I have read about the formation of natural atmospheric ozone on Wikipedia, where it is claiming that the ozone creation step requires an extra molecule in order to conserve momentum and energy:
$$\ce{O + O2 + A -> O3 + A}$$

I understand that the stated reason implies that it is impossible to have $$\ce{O + O2 -> O3},$$ but I do not understand why it cannot be:
$$\ce{O2 + O2 -> O3 + O}.$$

Specifically, why does the source claim that we have complete cleavage of the $\ce{O2}$ double-bond before the $\ce{O^{..}}$ radical attacks an $\ce{O2}$ molecule? Why is it not possible that after one bond is cleaved we have an $\ce{^.O−O^.}$ radical reacts with an $\ce{O2}$ molecule directly to form $\ce{O3}$ and an $\ce{O^{..}}$ radical? In particular:
$$\ce{O=O + ^.O−O^. -> O=O^+−O^- + O^{..}}$$

This not only solves the momentum+energy conservation problem, but also seems simpler. The other proposal requires 3 molecules to meet at the same time and does not really explain how the momentum/energy gets transferred.

  • $\begingroup$ Hmm I just thought about another possibility: Maybe ( O + O2 → O3 ) can conserve momentum and energy, by having the resulting molecule spin, but I'm not too sure.. $\endgroup$
    – user21820
    Commented Nov 8, 2023 at 8:00
  • $\begingroup$ It would be removing a wedge by a wedge, as there are 3 fundamental conservation laws (energy/mass, momentum, angular momentum). Generally, 2 molecules forming one need some inert object to pass energy there, respecting all these laws. $\endgroup$
    – Poutnik
    Commented Nov 8, 2023 at 8:15
  • $\begingroup$ @Poutnik: Why can't the bond lengths in a spinning molecule compensate for the extra energy? Anyway, that's a separate question from the one I'm asking haha.. $\endgroup$
    – user21820
    Commented Nov 8, 2023 at 8:17
  • $\begingroup$ Resulting O3 cannot pass extra energy to its spinning, it would break angular momentum conservation. // BTW, the molecule O2 is at the same time double bonded and biradical (counter intuitive state explained by Molecular Orbitals theory and manifested by oxygen paramagnetism. Substances with molecules without unpaired electrons are diamagnetic. $\endgroup$
    – Poutnik
    Commented Nov 8, 2023 at 8:21
  • $\begingroup$ @Poutnik: I think you missed my point: If the bond length is changed, the energy it stores is also changed. So we have an extra degree of freedom to compensate for the extra energy, no? By the way, I am aware that the Lewis structures are merely approximates, and MO theory is more accurate. But can it answer my question? $\endgroup$
    – user21820
    Commented Nov 8, 2023 at 9:05

2 Answers 2


The $\ce{O + O2 + M \to O3 + M }$ does appear written as such in many texts and is unusual in that other atom-diatom reactions that have been studied do not seem to need any $M$, such as $\ce{OH + Cl \to HCL +O}$, $\ce{ H +NO2 \to OH +NO}$, $\;\ce{F + H2 \to HF + H}$, $\;\ce{O +HCl\to OH + Cl}$ and many others. How a reaction proceeds depends both on the shape of the potential surface between the four species involved as the reaction evolves and on how much rotational and vibrational energy the diatomic has and the mutual approach velocity.

There could be two reasons for writing the reaction with $M$ given that a three body collision reaction is very unlikely to happen on statistical grounds and that is either that the nascent ozone has so much energy that it needs a collision with a third body to remove this energy to stabilise it, or that a transient complex is initially formed such as $\ce{O + M \to OM}$ or $\ce{O2 + M\to O2M}$ which then reacts with the remaining species to give products. The $\ce{I + I \to I2}$ reaction proceeds via an $IM$ intermediate, for example.

(Any linear momentum considerations can be accounted for by different velocities after reaction and changes in rotational energy levels for angular momentum conservation. The $\ce{O3 + O \to O2 + O2 }$ reaction has quite a large activation energy $\sim 18$ kJ/mol so the reverse reaction probably has a larger one making it even slower that this one).

