Question

Identify A, B, and C in the given reaction sequence: \[ \text{A} \xrightarrow{\text{HNO}_3} \text{Pb(NO}_3)_2 \xrightarrow{\text{H}_2\text{SO}_4} \text{B} \rightarrow \text{C (Yellow ppt)}. \]

• A. PbCl$_2$, PbSO$_4$, PbCrO$_4$
• B. PbS, PbSO$_4$, PbCrO$_4$
• C. PbS, PbSO$_4$, Pb(CH$_3$COO)$_2$
• D. PbCl$_2$, Pb(SO$_4$)$_2$, PbCrO$_4$

Hint:

The reaction sequence involves the formation of lead nitrate, lead sulfate, and lead chromate. This is a typical sequence for testing the presence of lead in reactions with chromates.

Answer 1

The Correct Option is B

1. Analyze the given reaction sequences:
   • The initial compound \(\text{A}\) reacts with nitric acid \(\text{HNO}_3\) to form lead nitrate \(\text{Pb(NO}_3)_2\).
   • Lead nitrate \(\text{Pb(NO}_3)_2\) further reacts with sulfuric acid \(\text{H}_2\text{SO}_4\) to form \(\text{B}\), which is lead sulfate \(\text{PbSO}_4\).
   • Compound \(\text{B}\) undergoes another transformation to give a yellow precipitate, indicating the formation of lead chromate \(\text{PbCrO}_4\), labeled as \(\text{C}\). 
2. Determine the identity of \(\text{A}\):
   • To yield lead nitrate when treated with nitric acid, \(\text{A}\) could most likely be lead sulfide \(\text{PbS}\). This is because \(\text{PbS} \xrightarrow{\text{HNO}_3} \text{Pb(NO}_3)_2\) can be balanced to show the transformation.
3. Summarize the full reaction pathway:
   • Step 1: \(\text{PbS} \xrightarrow{\text{HNO}_3} \text{Pb(NO}_3)_2\)
   • Step 2: \(\text{Pb(NO}_3)_2 \xrightarrow{\text{H}_2\text{SO}_4} \text{PbSO}_4\)
   • Step 3: \(\text{PbSO}_4 \rightarrow \text{PbCrO}_4 \, \text{(yellow ppt)}\)
4. Correct Option Evaluation:
   • The sequence through which \(\text{PbS}\) turns into \(\text{PbSO}_4\) and finally into the yellow precipitate \(\text{PbCrO}_4\) matches the option: "PbS, PbSO$_4$, PbCrO$_4$".
5. Elimination of Other Options:
   • Options involving \(\text{PbCl}_2\) and \(\text{Pb(SO}_4)_2\) do not fit the pathway as \(\text{PbS}\\) directly leads to \(\text{Pb(NO}_3)_2\) and \(\text{PbSO}_4\).

Thus, the correct sequence of reactions is captured by the option: PbS, PbSO$_4$, PbCrO$_4$.

Answer 2

Step 1: Identify compound A

The first reaction is:

A ⟶_HNO₃ Pb(NO₃)₂

This indicates that compound A reacts with nitric acid to form lead nitrate. Hence, A must be metallic lead (Pb).

So, A = Pb

Step 2: Identify compound B

Pb(NO₃)₂ reacts with H₂SO₄ to give compound B. The reaction is: \[ \text{Pb(NO}_3)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{PbSO}_4 \downarrow + 2\text{HNO}_3 \] PbSO₄ is a white precipitate formed by double displacement.

So, B = PbSO₄

Step 3: Identify compound C (Yellow precipitate)

Compound B is then treated to give a yellow precipitate C. The only classic yellow precipitate with lead is lead chromate (PbCrO₄), which is formed when Pb²⁺ reacts with chromate ions: \[ \text{Pb}^{2+} + \text{CrO}_4^{2-} \rightarrow \text{PbCrO}_4 \downarrow \, (\text{yellow ppt}) \] This confirms C is lead chromate.

So, C = PbCrO₄

Final Identifications:
A = Pb, B = PbSO₄, C = PbCrO₄