question_answer

The correct order of bond angles (smallest first) in\[{{H}_{2}}S,N{{H}_{3}},B{{F}_{3}}\]and\[Si{{H}_{4}}\]is

A) \[{{H}_{2}}S<Si{{H}_{4}}<N{{H}_{3}}<B{{F}_{3}}\]

B) \[N{{H}_{3}}<{{H}_{2}}S<Si{{H}_{4}}<B{{F}_{3}}\]

C) \[{{H}_{2}}S<N{{H}_{3}}<Si{{H}_{4}}<B{{F}_{3}}\]

D) \[{{H}_{2}}S<N{{H}_{3}}<B{{F}_{3}}<Si{{H}_{4}}\]

Correct Answer: C

Solution :

Species	Structure	lp	bp	VSEPR	Bon Bangla
\[{{H}_{2}}S\] \[N{{H}_{3}}\]		2 1	2 3	\[lp-lp\] \[lp-bp\] \[bp-bp\] \[lp-bp\] \[bp-bp\]	\[{{90}^{o}}\] \[{{107}^{o}}\]
\[B{{F}_{3}}\] \[Si{{H}_{4}}\]		0 0	3 4	\[bp-bp\] \[bp-bp\]	\[{{120}^{o}}\] \[{{109}^{o}}\]

Thus, bond angle\[{{H}_{2}}S<N{{H}_{3}}<Si{{H}_{4}}<B{{F}_{3}}\]