Paramagnetism is not shown by

Options

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

Answer

To determine which of the given options does not exhibit paramagnetism, we need to analyze the electronic configurations of each species and identify the presence of unpaired electrons. Paramagnetism occurs in species that have unpaired electrons in their molecular orbitals.

### Step-by-Step Solution:

1. **Understanding Paramagnetism**:
- A species is paramagnetic if it has unpaired electrons in its molecular orbitals. Conversely, if all electrons are paired, the species is diamagnetic.

2. **Analyzing O2 (Oxygen molecule)**:
- The electronic configuration of O2 is:
- Total electrons = 16 (8 from each oxygen atom).
- Filling the molecular orbitals:
- σ2s², σ*2s², σ2pz², π2py², π2px², π*2py¹, π*2px¹.
- In O2, there are 2 unpaired electrons in the π* orbitals.
- Conclusion: O2 is **paramagnetic**.

3. **Analyzing O2- (Oxygen anion)**:
- For O2-, we add one extra electron to O2:
- Total electrons = 17.
- Filling the molecular orbitals:
- σ2s², σ*2s², σ2pz², π2py², π2px², π*2py² (extra electron fills one of the π* orbitals).
- In O2-, there is 1 unpaired electron in the π* orbital.
- Conclusion: O2- is **paramagnetic**.

4. **Analyzing H2+ (Hydrogen cation)**:
- H2+ has 1 electron:
- Total electrons = 1.
- Filling the molecular orbitals:
- σ1s¹.
- There is 1 unpaired electron in the σ1s orbital.
- Conclusion: H2+ is **paramagnetic**.

5. **Analyzing O2-- (Dioxygen dianion)**:
- For O2--, we add two extra electrons to O2:
- Total electrons = 18.
- Filling the molecular orbitals:
- σ2s², σ*2s², σ2pz², π2py², π2px², π*2py², π*2px² (both π* orbitals are filled).
- All electrons are paired in O2--.
- Conclusion: O2-- is **diamagnetic**.

### Final Conclusion:
- The species that does not exhibit paramagnetism is **O2--** (dioxygen dianion), as it has no unpaired electrons.