`N_2` and `O_2` are converted into monoanions `N_2^-` and `O_2^-` respectively. Which of the following statements in wrong ?

Options

A. In `N_2` , the N-N bond weakens
B. In `O_2^-` , O-O bond length increases
C. In `O_2^-` , bond order decreases
D. `N_2^-` , becomes diamagnetic

Explanation

To solve the question regarding the conversion of \( N_2 \) and \( O_2 \) into their respective monoanions \( N_2^- \) and \( O_2^- \), we need to analyze the bond order and magnetic properties of these species. Let's break down the solution step by step.

### Step 1: Determine the Electron Configuration of \( N_2 \)

The molecular orbital configuration for \( N_2 \) (which has 14 electrons) is:

\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]

### Step 2: Calculate the Bond Order of \( N_2 \)

The bond order is calculated using the formula:

\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

For \( N_2 \):

- Bonding electrons = 10
- Antibonding electrons = 4

\[ \text{Bond Order} = \frac{10 - 4}{2} = 3.0 \]

### Step 3: Determine the Electron Configuration of \( N_2^- \)

When \( N_2 \) gains an electron to form \( N_2^- \), the configuration becomes:

\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \]

### Step 4: Calculate the Bond Order of \( N_2^- \)

For \( N_2^- \):

- Bonding electrons = 10
- Antibonding electrons = 5

\[ \text{Bond Order} = \frac{10 - 5}{2} = 2.5 \]

### Step 5: Analyze the Bond Strength and Length for \( N_2^- \)

Since the bond order decreased from 3.0 to 2.5, the bond strength decreases, and the bond length increases. This means that the bond becomes weaker in \( N_2^- \).

### Step 6: Determine the Electron Configuration of \( O_2 \)

The molecular orbital configuration for \( O_2 \) (which has 16 electrons) is:

\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \]

### Step 7: Calculate the Bond Order of \( O_2 \)

For \( O_2 \):

- Bonding electrons = 10
- Antibonding electrons = 6

\[ \text{Bond Order} = \frac{10 - 6}{2} = 2.0 \]

### Step 8: Determine the Electron Configuration of \( O_2^- \)

When \( O_2 \) gains an electron to form \( O_2^- \), the configuration becomes:

\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]

### Step 9: Calculate the Bond Order of \( O_2^- \)

For \( O_2^- \):

- Bonding electrons = 10
- Antibonding electrons = 7

\[ \text{Bond Order} = \frac{10 - 7}{2} = 1.5 \]

### Step 10: Analyze the Bond Strength and Length for \( O_2^- \)

The bond order decreases from 2.0 to 1.5, indicating that the bond strength decreases and the bond length increases in \( O_2^- \).

### Step 11: Determine the Magnetic Properties

- \( N_2 \) is diamagnetic (no unpaired electrons).
- \( N_2^- \) has one unpaired electron, making it paramagnetic.
- \( O_2 \) has two unpaired electrons, making it paramagnetic.
- \( O_2^- \) has two unpaired electrons, making it paramagnetic.

### Conclusion

The statement that \( N_2^- \) is diamagnetic is **incorrect**. Therefore, the wrong statement among the options is that \( N_2^- \) becomes diamagnetic.