I don't understand the chemistry here: fluorine is a halogen too.

I read that fluorine in the atmosphere readily forms HF, which is (somewhat) stable and doesn't catalytically break down ozone.

But, shouldn't loose chlorine in the atmosphere form HCl similarly?

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    $\begingroup$ Have a look at the well known chemistry of ozone depletion. If you replace the chlorine atoms in there with fluorine, you get compounds which are either more or far less stable. And HF is a very stable molecule in the gas phase. There is no liquid water in the upper stratosphere. $\endgroup$ – Karl Jan 7 at 20:03
  • $\begingroup$ chemistry.stackexchange.com/q/14230/102629 $\endgroup$ – cngzz1 Jan 7 at 23:57
  • $\begingroup$ Karl (+1): Your brief comment is actually quite accurate upon my further research. While F2 and F- do not readily lend themselves to radicalization (apparently the need for VUV light and .H presence), once activated highly active radicals are formed leading to stable compounds. $\endgroup$ – AJKOER Jan 8 at 13:02

TL;DR: The mechanism is different for fluorine and chlorine atoms. Fluorine atoms react and forms stable HF molecules while chlorine atoms turns into a radical by the action of UV which helps in the destruction of ozone.

Long answer:

Ozone depleting substance(ODS) are gases that take part in ozone depletion process. Most of the ODS are primarily chlorofluorocarbons(CFC) and halons that contain chlorine and bromine atoms respectively which reach the stratosphere and reacts with ozone and lead to ozone depletion. In this context, the chemical component responsible are chlorine radical (Cl·) and bromine radical (Br·). The ODS travel to the stratosphere without being destroyed in the troposphere due to their low reactivity and once in the stratosphere, the Cl and Br atoms are released from the parent compounds by the action of ultraviolet light, e.g.

$$\ce{CFCl3 ->[UV] Cl· + ·CFCl2}$$

Ozone is a highly reactive molecule that easily reduces to the more stable oxygen form by the catalytic action from the halogen radicals. Cl and Br atoms helps in destruction of ozone molecules through a variety of complex cycles. A simple cycle is as follows:

$$\ce{Cl· + O3 → ClO + O2}$$

$$\ce{ClO + O3 → Cl· + 2 O2}$$

Chlorine radical is released and the process is continued and the overall effect result in ozone depletion. Halon gases contains bromine and the process is similar to that of what CFCs does but the action is more potent and is more efficient in this ozone depletion process.

What about fluorine and iodine?

CFCs also contains fluorine atoms which gets released along with chlorine and bromine on photodissociation in the stratosphere. But this fluorine by itself does not contribute to ozone depletion. Fluorine atoms released are quickly sequestered into carbonyl compounds and subsequently into hydrogen fluoride by reacting with water and methane molecules which is very stable in the stratosphere. As such, CFCs are now being replaced with Hydrofluorocarbons (HFCs) according to Montreal protocol (but they do contribute to global warming).

Iodine containing organic molecules react so rapidly in the lower atmosphere that they do not reach the stratosphere in significant quantities and does not play role in ozone depletion.


  1. https://en.wikipedia.org/wiki/Ozone_depletion
  2. https://csl.noaa.gov/assessments/ozone/2010/twentyquestions/Q7.pdf
  3. https://hal.archives-ouvertes.fr/hal-00256296/document
  • $\begingroup$ Nice writeup, however, it technically despite its length (as compared to my answer, which was 1st presented with virtually the same argument) my reading notes the implicit citation of only UV light as the agent radicalizing chlorine, which technically does not impact fluorine (only VUV as my answer and sources notes in further conjunction with the hydrogen atom radical). The question relates to halogen differences. $\endgroup$ – AJKOER Jan 8 at 12:24
  • $\begingroup$ Also, you imply that F2 does not form similarly an active atomic fluorine radical, that is wrong. Read my source which, in a reaction chain, clearly cites the $\ce{.F}$ radical. As having presented a competing (actually correct) answer, I am, per professional courtesy only, not downgrading your response. $\endgroup$ – AJKOER Jan 8 at 12:42

Here is a depiction of how 'chlorine' destroys ozone per an education source:

CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle:

$\ce{.Cl + O3 → .ClO + O2 (step 1)}$

$\ce{.ClO +O. → .Cl + O2(step 2) }$

$\ce{O3 + O. → 2 O2(Overall reaction)}$

Chlorine is able to destroy so much of the ozone because it acts as a catalyst.

Note, the reaction cycle depends on the presence of chlorine as the chlorine atom radical.

However, with fluorine an apparent difference in the degree of relative stability as to quote a source work: Effect of VUV radiation and reactive hydrogen atoms on depletion of fluorine from polytetrafluoroethylene surface:

• Already only VUV radiation itself can cause depletion of fluorine.

• The strongest F depletion observed after a synergistic exposure to VUV and H atoms.

• Only H atoms without the help of VUV were not effective enough.

So, unlike chlorine and bromine, where UV radiation, itself, is apparently sufficient to induce $\ce{.Cl}$ and $\ce{.Br}$ formation, the stability of fluorine and its tendency to create stable $\ce{HF}$, for example, is likely the explanation as to why it is not usually cited in a corresponding ozone depletion cycle.


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