When \(\mathrm{FeCr_2O_4}\) is fused with \(\mathrm{Na_2CO_3}\) 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 coloured compound (C). An acidified solution of compound (C) oxidises iodide to (D). Identify (A), (B), (C) and (D).
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Solution and Explanation
Step 2: Identifying Compound (B). Upon acidification, \( \text{Na}_2\text{CrO}_4 \) is converted to potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)), which is orange in color. Thus, \( B = \text{K}_2\text{Cr}_2\text{O}_7 \).
Step 3: Identifying Compound (C). The reaction of \( \text{K}_2\text{Cr}_2\text{O}_7 \) with KCl leads to the formation of orange-colored \( \text{Cr}_2\text{O}_7^{2-} \) ions. Thus, \( C \) is an orange-colored chromium compound. Thus, \( C = \text{K}_2\text{Cr}_2\text{O}_7 \).
Step 4: Identifying Compound (D). An acidified solution of \( C \) oxidizes sodium sulfite (\( \text{Na}_2\text{SO}_3 \)) to sodium sulfate (\( \text{Na}_2\text{SO}_4 \)). This is a typical redox reaction where chromium compounds act as oxidizing agents.
Thus, \( D = \text{Na}_2\text{SO}_4 \).
So, the identities are:
\( A = \text{Na}_2\text{CrO}_4 \)
\( B = \text{K}_2\text{Cr}_2\text{O}_7 \)
\( C = \text{K}_2\text{Cr}_2\text{O}_7 \)
\( D = \text{Na}_2\text{SO}_4 \)
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