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Correct BF bond length follows the sequence?

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{BF}_{3} < {BF}_{2} - OH < {BF}_{2} - {NH}_{2} < {BF}_{4}^{-}

{BF}_{2} - {{NH}_{2} < {BF}_{2} - OH <}_{} {BF}_{3} < {BF}_{4}^{-}

{{BF}_{3} <}_{} {BF}_{4}^{-} <

{BF}_{2} - OH <_{} {BF}_{2} - {NH}_{2}

{BF}_{3} <_{} {BF}_{2} - {NH}_{2} <

{BF}_{2} - OH <_{} {BF}_{4}^{-}

answer is D.

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Detailed Solution

The planar molecule

{BF}_{3}

contains

{sp}^{2}

hybridised boron. The BF bond develops some double bond characteristics as a result of pipi back bonding (back donating).
Boron in

{sp}^{3}

has hybridised and has a tetrahedral structure in

{BF}_{4}^{-}

. It only has single bonds. The order of B–F bond length follows the sequence because double bonds have shorter bond lengths than single bonds and because N and O exhibit a higher electronegativity factor.

{BF}_{3} <_{} {BF}_{2} - {NH}_{2} <

{BF}_{2} - OH <_{} {BF}_{4}^{-}

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