Question

A white precipitate was formed when BaCl₂ was added to water extract of an inorganic salt. Further, a gas ‘X’ with characteristic odour was released when the formed white precipitate was dissolved in dilute HCl. The anion present in the inorganic salt is

• A. I^–
• B. \(SO^{2-}_3\)
• C. S^2–
• D. \(NO^{-}_2\)

Answer 1

The Correct Option is B

 To solve this question, we need to determine which anion, when combined with BaCl₂, forms a white precipitate that releases a gas with a characteristic odour when dissolved in dilute HCl. Let's analyze each given option:

1. Evidences and Reactions:
   • BaCl₂ is known to form a white precipitate with certain anions such as sulfates and sulfites.
   • When the white precipitate dissolves in dilute HCl and produces a gas with a characteristic odour, it indicates the release of a specific gas.
2. Anion Analysis:
   • \(SO^{2-}_3\) (Sulfite ion):
     • Reaction: \(BaCl_2 + SO^{2-}_3 \rightarrow BaSO_3 (s) + 2Cl^-\)
     • The white precipitate is BaSO₃ (barium sulfite).
     • Upon adding dilute HCl: \( BaSO_3 + 2HCl \rightarrow BaCl_2 + SO_2 (g) + H_2O\)
     • SO₂ gas is released, which has a characteristic sharp, pungent odour.
   • I^– (Iodide ion):
     • Iodides generally do not form precipitates with BaCl₂.
     • No reaction occurs that would form a white precipitate or release a gas with an odour.
   • S^2– (Sulfide ion):
     • Sulfides can precipitate but do not release a gas with a characteristic odour upon treatment with dilute HCl.
     • Instead, they typically form hydrogen sulfide gas (H₂S), which can be recognised by a rotten egg smell.
   • NO^-₂ (Nitrite ion):
     • Barium nitrite does not form a white precipitate with BaCl₂.
     • Therefore, it does not match the conditions described in the question.
3. Conclusion: Based on the analysis, the only anion that fits all conditions—forms a white precipitate with BaCl₂ and releases a characteristic smelling gas (SO₂) when treated with dilute HCl—is the sulfite ion, \(SO^{2-}_3\).

Hence, the correct answer is: \(SO^{2-}_3\).