A) \[BC{{l}_{3}}>B{{F}_{3}}BB{{r}_{3}}\]
B) \[BB{{r}_{3}}>BC{{l}_{3}}>B{{F}_{3}}\]
C) \[BB{{r}_{3}}>B{{F}_{3}}>BC{{l}_{3}}\]
D) \[B{{F}_{3}}>BC{{l}_{3}}>BB{{r}_{3}}\]
Correct Answer: B
Solution :
As the size of halogen atom increases, the acidic strength of boron halides increases. Thus, \[\text{B}{{\text{F}}_{\text{3}}}\] is the weakest Lewis acid. This is because of the \[p\pi -p\pi \] back bonding between the fully-filled unutilized 2p orbital?s of F and vacant 2p orbitals of boron which makes \[\text{B}{{\text{F}}_{\text{3}}}\] less electron deficient. Such back donation is not possible in case of \[\text{BC}{{\text{l}}_{\text{3}}}\] or \[\text{BB}{{\text{r}}_{\text{3}}}\] due to larger energy difference between their orbitals. Thus, these are more electron deficient. Since on moving down the group the energy difference increases, the Lewis acid character also increases. Thus, the tendency to behave as Lewis acid follows the order \[\text{BB}{{\text{r}}_{\text{3}}}>\text{BC}{{\text{l}}_{\text{3}}}\,>\text{B}{{\text{F}}_{\text{3}}}\]
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