Q.Suggest reasons why the B–F bond lengths in BF3 (130 pm) and BF4– (143 pm) differ.

Step 1 - BF3: back bonding is possible

Boron is $sp^2$ hybridised, forming three sigma B-F bonds, with one empty, unhybridised $2p$ orbital left over (perpendicular to the molecular plane). Each fluorine atom has filled $2p$ orbitals with lone pairs; one such filled F $2p$ orbital can overlap sideways with boron's empty $2p$ orbital, donating electron density into it ($p\pi$-$p\pi$ back bonding).

Step 2 - Effect on bond length in BF3

This back bonding gives each B-F bond partial double-bond character (in addition to the sigma bond), which shortens and strengthens the bond - the measured B-F bond length in BF3 is 130 pm, shorter than a "normal" single B-F bond would be expected to be.

Step 3 - BF4-: back bonding is no longer possible

When BF3 accepts a fourth $\text{F}^-$ ion (a Lewis base donating its electron pair) into boron's previously empty $2p$ orbital, boron rehybridises to $sp^3$ and now has no vacant orbital left for any pi back-donation. All four B-F bonds in $\text{BF}_4^-$ are therefore pure sigma (single) bonds only.

Step 4 - Result …

Identify boron's empty p orbital in BF3 and how it accepts pi back-donation from F lone pairs (partial double-bond character, shorter bond); note this orbital is consumed forming the 4th si …

• Attributing the bond-length change only to a change in hybridisation (sp2 to sp3) without explaining the underlying p-pi to p-pi back-bonding mechanism actually responsible.
• Getting the comparison backwards (thinking BF4- bonds should be shorter because there are "more bonds" or "more negative charge"). …