Question

When BC13 is treated with water, it hydrolyses and forms [B(OH)4]- only whereas
AIC13 in acidified aqueous solution forms [A1(H20).] ion, Explain what is the
hybridisation of boron and aluminium in these species?​

Answer 1

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