A) \[N{{H}_{3}}>N{{F}_{3}}>P{{F}_{3}}>P{{H}_{3}}\]
B) \[N{{H}_{3}}>P{{F}_{3}}>N{{F}_{3}}>P{{H}_{3}}\]
C) \[N{{F}_{3}}>N{{H}_{3}}>P{{H}_{3}}>P{{F}_{3}}\]
D) \[N{{H}_{3}}>P{{H}_{3}}>N{{F}_{3}}>P{{F}_{3}}\]
Correct Answer: A
Solution :
Bond angle of \[N{{H}_{3}}\] is greater than \[N{{F}_{3}}\]due to more electronegativity of F, the bond pair is more towards fluorine. Thus b.p.-b.p repulsions are less in\[N{{F}_{3}}.\] However, in the molecule \[P{{F}_{3}}.\]is expected to acquire partial double bond character due to resonance and this results in increase of the b.p.-b-p. repulsions to give a higher bond angle. The correct decreasing bond angle order is - \[\underset{(107{}^\circ )}{\mathop{N{{H}_{3}}}}\,\,\,>\,\,\underset{(102{}^\circ )}{\mathop{N{{F}_{3}}}}\,\,\,\,>\,\,\underset{(97{}^\circ )}{\mathop{P{{F}_{3}}}}\,\,\,>\,\,\underset{(94{}^\circ )}{\mathop{P{{H}_{3}}}}\,\]
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