In the reaction, `2S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I^(-)`. The eq. wt. of `Na_(2)S_(2)O_(3)` is equal to its:

Options

A. Mol. wt.
B. Mol. wt./2
C. `2xx"mol. w.t."`
D. Mol. w.t./6

Explanation

To find the equivalent weight of sodium thiosulfate (Na₂S₂O₃) in the given redox reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction provided is:
\[ 2S_{2}O_{3}^{2-} + I_{2} \rightarrow S_{4}O_{6}^{2-} + 2I^{-} \]

2. **Determine the Oxidation States**:
- In sodium thiosulfate (Na₂S₂O₃), the oxidation state of sulfur (S) can be determined.
- In \(S_{2}O_{3}^{2-}\), the average oxidation state of sulfur can be calculated. Let the oxidation state of sulfur be \(x\):
\[ 2x + 3(-2) = -2 \implies 2x - 6 = -2 \implies 2x = 4 \implies x = +2 \]
- Thus, the oxidation state of sulfur in \(S_{2}O_{3}^{2-}\) is +2.

3. **Calculate the Oxidation State in the Product**:
- In the product \(S_{4}O_{6}^{2-}\), we again let the oxidation state of sulfur be \(y\):
\[ 4y + 6(-2) = -2 \implies 4y - 12 = -2 \implies 4y = 10 \implies y = +2.5 \]
- Thus, the oxidation state of sulfur in \(S_{4}O_{6}^{2-}\) is +2.5.

4. **Determine the Change in Oxidation State**:
- For the reaction, we have 2 moles of \(S_{2}O_{3}^{2-}\) converting to \(S_{4}O_{6}^{2-}\).
- The change in oxidation state for one mole of sulfur is:
\[ \Delta \text{oxidation state} = 2.5 - 2 = 0.5 \]
- Since there are 4 sulfur atoms in \(S_{4}O_{6}^{2-}\) (from 2 moles of \(S_{2}O_{3}^{2-}\)), the total change for 2 moles of thiosulfate is:
\[ \text{Total change} = 4 \times 0.5 = 2 \]

5. **Calculate the n Factor**:
- The n factor is defined as the total change in oxidation state per mole of the compound. For 2 moles of \(S_{2}O_{3}^{2-}\), the change is 2, so for 1 mole:
\[ n = \frac{2}{2} = 1 \]

6. **Calculate the Equivalent Weight**:
- The equivalent weight (EW) is given by the formula:
\[ \text{EW} = \frac{\text{Molecular Weight}}{n} \]
- Since \(n = 1\), the equivalent weight of sodium thiosulfate is equal to its molecular weight.

### Conclusion:
Thus, the equivalent weight of \(Na_{2}S_{2}O_{3}\) is equal to its molecular weight.

### Answer:
The equivalent weight of \(Na_{2}S_{2}O_{3}\) is equal to its molecular weight.