The correct order of bond angles (smallest first) in $\mathrm{H}_{2}\mathrm{S}$, $\mathrm{NH}_{3}$, $\mathrm{BF}_{3}$ and $\mathrm{SiH}_{4}$ is

(C) $\mathrm{H}_{2}\mathrm{S}<\mathrm{NH}_{3}<\mathrm{SiH}_{4}<\mathrm{BF}_{3}$

(A) $\mathrm{H}_{2}\mathrm{S}<\mathrm{SiH}_{4}<\mathrm{NH}_{3}<\mathrm{BF}_{3}$

(B) $\mathrm{H}_{2}\mathrm{S}<\mathrm{NH}_{3}<\mathrm{BF}_{3}<\mathrm{SiH}_{4}$

(D) $\mathrm{NH}_{3}<\mathrm{H}_{2}\mathrm{S}<\mathrm{SiH}_{4}<\mathrm{BF}_{3}$

JEE Main · 2004 · 24 Apr 2004 · Offline · Chemical Bonding and Molecular Structure

The correct order of bond angles (smallest first) in

H_{2} S

NH_{3}

BF_{3}

and

SiH_{4}

Answer: (C)

Rank these by the geometry about the central atom and the number of lone pairs, since lone pairs compress bond angles most.

H X_{2} S

s p^{3}

with two lone pairs, forcing its angle down to about

9 2^{\circ}

NH X_{3}

s p^{3}

with one lone pair near

10 7^{\circ}

SiH X_{4}

s p^{3}

with no lone pair at the ideal

109. 5^{\circ}

BF X_{3}

s p^{2}

and flat at

12 0^{\circ}

. Smallest to largest, the order of bond angles is

H X_{2} S < NH X_{3} < SiH X_{4} < BF X_{3}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option