`N_(2) and O_(2)` are converted to monocations `N_(2)^(+) and O_(2)^(+)` respectively, which is wrong statement:

Options

A. In `N_(2)^(+)`, the N-N bond weakens
B. In `O_(2)^(+)`, the O-O bond order increase
C. In `O_(2)^(+)`, the paramagnetism decrease
D. `N_(2)^(+)` becomes diamagnetic

Answer

To determine which statement is wrong regarding the conversion of \(N_2\) and \(O_2\) to their respective monocations \(N_2^+\) and \(O_2^+\), we will analyze the bond order, bond strength, and magnetic properties of these species.

### Step 1: Calculate the bond order of \(N_2\)

1. **Determine the total number of electrons in \(N_2\)**:
- Each nitrogen atom has an atomic number of 7, contributing a total of 14 electrons for \(N_2\) (7 from each nitrogen).

2. **Fill the molecular orbital diagram**:
- The electron configuration for \(N_2\) is:
- \(\sigma_{1s}^2\), \(\sigma_{1s}^*^2\), \(\sigma_{2s}^2\), \(\sigma_{2s}^*^2\), \(\pi_{2p_x}^2\), \(\pi_{2p_y}^2\), \(\sigma_{2p_z}^2\)
- Count the bonding and antibonding electrons:
- Bonding: \(2 + 2 + 2 + 2 + 2 = 10\)
- Antibonding: \(2 + 2 = 4\)

3. **Calculate bond order**:
\[ \text{Bond order} = \frac{\text{Bonding electrons} - \text{Antibonding electrons}}{2} = \frac{10 - 4}{2} = 3 \]

### Step 2: Calculate the bond order of \(N_2^+\)

1. **Determine the total number of electrons in \(N_2^+\)**:
- \(N_2^+\) has one less electron, so it has 13 electrons.

2. **Fill the molecular orbital diagram**:
- The electron configuration for \(N_2^+\) is:
- \(\sigma_{1s}^2\), \(\sigma_{1s}^*^2\), \(\sigma_{2s}^2\), \(\sigma_{2s}^*^2\), \(\pi_{2p_x}^2\), \(\pi_{2p_y}^2\), \(\sigma_{2p_z}^1\)
- Count the bonding and antibonding electrons:
- Bonding: \(2 + 2 + 2 + 2 + 1 = 9\)
- Antibonding: \(2 + 2 = 4\)

3. **Calculate bond order**:
\[ \text{Bond order} = \frac{9 - 4}{2} = 2.5 \]

### Step 3: Calculate the bond order of \(O_2\)

1. **Determine the total number of electrons in \(O_2\)**:
- Each oxygen atom has an atomic number of 8, contributing a total of 16 electrons for \(O_2\).

2. **Fill the molecular orbital diagram**:
- The electron configuration for \(O_2\) is:
- \(\sigma_{1s}^2\), \(\sigma_{1s}^*^2\), \(\sigma_{2s}^2\), \(\sigma_{2s}^*^2\), \(\pi_{2p_x}^2\), \(\pi_{2p_y}^2\), \(\pi_{2p_x}^1\), \(\pi_{2p_y}^1\)
- Count the bonding and antibonding electrons:
- Bonding: \(2 + 2 + 2 + 2 + 2 = 10\)
- Antibonding: \(2 + 2 = 4\)

3. **Calculate bond order**:
\[ \text{Bond order} = \frac{10 - 4}{2} = 2 \]

### Step 4: Calculate the bond order of \(O_2^+\)

1. **Determine the total number of electrons in \(O_2^+\)**:
- \(O_2^+\) has one less electron, so it has 15 electrons.

2. **Fill the molecular orbital diagram**:
- The electron configuration for \(O_2^+\) is:
- \(\sigma_{1s}^2\), \(\sigma_{1s}^*^2\), \(\sigma_{2s}^2\), \(\sigma_{2s}^*^2\), \(\pi_{2p_x}^2\), \(\pi_{2p_y}^2\), \(\pi_{2p_x}^1\)
- Count the bonding and antibonding electrons:
- Bonding: \(2 + 2 + 2 + 2 + 1 = 9\)
- Antibonding: \(2 + 2 = 4\)

3. **Calculate bond order**:
\[ \text{Bond order} = \frac{9 - 4}{2} = 2.5 \]

### Step 5: Analyze the statements

1. **In \(N_2^+\)**, the \(N-N\) bond weakens: **True** (bond order decreases from 3 to 2.5).
2. **In \(O_2^+\)**, the \(O-O\) bond order increases: **False** (bond order remains the same at 2 for \(O_2\) and increases to 2.5 for \(O_2^+\)).
3. **In \(O_2^+\)**, the paramagnetism decreases: **True** (since \(O_2\) is paramagnetic due to unpaired electrons, and \(O_2^+\) has fewer unpaired electrons).

### Conclusion

The wrong statement is: **In \(O_2^+\), the \(O-O\) bond order increases**.