In the reaction,
`I_(2)+2S_(2)O_(3)^(2-) rarr 2I^(-)+S_(4)O_(6)^(2-)`.
Equivalent wieght of iodine will be equal to

Options

A. Its molecular mass
B. `(1)/(2)` of its molecular mass
C. `(1)/(4)` of its molecular mass
D. Twice its molecular mass

Explanation

To find the equivalent weight of iodine in the given reaction, we can follow these steps:

### Step 1: Write down the reaction
The reaction given is:
\[ I_2 + 2 S_2O_3^{2-} \rightarrow 2 I^{-} + S_4O_6^{2-} \]

### Step 2: Determine the change in oxidation state
In the reaction, iodine (I) is present in the molecular form \( I_2 \) with an oxidation state of 0. When it is converted to iodide ion \( I^{-} \), its oxidation state becomes -1.

The change in oxidation state for one iodine atom is:
\[ 0 \text{ (in } I_2\text{)} \rightarrow -1 \text{ (in } I^{-}\text{)} \]
Thus, the change in oxidation state is:
\[ \Delta \text{Oxidation State} = 0 - (-1) = 1 \]

### Step 3: Determine the n-factor
The n-factor for iodine in this reaction is the total change in oxidation state per mole of iodine. Since one molecule of \( I_2 \) produces two \( I^{-} \) ions, the n-factor for \( I_2 \) is:
\[ n = 2 \times (0 - (-1)) = 2 \]

### Step 4: Calculate the molecular weight of iodine
The molecular weight of iodine \( I_2 \) can be calculated as follows:
- The atomic weight of iodine (I) is approximately 127 g/mol.
- Therefore, the molecular weight of \( I_2 \) is:
\[ \text{Molecular Weight of } I_2 = 2 \times 127 = 254 \text{ g/mol} \]

### Step 5: Calculate the equivalent weight
The equivalent weight (EW) of a substance is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n \text{-factor}} \]

Substituting the values we found:
\[ \text{Equivalent Weight of } I_2 = \frac{254 \text{ g/mol}}{2} = 127 \text{ g/equiv} \]

### Conclusion
Thus, the equivalent weight of iodine \( I_2 \) is 127 g/equiv.

### Final Answer
The equivalent weight of iodine will be equal to half of its molecular weight.

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