Is O2 A Double Bond

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

Since an oxygen cantlet 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 non accept an octet effectually each oxygen, in fact, each oxygen would only accept vii electrons
• B would correspond to a structure with a double bond between the oxygen atoms, ii lonely pairs on each oxygen and no unpaired electrons on each oxygen; B does accept an octet around each oxygen, only it is not a biradical
• C would correspond to a structure with a triple bond betwixt the oxygen atoms, ane lonely pair on each oxygen and an unpaired electron on each oxygen; however C does not have an octet effectually 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 "look" it to count for much since the oxygens practice not have octets. This disability to clearly predict the biradical nature of $\ce{O_2}$ illustrates one of the failings of both Lewis structures and resonance theory.

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

Edit: response to OP'south comment

When I retrieve of triple bail I don't think of ii 3 electron bonds(which is what you lot drew). Rather I call up of three ii electron bonds(ane sigma bond and 2 pi bonds)

Structure C does represent iii two-electron bonds (non 2 three-electron bonds), that'due south merely how y'all 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 = half-dozen - 3 unshared - (ane/2 * half-dozen shared)= 0,

there is no formal accuse on oxygen in the "triple bond" structure and as I noted to a higher place, there are 9 electrons effectually it (not 5), which is incommunicable for oxygen.

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

Yes, that's correct.

Is O2 A Double Bond,

Source: https://chemistry.stackexchange.com/questions/15058/why-is-o2-a-biradical

Posted by: andersonwition70.blogspot.com

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