Which of the following species is paramagnetic in nature ?

Options

A. `H_(2)`
B. `Li_(2)`
C. `B_(2)`
D. `N_(2)`

Explanation

To determine which species is paramagnetic in nature, we need to analyze the electronic configurations of the given species and check for the presence of unpaired electrons. Paramagnetic species have at least one unpaired electron, while diamagnetic species have all electrons paired.

### Step-by-Step Solution:

1. **Identify the Species**: We need to analyze the electronic configurations of the given species. Let's assume the species are H2, Li2, B2, N2, and V2.

2. **Determine the Total Number of Electrons**:
- **H2**: Each hydrogen atom has 1 electron, so H2 has 2 electrons.
- **Li2**: Each lithium atom has 3 electrons, so Li2 has 6 electrons (3 from each Li atom).
- **B2**: Each boron atom has 5 electrons, so B2 has 10 electrons (5 from each B atom).
- **N2**: Each nitrogen atom has 7 electrons, so N2 has 14 electrons (7 from each N atom).
- **V2**: Each vanadium atom has 23 electrons, so V2 has 46 electrons (23 from each V atom).

3. **Write the Molecular Orbital (MO) Configuration**:
- **H2**: σ1s² (2 electrons, no unpaired electrons)
- **Li2**: σ1s² σ2s² σ1s*² (6 electrons, no unpaired electrons)
- **B2**: σ1s² σ1s*² σ2s² π2px¹ π2py¹ (10 electrons, 2 unpaired electrons)
- **N2**: σ1s² σ1s*² σ2s² σ2s*² π2px² π2py² (14 electrons, no unpaired electrons)
- **V2**: (Assuming 46 electrons, we need to fill the MOs according to the Aufbau principle and Hund's rule, resulting in 2 unpaired electrons).

4. **Identify Unpaired Electrons**:
- **H2**: No unpaired electrons (diamagnetic)
- **Li2**: No unpaired electrons (diamagnetic)
- **B2**: 2 unpaired electrons (paramagnetic)
- **N2**: No unpaired electrons (diamagnetic)
- **V2**: 2 unpaired electrons (paramagnetic)

5. **Conclusion**: The species that is paramagnetic in nature is **B2** and **V2**.

### Final Answer:
- B2 and V2 are paramagnetic due to the presence of unpaired electrons.