In which of the following species, bonding is taking place in excited state ?

Options

A. a.`BeCl_(2)(g)`
B. b. `C Cl_(4)`
C. c. `NF_(3)`
D. d. `PCl_(5)`

Explanation

To determine in which of the given species bonding occurs in an excited state, we need to analyze the electronic configurations of each species and identify whether any excitation of electrons is required for bonding.

### Step-by-Step Solution:

1. **Identify the Species and Their Ground State Configurations**:
- **Option A: BeCl2**
- Beryllium (Be) has the electronic configuration: 1s² 2s².
- In the ground state, Be has 2 electrons in the 2s orbital.
- **Option B: CCl4**
- Carbon (C) has the electronic configuration: 1s² 2s² 2p².
- In the ground state, C has 2 electrons in the 2s orbital and 2 in the 2p orbital.
- **Option C: NF3**
- Nitrogen (N) has the electronic configuration: 1s² 2s² 2p³.
- In the ground state, N has 2 electrons in the 2s orbital and 3 in the 2p orbital.
- **Option D: PCl5**
- Phosphorus (P) has the electronic configuration: 1s² 2s² 2p⁶ 3s² 3p³.
- In the ground state, P has 2 electrons in the 3s orbital and 3 in the 3p orbital.

2. **Analyze Excited States for Bond Formation**:
- **BeCl2**:
- In the excited state, one electron from the 2s orbital can be promoted to the 2p orbital, allowing for sp hybridization. This enables bonding with chlorine atoms.
- **CCl4**:
- In the excited state, one electron from the 2s orbital can be promoted to the 2p orbital, allowing for sp³ hybridization. This enables bonding with chlorine atoms.
- **NF3**:
- No excitation is needed as nitrogen can use its 3 unpaired electrons in the 2p orbital to bond with fluorine atoms.
- **PCl5**:
- In the excited state, one of the 3s electrons can be promoted to an empty 3d orbital, allowing for bonding with five chlorine atoms.

3. **Conclusion**:
- The species that requires excitation for bonding are BeCl2, CCl4, and PCl5. However, NF3 does not require excitation for bonding.

### Final Answer:
The correct option is **BeCl2, CCl4, and PCl5**.