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

• Correct answer
  A

  $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}, \mathrm{sp}^3 d^2$
• B

  $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{3+}, s p^3$
• C

  $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right]^{3+}, d s p^2$
• D

  $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}, d^2 s p^3$

Solution

Aluminium chloride in acidified aqueous solution forms an octahedral complex. Aqueous solution is mostly water and when the compound is dissolved in acidified aqueous solution, the water fills its vacancies and an octahedral complex, ' $\mathrm{A}^{\prime}$ which is $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is formed.

Thus, the hybridisation state of $\mathrm{Al}$ in $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ is $\mathrm{sp}^3 \mathrm{~d}^2(\mathrm{~B})$.

Hence, the correct answer is option: 'A'.