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 is the oxidation number of sulphur in `H_2SO_3 and H_2S_2O_4` ?

Answer

To determine the oxidation number of sulfur in the compounds \( H_2SO_3 \) and \( H_2S_2O_4 \), we can use the "X method" as described in the video transcript. Let's go through the calculations step by step.

### Step 1: Calculate the oxidation number of sulfur in \( H_2SO_3 \)

1. **Identify the known oxidation states**:
- Hydrogen (H) has an oxidation state of \( +1 \).
- Oxygen (O) has an oxidation state of \( -2 \).

2. **Set up the equation**:
- For \( H_2SO_3 \):
\[ 2(+1) + X + 3(-2) = 0 \]

3. **Substitute the known values**:
- This gives us:
\[ 2 + X - 6 = 0 \]

4. **Solve for X**:
- Rearranging the equation:
\[ X - 4 = 0 \implies X = +4 \]

Thus, the oxidation number of sulfur in \( H_2SO_3 \) is \( +4 \).

### Step 2: Calculate the oxidation number of sulfur in \( H_2S_2O_4 \)

1. **Identify the known oxidation states**:
- Again, Hydrogen (H) has an oxidation state of \( +1 \).
- Oxygen (O) has an oxidation state of \( -2 \).

2. **Set up the equation**:
- For \( H_2S_2O_4 \):
\[ 2(+1) + 2X + 4(-2) = 0 \]

3. **Substitute the known values**:
- This gives us:
\[ 2 + 2X - 8 = 0 \]

4. **Solve for X**:
- Rearranging the equation:
\[ 2X - 6 = 0 \implies 2X = 6 \implies X = +3 \]

Thus, the oxidation number of sulfur in \( H_2S_2O_4 \) is \( +3 \).

### Summary of Results
- The oxidation number of sulfur in \( H_2SO_3 \) is \( +4 \).
- The oxidation number of sulfur in \( H_2S_2O_4 \) is \( +3 \).