The power of halides of boron to act as Lewis acids decreases in the order :

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

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Solution

The correct option is A $B B r_{3} > B C l_{3} > B F_{3}$
We know that the electronegativity of halogens decreases from fluorine to iodine as we move down the group. Hence, according to this, $B F_{3}$ should be the most Lewis acidic due to the electron withdrawing nature of fluorine atoms.
However, the order is just the opposite due to back-bonding. Back bonding due to the lone pair of electrons of halogen atom increases electron density on $B$ centre, thereby, reducing its lewis acid character. This back bonding is more in case of $B F_{3}$ due to similar size of orbitals of boron and fluorine atoms, whereas it decreases down the group due to the increasing size of orbitals as we move from $F$ to $I$ .
The extent of back bonding follows the order $B B r_{3} > B C l_{3} > B F_{3}$ , as orbital match follows the order $2 p - 2 p > 3 p - 2 p > 4 p - 2 p$ .
Hence, the correct order is $B B r_{3} > B C l_{3} > B F_{3}$ .

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