JEE Main 2025 · 23 January · evening · Single correct · +4 / −1
Identify A, B and C in the given below reaction sequence
A, →(HNO₃), Pb(NO₃)₂, →(H₂SO₄), B; (1) Ammonium acetate; (2) Acetic acid; (3) K₂CrO₄, ↓; C; Yellow ppt
- (a) PbCl₂,PbSO₄,PbCrO₄
- (b) PbS,PbSO₄, Pb(CH₃COO)₂
- (c) PbS,PbSO₄,PbCrO₄
- (d) PbCl₂, Pb(SO₄)₂,PbCrO₄
Answer and solution
Answer (c) PbS,PbSO₄,PbCrO₄
Solution
Chemistry ma’am · 4 steps
Follow lead the way the lab does: black PbS in Group II, then white PbSO4, then yellow PbCrO4.
- Finding A. In salt analysis, Pb²⁺ is precipitated in Group II by H₂S as black PbS. That precipitate is dissolved in hot dilute nitric acid, which turns it into Pb(NO₃)₂. So A is PbS: 3PbS + 8HNO₃ → 3Pb(NO₃)₂ + 2NO + 3S + 4H₂O.
- Finding B. Adding dilute H₂SO₄ to the lead nitrate solution gives a white precipitate of PbSO₄. Lead is +2 here, so the formula is PbSO₄, not Pb(SO₄)₂. So B is PbSO₄.
- Finding C. PbSO₄ dissolves in ammonium acetate as soluble lead acetate. Acetic acid keeps the solution acidic, and K₂CrO₄ then throws down yellow lead chromate.
(CH₃COO)₂Pb + K₂CrO₄ → PbCrO₄ v + 2CH₃COOK
- So. A, B and C are PbS, PbSO₄ and PbCrO₄, which is option (c).
The idea to keep. For lead, remember the colour trail: black PbS dissolves in HNO₃, white PbSO₄ dissolves in ammonium acetate, and K₂CrO₄ gives yellow PbCrO₄.
Answer from the NTA final answer key. Solution by PrepWiser.