Which of the following is a correct set with respect to molecule, hybridization, and shape?

Options

A. `BeCl_(2),sp^(2)`, linear
B. `BeCl_(2),sp^(2)`, triangular planar
C. `BCl_(3),sp^(2)`, triangular planar
D. `BCl_(3),sp^(3)`, tetrahedral

Answer

To determine the correct set with respect to molecule, hybridization, and shape, we will analyze each molecule provided in the options.

### Step-by-Step Solution:

1. **Analyze BeCl2 (Beryllium Chloride)**:
- **Valence Electrons**: Beryllium has 2 electrons in its outermost shell.
- **Bond Formation**: Beryllium will form 2 bond pairs with 2 chlorine atoms.
- **Hybridization Calculation**:
- Number of bond pairs = 2
- Number of lone pairs = 0
- Hybridization = Number of bond pairs + Number of lone pairs = 2 + 0 = 2 (which corresponds to sp hybridization).
- **Shape**: The shape is linear due to the arrangement of the two bond pairs.

Conclusion: BeCl2 has sp hybridization and a linear shape.

2. **Analyze BCl3 (Boron Trichloride)**:
- **Valence Electrons**: Boron has 3 electrons in its outermost shell.
- **Bond Formation**: Boron will form 3 bond pairs with 3 chlorine atoms.
- **Hybridization Calculation**:
- Number of bond pairs = 3
- Number of lone pairs = 0
- Hybridization = Number of bond pairs + Number of lone pairs = 3 + 0 = 3 (which corresponds to sp² hybridization).
- **Shape**: The shape is trigonal planar due to the arrangement of the three bond pairs.

Conclusion: BCl3 has sp² hybridization and a trigonal planar shape.

3. **Evaluate the Options**:
- **Option A**: Incorrect (BeCl2 is sp and linear).
- **Option B**: Incorrect (BeCl2 is sp, not sp).
- **Option C**: Correct (BCl3 is sp² and trigonal planar).
- **Option D**: Incorrect (BCl3 cannot be sp³, it is sp²).

### Final Conclusion:
The correct set with respect to molecule, hybridization, and shape is **Option C: BCl3, sp², trigonal planar**.