The correct order of strengths of back-bonding and lewis acidic strengths in Boron trihalides respectively is:

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

and

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

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

and

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

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

and

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

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

and

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

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Solution

The correct option is C $B F_{3} > B C l_{3} > B B r_{3}$ and $B F_{3} < B C l_{3} < B B r_{3}$
Lewis acids are electron pair acceptors.
In $B F_{3}$ , though the partial positive charge on B atom is more due to the high electronegativity of flourine, the tendency to accept a lone pair is reduced because of the ability of the fluorine atoms to donate their own lone pair to the empty boron orbitals forming $p (π) - p (π)$ back bonding. $F$ forms the strongest back bond due to the effective overlap between $2 p (F)$ and $2 p (B)$ orbitals. So, the electron deficiency of boron gets compensated and the lewis acidic character decreases.
But as the size of the orbital increases, the extent of back-bonding decreases. This is due to the ineffective overlap as in case of $3 p a n d 4 p$ obitals of $C l a n d B r$ respectively with that of $2 p$ orbital of $B$ . So, the order of back bonding is $B F_{3} > B C l_{3} > B B r_{3}$ and hence that of lewis acidity is $B F_{3} < B C l_{3} < B B r_{3}$

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