The energy of σ2pz molecular orbital is greater than π2px and π2py molecular orbitals in nitrogen molecule. Write the complete sequence of energy levels in the increasing order of energy in the molecule. Compare the relative stability and the magnetic behavior of the following species.

N2, N2+, N2-, N22+


Electronic configuration of N-atom (Z=7) is 1s22s22px12py12pz1. Total number of electrons present in N2 molecule is 14, 7 from each N-atom. From the view of various rules for filling of molecular orbitals, the electronic configuration of N2 molecule will be
σ1s2, σ*1s2, σ2s2, σ*2s2, π2px2π2py2, σ2pz2
Comparative stude of the relative stability and the magnetic behaviour of the following species
(i) N2 molecule σ1s2, σ*1s2, σ2s2, σ*2s2, π2px2π2py2, σ2pz2
Here, Nb=10, Na=4
Hence, Bond order=12(Nb-Na)=12(10-4)=3
Hence, presence of no unpaired electron indicates it to be diamagnetic.
(ii) N2+ ions σ1s2, σ*1s2, σ2s2, σ*2s2, π2px2π2py2, σ2pz1
Here, Nb=9, Na=4
so that BO=12(9-4)=52=2.5
Further, as N2+ion has one unpaired electron in the σ(2p2) orbital, therefore, it is paramagnetic in nature.
(iii) N2- ions σ1s2, σ*1s2, σ2s2, σ*2s2, π2px2π2py2, σ2pz2, π*2px1
Here, Nb=10, Na=5
so that BO=12(10-5)=52=2.5
Again, as it has one unpaired electron in the π*(2px) orbital, therefore, it is paramagnetic.
(iv) N22+ ions σ1s2, σ*1s2, σ2s2, σ*2s2, π2px2π2py2
Here, Nb=8, Na=4
Hence, BO=12(8-4)=2
The presence of no unpaired electron indicates it to be diamagnetic in nature.