The species which does not show paramagnetism is .

Options

A. `O_(2)`
B. `O_(2)^(o+)`
C. `O_(2)^(2-)`
D. `H_(2)^(o+)`

Answer

To determine which species does not show paramagnetism, we need to identify the species that is diamagnetic, meaning it has no unpaired electrons. Let's analyze each option step by step.

### Step 1: Analyze O2
- **Molecular Electronic Configuration of O2**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z², π*2p_y¹, π*2p_z¹
- **Unpaired Electrons**:
- There are 2 unpaired electrons in the π* orbitals.
- **Conclusion**: O2 is paramagnetic.

### Step 2: Analyze O2+
- **Molecular Electronic Configuration of O2+**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z², π*2p_y¹, π*2p_z⁰
- **Unpaired Electrons**:
- There is 1 unpaired electron in the π*2p_y orbital.
- **Conclusion**: O2+ is paramagnetic.

### Step 3: Analyze O2−
- **Molecular Electronic Configuration of O2−**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z², π*2p_y², π*2p_z²
- **Unpaired Electrons**:
- All electrons are paired.
- **Conclusion**: O2− is diamagnetic.

### Step 4: Analyze H2+
- **Molecular Electronic Configuration of H2+**:
- The configuration is:
- σ1s¹
- **Unpaired Electrons**:
- There is 1 unpaired electron.
- **Conclusion**: H2+ is paramagnetic.

### Final Conclusion
- The species that does not show paramagnetism (is diamagnetic) is **O2− (option C)**.