Bond order of O₂
Bond order 2, paramagnetic. 16 electrons: Nb = 10, Na = 6.
Molecular orbitals
16 electrons2bond order
paramagnetic
paramagnetic
(Nb − Na) / 2 = (10 − 6) / 2
σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px¹ = π*2py¹)
Why this shape
- Each O has 6 valence electrons. It uses two in one double bond to O, so four are left as two lone pairs.
- Steric number = one sigma bond + two lone pairs = 3, so each O is sp² hybridised. A double or triple bond counts once: only its sigma bond uses a hybrid orbital.
- Three electron pairs point to the corners of a flat triangle (trigonal planar 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
12 valence electronsi
The Lewis structure pairs every electron, but O2 is paramagnetic with two unpaired electrons. Only molecular orbital theory explains that.
Formal charges: none
Bond order step by step
- O₂ has 16 electrons: 8 from each O.
- 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*2px¹ = π*2py¹)
- Bonding electrons Nb = 10, antibonding electrons Na = 6.
- Bond order = (Nb − Na) / 2 = (10 − 6) / 2 = 2.
- Two electrons are unpaired (in π*2px and π*2py), so O₂ is paramagnetic.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
Bond order of O2: quick answers
What is the bond order of O2?
2. Bond order = (Nb − Na) / 2 = (10 − 6) / 2 = 2. Bonding electrons Nb = 10, antibonding electrons Na = 6.
Is O2 paramagnetic or diamagnetic?
Paramagnetic. Two electrons are unpaired (in π*2px and π*2py), so O₂ is paramagnetic.
What is the molecular orbital configuration of O2?
σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px¹ = π*2py¹), with 16 electrons in all.