$ceO2$ has a dual bond in its normal form. The is $ceO=O$. There are no unpaired electron in this case are there due to the fact that there space 2 lone bag on each oxygen.

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However 1 resonance structure would it is in $ceO-O$ (result of homolytic cleavage of dual bond) where each $ceO$ is a cost-free radical (a negatively fee one in ~ that). If you have this in hydrogen that is likely going to type hydrogen peroxide.

You could also have $ceO-O$ wherein 1 is positive and the other is negative and this is additionally 2 complimentary radicals.

And ultimately there is $ceO#O$ where both oxygens space positively charged and are totally free radicals. Why room both positive charged? the is due to the fact that 3 bonds already to oxygen means 1 lone pair and also 5 electrons approximately oxygen is +1.

Is it due to the fact that of these resonance structures providing 2 complimentary radicals in $ceO2$ the $ceO2$ is considered a biradical?


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edited Aug 15 "14 in ~ 2:59
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We can attract the 3 Lewis structures (or the equivalent resonance structures) pictured below for $ceO_2$

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Since one oxygen atom has actually 6 electrons,

A would correspond to a structure with a solitary bond between the oxygen atoms, 2 lone pairs on every oxygen and an unpaired electron on every oxygen; yet A does not have an octet about each oxygen, in fact, every oxygen would only have actually 7 electronsB would certainly correspond to a structure with a twin bond in between the oxygen atoms, 2 lone bag on every oxygen and no unpaired electrons on every oxygen; B does have actually an octet about each oxygen, yet it is no a biradicalC would certainly correspond come a framework with a triple bond between the oxygen atoms, 1 lone pair on every oxygen and an unpaired electron on every oxygen; however C does not have actually an octet about each oxygen, in fact, each oxygen would have actually 9 electrons and also this would certainly be difficult for oxygen

So while structure A would suggest a biradical, we wouldn"t "expect" it to count for much due to the fact that the oxygens execute not have actually octets. This inability to clearly predict the biradical nature that $ceO_2$ illustrates one of the failings of both Lewis structures and also resonance theory.

In order to effectively predict the biradical nature the $ceO_2$ we must relocate up come molecular orbital theory. Listed below is the molecular orbital diagram because that $ceO_2$. As you can see the does predict that $ceO_2$ must be a biradical v an unpaired electron in every of its degenerate, highest occupied molecule orbitals.

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Edit: an answer to OP"s comment

When ns think of triple bond ns don"t think the 2 3 electron bonds(whichis what friend drew). Fairly I think of 3 2 electron bonds(1 sigma bondand 2 pi bonds)

Structure C does stand for 3 two-electron binding (not 2 three-electron bonds), that"s just exactly how you draw the Lewis structure.

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

See more: A) How Many Lone Pairs Are On The Central Atom Of Brf3? ? Hybridization Of Brf3 (Bromine Trifluoride)

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

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

there is no formal fee on oxygen in the "triple bond" structure and as I provided above, there room 9 electrons roughly it (not 5), i beg your pardon is difficult for oxygen.

I am assuming electron are mutual equally with fifty percent around 1 atom andhalf roughly the other(which is the basis because that formal charge