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Chemistry Question on The Valence Shell Electron Pair Repulsion (VSEPR) Theory

Compare the relative stability of the following species and indicate theirmagnetic properties; O2O_2, O2+O^+_2, O2−O_2^- (superoxide), O22−O_2^{2-}(peroxide)

Answer

There are 1616 electrons in a molecule of dioxygen, 88 from each oxygen atom. The electronic configuration of oxygen molecule can be written as:
[σ−(1s)]2[σ∗(1s)]2[σ(2s)]2[σ∗(2s)]2[σ(1pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1[π∗(2py)]1[\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 1s1s orbital of each oxygen atom is not involved in boding, the number of bonding electrons = 88
= NbN_b and the number of anti-bonding orbitals = 44 = NaN_a.
Bond order = \frac{1}{2}$$(N_b-N_a)
=12(8−4)\frac{1}{2}(8-4)
= 22
Similarly, the electronic configuration of O2+O_2^+ can be written as:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]1KK[\sigma(2s)]^2[\sigma^*(2s)]^2[\sigma(2p_z)]^2[\pi(2p_x)]^2[\pi(2p_y)]^2[\pi^*(2p_x)]^1

NbN_b = 88
NaN_a = 33
Bond order of O2+O_2^+ = 12(8−3)\frac{1}{2}(8-3)
=2.52.5
Electronic configuration of O2−O_2^- ion will be:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]2[π∗(2py)]1KK[\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
NbN_b = 88
NaN_a = 55
Bond order of O2−O_2^-=12(8−5)\frac{1}{2}(8-5)
= 1.51.5
Electronic configuration of O22−O_2^{2-} ion will be:

KK[σ(2s)]2[σ∗(2s)]2[σ(2pz)]2[π(2px)]2[π(2py)]2[π∗(2px)]2[π∗(2py)]2KK[\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 = 88
Na = 66
Bond order of O22−=12(8−6)O_2^{2-}=\frac{1}{2}(8-6)
=11
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 O2+>O2>O2−>O22−O_2^+>O_2>O_2^->O_2^{2-}.