In one of the following compounds, the oxidation number of sulphur is not a whole number

Options

A. `Na_(2)S_(4)O_(6)`
B. `H_(2)SO_(5)`
C. `H_(2)SO_(4)`
D. `Na_(2)S_(2)O_(3)`

Answer

To determine in which of the following compounds the oxidation number of sulfur is not a whole number, we will analyze each compound step by step.

### Step 1: Analyze Na2S2O3
1. **Identify the oxidation states:**
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation:**
- Let the oxidation state of sulfur (S) be x.
- The compound has 2 sodium atoms and 3 oxygen atoms.
- The equation can be set up as follows:
\[ 2(+1) + 2x + 3(-2) = 0 \]
- This simplifies to:
\[ 2 + 2x - 6 = 0 \]
\[ 2x - 4 = 0 \]
\[ 2x = 4 \implies x = 2 \]
- The oxidation state of sulfur in Na2S2O3 is +2, which is a whole number.

### Step 2: Analyze H2SO5
1. **Identify the oxidation states:**
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation:**
- Let the oxidation state of sulfur (S) be x.
- The equation can be set up as follows:
\[ 2(+1) + x + 5(-2) = 0 \]
- This simplifies to:
\[ 2 + x - 10 = 0 \]
\[ x - 8 = 0 \]
\[ x = 8 \]
- The oxidation state of sulfur in H2SO5 is +8, which is a whole number.

### Step 3: Analyze Na2S4O6
1. **Identify the oxidation states:**
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation:**
- Let the oxidation state of sulfur (S) be x.
- The equation can be set up as follows:
\[ 2(+1) + 4x + 6(-2) = 0 \]
- This simplifies to:
\[ 2 + 4x - 12 = 0 \]
\[ 4x - 10 = 0 \]
\[ 4x = 10 \implies x = 2.5 \]
- The oxidation state of sulfur in Na2S4O6 is +2.5, which is **not a whole number**.

### Conclusion
The compound in which the oxidation number of sulfur is not a whole number is **Na2S4O6**.