We know that when it comes to C its orbitals hybridize (they become sp3 , sp2 or sp) and the energy levels of an hybridized orbital of this kind is lower than that of the p orbital that formed it but higher than the s orbital so for example energetically speaking is more convenient for an atom to have its electrons (even if they are non-bonding and they don't do too much) in an hybridized orbital if possible.And I know that O and N also can hybridize so the energy barrier between their orbitals (the s and the p orbitals) is not that high to be unable to do this so why don't they just hybridize first in O2 or N2 and then form the molecular orbitals like is the case with C-C bonds , why in the case of O-O bonds we have no hybridization? I mean ok I get the perspective of being able to form a double bond but O2 can have a double bond even if it was sp2 hybridized and then all the non-bonding electrons would be in lower energy orbitals (the other sp2).

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    $\begingroup$ Hybridization occurs in our heads, not in atoms. $\endgroup$ Commented Mar 26, 2020 at 22:24
  • $\begingroup$ Ok , but hybridization is a concept based on a theoretical framework , so what I ask is how would this problem/question be solved using this concept. $\endgroup$ Commented Mar 26, 2020 at 22:52

1 Answer 1


The molecule $O_2$ does not have a double bond O=O. It is a biradical which has a single bond O-O, and two independant electrons.

Anyway in diatomic molecules, it is not necessary to consider hybridization. It does not help. Hybridization is an approach that has been invented by Pauling to explain the geometry of polyatomic molecules. Hybridization is not really necessary. It allows to easily describe the shape of the molecules, but the same result may be obtained without hybridization, although it is much more complicated to get to the same result.


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