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  • When BCl_3 is treated with water, it hydrolyses and forms [B[OH]_4]^- only whereas AlCl_3 in acidified aqueous solution forms [Al (H_2O)_6]^3+ ion.

When BCl3 is treated with water, it hydrolyses and forms [B[OH]_{}4]^{-} only whereasAlCl_{3} in acidified aqueous solution forms [Al (H_2O)_{6}]^{3+} ion. Explain what is the hybridisation of boron and aluminium in these species?

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BCl_{3} + 3H_{2}O\rightarrow B(OH)_{3} + 3HCl
B(OH)_{3} + H_{2}O\rightarrow [B(OH)_{4}]^{-} + H^{+}
B(OH)_{3} in order to complete its octet, accepts an electron pair as ( OH^{-}) to give [B(OH)_{4}]^{-}. Boron here has one 2s orbital and three 2p orbitals. Thus, hybridization of B is sp^{3} in [B(OH)_{4}]^{-}

AlCl_{3} + 6H_{2}O \rightarrow [Al(H_{2}O)_{6}]^{3+} + 3Cl^{-}

The 6 H2O molecules get attached with Al i.e. they donate 6 electron pairs to the 3s,3p and 3d orbital of Al^{3+} ion. Therefore, the hybridization of the Al atom in [Al(H2O)_{6}]^{3+}+  is sp^{3}d^{2}

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