How many of the following species have bond order of `2.5` ?
`N_(2)^(o+)` (ii) `N_(2)^(Theta) ` (iii) `O_(2)^(o+)` (iv) `O_(2)^(Theta)` (v) `NO` (vi) `CN` .

Explanation

To determine how many of the given species have a bond order of 2.5, we will analyze each species one by one using the molecular orbital theory (MOT). The bond order can be calculated using the formula:

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

Let's analyze each species:

1. **N₂⁺ (Nitrogen cation)**:
- The electron configuration of N₂ is: \( \sigma_{2s}^2 \sigma_{2s}^*^0 \sigma_{2p_z}^2 \pi_{2p_x}^1 \pi_{2p_y}^1 \)
- For N₂⁺, we remove one electron from the highest energy level, which is one of the π orbitals:
- Configuration: \( \sigma_{2s}^2 \sigma_{2s}^*^0 \sigma_{2p_z}^2 \pi_{2p_x}^1 \pi_{2p_y}^0 \)
- Bonding electrons = 5 (2 from σ2s, 2 from σ2p_z, 1 from π2p_x)
- Antibonding electrons = 0
- Bond Order = \( \frac{5 - 0}{2} = 2.5 \)

2. **N₂⁻ (Nitrogen anion)**:
- For N₂⁻, we add one electron to the highest energy level:
- Configuration: \( \sigma_{2s}^2 \sigma_{2s}^*^0 \sigma_{2p_z}^2 \pi_{2p_x}^1 \pi_{2p_y}^1 \)
- Bonding electrons = 6 (2 from σ2s, 2 from σ2p_z, 2 from π2p_x and π2p_y)
- Antibonding electrons = 1 (from π2p_x or π2p_y)
- Bond Order = \( \frac{6 - 1}{2} = 2.5 \)

3. **O₂⁺ (Oxygen cation)**:
- The electron configuration of O₂ is: \( \sigma_{2s}^2 \sigma_{2s}^*^0 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- For O₂⁺, we remove one electron from the highest energy level:
- Configuration: \( \sigma_{2s}^2 \sigma_{2s}^*^0 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- Bonding electrons = 6 (2 from σ2s, 2 from σ2p_z, 2 from π2p_x)
- Antibonding electrons = 1 (from π2p_y)
- Bond Order = \( \frac{6 - 1}{2} = 2.5 \)

4. **O₂⁻ (Oxygen anion)**:
- For O₂⁻, we add one electron to the highest energy level:
- Configuration: \( \sigma_{2s}^2 \sigma_{2s}^*^0 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- Bonding electrons = 6
- Antibonding electrons = 2 (1 from π2p_y and 1 from π2p_x)
- Bond Order = \( \frac{6 - 2}{2} = 2 \)

5. **NO (Nitric oxide)**:
- The electron configuration of NO is similar to O₂:
- Configuration: \( \sigma_{2s}^2 \sigma_{2s}^*^0 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- Bonding electrons = 6
- Antibonding electrons = 1 (from π2p_y)
- Bond Order = \( \frac{6 - 1}{2} = 2.5 \)

6. **CN (Cyanide ion)**:
- The electron configuration of CN is similar to N₂⁺:
- Configuration: \( \sigma_{2s}^2 \sigma_{2s}^*^0 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- Bonding electrons = 6
- Antibonding electrons = 1 (from π2p_y)
- Bond Order = \( \frac{6 - 1}{2} = 2.5 \)

Now, let's summarize the bond orders of each species:
- N₂⁺: Bond Order = 2.5
- N₂⁻: Bond Order = 2.5
- O₂⁺: Bond Order = 2.5
- O₂⁻: Bond Order = 2
- NO: Bond Order = 2.5
- CN: Bond Order = 2.5

### Conclusion:
The species with a bond order of 2.5 are: N₂⁺, N₂⁻, O₂⁺, NO, and CN. Therefore, there are **5 species** with a bond order of 2.5.