The oxidation state of sulphur in sodium tetrathionate (`Na_(2)S_(4)O_(6)`) is :

Explanation

To determine the oxidation state of sulfur in sodium tetrathionate (`Na2S4O6`), we can follow these steps:

### Step 1: Write the formula and identify the components
The formula of sodium tetrathionate is `Na2S4O6`. It contains:
- 2 sodium (Na) atoms
- 4 sulfur (S) atoms
- 6 oxygen (O) atoms

### Step 2: Assign oxidation states to known elements
- Sodium (Na) typically has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2, except in peroxides or when bonded to fluorine.

### Step 3: Set up the oxidation state equation
Let the oxidation state of sulfur be represented as \( x \). The overall charge of a neutral compound is 0. Therefore, we can set up the equation based on the contributions of each element:

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

### Step 4: Simplify the equation
Substituting the known values into the equation gives:

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

### Step 5: Solve for \( x \)
Now, simplify the equation:

\[ 4x - 10 = 0 \]
\[ 4x = 10 \]
\[ x = \frac{10}{4} = 2.5 \]

### Step 6: Conclusion
The oxidation state of sulfur in sodium tetrathionate (`Na2S4O6`) is +2.5.