The oxidation number of sulphur in ` S_2F_2, H_2S` respectively, are

Options

A. + 1 and - 2
B. +1 and-2
C. +1 and +2
D. +1 and-2

Answer

To determine the oxidation number of sulfur in the compounds \( S_2F_2 \) and \( H_2S \), we can follow these steps:

### Step 1: Determine the oxidation number of sulfur in \( S_2F_2 \)

1. **Identify the electronegativity of the elements**: Fluorine (F) is more electronegative than sulfur (S). The electronegativity of fluorine is -1 when it forms a bond.

2. **Set up the equation**: In \( S_2F_2 \), there are 2 sulfur atoms and 2 fluorine atoms. The overall charge of the molecule is neutral (0).
\[ 2x + 2(-1) = 0 \]
where \( x \) is the oxidation number of sulfur.

3. **Solve the equation**:
\[ 2x - 2 = 0 \\ 2x = 2 \\ x = +1 \]
Thus, the oxidation number of sulfur in \( S_2F_2 \) is **+1**.

### Step 2: Determine the oxidation number of sulfur in \( H_2S \)

1. **Identify the electronegativity of the elements**: Hydrogen (H) has an oxidation state of +1.

2. **Set up the equation**: In \( H_2S \), there are 2 hydrogen atoms and 1 sulfur atom. The overall charge of the molecule is neutral (0).
\[ 2(+1) + x = 0 \]
where \( x \) is the oxidation number of sulfur.

3. **Solve the equation**:
\[ 2 + x = 0 \\ x = -2 \]
Thus, the oxidation number of sulfur in \( H_2S \) is **-2**.

### Summary of Results
- The oxidation number of sulfur in \( S_2F_2 \) is **+1**.
- The oxidation number of sulfur in \( H_2S \) is **-2**.

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