A) \[{{H}_{2}}S<N{{H}_{3}}<Si{{H}_{4}}<B{{F}_{3}}\]
B) \[N{{H}_{3}}<\text{ }{{H}_{2}}S\text{ }<\text{ }Si{{H}_{4}}<\text{ }B{{F}_{3}}\]
C) \[{{H}_{2}}S\text{ }<\text{ }Si{{H}_{4}}<\text{ }N{{H}_{3}}<\text{ }B{{F}_{3}}\]
D) \[{{H}_{2}}S\text{ }<\text{ }N{{H}_{3}}<\text{ }B{{F}_{3}}<\text{ }Si{{H}_{4}}\]
Correct Answer: A
Solution :
\[B{{F}_{3}}>Si{{H}_{4}}>\overset{..}{\mathop{N}}\,{{H}_{3}}>{{H}_{2}}\overset{.\,.}{\mathop{S}}\,:\] \[120{}^\circ \text{ }109{}^\circ 28\text{ }\,\,\,107{}^\circ ~~92{}^\circ \]
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