Question

When BCl₃ is treated with water, it hydrolyses and forms [B[OH]₄]^– only whereas AlCl₃ in acidified aqueous solution forms [Al (H₂O)₆]³⁺ ion. Explain what is the hybridisation of boron and aluminium in these species?

Answer 1

BCl_3 + 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, hybridisation of B in [B(OH)4]^- is sp³.