrm{PF}_{3}<\mathrm{C} \ell \mathrm{F}_{3}$

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Chemical Bonding and Molecular Structure

The correct increasing order for bond angles among

BF_{3}, PF_{3}

and

C ℓ F_{3}

is :

Answer: (C)

_{3}

is trigonal planar (no lone pair on B), giving a bond angle of

12 0^{\circ}

.

PF

_{3}

is pyramidal (one lone pair on P), with bond angle reduced to about

9 7^{\circ}

.

ClF

_{3}

is T-shaped/see-saw (two lone pairs on Cl), with the F–Cl–F bond angle further compressed to about

87. 5^{\circ}

.

Increasing order:

ClF_{3} < PF_{3} < BF_{3}

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option