An aqueous solution of BCl3 is:

BCl3 + 3H2O→ B(OH)3 + 3HCl B(OH)3 + H2O→[B(OH)4]- + H+ B(OH)3 due to its incomplete octet accepts an electron pair (as OH-) to give [B(OH)4]-. Boron in this ion involves one 2s orbital and three 2p orbitals. Thus, hybridization of B in [B(OH)4]- is sp3.