Which of the following statements correctly describes the molecular orbital diagram for N_2?

• A. The highest occupied molecular orbital is a bonding pi orbital.
• B. The bond order of N_2 is 2.
• C. N_2 has two unpaired electrons in its molecular orbital diagram.
• D. The sigma_{2p} orbital is lower in energy than the pi_{2p} orbitals.

Answer: (A)

Verified step by step guidance

1. Recall the electronic configuration of the nitrogen atom: 1s² 2s² 2p³. Since N₂ is a diatomic molecule formed by two nitrogen atoms, the total number of valence electrons to consider is 10 (5 from each nitrogen).
2. Construct the molecular orbital (MO) diagram for N₂ by combining the atomic orbitals of the two nitrogen atoms. For molecules with atomic number less than 8 (like N₂), the ordering of the 2p orbitals is such that the \( \sigma_{2p} \) orbital is higher in energy than the \( \pi_{2p} \) orbitals.
3. Fill the molecular orbitals with the 10 valence electrons following the Aufbau principle (lowest energy orbitals filled first), Pauli exclusion principle, and Hund's rule. This will help determine which orbitals are occupied and whether electrons are paired or unpaired.
4. Calculate the bond order using the formula: \[ \text{Bond order} = \frac{(\text{number of bonding electrons}) - (\text{number of antibonding electrons})}{2} \]. This will help verify the strength and stability of the N₂ bond.
5. Analyze the highest occupied molecular orbital (HOMO) to identify its type (bonding or antibonding, sigma or pi) and check the number of unpaired electrons to confirm the magnetic properties of N₂.