Question

Compare the relative stability of the following species and indicate theirmagnetic properties; \(O_2\), \(O^+_2\), \(O_2^-\) (superoxide), \(O_2^{2-}\)(peroxide)

Answer 1

There are \(16\) electrons in a molecule of dioxygen, \(8\) from each oxygen atom. The electronic configuration of oxygen molecule can be written as:
\([\sigma-(1s)]^2[\sigma^*(1s)]^2[\sigma(2s)]^2[\sigma^*(2s)]^2[\sigma(1p_z)]^2[\pi(2p_x)]^2[\pi(2p_y)]^2[\pi^*(2p_x)]^1[π*(2p_y)]^1\)

Since the \(1s\) orbital of each oxygen atom is not involved in boding, the number of bonding electrons = \(8\) 
= \(N_b\) and the number of anti-bonding orbitals = \(4\) = \(N_a\).
Bond order = \(\frac{1}{2}\)\((N_b-N_a)\)
=\(\frac{1}{2}(8-4)\)
= \(2\)
Similarly, the electronic configuration of \(O_2^+\) can be written as:

\(KK[\sigma(2s)]^2[\sigma^*(2s)]^2[\sigma(2p_z)]^2[\pi(2p_x)]^2[\pi(2p_y)]^2[\pi^*(2p_x)]^1\)

\(N_b\) = \(8\)
\(N_a\) = \(3\)
Bond order of \(O_2^+\) = \(\frac{1}{2}(8-3)\)
=\(2.5\)
Electronic configuration of \(O_2^-\) ion will be: 

\(KK[\sigma(2s)]^2[\sigma^*(2s)]^2[\sigma(2p_z)]^2[\pi(2p_x)]^2[\pi(2p_y)]^2[\pi^*(2p_x)]^2[\pi^*(2p_y)]^1\)
\(N_b\) = \(8\) 
\(N_a\) = \(5\)
Bond order of \(O_2^-\)=\(\frac{1}{2}(8-5)\)
= \(1.5\)
Electronic configuration of \(O_2^{2-}\) ion will be:

\(KK[\sigma(2s)]^2[\sigma^*(2s)]^2[\sigma(2p_z)]^2[\pi(2p_x)]^2[\pi(2p_y)]^2[\pi^*(2p_x)]^2[\pi^*(2p_y)]^2\)
Nb = \(8\) 
Na = \(6\)
Bond order of \(O_2^{2-}=\frac{1}{2}(8-6)\)
=\(1\)
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 \(O_2^+>O_2>O_2^->O_2^{2-}\).