Q. The correct order of bond angles (smallest first) in $H_{2}S,\ NH_{3},\ BF_{3}$and$\ SiH_{4}$ is
A) $H_{2}S < SiH_{4} < NH_{3} < BF_{3}$
B) $NH_{3} < H_{2}S < SiH_{4} < BF_{3}$
C) $H_{2}S < NH_{3} < SiH_{4} < BF_{3}$
D) $H_{2}S < NH_{3} < BF_{3} < SiH_{4}$
Explanation
(c)
[table] Species | Structure | $lp$ | $bp$ | VSEPR | Bond angle ; $H_{2}S$
$NH_{3}$
$BF_{3}$
$SiH_{4}$ | [image:4_image109.emf][image:4_image110.emf]
[image:4_image111.emf][image:4_image112.emf] | 2
1
0
0 | 2
3
3
4 | $lp - lp$
$lp - bp$
$\ bp - bp$
$lp - bp\$
$bp - bp$
$bp - bp$
$bp - bp$ | $90{^\circ}$
$107{^\circ}$
$120{^\circ}$
$109{^\circ}28''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''$ [/table]
Thus, bond angle $H_{2}S < NH_{3} < SiH_{4} < BF_{3}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.