Question
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.
JEE Main
Chemistry
p Block Elements (Group 15, 16, 17 & 18)
2026
JEE Main 2026 (06 April Shift 2)
JEE Main Chemistry Question (2026) — Solution
Options
- A. NO and [ Fe(NO) ] SO _4
- B. NO_2 and [ Fe(NO _2)] SO _4
- C. N_2O and [ Fe(N _2 O )] SO _4
- D. N_2O_4 and [ Fe(N _2 O _4)] SO _4
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 and [ Fe(NO) ] SO _4
Related: Chemistry — p Block Elements (Group 15, 16, 17 & 18) · All PYQ Banks