Sulphur and rest of the elements of group `16` are less electronegative than oxygen, Therefore, their atoms cannot take electrons easily. They can acquire `ns^2 np^6` configuration by sharing two electrons with the atoms of other elements and thus, exhibit `+2` oxidation state in their compounds. In addition to this, their atoms have vacant d-orbitals in their valence shell to which electrons can be promoted from the `p` and s-orbitals of the same shell. As a result, they can show `+4` and `+ 6` oxidation states.
The oxidation state of sulphur in `Na_2 S_4 O_6` is

Options

A. `2//3`
B. `3//2`
C. `3//5`
D. `5//2`

Explanation

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

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

### Step 2: Set up the equation
Let the oxidation state of sulfur be \( X \). In the compound \( \text{Na}_2\text{S}_4\text{O}_6 \), we have:
- 2 sodium atoms contributing \( 2 \times (+1) = +2 \).
- 6 oxygen atoms contributing \( 6 \times (-2) = -12 \).
- 4 sulfur atoms contributing \( 4X \).

### Step 3: Write the equation for the overall charge
The overall charge of the compound is neutral (0), so we can write the equation:
\[ 4X + 2 - 12 = 0 \]

### Step 4: Simplify the equation
Rearranging the equation gives:
\[ 4X - 10 = 0 \]
\[ 4X = 10 \]

### Step 5: Solve for \( X \)
Dividing both sides by 4:
\[ X = \frac{10}{4} = 2.5 \]

### Conclusion
The oxidation state of sulfur in \( \text{Na}_2\text{S}_4\text{O}_6 \) is \( +2.5 \) or \( \frac{5}{2} \).