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JEE Mains · Chemistry · STD 12 - 4. d and f- block elements

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.

  1. A NO and \([\text{Fe(NO)}]\text{SO}_4\)
  2. B NO\(_2\) and \([\text{Fe(NO}_2)]\text{SO}_4\)
  3. C N\(_2\)O and \([\text{Fe(N}_2\text{O})]\text{SO}_4\)
  4. D N\(_2\)O\(_4\) and \([\text{Fe(N}_2\text{O}_4)]\text{SO}_4\)
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Answer & Solution

Correct Answer

(A) NO and \([\text{Fe(NO)}]\text{SO}_4\)

Step-by-step Solution

Detailed explanation

The brown ring test is used for the detection of nitrate (\(\text{NO}_3^-\)) or nitrite (\(\text{NO}_2^-\)) ions. In this test, nitric oxide (NO) gas is evolved which reacts with the freshly prepared ferrous sulphate (\(\text{FeSO}_4\)) solution to form a brown ring of nitroso…
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