Q.5 The superoxide ion, O2– , is produced by the reaction K+O 2→KO2 . The correct statement(s)
pertaining to oxygen and superoxide ion is (are):
(A) oxygen is paramagnetic and has two unpaired electrons
(B) the bond order in oxygen is 2
(C) the bond order in superoxide is 1.5
(D) the superoxide ion is not paramagnetic

Oxygen (O₂) is paramagnetic with a bond order of 2, while the superoxide ion (O₂⁻) in KO₂ has a bond order of 1.5 and is also paramagnetic. Options (A) and (C) are correct.

Molecular Orbital Diagrams

Oxygen molecule O₂ has 16 valence electrons. Its MO configuration is: KK (σ2s)² (σ2s)² (σ2pz)² (π2px)² (π2py)² (π2px)¹ (π2py)¹. Two unpaired electrons in the degenerate π orbitals confirm paramagnetism. Bond order = (10 bonding – 6 antibonding)/2 = 2.​​

Superoxide ion O₂⁻ gains one electron (17 total), filling as: KK (σ2s)² (σ2s)² (σ2pz)² (π2px)² (π2py)² (π2px)² (π2py)¹. One unpaired electron in π2py makes it paramagnetic. Bond order = (10 – 7)/2 = 1.5.​​

Option Analysis

• (A) Correct: O₂ shows paramagnetism due to two unpaired electrons.​

• (B) Incorrect: Bond order is 2, but the statement claims it as 2 (wait, this matches—recheck: yes, bond order is exactly 2). Wait, standard confirmation shows bond order 2, so actually correct? No—question says “bond order in oxygen is 2″—yes correct, but let’s verify if it’s multiple choice single or multi.​
  Wait, upon tool data, all sources confirm O₂ bond order 2.​​

The question is multiple correct type. But (B) is correct per data.

Standard answer for such CSIR questions: (A),(B),(C) correct, (D) wrong.

(D) Incorrect: O₂⁻ is paramagnetic, not diamagnetic.​

Oxygen molecule and superoxide ion properties are key for CSIR NET Life Sciences, especially in superoxide ion O₂⁻ bond order calculations using molecular orbital theory.

Oxygen Molecule Properties

O₂ exhibits paramagnetism from two unpaired electrons in π*2p orbitals. Bond order of 2 corresponds to a double bond, confirmed by MO diagram with 10 bonding and 6 antibonding electrons.​

Superoxide Formation

KO₂ forms via K + O₂ → KO₂, yielding yellow paramagnetic solid with O₂⁻ anion. The extra electron reduces bond strength.​

Bond Order Calculation

For superoxide ion O₂⁻ bond order: 17 electrons yield configuration ending π² π¹. Formula gives (10-7)/2 = 1.5. Weaker than O₂’s 2.​​

Magnetic Properties

Both species are paramagnetic: O₂ (2 unpaired e⁻), O₂⁻ (1 unpaired e⁻ in π*2py). KO₂ compound shows paramagnetism.​

Correct options: (A), (B), (C). (D) false as superoxide ion O₂⁻ is paramagnetic.​