Question
Question: The correct order of B-F bond length follows the sequence: (A) \(B{F_3}\) < \(B{F_2} - OH\) < \(B{...
The correct order of B-F bond length follows the sequence:
(A) BF3 < BF2−OH < BF2−NH2 < BF4−
(B) BF2−NH2 < BF2−OH < BF3 < BF4−
(C) BF3 < BF4− < BF2−OH < BF2−NH2
(D) BF3 < BF2−NH2 < BF2−OH < BF4−
Solution
BF3undergoes back bonding which leads to double bond character, while BF2−OH and BF2−NH2have electronegative atoms O and N involved of which O is more electronegative than N.
Complete step by step answer:
In BF3: It has a sp2 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 BF4−: It is sp3 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 BF3.
Now talking about BF2−OH and BF2−NH2: 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 BF2−OHas compared to BF2−NH2. 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 BF3, followed by BF2−OH, then BF2−NH2 and that of BF4− 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.