Paramagnetism is exhibited by :

Options

A. `N_(2)`
B. `O_(2)`
C. He
D. `O_(2)^(2-)`

Explanation

To determine which species exhibits paramagnetism, we need to analyze the electronic configurations of the given options and check for the presence of unpaired electrons. Here's the step-by-step solution:

### Step 1: Understand Paramagnetism
Paramagnetism occurs in species that have unpaired electrons. If all electrons are paired, the species is diamagnetic.

### Step 2: Analyze Each Option

#### Option 1: N2 (Nitrogen molecule)
1. **Electronic Configuration**: The electronic configuration for N2 is:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_x² π2p_y²
- Total electrons = 14
2. **Check for Unpaired Electrons**: All electrons are paired.
3. **Conclusion**: N2 is diamagnetic.

#### Option 2: O2 (Oxygen molecule)
1. **Electronic Configuration**: The electronic configuration for O2 is:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_x² π2p_y² π*2p_y¹ π*2p_z¹
- Total electrons = 16
2. **Check for Unpaired Electrons**: There are 2 unpaired electrons in the π*2p_y and π*2p_z orbitals.
3. **Conclusion**: O2 is paramagnetic.

#### Option 3: He (Helium)
1. **Electronic Configuration**: The electronic configuration for He is:
- σ1s²
- Total electrons = 2
2. **Check for Unpaired Electrons**: All electrons are paired.
3. **Conclusion**: He is diamagnetic.

#### Option 4: O2²⁻ (Superoxide ion)
1. **Electronic Configuration**: The electronic configuration for O2²⁻ is:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_x² π2p_y² π*2p_y² π*2p_z²
- Total electrons = 18
2. **Check for Unpaired Electrons**: All electrons are paired.
3. **Conclusion**: O2²⁻ is diamagnetic.

### Final Conclusion
Based on the analysis:
- **Paramagnetism is exhibited by O2 (Oxygen molecule)**, as it has unpaired electrons.

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