Molecular orbital diagram of F₂
Bond order 1, diamagnetic. 18 electrons: Nb = 10, Na = 8.
Molecular orbitals
18 electrons1bond order
diamagnetic
diamagnetic
(Nb − Na) / 2 = (10 − 8) / 2
σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px² = π*2py²)
Why this shape
- Each F has 7 valence electrons. It uses one in one single bond to F, so six are left as three lone pairs.
- Steric number = one sigma bond + three lone pairs = 4, so each F is sp³ hybridised.
- Four electron pairs point to the corners of a tetrahedron (tetrahedral electron geometry). With only two atoms, the molecule itself is a straight line.
- Two atoms always lie on a straight line, so there is no bond angle.
- Both atoms are the same, so the bond is non-polar.
Lewis structure
14 valence electronsFormal charges: none
Filling the diagram
- F₂ has 18 electrons: 9 from each F.
- For O₂, F₂ and their ions, σ2pz lies below the π2p pair and fills first.
- Configuration: σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px² = π*2py²)
- Bonding electrons Nb = 10, antibonding electrons Na = 8.
- Bond order = (Nb − Na) / 2 = (10 − 8) / 2 = 1.
- Every electron is paired, so F₂ is diamagnetic.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
MO diagram of F2: quick answers
What is the molecular orbital configuration of F2?
σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px² = π*2py²). For O₂, F₂ and their ions, σ2pz lies below the π2p pair and fills first.
What is the bond order of F2 from its MO diagram?
1: Nb = 10 and Na = 8, so (10 - 8) / 2 = 1. It is diamagnetic.
Why is σ2pz below π2p in F2?
From O2 onwards the 2s and 2p orbitals are far apart in energy and hardly mix, so σ2pz stays below the π2p pair.