The correct order of B-F bond length follows the sequence:-

B F_{3} < B F_{2} - O H < B F_{2} - N H_{2} < B F_{4}^{-}

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B F_{2} - N H_{2} < B F_{2} - O H < B F_{3} < B F_{4}^{-}

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B F_{3} < B F_{4}^{-} < B F_{2} - O H < B F_{2} - N H_{2}

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B F_{3} < B F_{2} - N H_{2} < B F_{2} - O H < B F_{4}^{-}

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Solution

The correct option is D $B F_{3} < B F_{2} - N H_{2} < B F_{2} - O H < B F_{4}^{-}$
In ${B F}_{3}$ , Boron is ${s p}^{2}$ hybridised and ${B F}_{3}$ is planar molecule. Due to $p i - p i$ back bonding (back donation) the $B F$ bond acquires some double bond character.
In ${B F}_{4}^{-}$ , Boron in ${s p}^{3}$ hybridised and has a tetrahedral shape. It has pure single bonds. Since, double bonds have smaller bond length than single bond and keep in mind the electronegativity factor exhibited by $N$ and $O$ the order of $B - F$ bond length follows sequence.
${B F}_{3} < {B F}_{2} - {N H}_{2} < {B F}_{2} - O H < {B F}_{4}^{-}$

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