`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, we will follow these steps:

### Step 1: Identify the Reaction
We start with the reaction:
\[ X(aq) + Na_2O_2 \rightarrow Y(aq) \]
Here, \( X \) is a sodium salt that reacts with sodium peroxide (\( Na_2O_2 \)) to form another sodium salt \( Y \).

### Step 2: Determine the Nature of X and Y
Since \( X \) and \( Y \) are both sodium salts, we need to identify what kind of anions they contain. Sodium peroxide is known to act as an oxidizing agent.

### Step 3: Consider Possible Anions for X
One common sodium salt that reacts with sodium peroxide is sodium sulfite (\( Na_2SO_3 \)). When sodium sulfite reacts with sodium peroxide, it gets oxidized to sodium sulfate (\( Na_2SO_4 \)):
\[ Na_2SO_3 + Na_2O_2 \rightarrow Na_2SO_4 \]

### Step 4: Identify Y
From the reaction, we can conclude that:
- \( X \) is sodium sulfite (\( Na_2SO_3 \))
- \( Y \) is sodium sulfate (\( Na_2SO_4 \))

### Step 5: Reaction with Barium Chloride
Next, we have:
\[ Y(aq) \overset{BaCl_2}{\rightarrow} Z \]
This indicates that when sodium sulfate reacts with barium chloride, it forms barium sulfate (\( BaSO_4 \)), which is insoluble in dilute hydrochloric acid (\( HCl \)):
\[ Na_2SO_4 + BaCl_2 \rightarrow BaSO_4 \downarrow + 2 NaCl \]

### Step 6: Conclude the Anion in X
Since \( X \) is sodium sulfite, the anion present in the salt \( X \) is:
\[ SO_3^{2-} \] (sulfite ion)

### Final Answer
The anion present in the salt \( X \) is \( SO_3^{2-} \) (sulfite).

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