The oxidation number of sulphur is `Na_(2)S_(2)O_(4)` is

Options

A. `1.5`
B. `2.5`
C. 3
D. 2

Explanation

To find the oxidation number of sulfur in the compound \( \text{Na}_2\text{S}_2\text{O}_4 \), we can follow these steps:

### Step 1: Identify the known oxidation states
- Sodium (Na) has an oxidation state of \( +1 \).
- Oxygen (O) typically has an oxidation state of \( -2 \).

### Step 2: Write the formula for the total oxidation states
The formula for the total oxidation states in a neutral compound is:
\[ \text{Total oxidation state} = 0 \]
For \( \text{Na}_2\text{S}_2\text{O}_4 \), we have:
- 2 sodium atoms: \( 2 \times (+1) = +2 \)
- 4 oxygen atoms: \( 4 \times (-2) = -8 \)
- Let the oxidation state of sulfur be \( X \). Since there are 2 sulfur atoms, this contributes \( 2X \).

### Step 3: Set up the equation
Now, we can set up the equation based on the total oxidation states:
\[ +2 + 2X - 8 = 0 \]

### Step 4: Solve for \( X \)
Rearranging the equation gives:
\[ 2X - 6 = 0 \]
\[ 2X = 6 \]
\[ X = 3 \]

### Conclusion
The oxidation number of sulfur in \( \text{Na}_2\text{S}_2\text{O}_4 \) is \( +3 \).

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