In the reaction `2Na_(a)S_(2)O_(3)+I_(2) rarr Na_(2)S_(4)O_(6)+ 2NaI`, the equivalent weight of `Na_(2)S_(2)O_(3) (mol. Wt. =M)` is equal to

Options

A. (a)`M`
B. (b)`M//2`
C. (c )`M//3`
D. (d)`M//4`

Explanation

To find the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the reaction

\[ 2\text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2\text{NaI} \]

we will follow these steps:

### Step 1: Determine the oxidation states
1. **Identify the oxidation states of sulfur in both compounds.**
- In \( \text{Na}_2\text{S}_2\text{O}_3 \):
- Let the oxidation state of sulfur be \( x \).
- The equation for the oxidation state is:
\[ 2(+1) + 2x + 3(-2) = 0 \]
Simplifying, we get:
\[ 2 + 2x - 6 = 0 \Rightarrow 2x - 4 = 0 \Rightarrow 2x = 4 \Rightarrow x = +2 \]
- In \( \text{Na}_2\text{S}_4\text{O}_6 \):
- Let the oxidation state of sulfur be \( y \).
- The equation for the oxidation state is:
\[ 2(+1) + 4y + 6(-2) = 0 \]
Simplifying, we get:
\[ 2 + 4y - 12 = 0 \Rightarrow 4y - 10 = 0 \Rightarrow 4y = 10 \Rightarrow y = +2.5 \]

### Step 2: Calculate the change in oxidation number
2. **Determine the change in oxidation number for sulfur:**
- The change in oxidation state from \( +2 \) to \( +2.5 \) is:
\[ \Delta = 2.5 - 2 = 0.5 \]

### Step 3: Calculate the N-factor
3. **Calculate the N-factor (number of equivalents) of \( \text{Na}_2\text{S}_2\text{O}_3 \):**
- Since there are 2 sulfur atoms in \( \text{Na}_2\text{S}_2\text{O}_3 \), the total change in oxidation number is:
\[ \text{N-factor} = 2 \times 0.5 = 1 \]

### Step 4: Calculate the equivalent weight
4. **Calculate the equivalent weight:**
- The equivalent weight is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{\text{N-factor}} \]
- Substituting the values:
\[ \text{Equivalent weight} = \frac{M}{1} = M \]

### Final Answer
Thus, the equivalent weight of \( \text{Na}_2\text{S}_2\text{O}_3 \) is \( M \).

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