`S_(8)+NaOH to Na_(2)S+Na_(2)S_(2)O_(3)`

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 solve the reaction \( S_8 + NaOH \rightarrow Na_2S + Na_2S_2O_3 \), we will analyze the oxidation states of the elements involved and determine the type of reaction taking place.

### Step 1: Identify the oxidation states of the elements in the reactants and products.

1. **Sulfur in \( S_8 \)**:
- Sulfur is in its elemental form, so its oxidation state is 0.

2. **Sodium in \( NaOH \)**:
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Hydrogen (H) has an oxidation state of +1.

### Step 2: Determine the oxidation states in the products.

1. **In \( Na_2S \)**:
- Sodium (Na) has an oxidation state of +1.
- To find the oxidation state of sulfur (S), we set up the equation:
\[ 2(+1) + x = 0 \implies x = -2 \]
- Thus, sulfur in \( Na_2S \) is -2.

2. **In \( Na_2S_2O_3 \)**:
- Sodium (Na) is +1.
- The overall charge of the compound is neutral, so we set up the equation:
\[ 2(+1) + 2x + 3(-2) = 0 \]
\[ 2 + 2x - 6 = 0 \implies 2x - 4 = 0 \implies 2x = 4 \implies x = +2 \]
- Thus, the oxidation state of sulfur in \( Na_2S_2O_3 \) is +2.

### Step 3: Analyze the changes in oxidation states.

- **Sulfur changes from 0 to -2** in \( Na_2S \) (reduction).
- **Sulfur changes from 0 to +2** in \( Na_2S_2O_3 \) (oxidation).

### Step 4: Determine the type of reaction.

Since one species (sulfur) is both oxidized and reduced in the reaction, this is classified as a **disproportionation reaction**.

### Conclusion:

The reaction \( S_8 + NaOH \rightarrow Na_2S + Na_2S_2O_3 \) is a **disproportionation reaction**.

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