Question

The stability order of $ \text{O}_{2} $ and its ions is:
(A) $ O_2^{2 + } > O_2^ + > {O_2} > O_2^ - > O_2^{2 - } $
(B) $ O_2^{2 + } = O_2^{2 - } > O_2^ + = O_2^ - > {O_2} $
(C) $ O_2^{2 + } = O_2^ + > O_2^{ + 2 - } = O_2^ - > {O_2} $
(D) $ O_2^{2 - } = O_2^ - > {O_2} = O_2^ + > O_2^{2 + } $

Answer 1

Hint: This problem can be solved by the concept of molecular orbital theory. After we get to know the configuration of each of the oxygen ions, we can find out the bond order. The bond order will give the details of the stability of each of these ions. Higher the bond order, more is the stability.

Formula Used
We will use the following formula to find out the bond order
 $ BO=\dfrac{1}{2}\left( {{N}_{b}}-{{N}_{a}} \right) $
Where
 $ BO $ is the bond order
 $ {{N}_{b}} $ is the number of bonding electrons
 $ {{N}_{a}} $ is the number of antibonding electrons.

Complete Step By Step Solution
Let us now write the configuration of oxygen molecule that is $ {{O}_{2}} $
 $ {{O}_{2}}\to {{\left( \sigma 1s \right)}_{2}}{{\left( \sigma *1s \right)}_{2}}{{\left( \sigma 2s \right)}_{2}}{{\left( \sigma *2s \right)}_{2}}{{\left( {{\sigma }^{2}}pz \right)}_{2}} $
 $ {{\left( \pi 2px \right)}_{2}}={{\left( {{\pi }^{2}}py \right)}_{2}}{{\left( \pi *2px \right)}_{1}}={{\left( \pi *2py \right)}_{1}} $
For $ {{O}_{2}}^{+} $ ion, the one electron is removed from the configuration, and it gains positive charge
For $ {{O}_{2}}^{-} $ ion, one electron is added to the configuration, and it gains negative charge
For $ {{O}_{2}}^{2-} $ ion, two electrons are added to the configuration, and its overall charge becomes $ 2- $
For $ {{O}_{2}}^{2+} $ ion, two electrons are removed from the configuration, and it gains overall charge of $ 2+ $
Now, we will find out the bond order of each of the oxygen ions
For $ {{O}_{2}} $ , we have
Bond order $ =\dfrac{1}{2}\left( 10-6 \right)=2 $
For $ {{O}_{2}}^{-} $ , we have
Bond order $ =\dfrac{1}{2}\left( 10-7 \right)=1.5 $
For $ {{O}_{2}}^{2-} $ , we have
Bond order $ =\dfrac{1}{2}\left( 10-8 \right)=1 $
For $ {{O}_{2}}^{+} $ , we have
Bond order $ =\dfrac{1}{2}\left( 10-5 \right)=2.5 $
For $ {{O}_{2}}^{2+} $ , we have
Bond order $ =\dfrac{1}{2}\left( 10-4 \right)=3 $
Now, we have calculated the bond order of all of the ions provided to us in the question
The stability of the ions depends upon its bond order
The stability of the ions will be more when the bond order is greater
So, the order of stability of the oxygen ions will be
 $ {{O}_{2}}^{2+}>{{O}_{2}}^{+}>{{O}_{2}}>{{O}_{2}}^{-}>{{O}_{2}}^{2-} $
Hence, the correct option is (A).

Note
The order of the bond shows the number of chemical bonds between a pair of atoms present. The order for the bond describes the bond's stability. The molecular orbital offers an easy understanding of a chemical bond's notion of the bond order. The degree of covalent bonds between the atoms is quantified.