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

\[\ce{BCl3 + 3H2O -> B(OH)3 + 3HCl}\]

\[\ce{B(OH)3 + H2O -> [B(OH)4]– + H+}\] 

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.

\[\ce{AlCl3 + 6H2O ->[HCl] [Al(H2O)6]^{3+} + 3Cl-}\]

Hence, hybridisation of Al is sp³d².