What is the oxidation state of S in `Na_2S_2` ?

Options

A. `+1`
B. `-2`
C. `-1`
D. 0

Answer

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

### Step-by-Step Solution:

1. **Identify the oxidation state of sodium (Na):**
Sodium is an alkali metal (Group 1) and has a fixed oxidation state of +1.

2. **Write the formula for the compound:**
The formula for the compound is \( \text{Na}_2\text{S}_2 \). This means there are 2 sodium (Na) atoms and 2 sulfur (S) atoms.

3. **Set up the equation for the total oxidation states:**
Let the oxidation state of sulfur be \( x \). The total oxidation state for the compound can be expressed as:
\[ 2(\text{oxidation state of Na}) + 2(\text{oxidation state of S}) = 0 \]
Substituting the known oxidation state of sodium:
\[ 2(+1) + 2(x) = 0 \]

4. **Simplify the equation:**
\[ 2 + 2x = 0 \]

5. **Solve for \( x \):**
\[ 2x = -2 \\ x = -1 \]

6. **Conclusion:**
The oxidation state of sulfur in \( \text{Na}_2\text{S}_2 \) is \(-1\).

### Final Answer:
The oxidation state of sulfur (S) in \( \text{Na}_2\text{S}_2 \) is \(-1\).