Bond order of O₂⁻
Bond order 1.5, paramagnetic. 17 electrons: Nb = 10, Na = 7.
Molecular orbitals
17 electrons1.5bond order
paramagnetic
paramagnetic
(Nb − Na) / 2 = (10 − 7) / 2
The atom columns show neutral atoms; the ion has 1 more electron in the molecule.
σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px² = π*2py¹)
Bond order step by step
- O₂⁻ has 17 electrons: 8 from each O, plus one for the negative charge.
- 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*2py¹)
- Bonding electrons Nb = 10, antibonding electrons Na = 7.
- Bond order = (Nb − Na) / 2 = (10 − 7) / 2 = 1.5.
- One electron is unpaired (in one of the π*2p orbitals), 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-?
1.5. Bond order = (Nb − Na) / 2 = (10 − 7) / 2 = 1.5. Bonding electrons Nb = 10, antibonding electrons Na = 7.
Is O2- paramagnetic or diamagnetic?
Paramagnetic. One electron is unpaired (in one of the π*2p orbitals), so O₂⁻ is paramagnetic.
What is the molecular orbital configuration of O2-?
σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px² = π*2py¹), with 17 electrons in all.