In which one of the following compounds does sulphur have the least oxidation number

Options

A. `SO_2`
B. `H_2SO_4`
C. `Na_2S_2O_3`
D. `Na_2S_4O_6`

Explanation

To determine in which compound sulfur has the least oxidation number, we need to analyze the oxidation states of sulfur in the given compounds. Let's break down the process step by step.

### Step 1: Identify the Compounds
The question provides several compounds. We will analyze the oxidation state of sulfur in each of these compounds. The compounds are:
1. \( H_2S \)
2. \( H_2SO_4 \)
3. \( Na_2H_2S_2O_3 \)
4. \( Na_2S_4O_6 \)

### Step 2: Calculate the Oxidation State of Sulfur in Each Compound

#### Compound 1: \( H_2S \)
- The oxidation state of hydrogen (H) is +1.
- Let the oxidation state of sulfur (S) be \( x \).
- The equation for the compound is:
\[ 2(+1) + x = 0 \implies 2 + x = 0 \implies x = -2 \]
- **Oxidation state of sulfur in \( H_2S \) is -2.**

#### Compound 2: \( H_2SO_4 \)
- The oxidation state of hydrogen (H) is +1, and oxygen (O) is -2.
- Let the oxidation state of sulfur (S) be \( x \).
- The equation for the compound is:
\[ 2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6 \]
- **Oxidation state of sulfur in \( H_2SO_4 \) is +6.**

#### Compound 3: \( Na_2H_2S_2O_3 \)
- The oxidation state of sodium (Na) is +1, hydrogen (H) is +1, and oxygen (O) is -2.
- Let the oxidation state of sulfur (S) be \( x \).
- The equation for the compound is:
\[ 2(+1) + 2(+1) + 2x + 3(-2) = 0 \implies 2 + 2 + 2x - 6 = 0 \implies 2x - 2 = 0 \implies x = +1 \]
- **Oxidation state of sulfur in \( Na_2H_2S_2O_3 \) is +1.**

#### Compound 4: \( Na_2S_4O_6 \)
- The oxidation state of sodium (Na) is +1, and oxygen (O) is -2.
- Let the oxidation state of sulfur (S) be \( x \).
- The equation for the compound is:
\[ 2(+1) + 4x + 6(-2) = 0 \implies 2 + 4x - 12 = 0 \implies 4x - 10 = 0 \implies x = +2.5 \]
- **Oxidation state of sulfur in \( Na_2S_4O_6 \) is +2.5.**

### Step 3: Compare the Oxidation States
Now we compare the oxidation states of sulfur in all the compounds:
- \( H_2S \): -2
- \( H_2SO_4 \): +6
- \( Na_2H_2S_2O_3 \): +1
- \( Na_2S_4O_6 \): +2.5

### Conclusion
The compound in which sulfur has the least oxidation number is **\( H_2S \)**, where sulfur has an oxidation state of **-2**.

### Final Answer
**The compound in which sulfur has the least oxidation number is \( H_2S \).**

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