In the compound `Na_(2)S_(2)O_(3)`, the oxidation state of sulphur is:

Options

A. `-2`
B. `+2`
C. `+4`
D. `+6`

Answer

To determine the oxidation state of sulfur in the compound \( \text{Na}_2\text{S}_2\text{O}_3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the elements and their usual oxidation states**:
- Sodium (\( \text{Na} \)) typically has an oxidation state of \( +1 \).
- Oxygen (\( \text{O} \)) typically has an oxidation state of \( -2 \).

2. **Set up the equation**:
- Let the oxidation state of the first sulfur atom be \( x \) and the second sulfur atom be \( -2 \) (as indicated in the video).
- The overall charge of the compound is neutral (0).

3. **Write the equation based on the oxidation states**:
\[ 2(\text{Na}) + x + (-2) + 3(-2) = 0 \]
- Here, we have:
- 2 sodium atoms contributing \( 2 \times (+1) = +2 \)
- One sulfur atom with oxidation state \( x \)
- One sulfur atom with oxidation state \( -2 \)
- Three oxygen atoms contributing \( 3 \times (-2) = -6 \)

4. **Substitute the values into the equation**:
\[ 2 + x - 2 - 6 = 0 \]

5. **Simplify the equation**:
\[ x - 6 = 0 \]

6. **Solve for \( x \)**:
\[ x = +6 \]

7. **Conclusion**:
- The oxidation state of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) is \( +6 \).

### Final Answer:
The oxidation state of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) is \( +6 \).