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.

Explanation

To determine the correct set with respect to molecule, hybridization, and shape, let's analyze the given molecules one by one.

### Step 1: Analyze BeCl2
- **Molecule**: BeCl2
- **Valence Electrons**: Beryllium (Be) has 2 valence electrons and forms 2 bonds with chlorine (Cl).
- **Hybridization**: Since BeCl2 has 2 bonding pairs and no lone pairs, the hybridization is **sp**.
- **Shape**: The shape of BeCl2 is **linear** due to the arrangement of the two bonding pairs around the central atom.

**Conclusion**: For BeCl2, hybridization is sp and shape is linear.

### Step 2: Analyze BCl2
- **Molecule**: BCl2
- **Valence Electrons**: Boron (B) has 3 valence electrons and forms 2 bonds with chlorine (Cl).
- **Hybridization**: Boron in BCl2 has 2 bonding pairs and no lone pairs, so the hybridization is **sp**.
- **Shape**: The shape of BCl2 is also **linear**.

**Conclusion**: For BCl2, hybridization is sp and shape is linear.

### Step 3: Analyze BCl3
- **Molecule**: BCl3
- **Valence Electrons**: Boron (B) has 3 valence electrons and forms 3 bonds with chlorine (Cl).
- **Hybridization**: Boron in BCl3 has 3 bonding pairs and no lone pairs, so the hybridization is **sp²**.
- **Shape**: The shape of BCl3 is **trigonal planar** due to the arrangement of the three bonding pairs around the central atom.

**Conclusion**: For BCl3, hybridization is sp² and shape is trigonal planar.

### Final Evaluation
- **BeCl2**: Hybridization is sp (correct), Shape is linear (correct).
- **BCl2**: Hybridization is sp (correct), Shape is linear (correct).
- **BCl3**: Hybridization is sp² (correct), Shape is trigonal planar (correct).

### Correct Set
The correct set is for **BCl3**: Hybridization is sp² and shape is trigonal planar.