If the moleucar wt of `Na_(2)S_(2)O_(3)` and `I_(2)` are `M_(1)` and `M_(2)` respectivly then what will be the equivalent wt of `Na_(2)S_(2)O_(3)` and `I_(2)` in the following reaction
`S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I^(-)`

Options

A. `M_(1)M_(2)`
B. `M_(1)M_(2)//2`
C. `2M_(1),M_(2)`
D. `Mlt_(1),2 M_(2)`

Explanation

To find the equivalent weight of `Na2S2O3` and `I2` in the given reaction, we will follow these steps:

### Step 1: Write the balanced reaction
The reaction given is:
\[ \text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^- \]

### Step 2: Determine the oxidation states
1. For `S2O3^{2-}`, the 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 `S2O3^{2-}` is +2.

2. In `S4O6^{2-}`, 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 `S4O6^{2-}` is +2.5.

### Step 3: Calculate the change in oxidation state
- The change in oxidation state for sulfur from `S2O3^{2-}` to `S4O6^{2-}` is:
\[ \Delta \text{Oxidation State} = +2.5 - (+2) = +0.5 \]
- Since there are 2 sulfur atoms in `S2O3^{2-}`, the total change is:
\[ 2 \times 0.5 = 1 \]

### Step 4: Determine the n-factor for `Na2S2O3`
The n-factor for `Na2S2O3` is the total change in oxidation state per molecule, which is 1.

### Step 5: Determine the n-factor for `I2`
1. In the reaction, `I2` is reduced to `2I^-`, which means:
- Each iodine atom goes from 0 in `I2` to -1 in `I^-`.
- The total change for `I2` is:
\[ 2 \text{ (for 2 electrons)} \]
Thus, the n-factor for `I2` is 2.

### Step 6: Calculate the equivalent weights
1. The equivalent weight of `Na2S2O3`:
\[ \text{Equivalent weight of } Na2S2O3 = \frac{M_1}{n_1} = \frac{M_1}{1} = M_1 \]

2. The equivalent weight of `I2`:
\[ \text{Equivalent weight of } I2 = \frac{M_2}{n_2} = \frac{M_2}{2} \]

### Step 7: Summary of equivalent weights
- The equivalent weight of `Na2S2O3` is \( M_1 \).
- The equivalent weight of `I2` is \( \frac{M_2}{2} \).

### Final Answer
Thus, the equivalent weights of `Na2S2O3` and `I2` in the reaction are:
- Equivalent weight of `Na2S2O3`: \( M_1 \)
- Equivalent weight of `I2`: \( \frac{M_2}{2} \)