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JEE Main20266 April 2026Evening ShiftChemistryp Block Elements (Group 15, 16, 17 & 18)Actual

Treatment of a gas ' X ' with a freshly prepared ferrous sulphate solution gives a compound ' Y ' as a brown ring. The compounds X and Y are.

Options

  1. ANO and [ Fe(NO) ] SO ₄
  2. BNO₂ and [ Fe(NO ₂)] SO ₄
  3. CN₂ O and [ Fe(N ₂ O )] SO ₄
  4. DN₂ O₄ and [ Fe(N ₂ O ₄)] SO ₄

Correct answer

A. NO and [ Fe(NO) ] SO ₄

Step-by-step solution

The brown ring test is used for the detection of nitrate ( NO ₃^- ) or nitrite ( NO ₂^- ) ions. In this test, nitric oxide (NO) gas is evolved which reacts with the freshly prepared ferrous sulphate ( FeSO ₄ ) solution to form a brown ring of nitroso ferrous sulphate. The gas X is NO. The reaction is given by: FeSO ₄ + NO [ Fe(NO) ] SO ₄ The actual composition of the brown ring complex is [ Fe(H ₂ O )₅( NO )] SO ₄ , which is often represented as [ Fe(NO) ] SO ₄ . Thus, X is NO and Y is [ Fe(NO) ] SO ₄ . Answer: NO

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