Bond order of O₂²⁻
Bond order 1, diamagnetic. 18 electrons: Nb = 10, Na = 8.
Molecular orbitals
18 electrons1bond order
diamagnetic
diamagnetic
(Nb − Na) / 2 = (10 − 8) / 2
The atom columns show neutral atoms; the ion has 2 more electrons in the molecule.
σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px² = π*2py²)
Bond order step by step
- O₂²⁻ has 18 electrons: 8 from each O, plus two 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*2py²)
- Bonding electrons Nb = 10, antibonding electrons Na = 8.
- Bond order = (Nb − Na) / 2 = (10 − 8) / 2 = 1.
- Every electron is paired, so O₂²⁻ is diamagnetic.
Class 11, Chapter 4 Chemical Bonding and Molecular Structure.
Bond order of O2 2-: quick answers
What is the bond order of O2 2-?
1. Bond order = (Nb − Na) / 2 = (10 − 8) / 2 = 1. Bonding electrons Nb = 10, antibonding electrons Na = 8.
Is O2 2- paramagnetic or diamagnetic?
Diamagnetic. Every electron is paired, so O₂²⁻ is diamagnetic.
What is the molecular orbital configuration of O2 2-?
σ1s² σ*1s² σ2s² σ*2s² σ2pz² (π2px² = π2py²) (π*2px² = π*2py²), with 18 electrons in all.