Question

Compare the relative stability of following species N2, N2+, N2-

Answer 1

Here we can take help from bond order

Bond Order=1/2( nb-na )

Bond order Of N2:

Total number of electrons=14

EC:- σ1s²  σ*1s²  σ2s²  σ*2s²  π2py²  [π2pz²  σ2px² ]

Therefore,

Bond Order=10-4/2=6/2=3

Bond Order of N2⁺:

Total no. of electrons 3

EC:- σ1s²  σ*1s²  σ2s²  σ*2s²  π 2py² [π2pz²  σ2px1 ]

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

Bond order of N2⁻:

Total no. of electrons =15

EC:- (σ1s)2 (σ1s*)2 (π 2py)2 (π2pz)2 (σ2s)2 (π2py*)1

Bond order=10-5/2=2.5

Higher bond order implies more stability

Therefore N2 most stable

N2+ and N2- have same bond order therefore we can say they have same energies

But since N2- has 1 more antibondig electron

therefore

N2+ is more stable than N2-

Final order N2 > N2+ > N2-

(although N2>N2+ = N2- is also correct as per class 11)