A) \[N{{H}_{3}}>PC{{l}_{3}}>BC{{l}_{3}}\]
B) \[BC{{l}_{3}}>N{{H}_{3}}>PC{{l}_{3}}\]
C) \[BC{{l}_{3}}>PC{{l}_{3}}>N{{H}_{3}}\]
D) \[PC{{l}_{3}}>BC{{l}_{3}}>N{{H}_{3}}\]
Correct Answer: B
Solution :
\[BC{{l}_{3}}\] is trigonal in shape where bond angle is \[120{}^\circ \]. In \[N{{H}_{3}}\] and \[PC{{l}_{3}}\] (both are having pyramidal geometry), though the central atom has equal number of lone pairs and bond pairs, the valence shell of P is relatively bigger, so electrons suffers lesser repulsion and thus bond angle is lower. Thus, the correct order of bond angle is \[BC{{l}_{3}}>N{{H}_{3}}>PC{{l}_{3}}\].
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