`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 determine the type of reaction represented by the equation \( I_2 + S_2O_3^{2-} \rightarrow I^{-} + S_4O_6^{2-} \), we will follow these steps:

### Step 1: Identify the oxidation states of each element in the reaction.

1. **For \( I_2 \)**: The oxidation state of iodine in its elemental form is 0.
2. **For \( I^{-} \)**: The oxidation state of iodine is -1.
3. **For \( S_2O_3^{2-} \)**: Let the oxidation state of sulfur be \( X \). The equation becomes:
\[ 2X + 3(-2) = -2 \implies 2X - 6 = -2 \implies 2X = 4 \implies X = +2 \]
So, sulfur in \( S_2O_3^{2-} \) has an oxidation state of +2.
4. **For \( S_4O_6^{2-} \)**: Let the oxidation state of sulfur be \( Y \). The equation becomes:
\[ 4Y + 6(-2) = -2 \implies 4Y - 12 = -2 \implies 4Y = 10 \implies Y = 2.5 \]
Thus, sulfur in \( S_4O_6^{2-} \) has an oxidation state of +2.5.

### Step 2: Analyze the changes in oxidation states.

- **Iodine**: Changes from 0 (in \( I_2 \)) to -1 (in \( I^{-} \)). This indicates a reduction (gain of electrons).
- **Sulfur**: Changes from +2 (in \( S_2O_3^{2-} \)) to +2.5 (in \( S_4O_6^{2-} \)). This indicates an oxidation (loss of electrons).

### Step 3: Classify the reaction.

Since one species (iodine) is being reduced while another species (sulfur) is being oxidized, this reaction is classified as a **redox reaction**. Specifically, because two different reactant molecules are involved, it is an **intermolecular redox reaction**.

### Conclusion

The type of reaction represented by the equation \( I_2 + S_2O_3^{2-} \rightarrow I^{-} + S_4O_6^{2-} \) is an **intermolecular redox reaction**.

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