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Question: The stability order of \( \text{O}_{2} \) and its ions is: (A) \( O_2^{2 + } > O_2^ + > {O_2} > O_...

The stability order of O2\text{O}_{2} and its ions is:
(A) O22+>O2+>O2>O2−>O22−O_2^{2 + } > O_2^ + > {O_2} > O_2^ - > O_2^{2 - }
(B) O22+=O22−>O2+=O2−>O2O_2^{2 + } = O_2^{2 - } > O_2^ + = O_2^ - > {O_2}
(C) O22+=O2+>O2+2−=O2−>O2O_2^{2 + } = O_2^ + > O_2^{ + 2 - } = O_2^ - > {O_2}
(D) O22−=O2−>O2=O2+>O22+O_2^{2 - } = O_2^ - > {O_2} = O_2^ + > O_2^{2 + }

Explanation

Solution

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=12(Nb−Na)BO=\dfrac{1}{2}\left( {{N}_{b}}-{{N}_{a}} \right)
Where
BOBO is the bond order
Nb{{N}_{b}} is the number of bonding electrons
Na{{N}_{a}} is the number of antibonding electrons.

Complete Step By Step Solution
Let us now write the configuration of oxygen molecule that is O2{{O}_{2}}
O2→(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)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}}
(π2px)2=(π2py)2(π∗2px)1=(π∗2py)1{{\left( \pi 2px \right)}_{2}}={{\left( {{\pi }^{2}}py \right)}_{2}}{{\left( \pi *2px \right)}_{1}}={{\left( \pi *2py \right)}_{1}}
For O2+{{O}_{2}}^{+} ion, the one electron is removed from the configuration, and it gains positive charge
For O2−{{O}_{2}}^{-} ion, one electron is added to the configuration, and it gains negative charge
For O22−{{O}_{2}}^{2-} ion, two electrons are added to the configuration, and its overall charge becomes 2−2-
For O22+{{O}_{2}}^{2+} ion, two electrons are removed from the configuration, and it gains overall charge of 2+2+
Now, we will find out the bond order of each of the oxygen ions
For O2{{O}_{2}} , we have
Bond order =12(10−6)=2=\dfrac{1}{2}\left( 10-6 \right)=2
For O2−{{O}_{2}}^{-} , we have
Bond order =12(10−7)=1.5=\dfrac{1}{2}\left( 10-7 \right)=1.5
For O22−{{O}_{2}}^{2-} , we have
Bond order =12(10−8)=1=\dfrac{1}{2}\left( 10-8 \right)=1
For O2+{{O}_{2}}^{+} , we have
Bond order =12(10−5)=2.5=\dfrac{1}{2}\left( 10-5 \right)=2.5
For O22+{{O}_{2}}^{2+} , we have
Bond order =12(10−4)=3=\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
O22+>O2+>O2>O2−>O22−{{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.