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
- ANO and [ Fe(NO) ] SO ₄
- BNO₂ and [ Fe(NO ₂)] SO ₄
- CN₂ O and [ Fe(N ₂ O )] SO 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