Which species does not exhibit paramagnetism ?

Options

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

Answer

To determine which species does not exhibit paramagnetism, we need to analyze the electronic configurations of the given species and check for the presence of unpaired electrons. A species is considered paramagnetic if it has unpaired electrons; if all electrons are paired, it is diamagnetic.

### Step-by-Step Solution:

1. **Identify the Species**: We need to analyze the following species: N₂, N₂⁺, O₂, O₂⁻, CO, and NO.

2. **Write the Electronic Configuration for Each Species**:
- **N₂**:
- Total electrons = 14
- Configuration: σ1s² σ*1s² σ2s² σ*2s² (2 electrons in bonding and antibonding 2s)
- Then: π2px² = π2py² (4 electrons in bonding 2p) and σ2pz² (2 electrons in bonding 2p).
- Result: All electrons are paired. (Diamagnetic)

- **N₂⁺**:
- Total electrons = 13 (1 electron removed from N₂)
- Configuration: σ1s² σ*1s² σ2s² σ*2s² π2px² = π2py² σ2pz¹
- Result: One unpaired electron in σ2pz. (Paramagnetic)

- **O₂**:
- Total electrons = 16
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² = π2py² π*2px¹ = π*2py¹
- Result: Two unpaired electrons in π*2px and π*2py. (Paramagnetic)

- **O₂⁻**:
- Total electrons = 17 (1 extra electron added to O₂)
- Configuration: Same as O₂ but with one more electron in π*2px.
- Result: One unpaired electron in π*2px. (Paramagnetic)

- **CO**:
- Total electrons = 18
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² = π2py²
- Result: All electrons are paired. (Diamagnetic)

- **NO**:
- Total electrons = 15
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² = π2py² σ2pz¹
- Result: One unpaired electron in σ2pz. (Paramagnetic)

3. **Conclusion**:
- The species that does not exhibit paramagnetism is **CO** because it has all paired electrons, making it diamagnetic.

### Final Answer:
**CO does not exhibit paramagnetism.**