Question

Which of the following options represents the correct bond order?
(a)- $O_{2}^{-}$ >${{O}_{2}}$ <$O_{2}^{+}$
(b)- $O_{2}^{-}$ <${{O}_{2}}$ >$O_{2}^{+}$
(c)- $O_{2}^{-}$ >${{O}_{2}}$ >$O_{2}^{+}$
(d)- $O_{2}^{-}$ <${{O}_{2}}$ <$O_{2}^{+}$

Answer 1

Hint: The bond order of the molecules or ions can be calculated by the formula: $B.O.=\dfrac{1}{2}({{N}_{b}}-{{N}_{a}})$, where ${{N}_{b}}$ is the number of electrons present in the bonding orbital and ${{N}_{a}}$ is the number of electrons present in the antibonding orbital.

Complete step by step answer:
The stability of a molecule is decided on the factor of bond order of that molecule and it is calculated by the formula:
$B.O.=\dfrac{1}{2}({{N}_{b}}-{{N}_{a}})$
Where ${{N}_{b}}$ is the number of electrons present in the bonding orbital and ${{N}_{a}}$ is the number of electrons present in the antibonding orbital.
In$O_{2}^{-}$, there is a negative charge on the oxygen atom that means that one electron is added to the molecule. The number of electrons in ${{O}_{2}}$ is 16 and adding 1 electron, the total number of the electrons in $O_{2}^{-}$ will be 17.
The configuration will be:
$\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\text{ }\sigma 2p_{z}^{2}\text{ }\pi 2p_{x}^{2}\text{ }\pi 2p_{y}^{2}\text{ }{{\pi }^{*}}2p_{x}^{2}\text{ }{{\pi }^{*}}2p_{y}^{1}$
So, the number of electrons in a bonding orbital is 10, and the number of electrons in an antibonding orbital is 7.
The bond order will be: $B.O.=\dfrac{1}{2}({{N}_{b}}-{{N}_{a}})$
$B.O.=\dfrac{1}{2}(10-7)=\dfrac{3}{2}=1.5$
So, the bond order is 1.5.
In${{O}_{2}}$, the number of electrons is 16.
The configuration will be:
$\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\text{ }\sigma 2p_{z}^{2}\text{ }\pi 2p_{x}^{2}\text{ }\pi 2p_{y}^{2}\text{ }{{\pi }^{*}}2p_{x}^{1}\text{ }{{\pi }^{*}}2p_{y}^{1}$
So, the number of electrons in a bonding orbital is 10, and the number of electrons in an antibonding orbital is 6. The bond order will be: $B.O.=\dfrac{1}{2}({{N}_{b}}-{{N}_{a}})$
$B.O.=\dfrac{1}{2}(10-6)=\dfrac{4}{2}=2$
So, the bond order is 2.
In $O_{2}^{+}$, there is a positive charge on the oxygen atom that means that one electron is removed from the molecule. The number of electrons in ${{O}_{2}}$ is 16 and removing 1 electron, the total number of the electrons in $O_{2}^{+}$ will be 15.
The configuration will be:
$\sigma 1{{s}^{2}}\text{ }{{\sigma }^{*}}1{{s}^{2}}\text{ }\sigma 2{{s}^{2}}\text{ }{{\sigma }^{*}}2{{s}^{2}}\text{ }\sigma 2p_{z}^{2}\text{ }\pi 2p_{x}^{2}\text{ }\pi 2p_{y}^{2}\text{ }{{\pi }^{*}}2p_{x}^{1}\text{ }$
So, the number of electrons in a bonding orbital is 10, and the number of electrons in an antibonding orbital is 5.
The bond order will be: $B.O.=\dfrac{1}{2}({{N}_{b}}-{{N}_{a}})$
$B.O.=\dfrac{1}{2}(10-5)=\dfrac{5}{2}=2.5$
So, the bond order is 2.5.
So, the order will be $O_{2}^{-}$ <${{O}_{2}}$ <$O_{2}^{+}$.

Therefore, the correct answer is an option (d)- $O_{2}^{-}$ <${{O}_{2}}$ <$O_{2}^{+}$.

Note: In the configuration of the molecule the electrons $\text{ }{{\pi }^{*}}2{{p}_{x}}\text{ and }{{\pi }^{*}}2{{p}_{y}}$ have equal energy, therefore, they are first given one-one electrons, only after that its pairing can be done.