Aluminium chloride in acidified aqueous solution forms a complex 'A', in which the hybridisation state of Al is 'B'. What are 'A' and 'B', respectively?

Solution & Explanation

\textbf{Answer:} (A) $[\mathrm{Al}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}$ with $s{p}^3{d}^2$ hybridisation. \textbf{Solution:} In acidified aqueous solution $\mathrm{AlCl}{}_3$ gives the hexaaqua ion $[\mathrm{Al}(\mathrm{H}{}_2\mathrm{O}){}_6]{}^{3+}$, so the coordination number is 6. Al(III) is $[\mathrm{Ne}]3\mathrm{s}{}^{0}3\mathrm{p}{}^{0}3\mathrm{d}{}^0$ — it has no $d$-electrons. To form 6 bonds it must use the empty outer $3d$ orbitals: one $3s$, three $3p$ and two $3d$ orbitals hybridise as $s{p}^3{d}^2$ (outer-orbital complex). Because the $d$-orbitals used are the outer ($3d$) set, the hybridisation is $s{p}^3{d}^2$ (not the inner $d^{2}s{p}^3$). Geometry is octahedral. Hence option (A).