The correct stability order for `N_2` and its given ions is :

Options

A. `N_(2) gt N_(2)^(+) = N_(2)^(-) gt N_(2)^(2-)`
B. `N_(2)^(+) gt N_(2)^(-) gt N_(2) gt N_(2)^(2-)`
C. `N_(2)^(-)gt N_(2)^(+) gt N_(2) gt N_(2)^(2-)`
D. `N_(2)^(2-) gt N_(2)^(-) = N_(2)^(+) gt N_(2)`

Explanation

To determine the correct stability order for \( N_2 \) and its ions (\( N_2^+ \), \( N_2^- \), and \( N_2^{2-} \)), we will calculate the bond order for each species and then compare them. The bond order is a measure of the stability of a molecule; a higher bond order indicates greater stability.

### Step 1: Calculate the bond order for \( N_2 \)

1. **Total Electrons**: \( N_2 \) has 14 electrons (7 from each nitrogen atom).
2. **Molecular Orbital Configuration**:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^*^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)
3. **Bond Order Calculation**:
- Bonding Electrons (nB) = 10 (from \( \sigma \) and \( \pi \) orbitals)
- Anti-bonding Electrons (nA) = 4 (from \( \sigma^* \) orbitals)
- Bond Order = \( \frac{nB - nA}{2} = \frac{10 - 4}{2} = 3 \)

### Step 2: Calculate the bond order for \( N_2^+ \)

1. **Total Electrons**: \( N_2^+ \) has 13 electrons (one less than \( N_2 \)).
2. **Molecular Orbital Configuration**:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^*^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^1 \) (one electron removed)
3. **Bond Order Calculation**:
- Bonding Electrons = 9
- Anti-bonding Electrons = 4
- Bond Order = \( \frac{9 - 4}{2} = 2.5 \)

### Step 3: Calculate the bond order for \( N_2^- \)

1. **Total Electrons**: \( N_2^- \) has 15 electrons (one more than \( N_2 \)).
2. **Molecular Orbital Configuration**:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^*^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^1 \) (one extra electron in anti-bonding)
3. **Bond Order Calculation**:
- Bonding Electrons = 10
- Anti-bonding Electrons = 5
- Bond Order = \( \frac{10 - 5}{2} = 2.5 \)

### Step 4: Calculate the bond order for \( N_2^{2-} \)

1. **Total Electrons**: \( N_2^{2-} \) has 16 electrons (two more than \( N_2 \)).
2. **Molecular Orbital Configuration**:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^*^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^1 \)
- \( \pi_{2p_y}^1 \) (two extra electrons in anti-bonding)
3. **Bond Order Calculation**:
- Bonding Electrons = 10
- Anti-bonding Electrons = 6
- Bond Order = \( \frac{10 - 6}{2} = 2 \)

### Step 5: Compare Bond Orders

- \( N_2 \): Bond Order = 3 (most stable)
- \( N_2^+ \): Bond Order = 2.5
- \( N_2^- \): Bond Order = 2.5
- \( N_2^{2-} \): Bond Order = 2 (least stable)

### Final Stability Order

The stability order from most stable to least stable is:
1. \( N_2 \)
2. \( N_2^+ \) = \( N_2^- \) (equal stability)
3. \( N_2^{2-} \)

### Answer

The correct stability order is:
\[ N_2 > N_2^+ = N_2^- > N_2^{2-} \]