Correct order of bond energy is :

Options

A. `N_2gtN_2^(+)gtN_2^(-)gtN_2^(2-)`
B. `N_2^(+)gtN_2^(-)gtN_2^(2-)gtN_2`
C. `N_2gtN_2^(-)=N_2^(+)gtN_2^(2-)`
D. `N_2^(-)gtN_2=N_2^(+)gtN_2^(2-)`

Explanation

To determine the correct order of bond energy for the nitrogen species \( N_2 \), \( N_2^+ \), \( N_2^- \), and \( N_2^{2-} \), we will calculate the bond order for each species and then relate bond order to bond energy.

### Step-by-Step Solution:

1. **Determine the Electron Configuration**:
- For \( N_2 \):
- Total electrons = 14
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \)
- For \( N_2^+ \):
- Total electrons = 13 (loss of one electron)
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1 \)
- For \( N_2^- \):
- Total electrons = 15 (gain of one electron)
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^*^1 \)
- For \( N_2^{2-} \):
- Total electrons = 16 (gain of two electrons)
- Configuration: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi_{2p_x}^*^2 \pi_{2p_y}^*^2 \)

2. **Calculate Bond Order**:
- **Bond Order Formula**:
\[ \text{Bond Order} = \frac{1}{2} \left( \text{Number of electrons in bonding orbitals} - \text{Number of electrons in anti-bonding orbitals} \right) \]
- For \( N_2 \):
- Bonding electrons = 10, Anti-bonding electrons = 4
- Bond Order = \( \frac{1}{2} (10 - 4) = 3 \)
- For \( N_2^+ \):
- Bonding electrons = 9, Anti-bonding electrons = 4
- Bond Order = \( \frac{1}{2} (9 - 4) = 2.5 \)
- For \( N_2^- \):
- Bonding electrons = 10, Anti-bonding electrons = 5
- Bond Order = \( \frac{1}{2} (10 - 5) = 2.5 \)
- For \( N_2^{2-} \):
- Bonding electrons = 10, Anti-bonding electrons = 6
- Bond Order = \( \frac{1}{2} (10 - 6) = 2 \)

3. **Relate Bond Order to Bond Energy**:
- Bond energy is directly proportional to bond order. Therefore, the higher the bond order, the higher the bond energy.
- The calculated bond orders are:
- \( N_2 \): 3
- \( N_2^+ \): 2.5
- \( N_2^- \): 2.5
- \( N_2^{2-} \): 2

4. **Order of Bond Energy**:
- From the bond orders, we can establish the order of bond energy:
\[ N_2 > N_2^+ = N_2^- > N_2^{2-} \]

5. **Final Answer**:
- The correct order of bond energy is:
\[ N_2 > N_2^+ = N_2^- > N_2^{2-} \]