Identify A, B and C in the given below reaction sequence {A} {{HNO3}} {Pb(NO3)2} {{H2SO4}} {B} [{(2) Acetic acid, (3) }…

Step 1: Identify compound A

Compound A reacts with hot dilute $\ce{HNO3}$ to form $\ce{Pb(NO3)2}$. The options suggest $\ce{PbS}$ or $\ce{PbCl2}$. Note that usually, black $\ce{PbS}$ is dissolved in dilute $\ce{HNO3}$ to get soluble lead nitrate: $\ce{3PbS + 8HNO3 -> 3Pb(NO3)2 + 2NO + 3S + 4H2O}$ So A is $\ce{PbS}$.

Step 2: Identify compound B

To the $\ce{Pb(NO3)2}$ solution, dilute $\ce{H2SO4}$ is added. A white precipitate of lead sulfate (B) is formed. $\ce{Pb(NO3)2 + H2SO4 -> PbSO4 v + 2HNO3}$ So B is $\ce{PbSO4}$.

Step 3: Analyze the reaction of B to form C

The white precipitate of $\ce{PbSO4}$ is soluble in a concentrated solution of ammonium acetate ($\ce{CH3COONH4}$) due to the formation of the soluble complex lead acetate: $\ce{PbSO4 + 2CH3COONH4 -> Pb(CH3COO)2 + (NH4)2SO4}$ When potassium chromate ($\ce{K2CrO4}$) is added to this solution in the presence of acetic acid, a yellow precipitate of lead chromate (C) is obtained. $\ce{Pb(CH3COO)2 + K2CrO4 -> PbCrO4 v + 2CH3COOK}$ So C is $\ce{PbCrO4}$.

Step 4: Conclusion

Matches found are: $\text{A} = \ce{PbS}$, $\text{B} = \ce{PbSO4}$, $\text{C} = \ce{PbCrO4}$. This corresponds to option (b).

Key Points to Remember:

• Black $\ce{PbS}$ is dissolved by boiling with dilute $\ce{HNO3}$.
• White $\ce{PbSO4}$ precipitate uniquely dissolves in ammonium acetate solution.
• Lead shows a characteristic yellow precipitate of $\ce{PbCrO4}$ upon reaction with chromate ions.