Accordingly MO theory (a) O is paramagnetic and bond order is greater than O2. (b) Ois paramagnetic and bond order is less than O2. (c) Ois diamagnetic and bond order is less than O2. (d) Ois diamagnetic and bond order is more than O2. Ans. (a) Sol. Electronic configuration of O2+ z x y x y 1s *1s 2s *2s 2p 2p 2p *2p *2p (i) B.O. for O2 = ½ × [10 – 6] B.O. for O2 = 2 (ii) B.O. for O2+ = ½ × [10 – 5] B.O. for O2+ = 2.5 This shows that O2+ is paramagnetic with an unpaired electron & its Bond order is greater than O2.