Question

In molecular orbital diagram for $\text{O}_{\text{2}}^{\text{+}}$ ion, the highest occupied orbital is:
(A) $\sigma \,\text{MO}\,\text{orbital}$
(B) $\text{ }\!\!\pi\!\!\text{ }\,\text{MO}\,\text{orbital}$
(C) ${{\text{ }\!\!\pi\!\!\text{ }}^{*}}\,\text{MO}\,\text{orbital}$
(D) ${{\sigma }^{*}}\,\text{MO}\,\text{orbital}$

Answer 1

Hint: Molecular orbital theory was put forward by Hund and Mulliken, which can be applied to explain the properties, that was not explained by Valence bond theory. This theory explained the paramagnetic nature of $\text{O}_{\text{2}}^{\text{+}}$ ion as per Valence bond theory it should be diamagnetic.
- Molecular orbital diagram is the diagrammatic representation of all the molecular orbital and electronic configuration of molecular orbitals in a molecule.
- According to this theory the electrons in a molecule are present in the various molecular orbits. There are two types of molecular orbital – bonding molecular orbital and antibonding molecular orbital.
- The number of molecular orbitals formed is equal to the number of combining atomic orbitals.

Complete Solution :
The bonding molecular orbital has lower energy and greater stability than the corresponding ant-bonding molecular orbitals. The molecular orbitals of a molecule are filled according to the Aufbau principle obeying the Pauli’s exclusion principle and Hund’s rule.
The order of energies of molecular orbitals for homonuclear diatomic molecule like $\text{O}_{\text{2}}^{{}}$, ${{\text{F}}_{\text{2}}}$ and \[\text{N}{{\text{e}}_{\text{2}}}\] is -$\text{ }\!\!\sigma\!\!\text{ 1s}$,${{\text{ }\!\!\sigma\!\!\text{ }}^{*}}\text{1s}$, $\text{ }\!\!\sigma\!\!\text{ 2s}$, ${{\text{ }\!\!\sigma\!\!\text{ }}^{*}}\text{2s}$, $\text{ }\!\!\sigma\!\!\text{ 2}{{\text{p}}_{z}}$, $\text{ }\!\!\pi\!\!\text{ 2}{{\text{p}}_{\text{x}}}=\,\text{ }\!\!\pi\!\!\text{ 2}{{\text{p}}_{\text{y}}}$, ${{\text{ }\!\!\pi\!\!\text{ }}^{*}}\text{2}{{\text{p}}_{\text{x}}}=\,{{\text{ }\!\!\pi\!\!\text{ }}^{*}}\text{2}{{\text{p}}_{\text{y}}}$

Where- $\text{ }\!\!\sigma\!\!\text{ }$,$\text{ }\!\!\pi\!\!\text{ }$ represents the bonding molecular orbital, while ${{\text{ }\!\!\sigma\!\!\text{ }}^{*}}$, ${{\text{ }\!\!\pi\!\!\text{ }}^{*}}$ represents the anti-bonding molecular orbital in a molecular orbital diagram.

- The electronic configuration of molecular orbital of $\text{O}_{\text{2}}^{\text{+}}$ ion, which contains $15\,{{e}^{-}}$ is formed by the loss of one electron from $\text{O}_{\text{2}}^{{}}$ molecule is -$\text{ }\!\!\sigma\!\!\text{ 1}{{\text{s}}^{2}}$,${{\text{ }\!\!\sigma\!\!\text{ }}^{*}}\text{1}{{\text{s}}^{2}}$, $\text{ }\!\!\sigma\!\!\text{ 2}{{\text{s}}^{2}}$, ${{\text{ }\!\!\sigma\!\!\text{ }}^{*}}\text{2}{{\text{s}}^{2}}$, $\text{ }\!\!\sigma\!\!\text{ 2p}_{\text{z}}^{\text{2}}$, $\text{ }\!\!\pi\!\!\text{ 2p}_{x}^{2}=\,\text{ }\!\!\pi\!\!\text{ 2p}_{y}^{2}$, ${{\text{ }\!\!\pi\!\!\text{ }}^{*}}\text{2p}_{\text{x}}^{\text{1}}=\,{{\text{ }\!\!\pi\!\!\text{ }}^{*}}\text{2p}_{\text{y}}^{\text{0}}$,${{\text{ }\!\!\sigma\!\!\text{ }}^{*}}\text{2p}_{\text{z}}^{\text{0}}$.
So, it is clear that in molecular orbital diagram for$\text{O}_{\text{2}}^{\text{+}}$ion, the highest occupied orbital is ${{\text{ }\!\!\pi\!\!\text{ }}^{*}}2{{\text{p}}_{\text{x}}}\,\text{anti-bonding}\,\text{MO}\,\text{orbital}$.
So, the correct answer is “Option C”.

Note: -Electronic configuration of molecular orbital must be according to Hund’s maximum multiplicity rule, according to which the orbital available in the subshell of a molecule are first filled singly with parallel spin electron before they begin to pair and subshell give maximum number of unpaired electron with parallel spin.