In which of the following processes magnetic moment and Bond order, both are changed.

Options

A. `NO to NO^+`
B. `O_2^+ to O_2`
C. `N_2 to N_2^-`
D. All of these

Answer

To determine in which of the following processes both the magnetic moment and bond order change, we will analyze the changes in molecular orbitals for the given species.

### Step-by-Step Solution:

1. **Understanding Magnetic Moment and Bond Order**:
- The magnetic moment (\( \mu \)) is determined by the number of unpaired electrons in a molecule. The formula is given by:
\[ \mu = \sqrt{n(n + 2)} \]
where \( n \) is the number of unpaired electrons.
- The bond order is calculated as:
\[ \text{Bond Order} = \frac{N_b - N_a}{2} \]
where \( N_b \) is the number of electrons in bonding molecular orbitals and \( N_a \) is the number of electrons in anti-bonding molecular orbitals.

2. **Analyzing the Given Processes**:
- **Process A: \( \text{NO} \) to \( \text{NO}^+ \)**:
- \( \text{NO} \) has 15 electrons, resulting in a bond order of 2.5 and a magnetic moment due to 1 unpaired electron.
- \( \text{NO}^+ \) has 14 electrons, resulting in a bond order of 3 and a magnetic moment of 0 (all electrons are paired).
- **Change**: Both bond order and magnetic moment change.

- **Process B: \( \text{NO}^+ \) to \( \text{NO}_2 \)**:
- \( \text{NO}_2 \) has 16 electrons, resulting in a bond order of 2 and a magnetic moment due to 2 unpaired electrons.
- **Change**: Both bond order and magnetic moment change.

- **Process C: \( \text{N}_2 \) to \( \text{N}_2^- \)**:
- \( \text{N}_2 \) has 14 electrons, resulting in a bond order of 3 and a magnetic moment of 0.
- \( \text{N}_2^- \) has 15 electrons, resulting in a bond order of 2.5 and a magnetic moment due to 1 unpaired electron.
- **Change**: Both bond order and magnetic moment change.

3. **Conclusion**:
- In all three processes (A, B, and C), both the magnetic moment and bond order change. Therefore, the correct answer is **D: All of the above**.