`X(aq)+Na_(2)O_(2) to Y(aq.) overset(BaCl_(2))to underset("Insoluble in dil. HCl")(Zdarr)`
X and Y are different sodium salts, then anion present in the salt (X) is:

Options

A. `Cr_(2)O_(7)^(2-)`
B. `C_(2)O_(4)^(2-)`
C. `SO_(3)^(2-)`
D. `SO_(4)^(2-)`

Answer

To solve the problem step by step, we need to analyze the reactions and the oxidation states involved.

### Step 1: Identify the reaction between X and Na2O2
We start with the reaction:
\[ X(aq) + Na_2O_2 \rightarrow Y(aq) \]

Here, \( Na_2O_2 \) (sodium peroxide) can act as both an oxidizing and reducing agent. This means that it can oxidize some species while reducing others.

### Step 2: Analyze the possible anions in X
We need to consider the possible anions that could be present in the salt \( X \). The options given are:
1. \( Cr_2O_7^{2-} \) (dichromate)
2. \( C_2O_4^{2-} \) (oxalate)
3. \( SO_3^{2-} \) (sulfite)
4. \( SO_4^{2-} \) (sulfate)

### Step 3: Determine the oxidation states
- For \( Cr_2O_7^{2-} \), chromium has an oxidation state of +6. This would be reduced to +3 when treated with \( Na_2O_2 \). The resulting \( Cr^{3+} \) would not form an insoluble salt with \( BaCl_2 \).

- For \( C_2O_4^{2-} \), carbon has an oxidation state of +3. When oxidized by \( Na_2O_2 \), it would become +4, resulting in \( CO_4^{2-} \) (which does not form an insoluble salt with \( BaCl_2 \)).

- For \( SO_3^{2-} \), sulfur has an oxidation state of +4. When oxidized by \( Na_2O_2 \), it would become \( SO_4^{2-} \) (sulfate) with +6 oxidation state. \( BaSO_4 \) is insoluble in dilute HCl, which fits our requirement.

- For \( SO_4^{2-} \), sulfur is already in the +6 oxidation state, and it cannot be oxidized further. It would be reduced to \( SO_3^{2-} \), which does not form an insoluble salt with \( BaCl_2 \).

### Step 4: Conclusion
Based on the analysis, the only anion that fits all the criteria is \( SO_3^{2-} \) (sulfite). When \( SO_3^{2-} \) is oxidized, it forms \( SO_4^{2-} \), which when treated with \( BaCl_2 \) produces \( BaSO_4 \), an insoluble salt in dilute HCl.

Thus, the anion present in the salt \( X \) is:
\[ \text{SO}_3^{2-} \]