Arrange the following halides of boron in the increasing order of acidic character. $B F_{3}, B C l_{3}, B B r_{3}, B I_{3}$ .

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Solution

$B F_{3} < B C l_{3} < B B r_{3} < B I_{3}$ . $B I_{3}$ is strongest lewis because of no-back bonding between empty orbital of $B$ and filled p-orbital of $I$ while $B F_{3}$ has strongest backbonding so weakest acid.

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