`I_(2)+S_(2)O_(3)^(2-) to I^(-)+S_(4)O_(6)^(2-)`

Options

A. For disproportionation reaction.
B. For comproportionation reaction.
C. For either intermolecular redox reaction or displacement reaction
D. For either thermal combination redox reaction or thermal decomposition redox reaction.

Answer

To analyze the reaction \( I_2 + S_2O_3^{2-} \rightarrow I^{-} + S_4O_6^{2-} \) and determine its type, we will follow these steps:

### Step 1: Determine the oxidation states of the reactants.
- For \( I_2 \), the oxidation state of iodine (I) is 0 because it is in its elemental form.
- For \( S_2O_3^{2-} \):
- Let the oxidation state of sulfur (S) be \( x \).
- The equation for the oxidation states is:
\[ 2x + 3(-2) = -2 \]
Simplifying this gives:
\[ 2x - 6 = -2 \implies 2x = 4 \implies x = +2 \]
- Thus, the oxidation state of sulfur in \( S_2O_3^{2-} \) is +2.

### Step 2: Determine the oxidation states of the products.
- For \( I^{-} \), the oxidation state of iodine is -1.
- For \( S_4O_6^{2-} \):
- Let the oxidation state of sulfur be \( y \).
- The equation for the oxidation states is:
\[ 4y + 6(-2) = -2 \]
Simplifying this gives:
\[ 4y - 12 = -2 \implies 4y = 10 \implies y = +2.5 \]
- Thus, the oxidation state of sulfur in \( S_4O_6^{2-} \) is +2.5.

### Step 3: Analyze the changes in oxidation states.
- Iodine changes from 0 (in \( I_2 \)) to -1 (in \( I^{-} \)), indicating a reduction.
- Sulfur changes from +2 (in \( S_2O_3^{2-} \)) to +2.5 (in \( S_4O_6^{2-} \)), indicating an oxidation.

### Step 4: Determine the type of reaction.
Since there is both a reduction (iodine) and an oxidation (sulfur) occurring simultaneously, this is classified as a redox reaction. Additionally, because the reaction involves different species (iodine and thiosulfate), it is also an intermolecular redox reaction.

### Conclusion:
The reaction \( I_2 + S_2O_3^{2-} \rightarrow I^{-} + S_4O_6^{2-} \) is an **intermolecular redox reaction**.

---