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 which ionization process results in an increased bond order and a change in magnetic behavior, we will analyze each of the given processes step by step.

### Step 1: Analyze N2 to N2+
- **N2 Configuration**: The electronic configuration of N2 is \(1s^2 2s^2 2p^2\) (total of 10 electrons).
- **Bonding and Antibonding Orbitals**:
- Bonding orbitals: \(\sigma_{1s}^2, \sigma_{2s}^2, \sigma_{2p_z}^2, \pi_{2p_x}^2, \pi_{2p_y}^2\)
- Antibonding orbitals: \(\sigma_{1s}^*, \sigma_{2s}^*, \pi_{2p_x}^*, \pi_{2p_y}^*\)
- **Bond Order Calculation**:
- Bond Order = (Number of electrons in bonding orbitals - Number of electrons in antibonding orbitals) / 2
- For N2: Bond Order = (8 - 0) / 2 = 4
- **Magnetic Behavior**: N2 is diamagnetic (no unpaired electrons).

- **N2+ Configuration**: For N2+, we remove one electron from the highest energy level, which is one of the \(\pi\) orbitals.
- **Bond Order Calculation**:
- For N2+: Bond Order = (7 - 0) / 2 = 3.5
- **Magnetic Behavior**: N2+ is paramagnetic (one unpaired electron).

### Step 2: Analyze C2 to C2+
- **C2 Configuration**: The electronic configuration of C2 is \(1s^2 2s^2 2p^2\) (total of 8 electrons).
- **Bond Order Calculation**:
- For C2: Bond Order = (6 - 0) / 2 = 3
- **Magnetic Behavior**: C2 is diamagnetic.

- **C2+ Configuration**: For C2+, we remove one electron from the \(\pi\) orbital.
- **Bond Order Calculation**:
- For C2+: Bond Order = (5 - 0) / 2 = 2.5
- **Magnetic Behavior**: C2+ is paramagnetic (one unpaired electron).

### Step 3: Analyze NO to NO+
- **NO Configuration**: The electronic configuration of NO is \(1s^2 2s^2 2p^4\) (total of 11 electrons).
- **Bond Order Calculation**:
- For NO: Bond Order = (7 - 1) / 2 = 3
- **Magnetic Behavior**: NO is paramagnetic (one unpaired electron).

- **NO+ Configuration**: For NO+, we remove one electron from the \(\pi^*\) orbital.
- **Bond Order Calculation**:
- For NO+: Bond Order = (6 - 0) / 2 = 3
- **Magnetic Behavior**: NO+ is diamagnetic (no unpaired electrons).

### Step 4: Analyze O2 to O2+
- **O2 Configuration**: The electronic configuration of O2 is \(1s^2 2s^2 2p^4\) (total of 16 electrons).
- **Bond Order Calculation**:
- For O2: Bond Order = (10 - 6) / 2 = 2
- **Magnetic Behavior**: O2 is paramagnetic (two unpaired electrons).

- **O2+ Configuration**: For O2+, we remove one electron from the \(\pi^*\) orbital.
- **Bond Order Calculation**:
- For O2+: Bond Order = (9 - 6) / 2 = 1.5
- **Magnetic Behavior**: O2+ is paramagnetic (still has unpaired electrons).

### Conclusion
- **N2 to N2+**: Bond order decreases, magnetic behavior changes (diamagnetic to paramagnetic).
- **C2 to C2+**: Bond order decreases, magnetic behavior changes (diamagnetic to paramagnetic).
- **NO to NO+**: Bond order remains the same, magnetic behavior changes (paramagnetic to diamagnetic).
- **O2 to O2+**: Bond order decreases, magnetic behavior remains the same (paramagnetic).

The only process where the bond order increases and the magnetic behavior changes is from **NO to NO+**.

### Final Answer
The correct option is **NO to NO+**.