Which one of the following is the correct order of decreasing bond enthalpies for the given species?

According to Molecular Orbital Theory (MOT), the strength of a bond is directly related to its bond order, which is defined as half the difference between the number of electrons in bonding orbitals ($N_b$) and antibonding orbitals ($N_a$) . The sources state that as the bond order increases, the bond enthalpy increases and the bond length decreases .

We can determine the bond orders for the given species as follows:

1. $N_2$: With 14 electrons, its configuration results in a bond order of 3 . Its experimental bond enthalpy is very high, at 946.0 kJ/mol .
2. $O_2$: With 16 electrons, its configuration results in a bond order of 2 . Its experimental bond enthalpy is 498 kJ/mol .
3. $O_2^-$ (Superoxide): This ion has 17 electrons. The additional electron enters an antibonding orbital, reducing the bond order to 1.5 .
4. $O_2^{2-}$ (Peroxide): This ion has 18 electrons. With two additional electrons in antibonding orbitals compared to $O_2$, the bond order is 1 .

Comparing these values, the order of decreasing bond order is $N_2 (3) > O_2 (2) > O_2^- (1.5) > O_2^{2-} (1)$. Therefore, the order of decreasing bond enthalpies follows the same sequence .