JEE Main 2026 Apr 6 Shift 2, Chemistry Q20: Hydrogen

JEE Main2026Apr 6, Shift 2Chemistry

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:

A
NO and $[Fe(NO)] SO_{4}$
B
NO $_{2}$ and $[Fe(NO_{2})] SO_{4}$
C
N $_{2}$ O and $[Fe(N_{2} O)] SO_{4}$
D
N $_{2}$ O $_{4}$ and $[Fe(N_{2} O_{4})] SO_{4}$

Solution

The classic brown-ring test for nitrate identifies NO. In the test, NO $_{3}^{-}$ is reduced to NO by Fe $^{2 +}$ in dil. H $_{2}$ SO $_{4}$ ; the NO then forms a brown nitrosyl complex with excess Fe $^{2 +}$ :

NO + FeSO $_{4}$ $\to$ $[Fe(H_{2} O)_{5} (NO)] SO_{4}$ , commonly written $[Fe(NO)] SO_{4}$ .

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