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Chemistry Question 70 – JEE-MAIN 2026

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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.

The description of a 'brown ring' formed with freshly prepared ferrous sulfate solution is characteristic of a specific qualitative test.

Step 1: Identify the Chemical Test✦ Active

The formation of a 'brown ring' when a gas 'X' is treated with a freshly prepared ferrous sulfate solution is the characteristic indication of the Brown Ring Test. This test is primarily used for the detection of nitrate ions (NO3−), but the question describes the reaction of a gas 'X' directly.

Step 2: Determine Gas 'X'○ Expand

In the Brown Ring Test, nitrate ions are reduced to nitric oxide (NO) gas. This NO gas then reacts with ferrous ions. Therefore, the gas 'X' that directly reacts with ferrous sulfate to form the brown ring must be nitric oxide.

NO3−+3Fe2++4H+→NO+3Fe3++2H2O
💡 Teacher's Secret Hint

Remember that the brown ring test involves the reduction of nitrate to NO gas.

Step 3: Determine Compound 'Y'○ Expand

The nitric oxide (NO) gas then reacts with excess ferrous ions (Fe2+) from the freshly prepared ferrous sulfate solution to form a brown colored complex. This complex is a nitrosoferrous sulfate complex, commonly represented as [Fe(H2O)5NO]2+ or simplified as [Fe(NO)]SO4. This complex is responsible for the brown ring.

[Fe(H2O)6]2++NO→[Fe(H2O)5NO]2++H2O
💡 Teacher's Secret Hint

The brown ring complex is an adduct of ferrous ion and nitric oxide.

Step 4: Match with Options○ Expand

Based on the analysis, gas 'X' is NO and compound 'Y' is [Fe(NO)]SO4. Comparing this with the given options, option 1 matches these compounds.

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