Which of the following sets of molecules contains the same number of lone pairs of electrons in the central atom ?

Options

A. `SO_(2),ClF_(3),BrF_(3)`
B. `SF_(4),NH_(3),O_(3)`
C. `ClF_(3),XeF_(2),H_(2)O`
D. `H_(2)O,SF_(2),NH_(3)`

Explanation

To determine which set of molecules contains the same number of lone pairs of electrons on the central atom, we will analyze each molecule step by step.

### Step 1: Analyze SO2
1. **Valence Electrons Calculation**:
- Sulfur (S) has 6 valence electrons.
- Oxygen (O) has 6 valence electrons, and there are 2 oxygen atoms.
- Total = 6 (S) + 2 × 6 (O) = 18 valence electrons.

2. **Hybridization**:
- Use the formula:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence Electrons} + \text{Monovalent Atoms} - \text{Charge} \right) \]
- For SO2:
\[ \text{Hybridization} = \frac{1}{2} (6 + 0 + 0) = 3 \quad \text{(sp}^2\text{)} \]
- Bond pairs = 2 (S=O bonds), Lone pairs = 1.

### Step 2: Analyze ClF3
1. **Valence Electrons Calculation**:
- Chlorine (Cl) has 7 valence electrons.
- Fluorine (F) has 7 valence electrons, and there are 3 fluorine atoms.
- Total = 7 (Cl) + 3 × 7 (F) = 28 valence electrons.

2. **Hybridization**:
- For ClF3:
\[ \text{Hybridization} = \frac{1}{2} (7 + 3 + 0) = 5 \quad \text{(sp}^3\text{d)} \]
- Bond pairs = 3, Lone pairs = 2.

### Step 3: Analyze BrF3
1. **Valence Electrons Calculation**:
- Bromine (Br) has 7 valence electrons.
- Fluorine (F) has 7 valence electrons, and there are 3 fluorine atoms.
- Total = 7 (Br) + 3 × 7 (F) = 28 valence electrons.

2. **Hybridization**:
- For BrF3:
\[ \text{Hybridization} = \frac{1}{2} (7 + 3 + 0) = 5 \quad \text{(sp}^3\text{d)} \]
- Bond pairs = 3, Lone pairs = 2.

### Step 4: Analyze SF4
1. **Valence Electrons Calculation**:
- Sulfur (S) has 6 valence electrons.
- Fluorine (F) has 7 valence electrons, and there are 4 fluorine atoms.
- Total = 6 (S) + 4 × 7 (F) = 34 valence electrons.

2. **Hybridization**:
- For SF4:
\[ \text{Hybridization} = \frac{1}{2} (6 + 4 + 0) = 5 \quad \text{(sp}^3\text{d)} \]
- Bond pairs = 4, Lone pairs = 1.

### Step 5: Analyze NH3
1. **Valence Electrons Calculation**:
- Nitrogen (N) has 5 valence electrons.
- Hydrogen (H) has 1 valence electron, and there are 3 hydrogen atoms.
- Total = 5 (N) + 3 × 1 (H) = 8 valence electrons.

2. **Hybridization**:
- For NH3:
\[ \text{Hybridization} = \frac{1}{2} (5 + 3 + 0) = 4 \quad \text{(sp}^3\text{)} \]
- Bond pairs = 3, Lone pairs = 1.

### Step 6: Analyze O3
1. **Valence Electrons Calculation**:
- Oxygen (O) has 6 valence electrons, and there are 3 oxygen atoms.
- Total = 3 × 6 = 18 valence electrons.

2. **Hybridization**:
- For O3:
\[ \text{Hybridization} = \frac{1}{2} (6 + 0 + 0) = 3 \quad \text{(sp}^2\text{)} \]
- Bond pairs = 2, Lone pairs = 1.

### Conclusion
From the analysis:
- SO2 has 1 lone pair.
- ClF3 has 2 lone pairs.
- BrF3 has 2 lone pairs.
- SF4 has 1 lone pair.
- NH3 has 1 lone pair.
- O3 has 1 lone pair.

The set of molecules that contains the same number of lone pairs on the central atom is **SF4, NH3, and O3**, all having 1 lone pair.