Correct order of dissociation energy of `N_(2)` and `N_(2)^(+)` is:

Options

A. `N_(2) gt N_(2)^(+)`
B. `N_(2) = N_(2)^(+)`
C. `N_(2)^(+) gt N_(2)`
D. None

Answer

To determine the correct order of dissociation energy of \( N_2 \) and \( N_2^+ \), we need to analyze the bonding and stability of both species.

### Step-by-Step Solution:

1. **Understanding the Molecular Structure of \( N_2 \)**:
- Nitrogen (\( N \)) has the electronic configuration \( 1s^2 2s^2 2p^3 \).
- In \( N_2 \), two nitrogen atoms share three pairs of electrons, forming a triple bond.
- This triple bond consists of one sigma bond and two pi bonds, making \( N_2 \) a very stable molecule.

2. **Stability of \( N_2 \)**:
- The bond order of \( N_2 \) is 3 (due to the three bonding pairs).
- All electrons in \( N_2 \) are paired, which contributes to its stability and high bond dissociation energy.

3. **Understanding the Molecular Structure of \( N_2^+ \)**:
- \( N_2^+ \) is formed by removing one electron from \( N_2 \).
- This results in one unpaired electron, making \( N_2^+ \) paramagnetic.
- The bond order of \( N_2^+ \) is still 2.5 (since one bonding electron is removed), but the presence of an unpaired electron reduces the overall stability.

4. **Comparing Dissociation Energies**:
- The bond dissociation energy is the energy required to break the bonds in a molecule.
- For \( N_2 \), the dissociation energy is high due to the strong triple bond and the stability of the molecule.
- For \( N_2^+ \), the dissociation energy is lower because the removal of an unpaired electron is easier than breaking a bond with paired electrons.

5. **Conclusion**:
- Since \( N_2 \) has a higher dissociation energy than \( N_2^+ \), the correct order of dissociation energy is:
\[ D(N_2) > D(N_2^+) \]

### Final Answer:
The correct order of dissociation energy is \( D(N_2) > D(N_2^+) \).