A) \[N{{H}_{3}}\]
B) \[{{H}_{2}}O\]
C) \[BC{{l}_{3}}\]
D) none of these
Correct Answer: D
Solution :
Key Idea: The bond angle of \[As{{H}_{3}}\] is smaller than that of \[N{{H}_{3}}\] because electronegativity of \[As\] is less than\[N\] \[\underset{Bond\,\,angle={{107}^{o}}}{\mathop{(a)N{{H}_{3}}}}\,\underset{Bond\,\,angle={{105}^{o}}}{\mathop{(b){{H}_{2}}O}}\,\underset{Bond\,\,angle={{120}^{o}}}{\mathop{(c)BC{{l}_{3}}}}\,\] \[\therefore \]Bond angle of \[As{{H}_{3}}\] is greater than all of them.
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