Question

Suggest reasons why the B–F bond lengths in $BF_{3}$ (130 pm) and $[BF_{4}]^{-}$ (143 pm) differ?

Answer 1

"The central atom “B”  in $B{ F }_{ 3 }$ undergoes $s{ p }^{ 2 }$ hybridization. Therefore, the shape of the molecule is planar. It has vacant 2p orbital and also F atom has three lone pair of electrons. In $B{ F }_{ 3 } molecule p\pi \quad -\quad p\pi$ bond is formed by overlapping of the empty orbital of boron atoms and one 2p - orbital of fluorine.  

In$[B{ F }_{ 4 }{ ] }^{ - }$, the central atom boron undergoes ${ sp }^{ 3 }$ hybridization. Therefore, it is a tetrahedral molecule.

In$[B{ F }_{ 4 }{ ] }^{ - }$, they do not have vacant orbital to accepts the electrons from the fluorine atoms.

And it have pure single bond.

Hence double bonds are shorter than single bonds, therefore, B -F bond length in$B{ F }_{ 3 }$ is shorter (130 pm) than B-F bond length (143 pm) in $[B{ F }_{ 4 }{ ] }^{ - }$."

Answer 2

Answer:

In BF3 ‘B’ is sp2 hybridised and in BF4– ‘B’ is sp3 hybridised. Thus, the difference in bond length is due to the state of hybridisation

Hope this helps you !!!!