The bond order in `N_(2) ^(-)` is

Options

A. 2
B. `2.5`
C. `1.5`
D. 3

Answer

To find the bond order of the ion \( N_2^- \), we can follow these steps:

### Step 1: Determine the total number of electrons
- The nitrogen atom has an atomic number of 7, which means it has 7 electrons.
- For \( N_2 \), there are 2 nitrogen atoms, giving us \( 7 + 7 = 14 \) electrons.
- Since \( N_2^- \) has an additional electron due to the negative charge, the total number of electrons becomes \( 14 + 1 = 15 \) electrons.

### Step 2: Write the molecular orbital (MO) configuration
- The molecular orbital configuration for \( N_2 \) is:
- \( \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \)
- For \( N_2^- \), we need to add one more electron. This additional electron will occupy the next available molecular orbital, which is the \( \sigma_{2p_z} \) orbital.
- Therefore, the MO configuration for \( N_2^- \) is:
- \( \sigma_{1s}^2 \sigma^*_{1s}^2 \sigma_{2s}^2 \sigma^*_{2s}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^3 \)

### Step 3: Count the bonding and antibonding electrons
- **Bonding electrons**:
- From the configuration, we have:
- \( \sigma_{1s}^2 \) contributes 2
- \( \sigma_{2s}^2 \) contributes 2
- \( \pi_{2p_x}^2 \) contributes 2
- \( \pi_{2p_y}^2 \) contributes 2
- \( \sigma_{2p_z}^3 \) contributes 3
- Total bonding electrons = \( 2 + 2 + 2 + 2 + 3 = 11 \)

- **Antibonding electrons**:
- From the configuration, we have:
- \( \sigma^*_{1s}^2 \) contributes 2
- \( \sigma^*_{2s}^2 \) contributes 2
- Total antibonding electrons = \( 2 + 2 = 4 \)

### Step 4: Calculate the bond order
- The bond order formula is given by:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]
- Substituting the values:
\[ \text{Bond Order} = \frac{(11 - 4)}{2} = \frac{7}{2} = 3.5 \]

### Conclusion
The bond order in \( N_2^- \) is \( 3.5 \).

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