Find the pair that has the same bond order with diamagnetic and paramagnetic properties respectively.

Options

A. `F_(2) and O_(2)`
B. `N_(2) and O_(2)^(2-)`
C. `Li_(2) and B_(2)`
D. `B_(2) and O_(2)`

Explanation

To solve the problem of finding a pair of molecules that have the same bond order with diamagnetic and paramagnetic properties respectively, we can follow these steps:

### Step 1: Understand the Definitions
- **Bond Order**: It is calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \]
- **Diamagnetic**: A species is diamagnetic if all its electrons are paired.
- **Paramagnetic**: A species is paramagnetic if it has unpaired electrons.

### Step 2: Analyze the Given Options
We will analyze the molecular configurations and properties of the given pairs:

1. **O2 and F2**
- **O2**: Has 16 electrons.
- Molecular orbital configuration: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^1 \pi_{2p_y}^1\)
- Bond order: \(\frac{1}{2}(10 - 6) = 2\) (Paramagnetic due to 2 unpaired electrons)
- **F2**: Has 18 electrons.
- Molecular orbital configuration: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2\)
- Bond order: \(\frac{1}{2}(10 - 8) = 1\) (Diamagnetic due to all paired electrons)

**Conclusion**: O2 and F2 do not have the same bond order.

2. **N2 and Li2**
- **N2**: Has 14 electrons.
- Molecular orbital configuration: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2\)
- Bond order: \(\frac{1}{2}(10 - 4) = 3\) (Diamagnetic due to all paired electrons)
- **Li2**: Has 6 electrons.
- Molecular orbital configuration: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2\)
- Bond order: \(\frac{1}{2}(2 - 0) = 1\) (Diamagnetic due to all paired electrons)

**Conclusion**: N2 and Li2 do not have the same bond order.

3. **B2 and F2**
- **B2**: Has 10 electrons.
- Molecular orbital configuration: \(\sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^1 \pi_{2p_y}^1\)
- Bond order: \(\frac{1}{2}(6 - 4) = 1\) (Paramagnetic due to 2 unpaired electrons)
- **F2**: As calculated before, has a bond order of 1 and is diamagnetic.

**Conclusion**: B2 and F2 do not have the same bond order.

4. **Li2 and B2**
- **Li2**: As calculated before, bond order = 1 (Diamagnetic).
- **B2**: As calculated before, bond order = 1 (Paramagnetic).

**Conclusion**: Li2 and B2 have the same bond order of 1, but different magnetic properties (Li2 is diamagnetic and B2 is paramagnetic).

### Final Answer
The pair that has the same bond order with diamagnetic and paramagnetic properties respectively is **Li2 (Diamagnetic) and B2 (Paramagnetic)**.