Which of the following pair is expected to have the same bond order? (i) O2 , N2 (ii) O2+, N2- (iii) O2-, N2+ (iv) O2-,N2-

This is a Long Answer Type Questions as classified in NCERT Exemplar Ans: Option (ii) (i) The electronic configuration of O2 is: σ1s2 σ∗1s2 σ2s2 σ∗2s2 s2pz2 π2p2y π2p2x π*2p1y π*2p1x Bond order = 1 2 ( Nb- Na) Bond order = 1 2 (10-6)=2.0 The electronic configuration of N2 : σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x, π2p2y σ2p2z Bond order = 1 2 (10-4)=3.0 So, the bond order of O2 , N2 will not be equal. (ii) The electronic configuration of O2+ is: σ1s2 σ∗1s2 σ2s2 σ∗2s2 s2pz2,π2p2y , π2p2x, π*2p1x Bond order = 1 2 (10-5)=2.5 The electronic configuration of N2- σ1s2 σ∗1s2 σ2s2 σ∗2s2 ,π2p2x , π2p2y, s2pz2 π*2p1x Bond order = 1 2 (10-5)=2.5 Hence, the bond order of O2+,N2- will be equal. (iii) The electronic configuration of O2 - is: σ1s2 σ∗1s2 σ2s2 σ∗2s2 s2pz2,π2p2y , π2p2x, π*2p2y π*2p1x Bond order = 1 2 (10-7)=1.5 The electronic configuration of N2+ is: σ1s2 σ*1s2 σ2s2 σ*2s2 π2p2x , π2p2y σ2p1z Bond order = 1 2 (9-4)=2.5 Hence, the bond order of O2-, N2+ will not be equal. (iv) The electronic configuration of O2 - is: σ1s2 σ∗1s2 σ2s2 σ∗2s2 s2pz2,π2p2y , π2p2x, π*2p2y π*2p1x Bond order = 1 2 (10-7)=1.5 The electronic configuration of N2- σ1s2 σ∗1s2 σ2s2 σ∗2s2 ,π2p2x , π2p2y, s2pz2 π*2p1x Bond order = 1 2 (10-5)=2.5 Hence, the bond order of O2 - , N2- will not be equal.