The oxidation number of sulphur in `Na_2S_4O_6` is .

Options

A. ` 1.5`
B. ` 2.5`
C. ` 3.0`
D. ` 2.0`

Explanation

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

### Step 1: Identify the oxidation states of known elements
In the compound \( \text{Na}_2\text{S}_4\text{O}_6 \):
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2.

### Step 2: Write the equation based on the total charge
The total charge of the compound is neutral (0). Therefore, we can set up the equation based on the oxidation states:

\[ \text{Total oxidation state} = (\text{Oxidation state of Na}) + (\text{Oxidation state of S}) + (\text{Oxidation state of O}) = 0 \]

### Step 3: Substitute the known values into the equation
Substituting the known oxidation states into the equation, we have:

\[ 2(+1) + 4(x) + 6(-2) = 0 \]

Where \( x \) is the oxidation state of sulfur.

### Step 4: Simplify the equation
Now, simplify the equation:

\[ 2 + 4x - 12 = 0 \]

This simplifies to:

\[ 4x - 10 = 0 \]

### Step 5: Solve for the oxidation state of sulfur
Now, solve for \( x \):

\[ 4x = 10 \]
\[ x = \frac{10}{4} = 2.5 \]

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

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