BCl₃ on hydrolysis forms

${\mathrm{BCl}}_3+3{\mathrm{H}}_2\mathrm{O}\to \mathrm{B}{\left(\mathrm{OH}\right)}_3+3\mathrm{HCl}$

$\mathrm{B}{\left(\mathrm{OH}\right)}_3+{\mathrm{H}}_2\mathrm{O}\to {\left[\mathrm{B}{\left(\mathrm{OH}\right)}_4\right]}^{-}+{\mathrm{H}}^{+}$

$\mathrm{B}{\left(\mathrm{OH}\right)}_3$ due to its incomplete octet accepts an electron pair (as ${\mathrm{OH}}^{-}$) to give ${\left[\mathrm{B}{\left(\mathrm{OH}\right)}_4\right]}^{-}$.

Boron in this ion involves one 2s orbital and three 2p orbitals. Thus, hybridization of B in ${\left[\mathrm{B}{\left(\mathrm{OH}\right)}_4\right]}^{-}$ is ${\mathrm{sp}}^3$ and its shape is tetrahedral.