Question

An inorganic compound W undergoes the following reactions: $ W + \text{Na}_2\text{CO}_3 \xrightarrow{\text{O}_2 / \text{heat}} X + H^+ \xrightarrow{} Y(s) $ $ Y(aq) + \text{KCl} (aq) \xrightarrow{} Z(s) $ Z appears in the form of orange crystals and is used as an oxidising agent in acid medium. Identify the compound W.

• A. \(\text{K}_2\text{Cr}_2\text{O}_4\)
• B. \(\text{FeCr}_2\text{O}_4\)
• C. \(\text{Cu}(\text{Cr}_2\text{O}_4)_2\)
• D. \(\text{Na}_2\text{Cr}_2\text{O}_7\)

Hint:

In problems involving oxidation states and colored compounds, identify the oxidation state of the central metal ion and consider the color of the resulting product. Chromium compounds in the +6 oxidation state typically appear orange or yellow, which can help identify the correct compound.

Answer 1

The Correct Option is B

The reaction sequence involves the following steps: 
1. The reaction of compound W with sodium carbonate (\(\text{Na}_2\text{CO}_3\)) in the presence of heat and oxygen gives X and \(H^+\). 
2. \(X\) then reacts with potassium chloride (\(\text{KCl}\)) in aqueous solution to form \(Y\), which is a solid compound. 
3. Finally, the compound \(Y\) leads to the formation of \(Z\), which appears as orange crystals and is an oxidizing agent in acidic conditions. 
Let's analyze the compounds: 
- The orange color and oxidizing nature suggest that \(Z\) is likely to be chromium(VI) compound, which typically exists in the form of orange or yellow crystals. 
- The most common orange chromium(VI) compound is \(\text{Cr}_2\text{O}_3\), and it is often produced from \(\text{FeCr}_2\text{O}_4\), a chromium-containing mineral. Thus, \(\text{FeCr}_2\text{O}_4\) fits the description of the compound W, as it can undergo oxidation to produce the chromium(VI) compound (which appears as orange crystals) and act as an oxidizing agent. \[ \text{FeCr_2\text{O}_4} \] 
Thus, the correct answer is \( B \).