Question:

On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is :

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The brown gas is \(\mathrm{NO_2}\), its dimer is \(\mathrm{N_2O_4}\), and \(\mathrm{N_2O_4}\) with NO gives the blue solid \(\mathrm{N_2O_3}\).
Updated On: Nov 5, 2026
  • +2
  • +3
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The Correct Option is B

Solution and Explanation

Step 1: \(\mathrm{2Pb(NO_3)_2 \xrightarrow{\Delta} 2PbO + 4NO_2 + O_2}\); the brown gas A is \(\mathrm{NO_2}\).
Step 2: On cooling, \(\mathrm{2NO_2 \rightleftharpoons N_2O_4}\), colourless solid/liquid B.
Step 3: \(\mathrm{N_2O_4 + 2NO \rightleftharpoons 2N_2O_3}\); \(\mathrm{N_2O_3}\) is a blue solid C.
Step 4: In \(\mathrm{N_2O_3}\): \(2x + 3(-2) = 0\), so \(x = +3\).
Answer: (2) +3
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