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

Options

A. `sigma_(2s)^(**)ltpi_(2py)=pi_(2py)lt sigma_(2pz)`
B. `sigma_(2s)^(**)ltpi_(2px)ltpi_(2py)lt sigma_(2pz)`
C. `sigma_(2s)^(**)ltpi_(2pz)ltpi_(2px)lt sigma_(2py)`
D. `P_(2px)ltP_(2py)lts_(2s)^(**)lts_(2pz)`

Explanation

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

### Step 1: Determine the number of electrons in \( N_2 \)
- Each nitrogen atom has 7 electrons. Since \( N_2 \) is a diatomic molecule, the total number of electrons is:
\[ 2 \times 7 = 14 \text{ electrons} \]

### Step 2: Write the molecular orbital configuration
- The molecular orbital configuration for \( N_2 \) can be derived from the combination of atomic orbitals. The order of filling for the molecular orbitals is as follows:
- \( \sigma_{1s}^2 \) (2 electrons)
- \( \sigma_{1s}^* \) (0 electrons)
- \( \sigma_{2s}^2 \) (2 electrons)
- \( \sigma_{2s}^* \) (0 electrons)
- \( \sigma_{2p_z}^2 \) (2 electrons)
- \( \pi_{2p_x}^2 \) and \( \pi_{2p_y}^2 \) (4 electrons total)
- \( \pi_{2p_x}^* \) and \( \pi_{2p_y}^* \) (0 electrons)
- \( \sigma_{2p_z}^* \) (0 electrons)

### Step 3: Fill the molecular orbitals according to the Aufbau principle
- The order of energy levels for the molecular orbitals in \( N_2 \) 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 order of energies
- The correct order of energies for the molecular orbitals of \( N_2 \) is:
\[ \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} \]
- This means that the \( \sigma_{2p_z} \) orbital is lower in energy than the degenerate \( \pi_{2p_x} \) and \( \pi_{2p_y} \) orbitals.

### Conclusion
- The correct order of energies of molecular orbitals for \( N_2 \) is:
\[ \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} \]