
Explain why ${O_2}$ molecule is paramagnetic in nature?
Answer
589.5k+ views
Hint: Try to recall the molecular orbital theory. By drawing the molecular orbital diagram of oxygen molecules, you can easily explain the paramagnetic nature of oxygen molecules. Also, it is known to you that for a molecule to be paramagnetic it must have at least an unpaired electron.
Complete step by step solution:
It is known to you that one of the greatest achievements of molecular orbital approach was that you could explain the paramagnetic nature of oxygen molecules which you could not explain by valence bond approach.
According to valence bond theory, oxygen molecule should like this:
The electronic configuration of an oxygen atom(Z=8) = $1{s^2}2{s^2}2{p^4}$. Valence bond theory considers that the atomic orbitals overlap to form bonds and due to this electron are paired in overlapping and it shows that oxygen is a diamagnetic species.
But experimentally it is found that oxygen molecules are paramagnetic in nature and this could be only explained by molecular orbital approach.
According to molecular orbital theory, the atomic orbitals of oxygen atoms overlap to form orbitals of the molecule which you can see in the below diagram.
The electronic configuration of Oxygen molecule (according to molecular orbital theory) is $(\sigma 1{s^2}\sigma *1{s^2})(\sigma 2{s^2}\sigma *2{s^2})(\sigma 2{p_z}^2)(\pi 2{p_x}^2 = \pi 2{p_y}^2)(\pi 2{p_x}^1 = \pi 2{p_y}^1)$.
where, $\sigma $ and $\pi $ indicates bonding molecular orbital
$\sigma *$ and $\pi *$ indicates antibonding molecular orbital.
From the above electronic configuration of the ${O_2}$ molecule, you can easily say that since, the last two electrons go into two separate, $\pi $ degenerate orbitals .
Therefore, you can easily say that ${O_2}$ has two unpaired electrons and is paramagnetic.
Note: It should be remembered that you can also find the bond order of oxygen molecules by using the electronic configuration of oxygen molecules.
Bond order is defined as “half the difference between the number of electrons in bonding orbitals and antibonding molecular orbitals”. It indicates number of bonds in the given compound
Complete step by step solution:
It is known to you that one of the greatest achievements of molecular orbital approach was that you could explain the paramagnetic nature of oxygen molecules which you could not explain by valence bond approach.
According to valence bond theory, oxygen molecule should like this:
The electronic configuration of an oxygen atom(Z=8) = $1{s^2}2{s^2}2{p^4}$. Valence bond theory considers that the atomic orbitals overlap to form bonds and due to this electron are paired in overlapping and it shows that oxygen is a diamagnetic species.
But experimentally it is found that oxygen molecules are paramagnetic in nature and this could be only explained by molecular orbital approach.
According to molecular orbital theory, the atomic orbitals of oxygen atoms overlap to form orbitals of the molecule which you can see in the below diagram.
The electronic configuration of Oxygen molecule (according to molecular orbital theory) is $(\sigma 1{s^2}\sigma *1{s^2})(\sigma 2{s^2}\sigma *2{s^2})(\sigma 2{p_z}^2)(\pi 2{p_x}^2 = \pi 2{p_y}^2)(\pi 2{p_x}^1 = \pi 2{p_y}^1)$.
where, $\sigma $ and $\pi $ indicates bonding molecular orbital
$\sigma *$ and $\pi *$ indicates antibonding molecular orbital.
From the above electronic configuration of the ${O_2}$ molecule, you can easily say that since, the last two electrons go into two separate, $\pi $ degenerate orbitals .
Therefore, you can easily say that ${O_2}$ has two unpaired electrons and is paramagnetic.
Note: It should be remembered that you can also find the bond order of oxygen molecules by using the electronic configuration of oxygen molecules.
Bond order is defined as “half the difference between the number of electrons in bonding orbitals and antibonding molecular orbitals”. It indicates number of bonds in the given compound
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