Assuming 2s - 2p mixing is NOT operative , the paramagnetic among the following is

Options

A. `Be_2`
B. `B_2`
C. `C_2`
D. `N_2`

Answer

To determine which of the given compounds is paramagnetic when assuming that 2s - 2p mixing is not operative, we will analyze the molecular orbital (MO) configurations of the compounds mentioned: Be2, B2, C2, and N2.

### Step-by-step Solution:

1. **Understanding Molecular Orbital Theory (MOT)**:
- In MOT, atomic orbitals combine to form molecular orbitals. The energy levels of these molecular orbitals depend on the atomic orbitals involved and their energies.
- For the second period elements, the order of molecular orbitals (without 2s - 2p mixing) is:
- σ(1s) < σ*(1s) < σ(2s) < σ*(2s) < σ(2p_z) < π(2p_x) = π(2p_y) < π*(2p_x) = π*(2p_y) < σ*(2p_z)

2. **Counting Electrons in Each Compound**:
- **Be2**: Beryllium has 4 electrons (2 from each Be atom).
- Configuration: σ(1s)² σ*(1s)² σ(2s)² σ*(2s)²
- Total electrons = 8 (All paired, hence no unpaired electrons).
- **B2**: Boron has 5 electrons (2 from each B atom).
- Configuration: σ(1s)² σ*(1s)² σ(2s)² σ*(2s)² σ(2p_z)²
- Total electrons = 10 (All paired, hence no unpaired electrons).
- **C2**: Carbon has 6 electrons (2 from each C atom).
- Configuration: σ(1s)² σ*(1s)² σ(2s)² σ*(2s)² σ(2p_z)² π(2p_x)¹ π(2p_y)¹
- Total electrons = 12 (2 unpaired electrons in π orbitals, hence it is paramagnetic).
- **N2**: Nitrogen has 7 electrons (2 from each N atom).
- Configuration: σ(1s)² σ*(1s)² σ(2s)² σ*(2s)² σ(2p_z)² π(2p_x)² π(2p_y)²
- Total electrons = 14 (All paired, hence no unpaired electrons).

3. **Identifying Paramagnetic Species**:
- A molecule is paramagnetic if it has unpaired electrons. From our analysis:
- Be2: 0 unpaired electrons (diamagnetic)
- B2: 0 unpaired electrons (diamagnetic)
- C2: 2 unpaired electrons (paramagnetic)
- N2: 0 unpaired electrons (diamagnetic)

### Conclusion:
The only compound that is paramagnetic among the given options is **C2**.