The decreasing order of power of boron halides to act as Lewis acids is :

B F_{3} > B C l_{3} > B B r_{3}

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B B r_{3} > B C l_{3} > B F_{3}

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B C l_{3} > B F_{3} > B B r_{3}

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B C I_{3} > B B r_{3} > B F_{3}

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Solution

The correct option is B $B B r_{3} > B C l_{3} > B F_{3}$
Answer:- (B) $B {B r}_{3} > B {C l}_{3} > B F_{3}$ .
Reason:-
It is due to $p π - p π$ back bonding.
All halides have lone pairs, they donate one lone pair to boron by overlapping with the p orbital of boron. An effective overlap between orbitals decreases due to the large size of p orbital of halides. Hence, electron deficiency of boron atom in its halides is - $B {B r}_{3}, > B {C L}_{3} > B F_{3}$ . A species which is more electron deficient will accept electron easily and hence will be more acidic.
Thus, (B) is the correct answer.

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Similar questions

The correct order of Lewis acidity of $B X_{3}$ molecule is

B B r_{3}, B C l_{3}

B F_{3}

B F_{3}, B C l_{3}

B B r_{3}

B F_{3}

B C l_{3}

B B r_{3}

B I_{3}

B F_{3} < B C l_{3} < B B r_{3} < B I_{3}

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