Test Index

p Block Elements Part 3

Section: Chemistry

Question : 2 of 81

Marks: +1, -0

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 6 Apr 2026 shift 2]

NO and $[\mathrm{Fe}(\mathrm{NO})]\ \mathrm{SO}_4$
$\mathrm{NO}_2$ and $[\mathrm{Fe}(\mathrm{NO}_2)]\ \mathrm{SO}_4$
$\mathrm{N}_2\ \mathrm{O}$ and $[\mathrm{Fe}(\mathrm{N}_2\ \mathrm{O})]\ \mathrm{SO}_4$
$\mathrm{N}_2\ \mathrm{O}_4$ and $[\mathrm{Fe}(\mathrm{N}_2\ \mathrm{O}_4)]\ \mathrm{SO}_4$

Validate

Solution:

Concept: The brown ring test is a qualitative test for the presence of nitrates.

Chemical Equation / Formula:

NO + Fe2+ + [Fe(CN)6]4- → [Fe(NO)(H2O)]2+ (brown ring)

Explanation: A brown ring is formed when nitric oxide (NO) reacts with ferrous sulfate in the presence of a source of iron(II) ions. This complex ion, [Fe(NO)(H2O)]2+, gives the characteristic brown color.

Shortcut: The brown ring test is specific to nitrates.

Answer: A. NO and

[\mathrm{Fe}(\mathrm{NO})]\ \mathrm{SO}_4

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