Why are diatomic oxygen molecules STILL reactive especially with metallic elements like sodium and copper even at room temperature?

You would think that since the two oxygen atoms already have the much needed 8 valence electrons when they bonded with each other that they wouldn't need to react with anything else but that doesn't seem to be the case.

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    $\begingroup$ Having a full octet for an atom is much like breathing for a man: important for sure, but it's not the only thing they need. $\endgroup$ Feb 16, 2018 at 12:49
  • $\begingroup$ If you take at loo kat the MO diagram for Oxygen, you will see that oxygen is a di-radical as we call it. It has two unpaired electrons, while the oxide (so a single oxygen atom now) would have all orbitals (s and p) filled which is more stable. $\endgroup$ Feb 16, 2018 at 16:29
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    $\begingroup$ I don't understand what you mean by "STILL reactive". "Still" suggests that you think the situation has been going on for too long ("I asked you to do X and you still haven't done it!") but that doesn't make sense here. What other scenario (even if hypothetical) are you comparing with? $\endgroup$ Feb 16, 2018 at 16:51
  • $\begingroup$ I personally think the weird case is not that O2 is reactive, but that N2 isn't. $\endgroup$
    – Joshua
    Feb 16, 2018 at 17:35
  • $\begingroup$ @Joshua it takes a lot to overcome a triple bond $\endgroup$
    – mbrig
    Feb 16, 2018 at 18:02

3 Answers 3


To estimate reactivity solely based on the the octet rule is a crude, (semi-)empirical approach in the first place. Atoms and molecules don't know anything about their number of electrons in shells, but the formation of molecules are based on the minimization of (free-)energy of the system. The octet rule gives a hint, for what kind of systems the potential energy may be small, but does not say anything about the relative energies of two systems, both fulfilling the octet rule.

To "exactly" predict the reactivity, you would have to solve the Schrödinger equation for both states of your systems, find the eigenvalue of the energies, consider entropic contributions. Furthermore you would need to "sample" all possible pathways of your reaction, to check if the energy barriers are small enough to be overcome at a given temperature. (To be really exact, you still would need to estimate entropic contributions)

Yet still a chemists' intuition often can predict reactivity very well. Concerning your specific problem, one possible qualitative explanation would be, that the octet rule is not fulfilled for copper, but only for oxygen, before the reaction happens. Taking a look on the structure of Copper(II) oxide on the other hand, shows that it fulfils the octet rule for both, copper and oxygen.(Ignoring d-electrons for copper)

Furthermore, the reaction enriches the electron density in the strongly electronegative oxygen, while reducing the electron density in the "electropositive" copper, which also stabalizes the structure of copper oxide.

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    $\begingroup$ I really can't agree with the explanation based on copper's octet being satisfied. A good model should have predictive power, and if the loss of the two 4s electrons is really thermodynamically favourable, then surely copper would react with - say - hydrogen to form $\ce{CuH2}$, or nitrogen to form $\ce{Cu3N2}$. $\endgroup$
    – orthocresol
    Feb 17, 2018 at 0:38
  • $\begingroup$ The octet rule has some predictive power. But basically it is as i said: u have to calculate the relative energies of concurent systems. The octet rule is widely applicable in organic chemistry, where u will (hardly) find themodynamic stable compounds, in which the octet rule is not satisfied. But on the other hand, not every compound, u can write down on paper satisfying the octet rule, will be easy to synthesize or stable in the presence of oxygen. $\endgroup$
    – vk_s
    Feb 17, 2018 at 8:26

The octet-rule is a very inadequate rule for understanding overall reactivity.

Ultimately you need to understand two things to know whether reactivity is likely: the thermodynamics of any possible reaction; and the kinetics of any possible reaction mechanism for the reaction.

The formation of many oxides is thermodynamically favourable. That's why many compounds burn in an atmosphere containing oxygen. The formation of sodium oxide, carbon dioxide and many metal oxides is thermodynamically favourable. Carbon-containg things will burn in air, metals will often react (quickly, like sodium, or slowly, like iron). Some will burn spontaneously like caesium; others will take a little encouragement.

The second factor that matters is whether there is an easy way for the oxidation reaction to happen. Aluminium, for example, is very reactive but rapidly forms a protective layer of strong aluminium oxide (alumina) on the surface preventing further reaction; iron is relatively reactive but won't rust easily unless contaminants are present; gold won't react at all. Carbon needs to be heated to start the reaction but will burn by itself when it does start.

Where reactions have easy to access mechanisms and some thermodynamic advantage, oxygen is very reactive.

Another thing to remember about oxygen is that it has some reaction mechanisms that are less than obvious and easier than expected for a diatomic molecule. Unlike nitrogen, for example, where thermodynamically favourable reactions are inhibited by its strong triple bond (and filled octet) an oxygen molecule is actually a diradical: despite the apparent filled octets you would expect from simple electron counting, the molecule has two unpaired electrons (this requires a bit of molecular orbital theory to explain and is one of the observations that simple electron octet counting doesn't explain well). Radicals tend to be more reactive than paired electron orbitals (or filled octets).

In short, an octet-counting view is too simplistic to explain oxygen's reactivity and you need more sophisticated ways of looking at the electronic structure. And don't forget that you also need to know the thermodynamics and the kinetics of potential reaction mechanisms to get any useful predictions of whether reactivity is likely.


Oxygen (O2) generally exists as diradicals i.e. each oxygen bonded to each other through single bonds and the remaining two electrons remains on each oxygen atoms as radicals. So this structural feature makes oxygen act as a strong oxidizing agent. Even in our body oxygen coordinates through its radical and this structural arrangement makes oxygen can act as a strong oxidants in many places.

  • $\begingroup$ Saying that O2 is a diradical rationalises the fact that it is reactive, but that doesn't make the statement a true fact!! $\endgroup$
    – Karl
    Apr 12, 2018 at 18:04

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