Which of the following correctly describes the molecular orbital energy diagram for the diatomic molecule He_2?

• A. Only the bonding (σ_{1s}) orbital is filled, resulting in a bond order of 1.
• B. Both bonding (σ_{1s}) and antibonding (σ^*_{1s}) orbitals are filled, resulting in a bond order of 0.
• C. Neither bonding nor antibonding orbitals are filled, resulting in a bond order of 0.
• D. Only the antibonding (σ^*_{1s}) orbital is filled, resulting in a bond order of -1.

Answer: (B)

Verified step by step guidance

1. Step 1: Determine the total number of electrons in the He_2 molecule. Each helium atom has 2 electrons, so He_2 has a total of 4 electrons.
2. Step 2: Recall the order of molecular orbitals for diatomic molecules formed from 1s atomic orbitals: the bonding orbital is \( \sigma_{1s} \) and the antibonding orbital is \( \sigma^*_{1s} \).
3. Step 3: Fill the molecular orbitals with the 4 electrons following the Aufbau principle and Pauli exclusion principle: first fill the bonding \( \sigma_{1s} \) orbital with 2 electrons, then fill the antibonding \( \sigma^*_{1s} \) orbital with the remaining 2 electrons.
4. Step 4: Calculate the bond order using the formula: \[ \text{Bond order} = \frac{(\text{number of electrons in bonding orbitals}) - (\text{number of electrons in antibonding orbitals})}{2} \]. Substitute the values: 2 electrons in bonding and 2 in antibonding orbitals.
5. Step 5: Interpret the bond order result. A bond order of 0 means no net bond is formed, which explains why He_2 is not a stable molecule.