The difference in the oxidation states of the two types of sulphur atoms in `Na_(2)S_(4)O_(6)` is :

Explanation

To find the difference in the oxidation states of the two types of sulfur atoms in the compound \( \text{Na}_2\text{S}_4\text{O}_6 \), we will follow these steps:

### Step 1: Identify the oxidation states of sulfur in the compound
In \( \text{Na}_2\text{S}_4\text{O}_6 \), there are two different types of sulfur atoms. We will denote them as \( \text{S}_1 \) and \( \text{S}_2 \).

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

### Step 3: Set up the equation for the overall charge
The overall charge of the compound is neutral (0). Therefore, we can set up the equation based on the oxidation states:
\[ 2(+1) + 4x + 6(-2) = 0 \]
where \( x \) is the oxidation state of sulfur.

### Step 4: Simplify the equation
Substituting the known values into the equation:
\[ 2 + 4x - 12 = 0 \]
This simplifies to:
\[ 4x - 10 = 0 \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):
\[ 4x = 10 \\ x = \frac{10}{4} = 2.5 \]

### Step 6: Identify the two types of sulfur
In \( \text{Na}_2\text{S}_4\text{O}_6 \), we have two types of sulfur:
- \( \text{S}_1 \) (the one with oxidation state +5)
- \( \text{S}_2 \) (the one with oxidation state +2)

### Step 7: Calculate the difference in oxidation states
Now we can calculate the difference in oxidation states:
\[ \text{Difference} = +5 - (+2) = 3 \]

### Final Answer
The difference in the oxidation states of the two types of sulfur atoms in \( \text{Na}_2\text{S}_4\text{O}_6 \) is **3**.

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