`Na_(2)S_(2)O_(3)` is oxidized by `I_(2)` to give

Options

A. `Na_(2)S`
B. `NaHSO_(3)`
C. `NaSO_(4)`
D. `Na_(2)S_(4)O_(6)`

Explanation

To solve the problem of what `Na₂S₂O₃` (sodium thiosulfate) produces when oxidized by `I₂` (iodine), we can follow these steps:

### Step 1: Write the Reactants
We start with the reactants given in the problem:
- Sodium thiosulfate: `Na₂S₂O₃`
- Iodine: `I₂`

### Step 2: Write the Reaction
When sodium thiosulfate reacts with iodine, it undergoes oxidation. The balanced chemical reaction can be written as:
\[ \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + \text{NaI} \]

### Step 3: Balance the Reaction
To ensure the reaction is balanced, we can check the number of atoms on both sides:
- On the left: 2 Na, 2 S, 3 O, and 2 I
- On the right: 2 Na (from Na₂S₄O₆ and NaI), 4 S (from Na₂S₄O₆), 6 O (from Na₂S₄O₆), and 1 I (from NaI)

After balancing, we find:
\[ 2 \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2 \text{NaI} \]

### Step 4: Determine Oxidation States
Next, we determine the oxidation states of sulfur in both compounds:
1. In `Na₂S₂O₃`:
- Let the oxidation state of S be \( x \).
- The equation is: \( 2x + 2(-2) + 3(-2) = 0 \)
- Solving gives \( x = +2 \).

2. In `Na₂S₄O₆`:
- Let the oxidation state of S be \( y \).
- The equation is: \( 4y + 6(-2) = 0 \)
- Solving gives \( y = +2.5 \).

### Step 5: Conclusion
From the reaction and the oxidation states, we conclude that sodium thiosulfate (`Na₂S₂O₃`) is oxidized by iodine (`I₂`) to produce sodium tetrathionate (`Na₂S₄O₆`) and sodium iodide (`NaI`).

### Final Answer
Thus, the final product of the oxidation of `Na₂S₂O₃` by `I₂` is:
\[ \text{Na}_2\text{S}_4\text{O}_6 \]

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