The reaction `S_(8)+12OH^(-)rarr4S^(2-)+2S_(2)O_(3)^(2-)+6H_(2)O` is

Options

A. Combination reaction
B. Decomposition reaction
C. Non-metal displacement
D. Disproportionation reaction

Answer

To determine the type of reaction represented by the equation:

\[ S_{8} + 12OH^{-} \rightarrow 4S^{2-} + 2S_{2}O_{3}^{2-} + 6H_{2}O \]

we will analyze the reaction step by step.

### Step 1: Identify the Reactants and Products
The reactants in this reaction are sulfur (S₈) and hydroxide ions (OH⁻). The products are sulfide ions (S²⁻), thiosulfate ions (S₂O₃²⁻), and water (H₂O).

### Step 2: Determine the Type of Reaction
We need to classify the reaction into one of the following categories: combination reaction, decomposition reaction, non-metal displacement reaction, or disproportionation reaction.

- **Combination Reaction**: This occurs when two or more substances combine to form a single product. Here, we have multiple products formed from the reactants, so it is not a combination reaction.

- **Decomposition Reaction**: This occurs when a single compound breaks down into two or more simpler products. Since we are not starting with a single compound that breaks down, this is not a decomposition reaction.

- **Non-metal Displacement Reaction**: This involves a non-metal displacing another non-metal in a compound. In this case, there is no displacement occurring, so this is not applicable.

- **Disproportionation Reaction**: This occurs when a single substance is both oxidized and reduced in the same reaction. We will check the oxidation states to confirm if this is the case.

### Step 3: Calculate Oxidation States
1. **Oxidation State of S₈**: The oxidation state of elemental sulfur (S₈) is 0.
2. **Oxidation State of S²⁻**: The oxidation state of sulfur in S²⁻ is -2.
3. **Oxidation State of S₂O₃²⁻**: Let’s calculate the oxidation state of sulfur in thiosulfate (S₂O₃²⁻):
- The equation for the oxidation state can be set up as follows:
\[ 2x + 3(-2) = -2 \]
- Solving this gives:
\[ 2x - 6 = -2 \implies 2x = 4 \implies x = +2 \]
- Therefore, the oxidation state of sulfur in S₂O₃²⁻ is +2.

### Step 4: Analyze the Changes in Oxidation States
- Sulfur in S₈ (0) is reduced to S²⁻ (-2).
- Sulfur in S₈ (0) is oxidized to S₂O₃²⁻ (+2).

Since sulfur is both oxidized and reduced in this reaction, it confirms that this is a **disproportionation reaction**.

### Conclusion
The reaction \( S_{8} + 12OH^{-} \rightarrow 4S^{2-} + 2S_{2}O_{3}^{2-} + 6H_{2}O \) is a **disproportionation reaction**.

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