Among `H_(2), He_(2)^(+), Li_(2), Be_(2), B_(2), C_(2), N_(2), O_(2)^(-)` and `F_(2)`, the number of diamagnetic species is
(Atomic numbers: `H=1, He=2, Li=3, Be=4, B=5, C=6, N=7, O=8, F=9`)

Explanation

To determine the number of diamagnetic species among the given diatomic molecules, we need to analyze the electronic configurations of each molecule and check for unpaired electrons. A diamagnetic species has all its electrons paired.

### Step-by-Step Solution

1. **Identify the Diatomic Molecules**: The given molecules are \( H_2, He_2^+, Li_2, Be_2, B_2, C_2, N_2, O_2^-, \) and \( F_2 \).

2. **Determine the Total Electrons for Each Molecule**:
- \( H_2 \): 2 electrons
- \( He_2^+ \): 3 electrons (4 total from 2 He atoms minus 1 for the positive charge)
- \( Li_2 \): 6 electrons (3 from each Li atom)
- \( Be_2 \): 8 electrons (4 from each Be atom)
- \( B_2 \): 10 electrons (5 from each B atom)
- \( C_2 \): 12 electrons (6 from each C atom)
- \( N_2 \): 14 electrons (7 from each N atom)
- \( O_2^- \): 17 electrons (8 from each O atom plus 1 for the negative charge)
- \( F_2 \): 18 electrons (9 from each F atom)

3. **Write the Molecular Orbital Configurations**:
- **\( H_2 \)**: \( \sigma_{1s}^2 \) → Diamagnetic (paired)
- **\( He_2^+ \)**: \( \sigma_{1s}^2 \sigma_{1s}^*^1 \) → Paramagnetic (1 unpaired)
- **\( Li_2 \)**: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \) → Diamagnetic (paired)
- **\( Be_2 \)**: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \) → Diamagnetic (paired)
- **\( B_2 \)**: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p}^1 \pi_{2p}^1 \) → Paramagnetic (2 unpaired)
- **\( C_2 \)**: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p}^2 \) → Diamagnetic (paired)
- **\( N_2 \)**: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p}^2 \pi_{2p}^2 \) → Diamagnetic (paired)
- **\( O_2^- \)**: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p}^2 \pi_{2p}^2 \pi_{2p}^1 \) → Paramagnetic (1 unpaired)
- **\( F_2 \)**: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p}^2 \pi_{2p}^2 \) → Diamagnetic (paired)

4. **Count the Diamagnetic Species**:
- Diamagnetic: \( H_2, Li_2, Be_2, C_2, N_2, F_2 \) → Total = 6
- Paramagnetic: \( He_2^+, B_2, O_2^- \) → Total = 3

### Final Answer
The number of diamagnetic species among the given diatomic molecules is **6**.