Which one of the following is the 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 solve the question regarding the correct set of molecule, hybridization, and shape, we will analyze the options provided in the context of molecular geometry and hybridization.

### Step-by-Step Solution:

1. **Identify the Molecules and Their Structures**:
- We need to consider the molecules mentioned in the options, which may include BeCl2, BCl3, etc.

2. **Determine the Valence Electrons**:
- For BeCl2: Beryllium (Be) has 2 valence electrons, and each chlorine (Cl) has 7 valence electrons. Thus, BeCl2 has a total of 2 + (2 × 7) = 16 valence electrons.
- For BCl3: Boron (B) has 3 valence electrons, and each chlorine (Cl) has 7 valence electrons. Thus, BCl3 has a total of 3 + (3 × 7) = 24 valence electrons.

3. **Identify the Hybridization**:
- For BeCl2: The central atom Be has 2 bonding pairs and no lone pairs. The steric number is 2, leading to sp hybridization.
- For BCl3: The central atom B has 3 bonding pairs and no lone pairs. The steric number is 3, leading to sp² hybridization.

4. **Determine the Shape**:
- For BeCl2: With sp hybridization and a steric number of 2, the shape is linear.
- For BCl3: With sp² hybridization and a steric number of 3, the shape is trigonal planar.

5. **Evaluate the Options**:
- Option A: BeCl2, sp², and linear (Incorrect: BeCl2 is sp and linear)
- Option B: BeCl2, sp², and trigonal planar (Incorrect: BeCl2 is sp and linear)
- Option C: BCl3, sp², and trigonal planar (Correct: BCl3 is sp² and trigonal planar)
- Option D: Any other molecule (not provided in the question)

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