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Draw the valence shell molecular orbital diagram of the oxygen molecule and predict its magnetic nature.


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Last updated date: 25th Jun 2024
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Answer
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Hint: The magnetic property of a molecule can be explained based on the molecular orbital theory. The molecule which does not contain the unpaired electron is known as the paramagnetic. The molecule which has all-electron paid-up does not contribute towards the magnetic property. It is diamagnetic in nature. To solve such a problem write down the MOT diagram of molecules.

Complete step by step solution:
Let’s first draw the MOT of the oxygen molecule. The electronic configuration of oxygen atom is as shown below,
$\text{ O = 1}{{\text{s}}^{\text{2}}}\text{ 2}{{\text{s}}^{\text{2}}}\text{ 2}{{\text{p}}_{\text{x}}}^{\text{2}}\text{=2}{{\text{p}}_{\text{y}}}^{\text{1}}\text{=2}{{\text{p}}_{\text{z}}}^{\text{1}}\text{ }$
Thus the oxygen molecule ${{\text{O}}_{\text{2}}}$ contains 16 electrons. The MOT is as shown below,
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First of all, we can write the molecular orbital configuration of ${{\text{O}}_{\text{2}}}$ the molecule. In a ${{\text{O}}_{\text{2}}}$ molecule, there are a total of 16 electrons. The molecular orbital configuration of ${{\text{O}}_{\text{2}}}$ the molecule is as follows:
\[\]\[\text{ }\!\!\sigma\!\!\text{ 1}{{\text{s}}^{\text{2}}}\text{,}{{\text{ }\!\!\sigma\!\!\text{ }}^{\text{*}}}\text{1}{{\text{s}}^{\text{2}}}\text{, }\!\!\sigma\!\!\text{ 2}{{\text{s}}^{\text{2}}}\text{, }{{\text{ }\!\!\sigma\!\!\text{ }}^{\text{*}}}\text{2}{{\text{s}}^{\text{2}}}\text{, }\!\!\sigma\!\!\text{ 2}{{\text{p}}^{\text{2}}}_{z}\text{, 2p}_{\text{x}}^{\text{2}}\text{ }\!\!\pi\!\!\text{ = 2p}_{\text{y}}^{\text{2}}\text{ }\!\!\pi\!\!\text{ , 2p}_{\text{x}}^{1}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{*}}}\text{=2p}_{\text{y}}^{1}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{*}}}\]
There are 10 bonding and 6 nonbonding electrons in the orbitals according to the molecular orbital configuration.
Therefore, $\text{Bond order =}\dfrac{\text{1}}{\text{2}}\left[ \text{Bonding-antibonding} \right]$
= $\dfrac{1}{2}\left[ 10-6 \right]=\dfrac{1}{2}\left( 4 \right)=2$
Thus, the bond order ${{\text{O}}_{\text{2}}}$ is 2.
We know that if a molecule has paired electrons in molecular orbits then it shows diamagnetic properties however if a molecule has unpaired electrons in its MO diagram then it exhibits paramagnetic properties.
Here, from the above MO diagram we observe two unpaired electrons in the $\text{ }\!\!\pi\!\!\text{ *2py }$ and $\text{ }\!\!\pi\!\!\text{ *2px }$molecular orbitals. Thus due to two unpaired electrons in the MO diagram oxygen molecule is paramagnetic in nature.

Note: Note that if we look at the Lewis dot structure we see that electrons in two oxygens are paired thus oxygen molecules should be diamagnetic.
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But remember that the Lewis dot structure of the oxygen molecule is misleading. Magnetic properties are well studied by the MOT diagram.