Question

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

• A. $[Al(H_2O)_6]^{3+}, sp^3d^2$
• B. $[Al(H_2O)_4]^{3+}, sp^3$
• C. $[Al(H_2O)_4]^{3+}, dsp^2$
• D. $[Al(H_2O)_6]^{3+}, d^2sp^3$

Answer 1

The Correct Option is A

To determine the complex formed by aluminium chloride in an acidified aqueous solution, it is important to understand the nature of aluminium as a metal ion and its typical coordination behavior.

Step 1: Understand the coordination behavior of Al^{3+}
The aluminium ion, Al^{3+}, is a small and highly charged cation. It tends to form coordination complexes with ligands, typically water molecules, because they can donate electron pairs to the aluminium ion.

Step 2: Formation of the complex
In an aqueous solution, the aluminium ion will coordinate with water molecules. Each water molecule acts as a ligand and donates an electron pair to the aluminium ion, filling its coordination sphere. The typical coordination number for Al^{3+} is 6, leading to the formation of a hexa-aqua complex:

[Al(H_2O)_6]^{3+}

Step 3: Hybridization of [Al(H_2O)_6]^{3+}
The next step is to determine the hybridization state of the aluminium ion in the complex. The hexa-coordination involves 6 ligands, suggesting octahedral geometry. For this geometry, the hybrid orbitals involved are sp^3d^2:

This involves the promotion of an electron from the 3s orbital to the 3d orbital in order to provide 6 equivalent hybrid orbitals for bonding, specifically under conditions that support such hybridization.

Conclusion
Thus, the correct complex and hybridization state for aluminium chloride in acidified aqueous solution is:

[Al(H_2O)_6]^{3+}, sp^3d^2