Update in response to questions

The energy the two species have for reaction is relative to their difference in velocity, this may be quite small relative to any dissociation or activation energy. At $400$ K the average thermal energy is only $\approx 265\;\mathrm{cm^{-1}}$. The ozone and oxygen vibrational (B=rotational) (all frequencies are in wavenumbers) O2, 2063 (B=1.4) and O3, 1110,705,1042 (B=3.5,0.4,0.39) so the molecules are vibrationally 'cool' and rotationally 'hot'. The O3 dissociates at $\approx$ 11700, i.e. a huge energy relative to thermal energy, but not to UV photon energy.

When then O+O2 species approach (or any A+B species) they experience one another's potential before collision (they are not hard spheres) this potential is, approximately, of a Lennard-Jones type so the species are attracted initially and move towards one another at a distance many times typical bond length. At close range repulsion can occur. A reaction may occur provided the energy is not too great or too small and the approach trajectory correct or the O and O2 will partially orbit and then separate. (see figure from https://chemistry-maths-book.com/chapter-11/). Typically this last event is the most likely.

If reaction can happen the transient species $\ce{OO2}$ forms. The presence of the extra O atom will weaken the O2 bond and energy is transferred into new vibrations (this will take only a few picoseconds) and reaction may be complete, most of the time however the transient species falls apart with a bit of energy having being transferred between the departing species. If an O3 molecule is formed it has all the energy of the collision but this is small compared to dissociation energy and so the molecule will be stable but initially be vibrationally and rotationally 'hot'. This will remain so until a non-reactive collision with an other species removes some energy and thermalises the new O3 molecule. How long this takes depends on the pressure, but at experimental pressures of a few torr, will occur within tens of nanoseconds


The impact parameter is the distance away from the line of centres. This diagram is for spheres, but should give the general idea even for a slightly no spherical molecule such as O2.

  • $\begingroup$ So you are claiming that with ( O + M → OM ), the extra energy can somehow be absorbed by the complex formation? I don't get it at all. We still have too much extra energy. (Take the reference frame centred at the centre of mass of the ( O + M ), and after complex formation it has 0 velocity...) $\endgroup$
    – user21820
    Commented Nov 8, 2023 at 15:40
  • $\begingroup$ The essential is one versus two product entities. The former needs M, the latter does not. All other examples have two products. 2 H -> H2 needs some M too. $\endgroup$
    – Poutnik
    Commented Nov 8, 2023 at 15:41
  • $\begingroup$ And by the way none of the reactions you mentioned in your first paragraph has the conservation problem, since there are enough products to balance the equations. $\endgroup$
    – user21820
    Commented Nov 8, 2023 at 15:41
  • 1
    $\begingroup$ @user21820 M is supposed to be inert, being needed for mechanical reasons to honor conservation laws. It is not some catalyst in chemical sense. It serves just as a ball to be pushed. $\endgroup$
    – Poutnik
    Commented Nov 8, 2023 at 15:43
  • 1
    $\begingroup$ @user21820, the OM complex can form quite easily, even if M is an atom OM has rotational and vibrational energy as well as translational kinetic energy. Momentum (a vector) will also be conserved. Experiment shows that molecules are more effective than atoms in forming a complex, more vibrational modes and more polarisability for electronic interaction . Such a (short lived) complex is the only possible explanation for the $\ce{I + I}$ reaction as observed by experiment, i.e. experiment shows that $\ce{I + M =IM, IM + I\to I2 + M}$ is the mechanism. (Porter Proc. Roy. Soc. v261, p28, 1961) $\endgroup$
    – porphyrin
    Commented Nov 8, 2023 at 19:08

Evaluating $\ce{2 O2 -> O3 + O}$ singlet oxygen hypothesis:

  • The molecule $\ce{O2}$ – also called triplet oxygen – has in its standard state a double bond, but it already is a biradical CH SE: why-is-o2-a-biradical. It is paramagnetic and has 2 unpaired electrons.
  • The next state with higher energy is the singlet oxygen, is paramagnetic and fits the classical idea about double-bonded biatomic molecule. It is formed by specific chemical reactions and is very reactive.
  • The opposite reaction $\ce{O3 + O -> 2 O2}$ is fast.
  • Singlet oxygen is not $\ce{\cdot O-O\cdot}$, but rather $\ce{O=O}$, while the triple oxygen is like $\ce{O÷O}$.
  • Single oxygen in upper atmosphere is not formed from triplet oxygen, but by photolysis of ozone.
  • Molecular entities can emit a photon, but it is a process much slower than the resonance break.
  • $\ce{O2->[UV]2 O}$ is supported by the spectroscopic evidence of oxygen UV spectrum, when absorption below about $\pu{240 nm}$ confirms the breaking of oxygen molecules to atoms, as the needed energy is equivalent to $\pu{241 nm}$. There are not shown any local maxima at higher wavelengths, suggesting some lower energy photon excited oxygen molecule.

