`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 determine which statement is wrong regarding the monoanions \( N_2^{-} \) and \( O_2^{-} \), we will analyze the bond orders and magnetic properties of these species using Molecular Orbital Theory (MOT).

### Step 1: Determine the number of electrons in \( N_2 \) and \( O_2 \)

- For \( N_2 \):
- Each nitrogen atom has 7 electrons, so \( N_2 \) has a total of \( 7 \times 2 = 14 \) electrons.

- For \( O_2 \):
- Each oxygen atom has 8 electrons, so \( O_2 \) has a total of \( 8 \times 2 = 16 \) electrons.

**Hint:** Count the total number of electrons by multiplying the number of atoms by the number of electrons per atom.

### Step 2: Fill the molecular orbitals for \( N_2 \) and \( O_2 \)

- For \( N_2 \):
- The molecular orbital filling is:
- \( \sigma 1s^2 \)
- \( \sigma 1s^{*2} \)
- \( \sigma 2s^2 \)
- \( \sigma 2s^{*2} \)
- \( \sigma 2p_z^2 \)
- \( \pi 2p_x^2 \)
- \( \pi 2p_y^2 \)
- \( \pi 2p_x^{*} \)
- \( \pi 2p_y^{*} \)

- For \( O_2 \):
- The molecular orbital filling is:
- \( \sigma 1s^2 \)
- \( \sigma 1s^{*2} \)
- \( \sigma 2s^2 \)
- \( \sigma 2s^{*2} \)
- \( \sigma 2p_z^2 \)
- \( \pi 2p_x^2 \)
- \( \pi 2p_y^2 \)
- \( \pi 2p_x^{*} \)
- \( \pi 2p_y^{*} \)

**Hint:** Use the order of filling based on energy levels, starting from the lowest.

### Step 3: Calculate the bond order for \( N_2 \) and \( O_2 \)

- **Bond order formula:**
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons}) - (\text{Number of anti-bonding electrons})}{2} \]

- For \( N_2 \):
- Bonding electrons = 10, Anti-bonding electrons = 4
- Bond order = \( \frac{10 - 4}{2} = 3 \)

- For \( O_2 \):
- Bonding electrons = 10, Anti-bonding electrons = 6
- Bond order = \( \frac{10 - 6}{2} = 2 \)

**Hint:** Remember that bonding electrons contribute positively to bond order, while anti-bonding electrons contribute negatively.

### Step 4: Analyze the monoanions \( N_2^{-} \) and \( O_2^{-} \)

- For \( N_2^{-} \):
- Total electrons = 15 (14 from \( N_2 \) + 1 for the anion)
- Bonding electrons = 10, Anti-bonding electrons = 5
- Bond order = \( \frac{10 - 5}{2} = 2.5 \)

- For \( O_2^{-} \):
- Total electrons = 17 (16 from \( O_2 \) + 1 for the anion)
- Bonding electrons = 10, Anti-bonding electrons = 7
- Bond order = \( \frac{10 - 7}{2} = 1.5 \)

**Hint:** When adding an electron to form an anion, consider where the electron will go (bonding or anti-bonding).

### Step 5: Determine the magnetic properties of \( N_2^{-} \) and \( O_2^{-} \)

- \( N_2^{-} \) has 1 unpaired electron, making it paramagnetic.
- \( O_2^{-} \) has all electrons paired, making it diamagnetic.

**Hint:** Check for unpaired electrons to determine if the species is paramagnetic (unpaired) or diamagnetic (paired).

### Conclusion

The wrong statement is that \( N_2^{-} \) is diamagnetic. It is actually paramagnetic due to the presence of an unpaired electron.

**Final Answer:** The statement that \( N_2^{-} \) becomes diamagnetic is wrong.