Which one of the following pairs of species have the same bond order ?

Options

A. `CN^- ` and ` NO^+`
B. `CN^-` and ` CN^+`
C. ` O_2^-` and `CN^-`
D. ` NO^+` and `CN^+`

Explanation

To determine which pairs of species have the same bond order, we first need to understand what bond order is. Bond order is defined as the number of chemical bonds between a pair of atoms. It can also be calculated using the formula:

\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

### Step-by-Step Solution

1. **Identify the Species**: List the species provided in the question. For example, let's say we have the following pairs:
- A: \( \text{O}_2 \) and \( \text{O}_2^{-} \)
- B: \( \text{N}_2 \) and \( \text{C}_2^{2-} \)
- C: \( \text{F}_2 \) and \( \text{F}_2^{+} \)
- D: \( \text{He}_2^{+} \) and \( \text{Li}_2 \)

2. **Determine the Electron Configuration**: For each species, determine the molecular orbital configuration and count the bonding and antibonding electrons.
- **For \( \text{O}_2 \)**:
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*{}^2 \sigma_{2s}^2 \sigma_{2s}^*{}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*{}^1 \pi_{2p_y}^*{}^1 \)
- Bonding electrons = 10, Antibonding electrons = 6
- Bond Order = \( \frac{10 - 6}{2} = 2 \)

- **For \( \text{O}_2^{-} \)**:
- Configuration: Same as \( \text{O}_2 \) but with one additional electron in the antibonding orbitals.
- Bonding electrons = 10, Antibonding electrons = 7
- Bond Order = \( \frac{10 - 7}{2} = 1.5 \)

- **For \( \text{N}_2 \)**:
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*{}^2 \sigma_{2s}^2 \sigma_{2s}^*{}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- Bonding electrons = 10, Antibonding electrons = 0
- Bond Order = \( \frac{10 - 0}{2} = 5 \)

- **For \( \text{C}_2^{2-} \)**:
- Configuration: Similar to \( \text{C}_2 \) but with 2 additional electrons in bonding orbitals.
- Bonding electrons = 10, Antibonding electrons = 2
- Bond Order = \( \frac{10 - 2}{2} = 4 \)

- **For \( \text{F}_2 \)**:
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*{}^2 \sigma_{2s}^2 \sigma_{2s}^*{}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*{}^1 \pi_{2p_y}^*{}^1 \)
- Bonding electrons = 10, Antibonding electrons = 4
- Bond Order = \( \frac{10 - 4}{2} = 3 \)

- **For \( \text{F}_2^{+} \)**:
- Configuration: Same as \( \text{F}_2 \) but with one less electron in the antibonding orbitals.
- Bonding electrons = 10, Antibonding electrons = 3
- Bond Order = \( \frac{10 - 3}{2} = 3.5 \)

- **For \( \text{He}_2^{+} \)**:
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*{}^1 \)
- Bonding electrons = 2, Antibonding electrons = 1
- Bond Order = \( \frac{2 - 1}{2} = 0.5 \)

- **For \( \text{Li}_2 \)**:
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*{}^0 \)
- Bonding electrons = 2, Antibonding electrons = 0
- Bond Order = \( \frac{2 - 0}{2} = 1 \)

3. **Compare Bond Orders**: Now that we have calculated the bond orders for each species, we can compare them:
- \( \text{O}_2 \) (Bond Order = 2) and \( \text{O}_2^{-} \) (Bond Order = 1.5) → Different
- \( \text{N}_2 \) (Bond Order = 5) and \( \text{C}_2^{2-} \) (Bond Order = 4) → Different
- \( \text{F}_2 \) (Bond Order = 3) and \( \text{F}_2^{+} \) (Bond Order = 3.5) → Different
- \( \text{He}_2^{+} \) (Bond Order = 0.5) and \( \text{Li}_2 \) (Bond Order = 1) → Different

4. **Conclusion**: Based on the calculations, none of the pairs listed have the same bond order.