JEE Main 2026 · 6 Apr · Shift 2 · Chemistry Q70

← JEE Main 2026 · 6 Apr · Shift 2

ChemistrySingle correct+4 / -1Question 70 of 75

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.

NO and

\ce{[Fe(NO)]SO4}

\ce{NO2}

and

\ce{[Fe(NO2)]SO4}

\ce{N2O}

and

\ce{[Fe(N2O)]SO4}

\ce{N2O4}

and

\ce{[Fe(N2O4)]SO4}

Answer and worked solution

Official answer: A

Brown-ring test. Nitrate is reduced to NO, which combines with $\ce{Fe^2+}$ :
$\ce{[Fe(H2O)6]^2+ + NO -> [Fe(H2O)5(NO)]^2+ + H2O}$
X is NO and Y is the nitrosyl iron(II) sulphate complex, written $\ce{[Fe(NO)]SO4}$ .

(A)

Topics:Anion analysis

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JEE Main 2026 · 6 Apr · Shift 2, Chemistry question 70