Question

The bond strength in $O_{2}^{+}$, ${{O}_{2}}$, $O_{2}^{-}$ and $O_{2}^{2-}$ follows the order –
a.) $O_{2}^{2-}>O_{2}^{+}>{{O}_{2}}>O_{2}^{+}$
b.) $O_{2}^{+}>{{O}_{2}}>O_{2}^{-}>O_{2}^{2-}$
c.) ${{O}_{2}}>O_{2}^{-}>O_{2}^{2-}>O_{2}^{+}$
d.) $O_{2}^{-}>O_{2}^{2-}>O_{2}^{+}>{{O}_{2}}$

Answer 1

Bond strength can be analyzed by calculating the bond order of the individual molecules. In order to find the bond order, count the number of total electrons present in the compound and use molecular orbital theory.

Complete step by step answer:
Let us analyze all the given compound and find its bond order.
The number of electrons in $O_{2}^{+}$= 15
The number of electrons in ${{O}_{2}}$= 16
The number of electrons in $O_{2}^{-}$ = 17
The number of electrons in $O_{2}^{2-}$ = 18
Now, let us write the electronic configuration of ${{O}_{2}}$according to Molecular Orbital Theory –
$\left( \text{ }\\!\\!\sigma\\!\\!\text{ 1}{{\text{s}}^{\text{2}}}\text{ }\\!\\!\sigma\\!\\!\text{ *1}{{\text{s}}^{\text{2}}} \right)\text{ }\left( \text{ }\\!\\!\sigma\\!\\!\text{ 2}{{\text{s}}^{\text{2}}}\text{ }\\!\\!\sigma\\!\\!\text{ *2}{{\text{s}}^{\text{2}}} \right)\text{ }\\!\\!\sigma\\!\\!\text{ 2p}_{z}^{2}\left( \text{ }\\!\\!\pi\\!\\!\text{ 2}{{\text{p}}_{\text{x}}}^{\text{2}}\text{= }\\!\\!\pi\\!\\!\text{ 2}{{\text{p}}_{\text{y}}}^{\text{2}} \right)\text{ }\\!\\!~\\!\\!\text{ }\left( \text{ }\\!\\!\pi\\!\\!\text{ *2}{{\text{p}}_{\text{x}}}^{1}\text{= }\\!\\!\pi\\!\\!\text{ *2}{{\text{p}}_{\text{y}}}^{1} \right)\text{ }\\!\\!~\\!\\!\text{ }$
where, σ and π indicate bonding molecular orbitals
σ* and π* indicate antibonding molecular orbitals.
As we can see, the number of Bonding electrons = 10
And, the number of Antibonding electrons = 6

As we know,
Bond Order = (Number of Bonding electrons – Number of Antibonding electrons) / 2
Therefore, Bond Order of BN = $\dfrac{6-2}{2}$ = 2.
We can calculate the bond of other elements similarly.
The bond order of $O_{2}^{+}$= 2.5
The bond order of in ${{O}_{2}}$= 2
The bond order of in $O_{2}^{-}$ = 1.5
The bond order of in $O_{2}^{2-}$ = 1

Arranging the following in order of bond stability, we get –
$O_{2}^{+}>{{O}_{2}}>O_{2}^{-}>O_{2}^{2-}$.
So, the correct answer is “Option B”.

Note: For stability of a compound, its number of bonding electrons should be more than the number of antibonding electrons.
A bond order equal to zero indicates that no bond exists, i.e. the compound doesn’t exist. The stability of compounds increases with increasing Bond order. Also, bond length decreases with increasing bond strength.