Paramagnetic molecules are molecules that have single electrons. When I draw the lewis structure of $\ce{O2}$, it appears to be a diamagnetic structure. What makes it paramagnetic?

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    $\begingroup$ Perhaps you could show us the structure you are drawing? I don't think that Lewis structures are going to help you here because they hold no information about the spin of electrons. $\endgroup$
    – bon
    Commented May 3, 2016 at 15:29
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    $\begingroup$ You are right about the single electrons. But note the plural. What if you have two unpaired electrons and your molecule is a biradical? ;-) $\endgroup$ Commented May 3, 2016 at 15:46
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    $\begingroup$ Is there even an answer on chem.SE explaining why O2 is paramagnetic? Everything I can find uses it as an example, but there's nothing explaining it... Edit: chemistry.stackexchange.com/a/39218/16683 The question isn't even about O2, sigh $\endgroup$ Commented May 3, 2016 at 16:00
  • $\begingroup$ You might look at these links:-physlink.com/Education/AskExperts/ae493.cfm ......... chemistrynotmystery.blogspot.in/2014/09/… $\endgroup$ Commented May 3, 2016 at 16:05

1 Answer 1


To understand the paramagnetic nature of $\ce{O2}$, we must first understand how atomic orbitals mix together to form molecular orbitals. In the diatomic molecules of the elements in the second period, a phenomenon known as $\mathrm{s}$-$\mathrm{p}$ mixing results in an increase in the energy of the $\sigma_\mathrm{2p_z}$ molecular orbital, and a decrease in the energy of the $\sigma_\mathrm{2s}$ orbital. This is also observed in the $\sigma^{*}_\mathrm{2p_z}$ and $\sigma^{*}_\mathrm{2s}$ orbitals.

The degree of $\mathrm{s}$-$\mathrm{p}$ mixing is determined by the energy gap between the $\mathrm{s}$ and $\mathrm{p}$ orbitals. The higher the energy gap, the less the orbitals mix. Because the $\mathrm{s}$-$\mathrm{p}$ gap increases across a period, $\mathrm{s}$-$\mathrm{p}$ mixing in $\ce{O2}$ and $\ce{F2}$ is actually so low that the $\sigma^{*}_\mathrm{2p_z}$ orbital is lower in energy than the $\pi_\mathrm{2p_{x,y}}$ orbital, unlike the preceding elements.

This can be seen here:

MO period 2


By constructing the molecular orbital diagram for $\ce{O2}$ and filling each orbital according to Hund's rule, it becomes evident that $\ce{O2}$ is a diradical, with two unpaired electrons of the same spin. This is what gives oxygen its paramagnetism.

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    $\begingroup$ This answer does not address the question's statement that the Lewis structure has no unpaired electrons. It would be nice if it clarified if and why Lewis is wrong in this case. $\endgroup$
    – Juan Perez
    Commented Aug 3, 2021 at 19:37

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