Sulphur shows a large number of oxidation states in its compounds such as `H_2S, SO_2, H_2SO_3,Na_2S_2O_3, H_2SO_4, H_2S_2O_7, H_2S_2O_4` and elemental S . In the reactions of sulphur compounds , the oxidation number of sulphur increases and decreases in different reactions and the compounds act as reducing and oxidising agents.
(i) `H_2S +HNO_3 rarr NO+S+H_2O`
(ii) `2Na_2S_2O_3+I_2rarr Na_2S_4O_6+2NaI`
(iii) `H_2SO_4+2HBr rarr 2H_2O+Br_2+SO_2`
(iv) `S_8+12OH^(-) rarr 4S^(2-) +2S_2O_3^(2-) +6H_2O`
What type of reaction is reaction (iv) ?

Answer

To determine the type of reaction for the given equation (iv) \( S_8 + 12OH^- \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O \), we will analyze the oxidation states of sulfur in the reactants and products.

### Step-by-Step Solution:

1. **Identify the reactants and products**:
The reaction is given as:
\[ S_8 + 12OH^- \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O \]

2. **Determine the oxidation state of sulfur in the reactants**:
In \( S_8 \), sulfur is in its elemental form, so the oxidation state is:
\[ \text{Oxidation state of } S \text{ in } S_8 = 0 \]

3. **Determine the oxidation state of sulfur in the products**:
- For \( S^{2-} \):
\[ \text{Oxidation state of } S = -2 \]
- For \( S_2O_3^{2-} \):
Let the oxidation state of sulfur in \( S_2O_3^{2-} \) be \( x \). The equation for the oxidation state 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 \( +2 \).

4. **Compare the oxidation states**:
- In the reactants, the oxidation state of sulfur is \( 0 \).
- In the products, sulfur has oxidation states of \( -2 \) (in \( 4S^{2-} \)) and \( +2 \) (in \( 2S_2O_3^{2-} \)).

5. **Identify the type of reaction**:
Since the same element (sulfur) undergoes both oxidation (increase in oxidation state from \( 0 \) to \( +2 \)) and reduction (decrease in oxidation state from \( 0 \) to \( -2 \)) in the same reaction, this is classified as a **disproportionation reaction**.

### Conclusion:
The type of reaction (iv) is a **disproportionation reaction**.