Which of the following order of energies of molecular orbitals of `N_(2)` is correct ?

Options

A. `(pi2p_(y)) lt (sigma 2p_(z)) lt (pi^(**) 2p_(x)) = (pi^(**)2p_(y))`
B. `(pi2p_(y)) gt (sigma 2p_(z)) gt (pi^(**) 2p_(x)) = (pi^(**)2p_(y))`
C. `(pi2p_(y)) lt (sigma 2p_(z)) gt (pi^(**) 2p_(x)) = (pi^(**)2p_(y))`
D. `(pi2p_(y)) gt (sigma 2p_(z)) lt (pi^(**) 2p_(x)) = (pi^(**)2p_(y))`

Explanation

To determine the correct order of energies of molecular orbitals for \( N_2 \), we can follow these steps:

### Step 1: Determine the Total Number of Electrons
Nitrogen (N) has an atomic number of 7, so \( N_2 \) has a total of \( 2 \times 7 = 14 \) electrons.

### Step 2: Fill the Molecular Orbitals
The molecular orbitals for \( N_2 \) are filled in the following order based on their energy levels:
1. \( \sigma_{1s} \) (2 electrons)
2. \( \sigma^*_{1s} \) (2 electrons)
3. \( \sigma_{2s} \) (2 electrons)
4. \( \sigma^*_{2s} \) (2 electrons)
5. \( \sigma_{2p_z} \) (2 electrons)
6. \( \pi_{2p_x} \) and \( \pi_{2p_y} \) (4 electrons total, 2 in each)
7. \( \pi^*_{2p_x} \) and \( \pi^*_{2p_y} \) (0 electrons)
8. \( \sigma^*_{2p_z} \) (0 electrons)

### Step 3: Write the Order of Energies
The order of energies of the molecular orbitals for \( N_2 \) is:
- \( \sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z} \)

### Step 4: Identify the Correct Option
Based on the filled molecular orbitals, we can conclude that the correct order of energies is:
\[ \sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z} \]

### Conclusion
The correct order of energies of molecular orbitals for \( N_2 \) is as stated above.