Which of the following molecule has a linear shape?

Options

A. `BeCl_(2)`
B. `BF_(3)`
C. `CH_(4)`
D. `C_(2)H_(4)`

Answer

To determine which molecule has a linear shape among the given options, we will analyze each molecule step by step, focusing on their structures, steric numbers, hybridization, and resulting geometries.

### Step 1: Analyze BECl2
1. **Structure**: Beryllium (Be) is the central atom and forms two bonds with chlorine (Cl).
2. **Valence Electrons**: Beryllium has 2 valence electrons, which allows it to form 2 sigma bonds.
3. **Steric Number Calculation**:
- Number of sigma bonds: 2 (Be-Cl bonds)
- Lone pairs: 0
- Steric number = 2 (2 sigma bonds + 0 lone pairs)
4. **Hybridization**: With a steric number of 2, the hybridization is sp.
5. **Geometry**: The geometry for sp hybridization is linear.

### Step 2: Analyze BF3
1. **Structure**: Boron (B) is the central atom and forms three bonds with fluorine (F).
2. **Valence Electrons**: Boron has 3 valence electrons, allowing it to form 3 sigma bonds.
3. **Steric Number Calculation**:
- Number of sigma bonds: 3 (B-F bonds)
- Lone pairs: 0
- Steric number = 3 (3 sigma bonds + 0 lone pairs)
4. **Hybridization**: With a steric number of 3, the hybridization is sp².
5. **Geometry**: The geometry for sp² hybridization is trigonal planar.

### Step 3: Analyze CH4
1. **Structure**: Carbon (C) is the central atom and forms four bonds with hydrogen (H).
2. **Valence Electrons**: Carbon has 4 valence electrons, allowing it to form 4 sigma bonds.
3. **Steric Number Calculation**:
- Number of sigma bonds: 4 (C-H bonds)
- Lone pairs: 0
- Steric number = 4 (4 sigma bonds + 0 lone pairs)
4. **Hybridization**: With a steric number of 4, the hybridization is sp³.
5. **Geometry**: The geometry for sp³ hybridization is tetrahedral.

### Step 4: Analyze C2H4
1. **Structure**: C2H4 has two carbon atoms, each bonded to two hydrogen atoms and to each other via a double bond.
2. **Valence Electrons**: Each carbon has 4 valence electrons.
3. **Steric Number Calculation**:
- For each carbon atom:
- Number of sigma bonds: 3 (2 C-H bonds + 1 C=C bond)
- Lone pairs: 0
- Steric number = 3 (3 sigma bonds + 0 lone pairs)
4. **Hybridization**: With a steric number of 3, the hybridization is sp² for both carbon atoms.
5. **Geometry**: The geometry for sp² hybridization is trigonal planar.

### Conclusion
After analyzing all the options:
- **BECl2** has a linear shape.
- **BF3** has a trigonal planar shape.
- **CH4** has a tetrahedral shape.
- **C2H4** has a trigonal planar shape.

Thus, the molecule with a linear shape is **BECl2**.