According to MOT, two atomic orbitals overlap resulting in the formation of molecular orbital formed. Number of atomic orbitals overlapping together is equal to the molecule orbital formed. The two atomic orbital thus formed by LCAO (linear combination of atomic orbital) in the phase or in the different phase are known as bonding and antibonding molecular orbitals respectively. The energy of bonding molecular orbital is lower than that of the pure atomic orbitals by an amount `Delta`. This known as the stabilization energy. The enerby of antibonding molecular orbital in increased by `Delta'` (destabilisation energy).
Q. Which among the following pairs contains both paramagnetic species.

Options

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

Answer

To determine which among the given pairs contains both paramagnetic species, we need to analyze the electronic configurations of the species in question. Paramagnetic species have unpaired electrons, while diamagnetic species have all paired electrons.

### Step-by-Step Solution:

1. **Identify the Species**: We need to evaluate the electronic configurations of the species mentioned in the question. The species we will analyze are O2, O2^-, and N2^-.

2. **Determine the Total Number of Electrons**:
- **O2**: Oxygen has an atomic number of 8. Therefore, O2 has 16 electrons (8 from each oxygen atom).
- **O2^-**: This species has one additional electron, giving it a total of 17 electrons.
- **N2^-**: Nitrogen has an atomic number of 7. Therefore, N2 has 14 electrons (7 from each nitrogen atom), and N2^- has 15 electrons.

3. **Construct the Molecular Orbital (MO) Diagram**:
- For O2 (16 electrons):
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² (bonding) π2p_x² = π2p_y² (bonding) π*2p_x² = π*2p_y² (antibonding)
- All electrons are paired in O2, hence it is **diamagnetic**.

- For O2^- (17 electrons):
- The additional electron goes into one of the antibonding orbitals (π*2p_x or π*2p_y).
- This results in one unpaired electron, making O2^- **paramagnetic**.

- For N2^- (15 electrons):
- The electronic configuration is σ1s² σ*1s² σ2s² σ*2s² σ2p_z² (bonding) π2p_x² = π2p_y² (bonding) π*2p_x¹ (one unpaired electron).
- N2^- has one unpaired electron, making it **paramagnetic**.

4. **Conclusion**:
- O2 is diamagnetic (all paired electrons).
- O2^- is paramagnetic (one unpaired electron).
- N2^- is paramagnetic (one unpaired electron).
- Therefore, the pair that contains both paramagnetic species is **O2^- and N2^-**.

### Final Answer:
The pair that contains both paramagnetic species is **O2^- and N2^-**.