Question:
Aluminium chloride in acidified aqueous solution gives a complex ion “X”. The hybridization of the central atom in X is:
Aluminium chloride in acidified aqueous solution gives a complex ion “X”. The hybridization of the central atom in X is:
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For octahedral complexes (coordination number 6), the hybridization of the central atom is generally sp\(^3\)d\(^2\).
Updated On: Jun 26, 2026
- sp\(^3\)
- sp\(^2\)d
- d\(^2\)sp\(^3\)
- sp\(^3\)d\(^2\)
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The Correct Option is D
Solution and Explanation
Step 1: Understand the coordination.
Aluminium chloride forms a complex ion [AlCl\(_4\)]\(^-\) in acidified aqueous solution. The aluminium atom is surrounded by six ligands when considering hydrolysis to [Al(H\(_2\)O)\(_6\)]\(^3+\) in acidic solution.
Step 2: Determine the coordination number.
The central Al atom is coordinated to six ligands → coordination number = 6.
Step 3: Relate coordination to hybridization.
For coordination number 6, the central atom uses one s, three p, and two d orbitals for bonding.
Step 4: Write the hybridization.
\[ \text{Hybridization} = sp^3d^2 \]
Step 5: Conclusion.
The central atom in the complex ion X has \(\boxed{sp^3d^2}\) hybridization.
Aluminium chloride forms a complex ion [AlCl\(_4\)]\(^-\) in acidified aqueous solution. The aluminium atom is surrounded by six ligands when considering hydrolysis to [Al(H\(_2\)O)\(_6\)]\(^3+\) in acidic solution.
Step 2: Determine the coordination number.
The central Al atom is coordinated to six ligands → coordination number = 6.
Step 3: Relate coordination to hybridization.
For coordination number 6, the central atom uses one s, three p, and two d orbitals for bonding.
Step 4: Write the hybridization.
\[ \text{Hybridization} = sp^3d^2 \]
Step 5: Conclusion.
The central atom in the complex ion X has \(\boxed{sp^3d^2}\) hybridization.
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