If Hund rule violate, then find the total number of species among following which whill be dimagnetic:
`B_(2), O_(2), N_(2)^(-), C_(2), NO, OF, N_(2)^(2-), BN`

Answer

To determine the total number of species that will be diamagnetic among the given molecules if Hund's rule is violated, we will analyze each species one by one.

### Step-by-Step Solution:

1. **B₂ (Boron Dimer)**:
- Total electrons = 10
- Molecular orbital filling:
- σ1s (2), σ*1s (2), σ2s (2), σ*2s (2), π2px (2), π2py (2)
- All electrons are paired.
- **Conclusion**: B₂ is **diamagnetic**.

2. **O₂ (Oxygen Dimer)**:
- Total electrons = 16
- Molecular orbital filling:
- σ1s (2), σ*1s (2), σ2s (2), σ*2s (2), σ2pz (2), π2px (2), π2py (2), π*2px (1), π*2py (1)
- If Hund's rule is violated, we would pair the electrons in the π* orbitals.
- All electrons are paired.
- **Conclusion**: O₂ is **diamagnetic**.

3. **N₂⁻ (Nitrogen Dimer with a negative charge)**:
- Total electrons = 15
- Molecular orbital filling:
- σ1s (2), σ*1s (2), σ2s (2), σ*2s (2), π2px (2), π2py (2), σ2pz (2), π*2px (1)
- If Hund's rule is violated, we would pair the electron in the π* orbital.
- All electrons are paired.
- **Conclusion**: N₂⁻ is **diamagnetic**.

4. **C₂ (Carbon Dimer)**:
- Total electrons = 12
- Molecular orbital filling:
- σ1s (2), σ*1s (2), σ2s (2), σ*2s (2), π2px (2), π2py (2)
- All electrons are paired.
- **Conclusion**: C₂ is **diamagnetic**.

5. **NO (Nitric Oxide)**:
- Total electrons = 15
- Molecular orbital filling:
- σ1s (2), σ*1s (2), σ2s (2), σ*2s (2), σ2pz (2), π2px (2), π2py (2), π*2px (1)
- If Hund's rule is violated, we would pair the unpaired electron.
- This would lead to one unpaired electron.
- **Conclusion**: NO is **paramagnetic**.

6. **OF (Oxygen Fluoride)**:
- Total electrons = 17
- Molecular orbital filling:
- σ1s (2), σ*1s (2), σ2s (2), σ*2s (2), σ2pz (2), π2px (2), π2py (2), π*2px (1), π*2py (1)
- If Hund's rule is violated, we would pair the unpaired electrons.
- This would lead to two unpaired electrons.
- **Conclusion**: OF is **paramagnetic**.

7. **N₂²⁻ (Nitrogen Dimer with a double negative charge)**:
- Total electrons = 16
- Molecular orbital filling:
- σ1s (2), σ*1s (2), σ2s (2), σ*2s (2), π2px (2), π2py (2), σ2pz (2), π*2px (2), π*2py (2)
- All electrons are paired.
- **Conclusion**: N₂²⁻ is **diamagnetic**.

8. **BN (Boron Nitride)**:
- Total electrons = 10
- Molecular orbital filling:
- σ1s (2), σ*1s (2), σ2s (2), σ*2s (2), σ2pz (2)
- All electrons are paired.
- **Conclusion**: BN is **diamagnetic**.

### Summary of Results:
- Diamagnetic species: B₂, O₂, N₂⁻, C₂, N₂²⁻, BN
- Paramagnetic species: NO, OF

### Total Count of Diamagnetic Species:
There are **6 diamagnetic species** among the given options.