JEE Main 2020 · 4 September · morning · Single correct · +4 / −1

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 :

  1. (a) +3
  2. (b) +4
  3. (c) +5
  4. (d) +2
Answer and solution

Answer (a) +3

Solution

Chemistry ma’am · 4 steps

Follow the gas through each step: brown NO2, colourless N2O4, then the blue solid N2O3.

  1. Brown gas A. Lead nitrate decomposes on heating: 2Pb(NO₃)₂ →(Δ) 2PbO + 4NO₂ + O₂. The brown gas A is NO₂.
  2. Colourless B. On cooling, NO₂ dimerises: 2NO₂ ⇌ N₂O₄. The colourless solid or liquid B is N₂O₄.
  3. Blue solid C. N₂O₄ reacts with NO: N₂O₄ + 2NO → 2N₂O₃. Dinitrogen trioxide, N₂O₃, is the blue solid C.

    2x + 3(-2) = 0 ⇒ x = +3

  4. So. Nitrogen in N₂O₃ has oxidation number +3, which is option (a).

The idea to keep. Learn the nitrogen oxides by colour: NO₂ is brown, N₂O₄ is colourless, N₂O₃ is blue. Once you name the compound, the oxidation number takes one line.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Redox Reactions

Asked about 90 times in JEE Main since 2020. This question is from 2020.