Question

Bond order of N2 is greater than that of N2+ but the bond order of O2 is less than that of O2+

Answer 1

Bond order = No. of electron in bonding MOs - No. of electron in antibonding orbital/2.

MO Configuration :

N2( Total 14e-) =σ1s2σ*1s2σ2s2σ*2s2(π2px2=π2py2)σ2pz2
thus Bond order = 10-4/2 =6/2= 3

N2+(Total 13e-) =σ1s2σ*1s2σ2s2σ*2s2(π2px2=π2py2)σ2pz1
thus Bond order = 9-4/2=5/2= 2.5

Hence bond order of N2 is greater than N2+

Similarly,

O2 (16e-) =σ1s2σ*1s2σ2s2σ*2s2σ2pz2(π2px2=π2py2)(π*2px1=π*2py1)
Bond order = (8-4)/2 = 4/2 = 2
the 1s orbital of each oxygen atom is not involved in boding and hence not considered while calculating B.O

O22+ (14e-) =σ1s2σ*1s2σ2s2σ*2s2σ2pz2(π2px2=π2py2)
Bond order = 8-3/2 = 5/2 = 2.5

Answer 2

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