According to MOT, two atomic orbitals overlap resulting in the formation of molecular orbital formed. Number of atomic orbitals overlapping together is equal to the molecule orbital formed. The two atomic orbital thus formed by LCAO (linear combination of atomic orbital) in the phase or in the different phase are known as bonding and antibonding molecular orbitals respectively. The energy of bonding molecular orbital is lower than that of the pure atomic orbitals by an amount `Delta`. This known as the stabilization energy. The enerby of antibonding molecular orbital in increased by `Delta'` (destabilisation energy).
Q . The bond order of `N_(2)^(-)` is equal to that of

Options

A. `O_(2)`
B. `O_(2)^(2-)`
C. `O_(2)^(+)`
D. None

Answer

To determine the bond order of \(N_2^-\) and find its equivalent in the given options, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Bond Order**:
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{(\text{Number of Bonding Electrons} - \text{Number of Antibonding Electrons})}{2} \]

2. **Determine the Electron Configuration of \(N_2^-\)**:
- The nitrogen atom (N) has an atomic number of 7, which means it has 7 electrons.
- For \(N_2\), the total number of electrons is \(7 + 7 = 14\).
- The negative charge (\(-\)) indicates that there is one additional electron, making it 15 electrons in total for \(N_2^-\).

3. **Fill the Molecular Orbitals**:
- The molecular orbital configuration for \(N_2\) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \]
- For \(N_2^-\), we add one more electron to the next available molecular orbital, which is the \(\pi_{2p_x}\) or \(\pi_{2p_y}\) (both are degenerate):
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^1 \]

4. **Count the Bonding and Antibonding Electrons**:
- Bonding Electrons:
- \(\sigma_{1s}^2\) = 2
- \(\sigma_{2s}^2\) = 2
- \(\sigma_{2p_z}^2\) = 2
- \(\pi_{2p_x}^2\) = 2
- \(\pi_{2p_y}^2\) = 2
- Total Bonding Electrons = \(2 + 2 + 2 + 2 + 2 = 10\)
- Antibonding Electrons:
- \(\sigma_{1s}^*^2\) = 2
- \(\sigma_{2s}^*^2\) = 2
- Total Antibonding Electrons = \(2 + 2 = 4\)

5. **Calculate the Bond Order**:
\[ \text{Bond Order} = \frac{(10 - 4)}{2} = \frac{6}{2} = 3 \]

6. **Compare with Given Options**:
- We need to find a molecule with a bond order of 2.5.
- From the options provided, we find that \(O_2^+\) has a bond order of 2.5.

### Final Answer:
The bond order of \(N_2^-\) is equal to that of \(O_2^+\).

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