With the help of molecular oribital theory explain Why `N_2` has greater dissociation energy than `N_2^+` ?

Explanation

To explain why \( N_2 \) has greater dissociation energy than \( N_2^+ \) using molecular orbital theory, we will follow these steps:

### Step 1: Determine the Electronic Configuration of \( N_2 \)

1. **Count the Total Electrons**: Nitrogen (N) has 7 electrons, so \( N_2 \) has \( 7 + 7 = 14 \) electrons.
2. **Fill the Molecular Orbitals**: The molecular orbital configuration for \( N_2 \) is:
- \( \sigma_{1s}^2 \)
- \( \sigma^*_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^*_{2s}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)

### Step 2: Calculate the Bond Order of \( N_2 \)

1. **Identify Bonding and Antibonding Electrons**:
- **Bonding Electrons**: \( 2 (\sigma_{1s}) + 2 (\sigma_{2s}) + 2 (\pi_{2p_x}) + 2 (\pi_{2p_y}) + 2 (\sigma_{2p_z}) = 10 \)
- **Antibonding Electrons**: \( 2 (\sigma^*_{1s}) + 2 (\sigma^*_{2s}) = 4 \)
2. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{\text{Bonding Electrons} - \text{Antibonding Electrons}}{2} = \frac{10 - 4}{2} = 3 \]

### Step 3: Determine the Electronic Configuration of \( N_2^+ \)

1. **Count the Total Electrons**: \( N_2^+ \) has one less electron, so it has \( 13 \) electrons.
2. **Fill the Molecular Orbitals**: The molecular orbital configuration for \( N_2^+ \) is:
- \( \sigma_{1s}^2 \)
- \( \sigma^*_{1s}^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma^*_{2s}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^1 \) (only one electron in the \( \sigma_{2p_z} \) orbital)

### Step 4: Calculate the Bond Order of \( N_2^+ \)

1. **Identify Bonding and Antibonding Electrons**:
- **Bonding Electrons**: \( 2 (\sigma_{1s}) + 2 (\sigma_{2s}) + 2 (\pi_{2p_x}) + 2 (\pi_{2p_y}) + 1 (\sigma_{2p_z}) = 9 \)
- **Antibonding Electrons**: \( 2 (\sigma^*_{1s}) + 2 (\sigma^*_{2s}) = 4 \)
2. **Calculate Bond Order**:
\[ \text{Bond Order} = \frac{9 - 4}{2} = \frac{5}{2} = 2.5 \]

### Step 5: Compare the Bond Orders and Dissociation Energies

1. **Bond Order Comparison**:
- \( N_2 \) has a bond order of \( 3 \).
- \( N_2^+ \) has a bond order of \( 2.5 \).
2. **Dissociation Energy Relation**: Higher bond order indicates stronger bonds and thus greater dissociation energy.
- Since \( N_2 \) has a higher bond order than \( N_2^+ \), it follows that \( N_2 \) has greater dissociation energy.

### Conclusion

Thus, \( N_2 \) has greater dissociation energy than \( N_2^+ \) because it has a higher bond order (3 compared to 2.5), which correlates with stronger bonding and higher energy required to break the bond.

---