Question

When \( \text{FeCr}_2\text{O}_4 \) is fused with \( \text{Na}_2\text{CO}_3 \) in the presence of air, it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange-colored compound (C). An acidified solution of compound (C) oxidizes \( \text{Na}_2\text{SO}_3 \) to (D). Identify (A), (B), (C), and (D).

Hint:

Chromates and dichromates are strong oxidizing agents. Chromates are yellow in alkaline conditions, while dichromates are orange in acidic conditions.

Answer 1

To solve the problem, we need to identify compounds (A), (B), (C), and (D) in the given reaction sequence starting from the fusion of \( \text{FeCr}_2\text{O}_4 \) with \( \text{Na}_2\text{CO}_3 \).

1. Reaction of \( \text{FeCr}_2\text{O}_4 \) with \( \text{Na}_2\text{CO}_3 \) in Air:
\( \text{FeCr}_2\text{O}_4 \) contains Cr in the +3 state. When fused with \( \text{Na}_2\text{CO}_3 \) in the presence of air (oxygen), Cr³⁺ is oxidized to Cr⁶⁺, forming sodium chromate (\( \text{Na}_2\text{CrO}_4 \)), a yellow solution. The reaction is:

\( 2\text{FeCr}_2\text{O}_4 + 4\text{Na}_2\text{CO}_3 + 7/2\text{O}_2 \rightarrow 2\text{Na}_2\text{CrO}_4 + 2\text{NaCrO}_2 + \text{Fe}_2\text{O}_3 + 4\text{CO}_2 \).
Compound (A) is \( \text{Na}_2\text{CrO}_4 \).

2. Acidification of Compound (A):
Acidifying \( \text{Na}_2\text{CrO}_4 \) converts chromate (\( \text{CrO}_4^{2-} \)) to dichromate (\( \text{Cr}_2\text{O}_7^{2-} \)) via:

\( 2\text{Na}_2\text{CrO}_4 + 2\text{H}^+ \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{Na}^+ + \text{H}_2\text{O} \).
Compound (B) is \( \text{Na}_2\text{Cr}_2\text{O}_7 \).

3. Reaction of Compound (B) with KCl:
\( \text{Na}_2\text{Cr}_2\text{O}_7 \) reacts with KCl to form potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)), an orange-colored compound:

\( \text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl} \).
Compound (C) is \( \text{K}_2\text{Cr}_2\text{O}_7 \).

4. Oxidation of \( \text{Na}_2\text{SO}_3 \) by Acidified Compound (C):
Acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \) acts as an oxidizing agent, converting \( \text{Na}_2\text{SO}_3 \) (where S is +4) to \( \text{Na}_2\text{SO}_4 \) (where S is +6):

\( \text{K}_2\text{Cr}_2\text{O}_7 + 4\text{H}_2\text{SO}_4 + 3\text{Na}_2\text{SO}_3 \rightarrow \text{Cr}_2(\text{SO}_4)_3 + \text{K}_2\text{SO}_4 + 3\text{Na}_2\text{SO}_4 + 4\text{H}_2\text{O} \).
Compound (D) is \( \text{Na}_2\text{SO}_4 \).

Final Answer:
(A) is \( \text{Na}_2\text{CrO}_4 \), (B) is \( \text{Na}_2\text{Cr}_2\text{O}_7 \), (C) is \( \text{K}_2\text{Cr}_2\text{O}_7 \), and (D) is \( \text{Na}_2\text{SO}_4 \).