A) \[N{{H}_{3}}\]
B) \[PC{{l}_{3}}\]
C) \[BC{{l}_{3}}\]
D) \[Cl{{F}_{3}}\]
Correct Answer: C
Solution :
\[BC{{l}_{3}}\] has not the complete octet in boron (B), hence, it is electron deficient compound \[Cl\dot{\times }\overset{Cl}{\mathop{\overset{{\dot{\times }}}{\mathop{B}}\,}}\,\,\dot{\times }Cl\]
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