The average oxidation state of sulphur in `Na_2S_4O_6` is

Options

A. `+2.5`
B. `+2`
C. `+3.0`
D. `+3.5`

Answer

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

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

### Step 2: Write the formula for the total oxidation state
The total oxidation state of the compound must equal zero since it is a neutral compound. We can express this as:
\[ \text{Total oxidation state} = (\text{Oxidation state of Na}) + (\text{Oxidation state of S}) + (\text{Oxidation state of O}) = 0 \]

### Step 3: Set up the equation
Let \( x \) be the oxidation state of sulfur (S). The formula \( \text{Na}_2\text{S}_4\text{O}_6 \) contains:
- 2 sodium atoms: \( 2 \times (+1) = +2 \)
- 4 sulfur atoms: \( 4 \times x = 4x \)
- 6 oxygen atoms: \( 6 \times (-2) = -12 \)

Putting this all together, we have:
\[ 2 + 4x - 12 = 0 \]

### Step 4: Solve for \( x \)
Now, we can simplify the equation:
\[ 4x - 10 = 0 \]
\[ 4x = 10 \]
\[ x = \frac{10}{4} = 2.5 \]

### Step 5: Conclusion
The average oxidation state of sulfur in \( \text{Na}_2\text{S}_4\text{O}_6 \) is \( 2.5 \).

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