In which of the following diatomjic molecules/ions is the bond order of each molecule/ion=2.5?

Options

A. `N_(2) ^(+), NO, O _(2) ^(+)`
B. `O_(2) ^(+), NO,CN^(-)`
C. `N_(2) ^(+), NO, CN^(-)`
D. `cn^(-), n_(2) ^(+), n_(2)`

Explanation

To determine which diatomic molecules or ions have a bond order of 2.5, we will use Molecular Orbital Theory (MOT) to calculate the bond order for each given species. The bond order can be calculated using the formula:

\[ \text{Bond Order} = \frac{1}{2} (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \]

### Step-by-Step Solution:

1. **Identify the species and their total number of electrons:**
- **N₂⁺:** Each nitrogen atom has 7 electrons. For N₂, there are 14 electrons. Since it is N₂⁺, we lose 1 electron, resulting in 13 electrons.
- **NO:** Nitrogen has 7 electrons and oxygen has 8 electrons, resulting in 15 electrons.
- **O₂⁺:** Each oxygen atom has 8 electrons. For O₂, there are 16 electrons. Since it is O₂⁺, we lose 1 electron, resulting in 15 electrons.
- **CN⁻:** Carbon has 6 electrons and nitrogen has 7 electrons, resulting in 13 electrons. Since it is CN⁻, we gain 1 electron, resulting in 14 electrons.

2. **Construct the Molecular Orbital (MO) diagrams for each species:**
- **N₂⁺ (13 electrons):**
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)
- \(\sigma_{2p_z}^1\)
- Total bonding electrons = 9, antibonding electrons = 2.
- Bond Order = \(\frac{1}{2}(9 - 2) = \frac{7}{2} = 2.5\).

- **NO (15 electrons):**
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)
- \(\pi^*_{2p_x}^1\)
- Total bonding electrons = 10, antibonding electrons = 2.
- Bond Order = \(\frac{1}{2}(10 - 2) = \frac{8}{2} = 2.5\).

- **O₂⁺ (15 electrons):**
- Similar to NO, we have 15 electrons.
- Total bonding electrons = 10, antibonding electrons = 2.
- Bond Order = \(\frac{1}{2}(10 - 2) = \frac{8}{2} = 2.5\).

- **CN⁻ (14 electrons):**
- \(\sigma_{1s}^2\)
- \(\sigma^*_{1s}^2\)
- \(\sigma_{2s}^2\)
- \(\sigma^*_{2s}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)
- \(\sigma_{2p_z}^2\)
- Total bonding electrons = 10, antibonding electrons = 2.
- Bond Order = \(\frac{1}{2}(10 - 2) = \frac{8}{2} = 3\).

3. **Conclusion:**
The species with a bond order of 2.5 are:
- N₂⁺
- NO
- O₂⁺