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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?
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Solution
\[\ce{BCl3 + 3H2O -> B(OH)3 + 3HCl}\]
\[\ce{B(OH)3 + H2O -> [B(OH)4]– + H+}\]
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.
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