Identify A, B, and C in the given reaction sequence:
\[
\text{A} \xrightarrow{\text{HNO}_3} \text{Pb(NO}_3)_2 \xrightarrow{\text{H}_2\text{SO}_4} \text{B} \rightarrow \text{C (Yellow ppt)}.
\]
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- PbCl$_2$, PbSO$_4$, PbCrO$_4$
- PbS, PbSO$_4$, PbCrO$_4$
- PbS, PbSO$_4$, Pb(CH$_3$COO)$_2$
- PbCl$_2$, Pb(SO$_4$)$_2$, PbCrO$_4$
The Correct Option is B
Approach Solution - 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 \(\text{C}\).
- 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.
- 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: \(\text{PbSO}_4 \rightarrow \text{PbCrO}_4 \, \text{(yellow ppt)}\)
- 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$_4$".
- 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$_4$.
Approach Solution -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
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