oxygen UV spectrum from researchgate.net

The third object mechanism:

Reminding the known:
In reference frame of zero total momentum, without $\ce{M}$, resulting $\ce{O3}$ from assumed $\ce{O2}$ + $\ce{O}$ collision has zero momentum and zero translational energy. Due angular momentum conservation, it has comparable rotational energy as original molecule $\ce{O2}$. Plus lower potential energy due bonding. Something does not fit. It is like when a skateboarder runs through an U-ramp and ends up on the other side free. Molecules do one vibrational period, bounce and part away.

If there is any inert entity $\ce{M}$, both $\ce{M}$ and $\ce{O3}$ are parting with the same but opposite sufficient momentum, according to conversion of energy released by bonding to translational energy. Like we when a skateboarder hits something and does not make it to the other side, being "bonded to the U ramp".

When there are two products, one plays simultaneously the role of $\ce{M}$, like in case $\ce{NO + O3 -> NO2 + O2}$.

Involving $\ce{M}$ does not mean that collisions of $\ce{O}$, $\ce{O2}$ and $\ce{M}$ occur at the same time point. It means one collides when the other two are still attached during resonance meeting before parting.

Resonance can be understood by 2 hydrogen atoms, approaching each other. They reach maximum kinetic and minimum potential energy at the distance equal nuclei distance of a hydrogen molecule. But the motion continues, kinetic energy decreases to zero, reaching maximum potential energy. Then the motion continues like if reversed in time. In fact, in is one vibrational period on the maximum energy level with one side amplitude in infinity (and then some).

Without existence of resonance, there would be no carbon in universe and no life based on carbon compounds. In fact there are even 2 resonances in row:
$\ce{2 ^4He <<=> ^8Be^{*}<<=>[+^4He]^{12}C^{*}->^{12}C + \gamma }$. These resonances have many orders shorter life than resonances of molecular entities. See Beryllium-8 and Triple-alpha_process.

As a curiosity, see Atomic hydrogen welding.

  • $\begingroup$ I seems that you did not read what I wrote in my comments objecting to your claim. I did not ask about M and O3. I asked about M and O, which you had claimed could combine into a complex. I stated that it faces the same issue, ( M + O ) has positive energy but zero momentum after combining, so all that energy must go somewhere. You must explain why ( M + O → MO ) is significantly different from ( O + O2 → O3 ), otherwise you have explained nothing at all. $\endgroup$
    – user21820
    Commented Nov 10, 2023 at 3:34
  • $\begingroup$ I did not claim M and O form complex. I need not to explain why something happens if I think it does not happen. IIRC it was Porphyrin's idea. BTW, there is also the difference forming product and forming a resonance that falls apart unless passes energy to other object . Like $\ce{2 H <<=> H2^{*}->[M]H2}$ $\endgroup$
    – Poutnik
    Commented Nov 10, 2023 at 5:25
  • $\begingroup$ M forms a resonance with O or O2, that falls apart, unless encounters O2 or O, forming O3 which together with M part with momentum and kinetic energy from the reaction. Direct O and O2 form a resonance O3, that falls apart, unless colliding with M, passing it energy and then not falling apart. $\endgroup$
    – Poutnik
    Commented Nov 10, 2023 at 5:36
  • $\begingroup$ Please re-read; I asked clearly about ( O + M ) and you answered incorrectly. Since you clearly misread, your answer here is not useful since it does not answer my question and also does not address my objection to the "OM complex formation" idea. Your last comment (claiming resonance rather than complex formation) may be viable but still does not answer my question (i.e. why not O2 + O2 → O3 + O). $\endgroup$
    – user21820
    Commented Nov 10, 2023 at 6:08
  • 3
    $\begingroup$ Do not you think you have slided to a confrontational instead of cooperative approach? / Try to put more thinking into the topic. $\endgroup$
    – Poutnik
    Commented Nov 10, 2023 at 7:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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