Which of the following molecules does NOT contain a lone pair of electron ?

Options

A. `NH_(3)`
B. `H_(2)O`
C. `PF_(5)`
D. `ClF_(3)`

Answer

To determine which of the given molecules does NOT contain a lone pair of electrons, we will analyze each molecule's hybridization and the presence of lone pairs.

### Step 1: Analyze NH3 (Ammonia)
1. **Identify the central atom**: Nitrogen (N).
2. **Count the valence electrons**: Nitrogen has 5 valence electrons.
3. **Count the univalent species**: There are 3 hydrogen (H) atoms bonded to nitrogen.
4. **Calculate N**:
\[ N = \frac{1}{2} \times (\text{valence electrons of N} + \text{number of univalent species}) \]
\[ N = \frac{1}{2} \times (5 + 3) = \frac{8}{2} = 4 \]
5. **Determine hybridization**: Since N = 4, the hybridization is sp³.
6. **Determine lone pairs**: Nitrogen forms 3 bonds with hydrogen, leaving 2 electrons as a lone pair.

**Conclusion**: NH3 has 1 lone pair.

### Step 2: Analyze H2O (Water)
1. **Identify the central atom**: Oxygen (O).
2. **Count the valence electrons**: Oxygen has 6 valence electrons.
3. **Count the univalent species**: There are 2 hydrogen atoms bonded to oxygen.
4. **Calculate N**:
\[ N = \frac{1}{2} \times (6 + 2) = \frac{8}{2} = 4 \]
5. **Determine hybridization**: Since N = 4, the hybridization is sp³.
6. **Determine lone pairs**: Oxygen forms 2 bonds with hydrogen, leaving 4 electrons as 2 lone pairs.

**Conclusion**: H2O has 2 lone pairs.

### Step 3: Analyze PF5 (Phosphorus Pentafluoride)
1. **Identify the central atom**: Phosphorus (P).
2. **Count the valence electrons**: Phosphorus has 5 valence electrons.
3. **Count the univalent species**: There are 5 fluorine atoms bonded to phosphorus.
4. **Calculate N**:
\[ N = \frac{1}{2} \times (5 + 5) = \frac{10}{2} = 5 \]
5. **Determine hybridization**: Since N = 5, the hybridization is sp³d.
6. **Determine lone pairs**: Phosphorus forms 5 bonds with fluorine, using all its valence electrons, leaving no lone pairs.

**Conclusion**: PF5 has 0 lone pairs.

### Step 4: Analyze ClF3 (Chlorine Trifluoride)
1. **Identify the central atom**: Chlorine (Cl).
2. **Count the valence electrons**: Chlorine has 7 valence electrons.
3. **Count the univalent species**: There are 3 fluorine atoms bonded to chlorine.
4. **Calculate N**:
\[ N = \frac{1}{2} \times (7 + 3) = \frac{10}{2} = 5 \]
5. **Determine hybridization**: Since N = 5, the hybridization is sp³d.
6. **Determine lone pairs**: Chlorine forms 3 bonds with fluorine, leaving 4 electrons as 2 lone pairs.

**Conclusion**: ClF3 has 2 lone pairs.

### Final Answer
The molecule that does NOT contain a lone pair of electrons is **PF5**.