A) \[N{{H}_{3}}\]
B) \[Be{{F}_{2}}\]
C) \[{{H}_{2}}O\]
D) \[C{{H}_{4}}\]
Correct Answer: C
Solution :
Due to \[lp-lp\] repulsions, bond angle in \[{{H}_{2}}O\] is lower \[({{104}^{o}}{{.5}^{o}})\] than that in \[N{{H}_{3}}\]\[({{107}^{o}})\] and \[C{{H}_{4}}({{109}^{o}}2{8}')\]. \[Be{{F}_{2}}\] on the other hand, has sp-hybridization and hence has a bond angle of \[{{180}^{o}}\].
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