Question

The hybridized state of Al\(^{3+}\) in the complex ion formed when AlCl\(_3\) is treated with aqueous acid is

• A. sp$^3$
• B. dsp$^2$
• C. sp$^3$d$^2$
• D. sp$^2$d
• E. sp$^2$

Hint:

For main group elements:
• Avoid assigning $d$ hybridization unless necessary
• Use simplified hybridization models

Answer 1

The Correct Option is A

Concept: When AlCl$_3$ reacts with water, it forms a hydrated complex: \[ AlCl_3 + 6H_2O \rightarrow [Al(H_2O)_6]^{3+} \]

Step 1: Identify central metal ion \[ \text{Central ion = Al}^{3+} \] Electronic configuration: \[ Al: 1s^2 2s^2 2p^6 3s^2 3p^1 \] \[ Al^{3+}: 1s^2 2s^2 2p^6 \] No valence electrons left in outer shell

Step 2: Coordination number \[ [Al(H_2O)_6]^{3+} \Rightarrow 6 \text{ ligands} \] So: \[ \text{Coordination number = 6} \]

Step 3: Expected geometry \[ \text{Coordination number 6} \Rightarrow \text{Octahedral geometry} \]

Step 4: Hybridization analysis For octahedral: \[ sp^3d^2 \] BUT IMPORTANT:
• Al is a main group element
• It does not effectively use d-orbitals
• Bonding is better described using outer orbitals only Thus simplified hybridization: \[ sp^3 \]

Step 5: Key reasoning
• Transition metals use $d$ orbitals
• Main group metals like Al use only $s$ and $p$
• Hence actual bonding description simplifies to $sp^3$ Final Answer: \[ \boxed{sp^3} \]