UPSEE Solved Model Paper 1

N2 = 𝜎1s2,𝜎∗1s2 , 𝜎2s2,𝜎∗2s2 , 𝜋2px2,𝜋2py2,𝜎2pz2
N2+ = 𝜎1s2,𝜎∗1s2,𝜎2s2 , 𝜎∗2s2 , 𝜋2px2,𝜋2py2,𝜎2pz1
Bond order of N2 = 1/2(8-2) = 3
Bond order of N2+ = 1/2(7-2) = 2.5
When N2+ is formed from N2, the electron is lost from the bonding molecular orbital and thus the bond order decreases. Consequently, the bond distance increases
O2 = 𝜎1s2,𝜎∗1s2,𝜎2s2 , 𝜎∗2s2 , 𝜎2p𝑧2,𝜋2px2 , 𝜋2py2,𝜋∗2px1,𝜋∗2py1
O2+ = 𝜎1s2,𝜎∗1s2 , 𝜎2s2,𝜎∗2s2 , 𝜎2p𝑧2,𝜋2px2 , 𝜋2py2,𝜋∗2px1,𝜋∗2py0
Bond order of O2 = 1/2(8-4) = 2
Bond order of O2+ = 1/2(8-3) = 2.5
When O2+ is formed, the electron is lost from an antibonding molecular orbital. Obviously, the bond order increases and the bond distance decreases.