Solution;
In $\mathrm{BF}_{3}$, the length of the $\mathrm{B}-\mathrm{F}$ bond is more limited than that of the $\mathrm{B}-\mathrm{F}$ bond in $\mathrm{BF}_{4}^{-} \cdot \mathrm{BF}_{3}$ is an animal categories that has an electron inadequacy. The fluorine and boron particles go through pn-pn back-holding with an empty p-orbital on boron to eliminate this inadequacy. This enriches the $B-F$ bond with a twofold bond character.
This twofold bond character causes a shortening of the bond length by $\mathrm{BF}_{3}(130 \mathrm{pm})$. Be that as it may, an adjustment of hybridization from $\mathrm{sp}^{2}$ (in $\mathrm{BF}_{3}$ ) to $\mathrm{sp}^{3}$ (in $\left.\mathrm{BF}_{4}^{-}\right)$occurs when $\mathrm{BF}_{3}$ organizes with the fluoride particle. Boron presently frames 4 bonds and the personality of the twofold bond is lost. This addresses the length of a $143 \mathrm{pm}$ B-F bond in $\mathrm{BF}_{4}{ }^{-}$ion.
