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
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(whichis what you drew). Rather I think of 3 2 electron bonds(1 sigma bondand 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 5electrons 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 andhalf around the other(which is the basis for formal charge
When I think of triple bond I dont 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.More
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. Thats 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.More
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 More
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,
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
Structure C does represent 3 two-electron bonds (not 2 three-electron bonds), that's just how you draw the Lewis structure.
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.
Yes, that's correct.
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,
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
Structure C does represent 3 two-electron bonds (not 2 three-electron bonds), that's just how you draw the Lewis structure.
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.
Yes, that's correct.
More
VOTE
VOTE
VOTE
VOTE
VOTE
VOTE