The number of molecules or ions having bond order 2.5 among `O_(2)^(+), CN, NO, N_(2)^(+), CO^(+), NO^(+), O_(2)^(-), CN^(-), N_(2),` is

Explanation

To determine the number of molecules or ions having a bond order of 2.5 among the given species \( O_2^+, CN, NO, N_2^+, CO^+, NO^+, O_2^-, CN^-, N_2 \), we will calculate the bond order for each species using the formula:

\[ \text{Bond Order} = \frac{1}{2} \left( N_b - N_a \right) \]

where \( N_b \) is the number of electrons in bonding molecular orbitals and \( N_a \) is the number of electrons in anti-bonding molecular orbitals.

### Step 1: Calculate Bond Order for Each Species

1. **For \( O_2^+ \)**:
- Total electrons = 15 (O2 has 16, remove 1 for \( O_2^+ \))
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \)
- \( N_b = 10 \), \( N_a = 5 \)
- Bond Order = \( \frac{1}{2} (10 - 5) = 2.5 \)

2. **For \( CN \)**:
- Total electrons = 13 (C has 6, N has 7)
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1 \)
- \( N_b = 9 \), \( N_a = 4 \)
- Bond Order = \( \frac{1}{2} (9 - 4) = 2.5 \)

3. **For \( NO \)**:
- Total electrons = 15 (N has 7, O has 8)
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- \( N_b = 10 \), \( N_a = 5 \)
- Bond Order = \( \frac{1}{2} (10 - 5) = 2.5 \)

4. **For \( N_2^+ \)**:
- Total electrons = 13 (N2 has 14, remove 1 for \( N_2^+ \))
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1 \)
- \( N_b = 8 \), \( N_a = 5 \)
- Bond Order = \( \frac{1}{2} (8 - 5) = 1.5 \)

5. **For \( CO^+ \)**:
- Total electrons = 13 (C has 6, O has 8, remove 1 for \( CO^+ \))
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- \( N_b = 10 \), \( N_a = 3 \)
- Bond Order = \( \frac{1}{2} (10 - 3) = 3.5 \)

6. **For \( NO^+ \)**:
- Total electrons = 14 (NO has 15, remove 1 for \( NO^+ \))
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- \( N_b = 9 \), \( N_a = 5 \)
- Bond Order = \( \frac{1}{2} (9 - 5) = 2 \)

7. **For \( O_2^- \)**:
- Total electrons = 17 (O2 has 16, add 1 for \( O_2^- \))
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \)
- \( N_b = 10 \), \( N_a = 6 \)
- Bond Order = \( \frac{1}{2} (10 - 6) = 2 \)

8. **For \( CN^- \)**:
- Total electrons = 14 (CN has 13, add 1 for \( CN^- \))
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \)
- \( N_b = 10 \), \( N_a = 4 \)
- Bond Order = \( \frac{1}{2} (10 - 4) = 3 \)

9. **For \( N_2 \)**:
- Total electrons = 14 (N2 has 14)
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- \( N_b = 10 \), \( N_a = 4 \)
- Bond Order = \( \frac{1}{2} (10 - 4) = 3 \)

### Step 2: Count the Species with Bond Order of 2.5
The species with a bond order of 2.5 are:
- \( O_2^+ \)
- \( CN \)
- \( NO \)
- \( N_2^+ \)

Thus, the total number of molecules or ions having a bond order of 2.5 is **4**.

### Final Answer
The number of molecules or ions having bond order 2.5 is **4**.