The species having bond angle of `120^(@)` is

Options

A. `PH_(3)`
B. `ClF_(3)`
C. `NCl_(3)`
D. `BCl_(3) `

Explanation

To determine the species that has a bond angle of 120 degrees, we need to analyze the hybridization of the given options and their corresponding bond angles.

### Step-by-Step Solution:

1. **Understanding Hybridization**:
- Hybridization is the mixing of atomic orbitals to form new hybrid orbitals. The type of hybridization determines the geometry and bond angles of the molecule.
- The common types of hybridization are:
- **sp**: Linear geometry, bond angle = 180°
- **sp²**: Trigonal planar geometry, bond angle = 120°
- **sp³**: Tetrahedral geometry, bond angle = 109.5°

2. **Analyzing the Given Options**:
- The question provides four species: PCl3, ClF3, NCl3, and BCl3.
- We need to determine the hybridization for each of these species.

3. **Determining Hybridization**:
- **PCl3 (Phosphorus trichloride)**:
- Phosphorus has 5 valence electrons and forms three bonds with chlorine, leaving one lone pair.
- Hybridization: sp³ (due to one lone pair and three bonding pairs). Bond angle ≈ 107° (not 120°).
- **ClF3 (Chlorine trifluoride)**:
- Chlorine has 7 valence electrons and forms three bonds with fluorine, leaving two lone pairs.
- Hybridization: sp³d (due to two lone pairs and three bonding pairs). Bond angle < 90° (not 120°).
- **NCl3 (Nitrogen trichloride)**:
- Nitrogen has 5 valence electrons and forms three bonds with chlorine, leaving one lone pair.
- Hybridization: sp³ (due to one lone pair and three bonding pairs). Bond angle ≈ 107° (not 120°).
- **BCl3 (Boron trichloride)**:
- Boron has 3 valence electrons and forms three bonds with chlorine, with no lone pairs.
- Hybridization: sp² (due to three bonding pairs). Bond angle = 120°.

4. **Conclusion**:
- Among the given options, **BCl3** has a bond angle of 120 degrees due to its sp² hybridization and trigonal planar geometry.

### Final Answer:
The species having a bond angle of 120 degrees is **BCl3**.