JEE Advanced 2016 Paper 2

(A) The molecular orbital energy configuration of C22– is
σ1s2,σ1s*2σ2s2,σ2s*2,π2px2=π2py2,σ2pz2
In the MO of C22– there is no unpaired electron hence it is diamagnetic
(B) Bond order of O22+ is 3 and O2 is 2 therefore bond length of O2 is greater than O22+
The molecular orbital energy configuration of N2+ is

Bond order of N2+=

1

2

(9−4)
=2.5
σ1s2,σ1s*2σ2s2,σ2s*2,π2px2=π2py2,σ2pz2,π2px*1=π2py*
Bond order of N2−=

1

2

(10−5)
=2.5
He2+has less energy as compare to two isolated He atoms