In which of the following ionisation processes, the bond order has increased and the magnetic behaviour has changed?

Options

A. `N_(2)toN_(2)^(+)`
B. `C_(2)to C_(2)^(+)`
C. `NO to NO^(+)`
D. `O_(2) to O_(2)^(+)`

Answer

To solve the question regarding the ionization processes where the bond order has increased and the magnetic behavior has changed, we will analyze each option step by step.

### Step 1: Understanding Bond Order and Magnetic Behavior
- **Bond Order** is calculated using the formula:
\[ \text{Bond Order} = \frac{(N_b - N_a)}{2} \]
where \(N_b\) is the number of electrons in bonding molecular orbitals and \(N_a\) is the number of electrons in antibonding molecular orbitals.

- **Magnetic Behavior**:
- **Paramagnetic**: Species with unpaired electrons.
- **Diamagnetic**: Species with all electrons paired.

### Step 2: Analyze Each Option

#### Option 1: \(N_2\) to \(N_2^+\)
1. **Electrons in \(N_2\)**:
- Total electrons = 14
- Configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2\)
- Bond Order Calculation:
- Bonding = 10 (2 from each of \(\sigma_{1s}\), \(\sigma_{2s}\), \(\pi_{2p_x}\), \(\pi_{2p_y}\), \(\sigma_{2p_z}\))
- Antibonding = 4 (2 from \(\sigma^*_{1s}\), 2 from \(\sigma^*_{2s}\))
- Bond Order = \(\frac{10 - 4}{2} = 3\)

2. **Electrons in \(N_2^+\)**:
- Total electrons = 13
- Configuration: Same as \(N_2\) but one less electron in \(\sigma_{2p_z}\).
- Bond Order Calculation:
- Bonding = 9
- Antibonding = 4
- Bond Order = \(\frac{9 - 4}{2} = 2.5\)

3. **Magnetic Behavior**:
- \(N_2\) is diamagnetic; \(N_2^+\) is paramagnetic.
- **Conclusion**: Bond order decreased, so this option is incorrect.

#### Option 2: \(C_2\) to \(C_2^+\)
1. **Electrons in \(C_2\)**:
- Total electrons = 12
- Configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2\)
- Bond Order Calculation:
- Bonding = 8
- Antibonding = 4
- Bond Order = \(\frac{8 - 4}{2} = 2\)

2. **Electrons in \(C_2^+\)**:
- Total electrons = 11
- Configuration: Same as \(C_2\) but one less electron in the \(\pi\) orbitals.
- Bond Order Calculation:
- Bonding = 7
- Antibonding = 4
- Bond Order = \(\frac{7 - 4}{2} = 1.5\)

3. **Magnetic Behavior**:
- \(C_2\) is diamagnetic; \(C_2^+\) is paramagnetic.
- **Conclusion**: Bond order decreased, so this option is incorrect.

#### Option 3: \(NO\) to \(NO^+\)
1. **Electrons in \(NO\)**:
- Total electrons = 15
- Configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi^*_{2p_x}^1\)
- Bond Order Calculation:
- Bonding = 10
- Antibonding = 4
- Bond Order = \(\frac{10 - 4}{2} = 3\)

2. **Electrons in \(NO^+\)**:
- Total electrons = 14
- Configuration: Same as \(NO\) but one less electron in the \(\pi^*\) orbital.
- Bond Order Calculation:
- Bonding = 10
- Antibonding = 3
- Bond Order = \(\frac{10 - 3}{2} = 3.5\)

3. **Magnetic Behavior**:
- \(NO\) is paramagnetic; \(NO^+\) is diamagnetic.
- **Conclusion**: Bond order increased from 3 to 3.5, and magnetic behavior changed from paramagnetic to diamagnetic. This is the correct answer.

### Final Answer
The correct option is **Option 3: \(NO\) to \(NO^+\)**, where the bond order has increased and the magnetic behavior has changed.