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

Explanation

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

### Step 1: Identify the oxidation states of known elements
In sodium tetrathionate, we know:
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

### Step 2: Set up the equation for the oxidation states
Let the oxidation state of sulfur (S) be represented as \( x \). The formula for sodium tetrathionate is Na₂S₄O₆, which contains:
- 2 sodium atoms
- 4 sulfur atoms
- 6 oxygen atoms

The overall charge of the compound is neutral (0). Therefore, we can set up the equation based on the sum of the oxidation states:

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

### Step 3: Simplify the equation
Substituting the values into the equation gives:

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

### Step 4: Combine like terms
Now, combine the constants:

\[ 4x - 10 = 0 \]

### Step 5: Solve for \( x \)
Now, isolate \( x \):

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

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

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