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The correct order of B-F bond length follows the sequence:
(A) $B{F_3}$ < $B{F_2} - OH$ < $B{F_2} - N{H_2}$ < $B{F_4}^ - $
(B) $B{F_2} - N{H_2}$ < $B{F_2} - OH$ < $B{F_3}$ < $B{F_4}^ - $
(C) $B{F_3}$ < $B{F_4}^ - $ < $B{F_2} - OH$ < $B{F_2} - N{H_2}$
(D) $B{F_3}$ < $B{F_2} - N{H_2}$ < $B{F_2} - OH$ < $B{F_4}^ - $

Answer
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Hint: $B{F_3}$undergoes back bonding which leads to double bond character, while $B{F_2} - OH$ and $B{F_2} - N{H_2}$have electronegative atoms O and N involved of which O is more electronegative than N.

Complete step by step answer:
> In $B{F_3}$: It has a $s{p^2}$ hybridised structure and so a trigonal planar structure. B has vacant 2p orbital while F atoms have completely filled unutilised 2p orbitals. Since both of these orbitals are of the same energy level (2p), pπ-pπ back bonding occurs between B and F. In this a lone pair of electrons from the unutilised completely filled orbital of F is transported to the vacant 2p orbital of B atom. This type of back bonding causes the bond between B and F to develop a double bond character. This causes the bond length B-F to be shorter than normal.
Hence it has the shortest bond length here.
> In $B{F_4}^ - $: It is $s{p^3}$ hybridised with a tetrahedral shape and B atom here does not have any vacant 2p orbital left and so there is no scope for back bonding. So, all the B-F bond lengths are purely single bonds.
So, they have a higher bond length than $B{F_3}$.
> Now talking about $B{F_2} - OH$ and $B{F_2} - N{H_2}$: we know that electronegativity of O is much more than that of N.
Electronegativity has a direct effect on the bond length between 2 atoms. A more electronegative atom has a higher tendency to pull the pair of electrons forming the bond towards itself. This causes the bond length to become shorter than usual and since O is more electronegative than N, it attracts the bonding electrons more and so the bond length of B-F is shorter in $B{F_2} - OH$as compared to $B{F_2} - N{H_2}$. But still this bond length is more than double bond length but less than that of single bond.
> So the bond length is smallest in $B{F_3}$, followed by $B{F_2} - OH$, then $B{F_2} - N{H_2}$ and that of $B{F_4}^ - $ is the longest.

Hence the correct option is: (A).

Note: Back bonding occurs only if one of the atoms has a vacant orbital and other atom has a lone pair of electrons to transfer. Both of these should involve orbitals of the same energy only.
Also a more electronegative atom attracts the bonding electron pair towards it more and so has a shorter bond length as compared to lesser electronegative atoms.