`N_(2) and O_(2)` are converted into mono anions, `N_(2)^(-)` and `O_(2)^(-)` respectively. Which of the following is wrong?

Options

A. In `N_(2)^(-)`, the `N-N` bond weakens
B. In `O_(2)^(-)`, the `O-O` bond length increases
C. In `O_(2)^(-)`, the bond order decreases
D. `N_(2)^(-)` becomes diamagnetic.

Explanation

To solve the question regarding the monoanions \( N_2^- \) and \( O_2^- \), we need to analyze the bond order and magnetic properties of these species based on Molecular Orbital Theory (MOT).

### Step-by-Step Solution:

1. **Identify the Total Number of Electrons**:
- For \( N_2 \): Each nitrogen atom contributes 7 electrons, so \( N_2 \) has \( 7 + 7 = 14 \) electrons.
- For \( O_2 \): Each oxygen atom contributes 8 electrons, so \( O_2 \) has \( 8 + 8 = 16 \) electrons.

2. **Determine the Electron Configuration for \( N_2^- \)**:
- Adding one electron to \( N_2 \) gives \( N_2^- \) a total of 15 electrons.
- The molecular orbital configuration for \( N_2 \) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \]
- The additional electron goes into the \( \pi_{2p_x}^* \) or \( \pi_{2p_y}^* \) orbital, making it paramagnetic due to the presence of one unpaired electron.

3. **Calculate the Bond Order for \( N_2^- \)**:
- Bond order is calculated as:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]
- For \( N_2^- \):
- Bonding electrons = 10 (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} \))
- Antibonding electrons = 5 (from \( \sigma_{1s}^*, \sigma_{2s}^*, \pi_{2p_x}^*, \pi_{2p_y}^* \))
- Thus, Bond Order = \( \frac{10 - 5}{2} = 2.5 \).

4. **Determine the Electron Configuration for \( O_2^- \)**:
- Adding one electron to \( O_2 \) gives \( O_2^- \) a total of 17 electrons.
- The molecular orbital configuration for \( O_2 \) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^1 \]
- The additional electron goes into the \( \pi_{2p_z}^* \) orbital, making it diamagnetic since all electrons are paired.

5. **Calculate the Bond Order for \( O_2^- \)**:
- For \( O_2^- \):
- Bonding electrons = 10 (same as \( O_2 \))
- Antibonding electrons = 6 (5 from \( O_2 \) plus 1 additional from \( O_2^- \))
- Thus, Bond Order = \( \frac{10 - 6}{2} = 2 \).

6. **Evaluate the Statements**:
- **Statement A**: Bond order decreases, bond length increases, bond weakens. (Correct)
- **Statement B**: Bond order decreases for \( O_2^- \). (Correct)
- **Statement C**: \( N_2^- \) becomes diamagnetic. (Incorrect, it is paramagnetic)
- **Statement D**: \( O_2^- \) is diamagnetic. (Correct)

### Conclusion:
The wrong statement is **C**: \( N_2^- \) becomes diamagnetic.