Chemistry · Chemical Bonding · NEET
O2+ = 2.5, O2 = 2, O2- = 1.5, O2^2- = 1. Neutral O2 has 16 electrons and bond order 2. O2+ has one electron less (15 e), so 2.5. O2- (superoxide) has one extra electron (17 e), so 1.5. O2^2- (peroxide) has two extra electrons (18 e), so 1. Each electron you add or remove changes the bond order by 0.5 because it goes into an antibonding orbital.
Bond order = 1/2 (bonding electrons - antibonding electrons). For O2 (16 e), the MO filling gives 10 bonding and 6 antibonding electrons: 1/2 (10 - 6) = 2. The last electrons of O2 sit in the two pi* (antibonding) orbitals. So removing an electron (O2+) takes it OUT of an antibonding orbital, raising bond order to 2.5. Adding electrons (O2-, O2^2-) puts them INTO pi* orbitals, lowering the bond order.
Superoxide is O2- with a single negative charge and bond order 1.5 (17 electrons, one unpaired electron, so paramagnetic). Peroxide is O2^2- with a double negative charge and bond order 1 (18 electrons, all paired, so diamagnetic). Easy way to remember: 'super' = higher bond order (1.5), 'per' with 2 charges = lower bond order (1).
O2 has its two highest-energy electrons in antibonding (pi*) orbitals. When you remove one electron to form O2+, you take it away from an antibonding orbital. Removing an antibonding electron makes the molecule stronger, so bond order goes UP from 2 to 2.5. That is why O2+ has a shorter, stronger bond than O2.
Higher bond order means shorter and stronger bond. So bond LENGTH order is opposite to bond order: O2+ < O2 < O2- < O2^2- (peroxide has the longest bond). Bond STRENGTH (bond enthalpy) follows bond order: O2+ > O2 > O2- > O2^2-. NEET loves asking you to arrange these, so learn both directions.
O2 has 2 unpaired electrons (paramagnetic). O2+ has 1 unpaired electron (paramagnetic). O2- (superoxide) has 1 unpaired electron (paramagnetic). Only O2^2- (peroxide) has all electrons paired, so it is diamagnetic. Trap: many students wrongly call O2+ diamagnetic - it is NOT (a real NEET 2022 answer).
Which amongst the following is an incorrect statement? (A) The bond orders of O2+, O2, O2- and O2^2- are 2.5, 2, 1.5 and 1, respectively. (B) C2 molecule has four electrons in its two degenerate pi molecular orbitals. (C) H2+ ion has one electron. (D) O2+ ion is diamagnetic.
Which one of the following pairs of species have the same bond order?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Start at O2 = 2. For O2+ add 0.5 (remove an electron) to get 2.5. For O2- subtract 0.5 to get 1.5. For O2^2- subtract another 0.5 to get 1. The pattern is 2.5, 2, 1.5, 1 in the order O2+, O2, O2-, O2^2-.
Peroxide O2^2- is diamagnetic. It has 18 electrons, and its two pi* orbitals are completely filled with paired electrons, so there are no unpaired electrons. Its bond order is 1.
O2+ has the strongest and shortest bond because it has the highest bond order (2.5). Bond strength order is O2+ > O2 > O2- > O2^2-, and bond length is the reverse.
O2 has 16 electrons, O2+ has 15 (one removed), O2- (superoxide) has 17 (one added), and O2^2- (peroxide) has 18 (two added). Each change of one electron shifts the bond order by 0.5.
In O2 the highest orbitals available are antibonding (pi*). Extra electrons added to make O2- and O2^2- must go into these antibonding orbitals. Antibonding electrons weaken the bond, so the bond order goes down.
