Aluminium chloride in acidified aqueous solution forms an ion having geometry

Solution

Chemistry ma’am · 3 steps

In acidified water, Al³⁺ is hydrated: six water molecules surround it, so the ion is octahedral.

1. What forms. In acidified water, AlCl₃ gives free Al³⁺ ions. The acid stops them from hydrolysing to Al(OH)₃. The small, highly charged Al³⁺ pulls in water molecules and forms the hydrated ion [Al(H₂O)₆]³⁺.

   AlCl₃ + 6H₂O → [Al(H₂O)₆]³⁺ + 3Cl⁻

2. Count electron pairs. Each water molecule donates one lone pair to aluminium, so Al has 6 bond pairs and 0 lone pairs. By VSEPR, six bond pairs spread out as far apart as possible, which gives an octahedron (sp³d² hybridisation).
3. So. The ion [Al(H₂O)₆]³⁺ is octahedral, which is option (b).

The idea to keep. Ask which ion actually exists in that medium before you apply VSEPR. In water, Al³⁺ is [Al(H₂O)₆]³⁺, which is octahedral. In strong base it is [Al(OH)₄]⁻, which is tetrahedral.

Answer from the NTA final answer key. Solution by PrepWiser.