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$\ce{O2}$ 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?

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    $\begingroup$ Some of your proposed resonance structures are wrong. Both atoms in $\ce{O2}$ cannot have charge of the same sign, as that would violate the neutrality of the oxygen molecule. $\endgroup$
    – Nicolau Saker Neto
    Aug 9, 2014 at 22:07
  • $\begingroup$ but a molecule does not have to be neutral. Otherwise methoxide anions and hydroxide anions would not exist and lots more polyatomic ions wouldn't exist. Also O can have more than 2 bonds or less than 2 bonds. $\endgroup$
    – Caters
    Aug 9, 2014 at 22:12
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    $\begingroup$ Discrete polyatomic entities are only called molecules if they have no charge, otherwise they are termed ions. You are correct that polyatomic ions exist, but dioxygen, the most stable allotrope of oxygen and the molecule found in air, is not one of them. When we say oxygen gas or oxygen molecules, we are talking about neutral $\ce{O2}$ (or I suppose you could explicitly write $\ce{O_{2}^{0}}$). While $\ce{O2^{+}}$, $\ce{O2^{-}}$ and $\ce{O2^{2-}}$ are all known, they are not called oxygen, but dioxygenyl, superoxide and peroxide, respectively. $\endgroup$
    – Nicolau Saker Neto
    Aug 9, 2014 at 22:22
  • $\begingroup$ There are such things as "molecular ions." $\endgroup$
    – Dissenter
    Aug 15, 2014 at 3:32
  • $\begingroup$ yeah like the carbanion resonance structure of a carbonyl compound and the oxyanion resonance structure of a carbonyl compound. It is still considered to be that 1 molecule(whatever the carbonyl compound is) but now there are ions in it. This is really what is called a zwitterion because if carbon is positive O is negative and if C is negative O is positive. $\endgroup$
    – Caters
    Sep 28, 2014 at 16:03

2 Answers 2

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

enter image description here

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.

enter image description here

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.

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  • $\begingroup$ 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) This type of triple bond would make the oxygen positive with 5 electrons around it. Also oxygen when it is single bonded to only 1 atom has 3 lone pairs of electrons making there be 7 electrons around it(and thus a -1 charge). I am assuming electrons are shared equally with half around 1 atom and half around the other(which is the basis for formal charge), not necessarily that each one has an octet. $\endgroup$
    – Caters
    Aug 9, 2014 at 23:25
  • $\begingroup$ To best answer your comment, I placed an edit in my post. $\endgroup$
    – ron
    Aug 9, 2014 at 23:38
  • $\begingroup$ there is formal charge on the oxygen. I mean look at C=-O. There are 5 electrons around this oxygen because of 3 2 electron bonds and 1 lone pair. Same thing goes for the O=-O. I often use a slightly condensed lewis dot structure where I represent bonds with lines between atoms and nonbonding electrons with dots(most of the time just lone pairs). $\endgroup$
    – Caters
    Aug 9, 2014 at 23:43
  • $\begingroup$ No, for the oxygen in C-triple bond-O (carbon monoxide right?) there 8 electrons around it; 2 in the lone pair and the 6 in the triple bond. That's how you count electrons to determine how many electrons are around an atom, for example when you are trying to count and see if there is an octet. $\endgroup$
    – ron
    Aug 9, 2014 at 23:48
  • $\begingroup$ I said that the single bonded oxygen and the triple bonded oxygen are free radicals because if the oxygen in this state is separated each O has an odd number of valence electrons which means that it has unpaired electrons which means it is a free radical and is extremely reactive(Even more than the oxide anion is) $\endgroup$
    – Caters
    Aug 9, 2014 at 23:49
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There is no way to adequately describe $\ce{O2}$ molecule using naive valence bond theory. You need to learn Molecular Orbital Theory. Relevant links

http://en.wikipedia.org/wiki/Molecular_orbital_theory http://en.wikipedia.org/wiki/Molecular_orbital_diagram#Dioxygen

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    $\begingroup$ Apparently even in the first article where Pauling described valence bond theory, he realized the problem with dioxygen and was careful to describe the bonding in $\ce{O2}$ as involving three-electron bonds, which would make the molecule paramagnetic. I have also read that modern valence bond theory finds no difficulty in describing $\ce{O2}$, though I don't know if it still invokes the three-electron bond. $\endgroup$
    – Nicolau Saker Neto
    Aug 9, 2014 at 22:13
  • $\begingroup$ O2 is paramagnetic, so at least that's correct. You don't notice it unless you have liquid O2 and a big magnet though. $\endgroup$
    – user137
    Aug 9, 2014 at 23:04
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    $\begingroup$ @NicolauSakerNeto I believe in modern VB theory you also give the overall spin state of the wave function, in this case you will start with two unpaired one from scratch. The three electron bond is not used any more (to my knowledge), they most likely use some kind of hybrid with MO theory to describe the individual resonance structures. $\endgroup$
    – Martin - マーチン ♦
    Aug 14, 2014 at 10:50

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