`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 monoanions \(N_2^-\) and \(O_2^-\), we need to analyze the electronic configurations of both molecules and determine which statement among the options provided is incorrect. Let's go through the steps systematically.

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

1. **Count the Total Electrons**:
- \(N_2\) has 2 nitrogen atoms, each contributing 7 electrons, giving a total of \(14\) electrons.

2. **Build the Molecular Orbital Diagram**:
- The molecular orbital (MO) configuration for \(N_2\) is:
- \(\sigma_{1s}^2\)
- \(\sigma_{1s}^*^2\)
- \(\sigma_{2s}^2\)
- \(\sigma_{2s}^*^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)

Thus, the electronic configuration of \(N_2\) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]

### Step 2: Determine the Electronic Configuration of \(N_2^-\)

1. **Add One Electron**:
- For \(N_2^-\), we add one electron, making a total of \(15\) electrons.

2. **Update the Molecular Orbital Configuration**:
- The additional electron will go into the next available molecular orbital, which is \(\pi_{2p_x}^*\) or \(\pi_{2p_y}^*\) (both are degenerate).
- Therefore, the configuration for \(N_2^-\) becomes:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \text{ or } \pi_{2p_y}^*^1 \]

### Step 3: Determine the Electronic Configuration of \(O_2\)

1. **Count the Total Electrons**:
- \(O_2\) has 2 oxygen atoms, each contributing 8 electrons, giving a total of \(16\) electrons.

2. **Build the Molecular Orbital Diagram**:
- The molecular orbital configuration for \(O_2\) is:
- \(\sigma_{1s}^2\)
- \(\sigma_{1s}^*^2\)
- \(\sigma_{2s}^2\)
- \(\sigma_{2s}^*^2\)
- \(\sigma_{2p_z}^2\)
- \(\pi_{2p_x}^2\)
- \(\pi_{2p_y}^2\)

Thus, the electronic configuration of \(O_2\) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]

### Step 4: Determine the Electronic Configuration of \(O_2^-\)

1. **Add One Electron**:
- For \(O_2^-\), we add one electron, making a total of \(17\) electrons.

2. **Update the Molecular Orbital Configuration**:
- The additional electron will go into one of the degenerate \(\pi_{2p_x}^*\) or \(\pi_{2p_y}^*\) orbitals.
- Therefore, the configuration for \(O_2^-\) becomes:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \text{ or } \pi_{2p_y}^*^1 \]

### Step 5: Analyze the Statements

1. **Bond Order Calculation**:
- **For \(N_2\)**: Bond order = \(\frac{(8 - 2)}{2} = 3\)
- **For \(N_2^-\)**: Bond order = \(\frac{(8 - 3)}{2} = 2.5\)
- **For \(O_2\)**: Bond order = \(\frac{(10 - 6)}{2} = 2\)
- **For \(O_2^-\)**: Bond order = \(\frac{(10 - 7)}{2} = 1.5\)

2. **Evaluate Each Statement**:
- **Statement 1**: In \(N_2^-\), the bond weakens due to the addition of an electron to an anti-bonding orbital. (True)
- **Statement 2**: The bond length of \(O_2^-\) increases compared to \(O_2\) due to a decrease in bond order. (True)
- **Statement 3**: \(N_2^-\) is diamagnetic. (False, it is paramagnetic due to the presence of an unpaired electron)
- **Statement 4**: The bond order of \(O_2^-\) is less than that of \(O_2\). (True)

### Conclusion

The incorrect statement is **Statement 3**: \(N_2^-\) is diamagnetic.