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 \[\text{As}{{\text{H}}_{\text{3}}}\]is smaller than that of \[\text{N}{{\text{H}}_{3}}\]because electronegativity of As is less than N. [a]\[\underset{\text{Bond}\,\text{angle}\,\,\text{=}\,\,\text{10}{{\text{7}}^{\text{o}}}}{\mathop{N{{H}_{3}}}}\,\] [b]\[\underset{\text{Bond}\,\text{angle}\,\,=\,{{105}^{o}}}{\mathop{{{H}_{2}}O}}\,\] [c] \[\underset{\text{Bond}\,\text{angle}\,=\,{{120}^{o}}}{\mathop{~BC{{l}_{3}}}}\,\] \[\therefore \] Bond angle of \[\text{As}{{\text{H}}_{\text{3}}}\]is greater than all of them.
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