`N_(2) and O_(2)` are converted into monocations, `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 order increases
C. In `O_(2)^(+)`, the paramagnetism decreases
D. `N_(2)^(+)` becomes diamagnetic

Answer

To solve the question regarding the conversion of \(N_2\) and \(O_2\) into their respective monocations \(N_2^+\) and \(O_2^+\), we need to analyze the statements provided and determine which one is incorrect.

### Step-by-Step Solution:

1. **Understanding the Electron Configuration of \(N_2\)**:
- The total number of electrons in \(N_2\) is \(14\) (7 from each nitrogen atom).
- The electron configuration can be filled as follows:
- \(\sigma 1s^2\)
- \(\sigma^* 1s^2\)
- \(\sigma 2s^2\)
- \(\sigma^* 2s^2\)
- \(\pi 2p_x^2\)
- \(\pi 2p_y^2\)
- \(\sigma 2p_z^2\)

2. **Calculating the Bond Order of \(N_2\)**:
- Bond order = \( \frac{( \text{Number of bonding electrons} - \text{Number of anti-bonding electrons})}{2} \)
- Bonding electrons = \(8\) (from \(\sigma\) and \(\pi\) orbitals)
- Anti-bonding electrons = \(4\) (from \(\sigma^*\) orbitals)
- Therefore, bond order = \( \frac{(8 - 4)}{2} = 2\).

3. **Analyzing \(N_2^+\)**:
- For \(N_2^+\), we remove one electron from the highest energy orbital:
- Total electrons = \(13\).
- The filling will be:
- \(\sigma 1s^2\)
- \(\sigma^* 1s^2\)
- \(\sigma 2s^2\)
- \(\sigma^* 2s^2\)
- \(\pi 2p_x^2\)
- \(\pi 2p_y^2\)
- \(\sigma 2p_z^1\)
- Bond order for \(N_2^+\) = \( \frac{(7 - 2)}{2} = 2.5\).

4. **Understanding the Electron Configuration of \(O_2\)**:
- The total number of electrons in \(O_2\) is \(16\) (8 from each oxygen atom).
- The electron configuration is:
- \(\sigma 1s^2\)
- \(\sigma^* 1s^2\)
- \(\sigma 2s^2\)
- \(\sigma^* 2s^2\)
- \(\sigma 2p_z^2\)
- \(\pi 2p_x^2\)
- \(\pi 2p_y^2\)
- \(\pi^* 2p_x^1\)
- \(\pi^* 2p_y^1\)

5. **Calculating the Bond Order of \(O_2\)**:
- Bonding electrons = \(10\) (from \(\sigma\) and \(\pi\) orbitals)
- Anti-bonding electrons = \(6\) (from \(\sigma^*\) and \(\pi^*\) orbitals)
- Therefore, bond order = \( \frac{(10 - 6)}{2} = 2\).

6. **Analyzing \(O_2^+\)**:
- For \(O_2^+\), we remove one electron from the highest energy orbital:
- Total electrons = \(15\).
- The filling will be:
- \(\sigma 1s^2\)
- \(\sigma^* 1s^2\)
- \(\sigma 2s^2\)
- \(\sigma^* 2s^2\)
- \(\sigma 2p_z^2\)
- \(\pi 2p_x^2\)
- \(\pi 2p_y^2\)
- \(\pi^* 2p_x^1\)
- Bond order for \(O_2^+\) = \( \frac{(10 - 5)}{2} = 2.5\).

7. **Evaluating the Statements**:
- **Statement 1**: In \(N_2^+\), the N-N bond weakens. **(True)**
- **Statement 2**: In \(O_2^+\), the bond order increases. **(True)**
- **Statement 3**: In \(O_2^+\), the paramagnetism decreases. **(True)**
- **Statement 4**: \(N_2^+\) becomes diamagnetic. **(False)** - \(N_2^+\) is paramagnetic due to the presence of unpaired electrons.

### Conclusion:
The incorrect statement is: **"N2 positive becomes diamagnetic."**