Respective Bond Order for the following `N_(2),N_(2)^(+),N_(2)^(-)`

Options

A. `2.5, 2.5 and 3` respectively
B. `2,2.5 and 3` respectively
C. `3,2.5 and 3` respectively
D. `2.5,2.5 and 2.5` respectively.

Explanation

To find the respective bond order for the molecules \( N_2 \), \( N_2^+ \), and \( N_2^- \), we will follow these steps:

### Step 1: Understand Bond Order Calculation
The bond order can be 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 (BMOs) and \( N_a \) is the number of electrons in antibonding molecular orbitals (ABMOs).

### Step 2: Determine the Electron Configuration for \( N_2 \)
1. **Total Electrons**: \( N_2 \) has 14 electrons (7 from each nitrogen atom).
2. **Molecular Orbital Filling**:
- Fill the orbitals in the following order:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^*^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- Total in bonding: \( 2 + 2 + 2 + 2 + 2 + 2 = 10 \)
- Total in antibonding: \( 2 + 2 = 4 \)

3. **Calculate Bond Order**:
\[ \text{Bond Order for } N_2 = \frac{(10 - 4)}{2} = \frac{6}{2} = 3 \]

### Step 3: Determine the Electron Configuration for \( N_2^+ \)
1. **Total Electrons**: \( N_2^+ \) has 13 electrons (one less than \( N_2 \)).
2. **Molecular Orbital Filling**:
- Remove one electron from the highest energy orbital, which is \( \sigma_{2p_z} \):
- Configuration:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^*^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^1 \)
- Total in bonding: \( 10 - 1 = 9 \)
- Total in antibonding: \( 4 \)

3. **Calculate Bond Order**:
\[ \text{Bond Order for } N_2^+ = \frac{(9 - 4)}{2} = \frac{5}{2} = 2.5 \]

### Step 4: Determine the Electron Configuration for \( N_2^- \)
1. **Total Electrons**: \( N_2^- \) has 15 electrons (one more than \( N_2 \)).
2. **Molecular Orbital Filling**:
- Add one electron to the highest energy orbital, which is \( \pi_{2p_x} \) or \( \pi_{2p_y} \):
- Configuration:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^*^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^*^1 \) (or \( \pi_{2p_y}^*^1 \))
- Total in bonding: \( 10 \)
- Total in antibonding: \( 5 \)

3. **Calculate Bond Order**:
\[ \text{Bond Order for } N_2^- = \frac{(10 - 5)}{2} = \frac{5}{2} = 2.5 \]

### Final Results
- \( N_2 \): Bond Order = 3
- \( N_2^+ \): Bond Order = 2.5
- \( N_2^- \): Bond Order = 2.5

### Conclusion
The respective bond orders for \( N_2 \), \( N_2^+ \), and \( N_2^- \) are 3, 2.5, and 2.5.