The oxidation number of `S` in `Na_(2)S_(4)O_(6)` is

Options

A. `+0.5`
B. `2.5`
C. `+4`
D. `+6`

Explanation

To find the oxidation number of sulfur (S) in the compound \( \text{Na}_2\text{S}_4\text{O}_6 \), we can follow these steps:

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

### Step 2: Set up the equation
The total charge of the compound is neutral (0). Therefore, we can set up the equation based on the oxidation states:
\[ \text{Total oxidation state} = 0 \]
This can be expressed as:
\[ 2(\text{oxidation state of Na}) + 4(\text{oxidation state of S}) + 6(\text{oxidation state of O}) = 0 \]
Substituting the known values:
\[ 2(+1) + 4x + 6(-2) = 0 \]

### Step 3: Simplify the equation
Now, simplify the equation:
\[ 2 + 4x - 12 = 0 \]
Combine like terms:
\[ 4x - 10 = 0 \]

### Step 4: Solve for x
Now, isolate \( x \):
\[ 4x = 10 \]
\[ x = \frac{10}{4} = 2.5 \]

### Step 5: Interpret the result
The oxidation number of sulfur in \( \text{Na}_2\text{S}_4\text{O}_6 \) is \( +2.5 \). This indicates that the sulfur atoms in this compound have mixed oxidation states, likely with three sulfur atoms at \( +2 \) and one sulfur atom at \( +4 \).

### Final Answer
The oxidation number of sulfur in \( \text{Na}_2\text{S}_4\text{O}_6 \) is \( +2.5 \).

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