The correct bond order among the following species is :

Options

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

Answer

To determine the correct bond order among the species \(N_2\), \(N_2^+\), \(N_2^-\), and \(N_2^{2-}\), we will use the molecular orbital theory. Here’s a step-by-step solution:

### Step 1: Determine the number of electrons in each species
- **For \(N_2\)**: Each nitrogen atom has 7 electrons, so \(N_2\) has \(7 \times 2 = 14\) electrons.
- **For \(N_2^+\)**: This species has one less electron than \(N_2\), so it has \(14 - 1 = 13\) electrons.
- **For \(N_2^-\)**: This species has one more electron than \(N_2\), so it has \(14 + 1 = 15\) electrons.
- **For \(N_2^{2-}\)**: This species has two more electrons than \(N_2\), so it has \(14 + 2 = 16\) electrons.

### Step 2: Write the molecular orbital configuration for each species
- **For \(N_2\)** (14 electrons):
- Configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2\)

- **For \(N_2^+\)** (13 electrons):
- Configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^1\)

- **For \(N_2^-\)** (15 electrons):
- Configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi^*_{2p_x}^1\)

- **For \(N_2^{2-}\)** (16 electrons):
- Configuration: \(\sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \pi^*_{2p_x}^2\)

### Step 3: Calculate the bond order for each species
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} \left( \text{Number of electrons in bonding MOs} - \text{Number of electrons in antibonding MOs} \right) \]

- **For \(N_2\)**:
- Bonding electrons: 10 (from \(\sigma_{1s}\), \(\sigma_{2s}\), \(\pi_{2p_x}\), \(\pi_{2p_y}\), \(\sigma_{2p_z}\))
- Antibonding electrons: 4 (from \(\sigma^*_{1s}\) and \(\sigma^*_{2s}\))
- Bond Order: \(\frac{1}{2} (10 - 4) = \frac{6}{2} = 3\)

- **For \(N_2^+\)**:
- Bonding electrons: 9 (same as above but with one less in \(\sigma_{2p_z}\))
- Antibonding electrons: 4
- Bond Order: \(\frac{1}{2} (9 - 4) = \frac{5}{2} = 2.5\)

- **For \(N_2^-\)**:
- Bonding electrons: 10
- Antibonding electrons: 5 (including one from \(\pi^*_{2p_x}\))
- Bond Order: \(\frac{1}{2} (10 - 5) = \frac{5}{2} = 2.5\)

- **For \(N_2^{2-}\)**:
- Bonding electrons: 10
- Antibonding electrons: 6 (including two from \(\pi^*_{2p_x}\))
- Bond Order: \(\frac{1}{2} (10 - 6) = \frac{4}{2} = 2\)

### Step 4: Summarize the bond orders
- \(N_2\): Bond Order = 3
- \(N_2^+\): Bond Order = 2.5
- \(N_2^-\): Bond Order = 2.5
- \(N_2^{2-}\): Bond Order = 2

### Conclusion
The correct order of bond strength from highest to lowest is:
1. \(N_2\) (Bond Order = 3)
2. \(N_2^+\) and \(N_2^-\) (Bond Order = 2.5)
3. \(N_2^{2-}\) (Bond Order = 2)

Thus, the correct bond order among the given species is \(N_2 > N_2^+ = N_2^- > N_2^{2-}\).