As far as I know, Lewis Structures are convenient for representing molecules, but give no insight on the actual electron orbitals.

For example: The Lewis Structure of $\ce{O_2}$ does not reflect paramagnetism. Therefore, we use MO theory to explain it.

Then, why are Lewis Structures used to show if a molecule is a radical?

For example: hydroxyl is drawn as $\ce{.OH}$ to reflect the unpaired electron. That is a Lewis Structure.

Why do Lewis Structures work in this case?

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    $\begingroup$ Look at the history of atomic models. $\endgroup$ May 4, 2016 at 5:19

1 Answer 1


Lewis structures are excellent for showing radical character for an odd number of electrons.

The reason they fail for molecular oxygen is that they Lewis representation carries no intrinsic information about the molecular spin, and as such it cannot distinguish the unpaired electrons in molecular oxygen's antibonding orbitals as spin matched (triplet) or spin opposed (singlet).

However, when there are an odd number of electrons, it is clear there will be one unpaired spin, and it is not relevant whether what the spin orientation is for these cases.

That being said, and somewhat outside the scope of this question, Lewis structures cannot distinguish between doublet and quartet states.

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    $\begingroup$ Once you've counted your electrons and ended up with an odd number, you don't need Lewis structures (or anything else, for that matter) to conclude that you have a radical on your hands. $\endgroup$ May 4, 2016 at 7:50

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