The average oxidation states of sulphur in `Na_2S_2O_3 and Na_2S_4O_6` are respectively.

Options

A. `+5 & +2`
B. `+2 & +2.5`
C. `+5 & 2.5`
D. `+2 & +4`

Explanation

To find the average oxidation states of sulfur in the compounds \( \text{Na}_2\text{S}_2\text{O}_3 \) and \( \text{Na}_2\text{S}_4\text{O}_6 \), we will follow these steps:

### Step 1: Calculate the oxidation state of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \)

1. **Assign oxidation states:**
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of sulfur (S) be \( x \).

2. **Set up the equation:**
The compound has 2 sodium atoms, 2 sulfur atoms, and 3 oxygen atoms. The overall charge of the compound is neutral (0).
\[ 2(+1) + 2(x) + 3(-2) = 0 \]

3. **Simplify the equation:**
\[ 2 + 2x - 6 = 0 \]
\[ 2x - 4 = 0 \]

4. **Solve for \( x \):**
\[ 2x = 4 \implies x = +2 \]

### Step 2: Calculate the oxidation state of sulfur in \( \text{Na}_2\text{S}_4\text{O}_6 \)

1. **Assign oxidation states:**
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of sulfur (S) be \( y \).

2. **Set up the equation:**
The compound has 2 sodium atoms, 4 sulfur atoms, and 6 oxygen atoms. The overall charge of the compound is neutral (0).
\[ 2(+1) + 4(y) + 6(-2) = 0 \]

3. **Simplify the equation:**
\[ 2 + 4y - 12 = 0 \]
\[ 4y - 10 = 0 \]

4. **Solve for \( y \):**
\[ 4y = 10 \implies y = \frac{10}{4} = 2.5 \]

### Step 3: State the average oxidation states

- The average oxidation state of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) is \( +2 \).
- The average oxidation state of sulfur in \( \text{Na}_2\text{S}_4\text{O}_6 \) is \( +2.5 \).

### Final Answer:

The average oxidation states of sulfur in \( \text{Na}_2\text{S}_2\text{O}_3 \) and \( \text{Na}_2\text{S}_4\text{O}_6 \) are \( +2 \) and \( +2.5 \) respectively.

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