The trichlorides of group 13 being covalent in nature, form tetrahedral [M(OH)4]–,on hydrolysis in water.
When BCl3 is treated with water, it hydrolyses and forms [B[OH]4]–
BCl3 + 3H2O→ B(OH)3 + 3HCl
B(OH)3 + H2O→[B(OH)4]- + H+
The Boron in the [B[OH]4]– ion involves one 2s orbital and 3 2p orbitals.
Thus, the hybridization of the Boron atom is sp3.