Compare the relative stability of the following species and indicate their magnetic properties;
Question:

Compare the relative stability of the following species and indicate their magnetic properties;

$\mathrm{O}_{2}, \mathrm{O}_{2}^{+}, \mathrm{O}_{2}^{-}$(superoxide), $\mathrm{O}_{2}^{2-}$ (peroxide)

Solution:

There are 16 electrons in a molecule of dioxygen, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:

Since the 1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 8 = Nand the number of anti-bonding orbitals = 4 = Na.

Bond order $=\frac{1}{2}\left(N_{\mathrm{b}}-N_{\mathrm{a}}\right)$

$=\frac{1}{2}(8-4)$

$=2$

Similarly, the electronic configuration of $\mathrm{O}_{2}^{+}$can be written as:

Nb = 8

Na = 3

Bond order of $\mathrm{O}_{2}^{+}=\frac{1}{2}(8-3)$

= 2.5

Electronic configuration of $\mathrm{O}_{2}^{-}$ion will be:

Nb = 8

Na = 5

Bond order of $\mathrm{O}_{2}^{-}=\frac{1}{2}(8-5)$

= 1.5

Electronic configuration of $\mathrm{O}_{2}^{2-}$ ion will be:

Nb = 8

Na = 6

Bond order of $\mathrm{O}_{2}^{2-}=\frac{1}{2}(8-6)$

= 1

Bond dissociation energy is directly proportional to bond order. Thus, the higher the bond order, the greater will be the stability. On this basis, the order of stability is

$\mathrm{O}_{2}^{+}>\mathrm{O}_{2}>\mathrm{O}_{2}^{-}>\mathrm{O}_{2}^{2-}$