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

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The $B-F$ bond length in $BF_{3}$ is shorter than the $B-F$ bond length in $BF_{4}^{-}$.
$BF_{3}$ is an electron-deficient species. With a vacant $p$-orbital on boron, the fluorine and boron atoms undergo $p\pi-p\pi$ back-bonding to remove this deficiency. This imparts a partial double bond character to the $B-F$ bond.
This double-bond character causes the bond length to shorten in $BF_{3}$ $(130 \ pm)$.
However, when $BF_{3}$ coordinates with the fluoride ion, a change in hybridisation from $sp^{2}$ (in $BF_{3}$) to $sp^{3}$ (in $BF_{4}^{-}$) occurs.
Boron now forms $4 \sigma$ bonds and the double-bond character is lost. This accounts for a $B-F$ bond length of $143 \ pm$ in the $BF_{4}^{-}$ ion.

Explore More

Similar Questions

Which molecule has $sp^3$ hybridization of the central atom?

The bond length between an $sp^{3}$ hybridized carbon atom and another carbon atom is minimum in:

With which of the given pairs does $CO_2$ resemble in terms of structure?

Which among the following molecules have $sp^3d$ hybridisation with one lone pair of electrons on the central atom?
$(i)\ SF_4$
$(ii)\ [PCl_4]^+$
$(iii)\ XeO_2F_2$
$(iv)\ ClOF_3$

The hybridisation of the central atom in $NF_3$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo