Which of the following species exhibits the diamagnetic behaviour ?

Options

A. NO
B. `O_(2)^(2-)`
C. `O_(2)^(+)`
D. `O_(2)`

Explanation

To determine which of the given species exhibits diamagnetic behavior, we need to analyze each option based on the presence of unpaired electrons. Diamagnetic substances have all their electrons paired, resulting in a net magnetic moment of zero.

### Step-by-Step Solution:

1. **Understand Diamagnetic Behavior**:
- A species is diamagnetic if it has no unpaired electrons, meaning all electrons are paired in their respective orbitals.

2. **Analyze Each Option**:
- We will analyze the molecular orbital configurations for each species provided in the options.

3. **Option A: NO (Nitric Oxide)**:
- Total electrons = 7 (N) + 8 (O) = 15 electrons.
- Filling the molecular orbitals according to the molecular orbital theory, we find that there is 1 unpaired electron in NO.
- Conclusion: NO is **not** diamagnetic.

4. **Option B: O2²⁻ (Oxide Ion)**:
- Total electrons = 16 (O) + 2 (2 extra electrons) = 18 electrons.
- Filling the molecular orbitals, we find that all electrons are paired.
- Conclusion: O2²⁻ **is** diamagnetic.

5. **Option C: O2⁺ (Oxygen Cation)**:
- Total electrons = 16 (O) - 1 (1 electron removed) = 15 electrons.
- Filling the molecular orbitals, we find that there is 1 unpaired electron in O2⁺.
- Conclusion: O2⁺ is **not** diamagnetic.

6. **Option D: O2 (Oxygen)**:
- Total electrons = 16 electrons.
- Filling the molecular orbitals, we find that there are 2 unpaired electrons in O2.
- Conclusion: O2 is **not** diamagnetic.

7. **Final Conclusion**:
- The only species that exhibits diamagnetic behavior is **Option B: O2²⁻**.

### Summary:
- **Diamagnetic Species**: O2²⁻ (Option B)
- **Not Diamagnetic**: NO (Option A), O2⁺ (Option C), O2 (Option D)