Question

compare the relative stability of O2 minus and N2 + and comment on their magnetic behaviour​

Answer 1

electronic configuration of O2-(17 electrons)

σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px²≈π*2Py¹)

Na = 7, Nb = 10

now, B.O = 1/2[10 - 7] =1.5  

It has one unpaired electron so, it is paramagnetic

Bond Order of N2⁺:

Total number of electrons:13

Electronic Configuration=:σ1s²  σ*1s²  σ2s²  σ*2s²  π 2py² [π2pz²  σ2px1 ]

Bond Order=[9-4]/2=5/2=2.5 it is paramagnetic

because of bond order n2+ is more stable than O2-

hope it helps.............

Answer 2

Answer:

electronic configuration of O2-(17 electrons)

σ1s²,σ*1s², σ2s²,σ*2s², (π2px²≈π2Py²),(π*2Px²≈π*2Py¹)

Na = 7, Nb = 10

now, B.O = 1/2[10 - 7] =1.5  

It has one unpaired electron so, it is paramagnetic

Bond Order of N2⁺:

Total number of electrons:13

Electronic Configuration=:σ1s²  σ*1s²  σ2s²  σ*2s²  π 2py² [π2pz²  σ2px1 ]

Bond Order=[9-4]/2=5/2=2.5 it is paramagnetic

because of bond order n2+ is more stable than O2-