$2{\mathrm{BF}}_3+6\mathrm{NaH}\to \underset{\left(\mathrm{A}\right)}{{\mathrm{B}}_2{\mathrm{H}}_6}+6\mathrm{NaF}$

In this reaction hydride ions replace the fluoride ions. and diborane is formed, which is a Lewis acid due to incomplete octet. It breaks symmetrically with bases like tertiary amines.

${\mathrm{B}}_2{\mathrm{H}}_6+2{\mathrm{Me}}_3\mathrm{N}\to \underset{\left(\mathrm{B}\right)}{2{\left({\mathrm{Me}}_3\mathrm{N}\right)}^{+}}{\left({\mathrm{BH}}_3\right)}^{-}$ 

In this product geometry around Boron will be tetrahedral. And it will have ${\mathrm{sp}}^3$ hybridisation.