stability - Why is O2 a biradical?

$\ce{O2}$ has a double bond in its normal form. That is $\ce{O=O}$. There are no unpaired electrons in this case are there since there are 2 lone pairs on each oxygen.

However 1 resonance structure would be $\ce{O-O}$ (result of homolytic cleavage of double bond) where each $\ce{O}$ is a free radical (a negatively charged one at that). If you have this in hydrogen it is likely going to form hydrogen peroxide.

You could also have $\ce{O-O}$ where 1 is positive and the other is negative and this is also 2 free radicals.

And finally there is $\ce{O#O}$ where both oxygens are positively charged and are free radicals. Why are both positively charged? It is because 3 bonds already to oxygen means 1 lone pair and 5 electrons around oxygen is +1.

Is it because of these resonance structures giving 2 free radicals in $\ce{O2}$ that $\ce{O2}$ is considered a biradical?

Answer

We can draw the 3 Lewis structures (or the corresponding resonance structures) pictured below for $\ce{O_2}$

Since an oxygen atom has 6 electrons,

• A would correspond to a structure with a single bond between the oxygen atoms, 2 lone pairs on each oxygen and an unpaired electron on each oxygen; however A does not have an octet around each oxygen, in fact, each oxygen would only have 7 electrons


• B would correspond to a structure with a double bond between the oxygen atoms, 2 lone pairs on each oxygen and no unpaired electrons on each oxygen; B does have an octet around each oxygen, but it is not a biradical


• C would correspond to a structure with a triple bond between the oxygen atoms, 1 lone pair on each oxygen and an unpaired electron on each oxygen; however C does not have an octet around each oxygen, in fact, each oxygen would have 9 electrons and this would be impossible for oxygen

So while structure A would indicate a biradical, we wouldn't "expect" it to count for much since the oxygens do not have octets. This inability to clearly predict the biradical nature of $\ce{O_2}$ illustrates one of the failings of both Lewis structures and resonance theory.

In order to correctly predict the biradical nature of $\ce{O_2}$ we must move up to molecular orbital theory. Below is the molecular orbital diagram for $\ce{O_2}$. As you can see it does predict that $\ce{O_2}$ should be a biradical with an unpaired electron in each of its degenerate, highest occupied molecular orbitals.

Edit: response to OP's comment

When I think of triple bond I don't think of 2 3 electron bonds(which is what you drew). Rather I think of 3 2 electron bonds(1 sigma bond and 2 pi bonds)

Structure C does represent 3 two-electron bonds (not 2 three-electron bonds), that's just how you draw the Lewis structure.

This type of triple bond would make the oxygen positive with 5 electrons around it.

No, the formal charge on the oxygen in structure C is

Z = 6 - 3 unshared - (1/2 * 6 shared)= 0,

there is no formal charge on oxygen in the "triple bond" structure and as I noted above, there are 9 electrons around it (not 5), which is impossible for oxygen.

I am assuming electrons are shared equally with half around 1 atom and half around the other(which is the basis for formal charge

Yes, that's correct.