I recently saw a quote from an expert which said it takes three atoms to form a molecule. It seemed clear from the context that they were including (and in fact referring specifically to) molecules made up of two of the same atom. So to make one molecule of O2 you need three O atoms to participate in forming the molecular bond, you can't just put two O atoms in a box and make O2. Is this really true? If so, why?

  • 3
    $\begingroup$ Which expert? Where did you see this quote? $\endgroup$ – ACuriousMind Feb 27 '20 at 17:41
  • $\begingroup$ If it’s possible to break an $O_2$ molecule without a third $O$, it should be possible to make an $O_2$ molecule without a third $O$ $\endgroup$ – Superfast Jellyfish Feb 27 '20 at 18:54
  • $\begingroup$ You might get some other interesting examples/answers if you posted this in the Chemistry stackexchange. $\endgroup$ – Charlie Feb 27 '20 at 19:27
  • $\begingroup$ @FellowTraveller One can break an O$_2$ molecule by photoexcitation to an antibonding potential surface, but then the atoms leave with a lot of excess kinetic energy. That is not the initial state of the question. $\endgroup$ – user137289 Feb 27 '20 at 20:51
  • $\begingroup$ To be clear, your "expert" is saying that three are needed to FORM a molecule. Certainly, that expert is not saying that the resulting molecule must have three atoms, and by some extension, that only things with three atoms are true molecules. $\endgroup$ – Jeffrey J Weimer Feb 27 '20 at 22:10

There are different processes of molecular formation. Each of them has its own efficiency depending on the environmental conditions and the kind of molecules and reagents. A quite efficient mechanism is that of exchange reactions: a molecule $AB$ interacts with molecule $CD$ to end up with a molecule $AC$ and a molecule $BD$. In the case of diatomic molecules (like O$_2$) another efficient mechanism requires two approaching atoms and a third atom whose role is only to "extract" the excess energy with respect to the bound molecule.

Finally, there is a third mechanism, usually less efficient, but sometimes the only available, which is the radiative association, i.e. the case of two approaching atoms and the release of the excess energy through photon emission. A situation where such a mechanism dominates (du to the low density) is the molecular formation in interstellar clouds. In usual lab conditions triple collisions are more efficient and this is the probable reason for the statement you have read.

  • $\begingroup$ OK, thank you! I thought that energy conservation might be the key, but it wasn't clear to me why the extra energy couldn't simply be radiated away. Does the second mechanism you listed have a standard name? $\endgroup$ – kingskot Feb 27 '20 at 18:54
  • $\begingroup$ @kingskot Sometimes it is called three atom process. $\endgroup$ – GiorgioP Feb 27 '20 at 19:01
  • $\begingroup$ Radiative association would only be efficient for atoms of two different elements. $\endgroup$ – user137289 Feb 27 '20 at 20:45
  • 1
    $\begingroup$ @Pieter yes, of course. I have not numbers, but I would expect a very low efficiency for formation of O$_2$. $\endgroup$ – GiorgioP Feb 27 '20 at 21:34

Two oxygen atoms would just scatter off each other. Yes, there are potential surfaces with a minimum where a normal O$_2$ molecule sits in a bound state.

But two oxygen atoms have no way to get rid of their energy. If they are on an attracting potential, they will accelerate toward each other, move through the minimum, and bounce back. An elastic collision.

It needs a third to absorb energy.

  • 7
    $\begingroup$ Why can't the following process happen? 1) O and O accelerate toward each other, getting close enough that they momentarily form a highly-excited molecular state. 2) Highly-excited molecular state radiates energy as it transitions to lower states. 3) Eventually, you're left with O2 in the ground state. $\endgroup$ – probably_someone Feb 27 '20 at 18:24
  • $\begingroup$ @probably_someone The oxygen-oxygen bond is not a dipole. It does not radiate when the distance between the nuclei changes. $\endgroup$ – user137289 Feb 27 '20 at 20:42
  • $\begingroup$ Brilliant! +1 Thank you. How would your statement change for collision of N + O, a dipolar molecule that theoretically could radiate away the bond formation energy? $\endgroup$ – Jeffrey J Weimer Feb 27 '20 at 22:05
  • $\begingroup$ @Pieter Doesn't it have a quadrupole moment at least, though? $\endgroup$ – probably_someone Feb 27 '20 at 22:17
  • $\begingroup$ @JeffreyJWeimer As in Giorgio's answer, then radiative association is a possibility. And when the density is very very low as in interstellar clouds, it will be the way such molecules are formed. $\endgroup$ – user137289 Feb 27 '20 at 22:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.