The $\mathrm{B}-\mathrm{F}$ bond length in $\mathrm{BF}_{3}$ is shorter than the $\mathrm{B}-\mathrm{F}$ bond length in $\mathrm{BF}_{4}^{-} . \mathrm{BF}_{3}$ 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 double bond character to the $\mathrm{B}-\mathrm{F}$ bond.

This double-bond character causes the bond length to shorten in $\mathrm{BF}_{3}(130 \mathrm{pm})$. However, when $\mathrm{BF}_{3}$ coordinates with the fluoride ion, a change in hybridisation from $s p^{2}$ (in $\mathrm{BF}_{3}$ ) to $s p^{3}$ (in $\mathrm{BF}_{4}^{-}$) occurs.

Boron now forms $4 \sigma$ bonds and the double-bond character is lost. This accounts for a $B-F$ bond length of $143 \mathrm{pm}$ in $\mathrm{BF}_{4}^{-}$ion.