Ncert Exercises & Solutions

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

$O_{2}, O_{2}^{+}, O_{2}^{-}$ (superoxide), $O_{2}^{2 -}$ (peroxide)

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

{[σ - (1 s)]}^{2} {[σ^{*} (1 s)]}^{2} {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[σ (1 p_{z})]}^{2} {[π (2 p_{x})]}^{2} {[π (2 p_{y})]}^{2} {[π^{*} (2 p_{x})]}^{1} {[π^{*} (2 p_{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:

K K {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[σ (2 p_{z})]}^{2} {[π (2 p_{x})]}^{2} {[π (2 p_{y})]}^{2} {[π^{*} (2 p_{x})]}^{1}

_{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:

K K {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[σ (2 p_{z})]}^{2} {[π (2 p_{x})]}^{2} {[π (2 p_{y})]}^{2} {[π^{*} (2 p_{x})]}^{2} {[π^{*} (2 p_{y})]}^{1}

_{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:

K K {[σ (2 s)]}^{2} {[σ^{*} (2 s)]}^{2} {[σ (2 p_{z})]}^{2} {[π (2 p_{x})]}^{2} {[π (2 p_{y})]}^{2} {[π^{*} (2 p_{x})]}^{2} {[π^{*} (2 p_{y})]}^{2}

_{b}

= 8

_{a}

= 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 -}

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