Draw the molecular orbital diagram for the fluorine molecule (F2). Calculate the bond order and provide a concise statement to describe whether the bond order aligns with what you would predict based on the Lewis structure of F2.
The Correct Answer and Explanation is:
To answer this question, we will draw the molecular orbital (MO) diagram for the fluorine molecule (F₂), calculate its bond order, and compare it to the prediction from the Lewis structure.
Molecular Orbital Diagram for F₂
Fluorine has 9 electrons. In F₂, each atom contributes 9 electrons, making a total of 18 electrons. Only the valence electrons (2s and 2p orbitals) are considered in the MO diagram, which totals 14 valence electrons (7 from each atom).
Order of Molecular Orbitals for F₂ (Z ≥ 8):
- σ(2s)
- σ*(2s)
- σ(2pₓ)
- π(2p_y), π(2p_z)
- π*(2p_y), π*(2p_z)
- σ*(2pₓ)
Electron Filling:
- σ(2s): 2 electrons
- σ*(2s): 2 electrons
- σ(2pₓ): 2 electrons
- π(2p_y), π(2p_z): 4 electrons
- π*(2p_y), π*(2p_z): 4 electrons
- σ*(2pₓ): 0 electrons
All 14 valence electrons are now placed.
Bond Order Calculation
Bond order = (Number of bonding electrons – Number of antibonding electrons) / 2
- Bonding electrons = 2 (σ2s) + 2 (σ2pₓ) + 4 (π2p) = 8
- Antibonding electrons = 2 (σ2s) + 4 (π2p) = 6
Bond order = (8 – 6) / 2 = 1
Comparison with Lewis Structure
The Lewis structure of F₂ shows a single bond between the two fluorine atoms, which is consistent with a bond order of 1. The molecular orbital diagram confirms this, showing one net bond between the atoms after canceling out the bonding and antibonding effects. Thus, the MO theory and Lewis structure both predict the same type of bonding in F₂. This reinforces the reliability of both models for simple diatomic molecules like fluorine.
