Solution:
The molecular orbital configurations of
O2+ , O2− , O22− and O2 are
O2+ = σ1s2 , σ* 1s2,σ2s2 , σ* 2s2,σ2pz2,π2px2
~ π2py2,π*2px1 ~ π*2py0
O2− = σ1s2,σ*1s2,σ2s2,σ*2s2 , σ2pz2,π2px2
~ π2py2,π*2px2 ~ π*2py1
O22− = σ1s2,σ*1s2 , σ2s2,σ*2s2 , σ2pz2,π2px2
~ π2py2,π*2px2 ~ π*2py2
O2 = σ1s2,σ*1s2 , σ2s2,σ*2s2 , σ2pz2,π2px2
~ π2py2,π*2px1 ~ π*2py1
and the electronic configuration of O and O+ are
O = 1s2,2s2,2px2,2py1,2pz1
O+ = 1s2,2s2,2px1,2py1,2pz1
As O2+,O2,O2−,O and O+ have unpaired electrons, hence are paramagnetic.
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