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
C. `BCl_3, sp^3`, triangular planar
D. `BCl_3, sp^3`, tetrahedral

Explanation

To solve the question regarding the correct set of molecule, hybridization, and shape for BeCl2 and BeCl3, we will analyze each molecule step by step.

### Step 1: Analyze BeCl2
1. **Determine the Valence Electrons**:
- Beryllium (Be) has 2 valence electrons (configuration: 1s² 2s²).
- Each chlorine (Cl) has 7 valence electrons (configuration: 1s² 2s² 2p⁶ 3s² 3p⁵).

2. **Lewis Structure**:
- Beryllium forms two covalent bonds with two chlorine atoms, using its 2 valence electrons.
- The Lewis structure shows Be in the center with two Cl atoms bonded to it.

3. **Octet Rule**:
- Each chlorine atom achieves an octet (8 electrons) by sharing one electron with beryllium.
- Beryllium, however, only has 4 electrons around it (2 from the bonds), making it electron deficient.

4. **Steric Number**:
- The steric number is calculated as the number of bonded atoms plus the number of lone pairs.
- For BeCl2: 2 bonded atoms + 0 lone pairs = 2.

5. **Hybridization**:
- The hybridization corresponding to a steric number of 2 is sp.

6. **Shape**:
- With no lone pairs, the shape of BeCl2 is linear.

### Step 2: Analyze BeCl3
1. **Determine the Valence Electrons**:
- Boron (B) has 3 valence electrons (configuration: 1s² 2s² 2p¹).
- Each chlorine (Cl) has 7 valence electrons.

2. **Lewis Structure**:
- Boron forms three covalent bonds with three chlorine atoms, using all 3 of its valence electrons.
- The Lewis structure shows B in the center with three Cl atoms bonded to it.

3. **Octet Rule**:
- Each chlorine achieves an octet by sharing one electron with boron.
- Boron, however, only has 6 electrons around it, making it another electron deficient compound.

4. **Steric Number**:
- For BeCl3: 3 bonded atoms + 0 lone pairs = 3.

5. **Hybridization**:
- The hybridization corresponding to a steric number of 3 is sp².

6. **Shape**:
- With no lone pairs, the shape of BeCl3 is trigonal planar.

### Conclusion
- For BeCl2:
- Hybridization: sp
- Shape: Linear
- For BeCl3:
- Hybridization: sp²
- Shape: Trigonal planar

### Final Answer
The correct sets are:
- BeCl2: sp hybridization and linear shape.
- BeCl3: sp² hybridization and trigonal planar shape.