Ozone Layer and CFCs Revision Notes | A Level AQA Chemistry

1. Why the Ozone Layer Matters

Picture the Earth wrapped in an invisible chemical shield, roughly 15–35 km above the surface in the stratosphere. That shield is the ozone layer — and without it, the ultraviolet (UV) radiation streaming from the Sun would reach the ground at intensities capable of causing DNA damage, skin cancer, and ecosystem disruption.

Ozone, $O_3$, is a molecule made of three oxygen atoms. It forms naturally in the stratosphere when high-energy UV radiation splits ordinary oxygen molecules ($O_2$) into individual oxygen atoms, which then collide with other $O_2$ molecules. The key point for AQA is what ozone does once it exists: it absorbs ultraviolet radiation, converting that energy into heat rather than allowing it to pass through to the Earth's surface.

The absorption process works because UV photons carry exactly the right energy to break the bonds within $O_3$. When an ozone molecule absorbs a UV photon, it dissociates — and in doing so, it removes that UV energy from the radiation reaching the lower atmosphere. This is the mechanism by which the ozone layer acts as a protective filter.

2. CFCs — Stable Below, Destructive Above

Chlorofluorocarbons (CFCs) were once considered ideal industrial chemicals. They are non-toxic, non-flammable, and chemically unreactive at ground level — properties that made them popular as refrigerants, aerosol propellants, and foam-blowing agents throughout the mid-twentieth century. The problem is precisely what makes them so stable at ground level: they do not break down in the lower atmosphere, so they drift upward into the stratosphere intact.

A typical CFC such as trichlorofluoromethane ($CCl_3F$, also known as CFC-11) contains C–Cl bonds. At ground level, there is insufficient energy to break these bonds. In the stratosphere, however, Higher-energy, shorter-wavelength UV radiation (especially UV-C) is present in the stratosphere before being absorbed, so it can break C-Cl bonds in CFCs and generate chlorine radicals. — and UV photons carry enough energy to cause the C–Cl bond to break.

This bond-breaking process is called homolytic fission. Unlike heterolytic fission (where both electrons go to one atom), homolytic fission splits the bonding electron pair equally — one electron to each atom. The result is two species, each carrying an unpaired electron. These are called radicals, and they are extremely reactive.

The equation for this process can be written as:

Notice that the dot notation ($Cl\bullet$) is used to indicate the unpaired electron on the radical. AQA mark schemes expect this notation — omitting the dot is a common way to lose marks in mechanism questions.

3. Why C–Cl and Not C–F?

CFCs contain both C–Cl and C–F bonds, so a reasonable question is: why does the C–Cl bond break preferentially? The answer lies in bond enthalpy. The C–F bond has a mean bond enthalpy of approximately 485 kJ mol$^{-1}$, whereas the C–Cl bond has a mean bond enthalpy of approximately 339 kJ mol$^{-1}$. The C–Cl bond requires less energy to break, so UV photons of the wavelengths present in the stratosphere are sufficient to cause its homolytic fission, but not sufficient to break the stronger C–F bond.

This bond enthalpy comparison also explains why hydrofluorocarbons (HFCs), which contain only C–F and C–H bonds but no C–Cl bonds, do not produce ozone-depleting chlorine radicals — though they do have other environmental concerns as greenhouse gases.

4. The Chain Reaction: How One Chlorine Radical Destroys Many Ozone Molecules

Here is where the environmental impact becomes alarming. A single chlorine radical does not destroy just one ozone molecule — it acts as a catalyst in a chain reaction, regenerating itself after each propagation cycle and destroying an estimated tens of thousands of ozone molecules before it is eventually removed from the stratosphere.

Worked example: The Ozone Depletion Chain Reaction

Describe the chain reaction by which a chlorine radical catalyses the destruction of ozone in the stratosphere.

1. $$CCl_3F \xrightarrow{UV} \cdot CCl_2F + Cl\cdot$$ — UV radiation causes homolytic fission of the C–Cl bond in a CFC, releasing a chlorine radical ($Cl\bullet$) into the stratosphere. This is the initiation step.
2. $$Cl\cdot + O_3 \rightarrow ClO\cdot + O_2$$ — The chlorine radical reacts with an ozone molecule ($O_3$), breaking it down into an oxygen molecule ($O_2$) and a chlorine monoxide radical ($ClO\bullet$). One ozone molecule is destroyed.
3. $$ClO\cdot + O \rightarrow Cl\cdot + O_2$$ — The $ClO\bullet$ radical reacts with a free oxygen atom (also present in the stratosphere), regenerating the chlorine radical and producing another $O_2$ molecule. The chlorine radical is now free to attack another ozone molecule.
4. $$O_3 + O \rightarrow 2O_2$$ — Adding the two propagation steps together and cancelling $Cl\bullet$ (which appears on both sides) gives the overall equation. Ozone is converted to ordinary oxygen, and the chlorine radical is not consumed — it acts as a catalyst.

Answer: Each chlorine radical catalyses the destruction of ozone via a two-step propagation cycle. The $Cl\bullet$ is regenerated at the end of each cycle, allowing a single radical to destroy thousands of $O_3$ molecules before termination occurs.

The catalytic nature of this process is what makes CFC emissions so damaging. Even small quantities of CFCs released into the atmosphere can generate enough chlorine radicals to cause significant, long-term depletion of the ozone layer. This is why international agreements such as the Montreal Protocol (1987) phased out CFC production globally.

5. Exam Technique and Common Mistakes

AQA questions on this topic tend to follow predictable patterns. You will be asked to either describe how chlorine radicals form, explain why the ozone layer is beneficial, or evaluate the environmental impact of CFCs. Each command word demands a different response structure — and confusing them is one of the most common ways students drop marks.

For a describe question about radical formation, focus on the sequence of events: UV radiation is absorbed by the CFC in the stratosphere; the C–Cl bond undergoes homolytic fission; each atom receives one electron from the bonding pair; a chlorine radical ($Cl\bullet$) is produced. Do not add 'because' — describe questions want the what, not the why.

For an explain question, you must include the scientific reasoning. For example: 'The C–Cl bond undergoes homolytic fission because UV radiation in the stratosphere provides sufficient energy to break this bond, which has a lower bond enthalpy than the C–F bond.' The word 'because' is your signal that you are explaining, not describing.

6. Bringing It Together — Spec Point Summary

Both AQA specification requirements for this topic connect through a single narrative: ozone is beneficial because it absorbs UV radiation, and CFCs undermine this protection by releasing chlorine radicals via UV-induced homolytic fission of C–Cl bonds in the stratosphere. Every exam question on this topic is testing one or both of these ideas.

• Ozone ($O_3$) in the upper atmosphere absorbs ultraviolet radiation, acting as a protective filter for the Earth's surface.
• CFCs are stable at ground level but reach the stratosphere intact, where UV radiation is intense enough to break C–Cl bonds.
• Homolytic fission of the C–Cl bond produces a chlorine radical ($Cl\bullet$) — each bonding atom receives one electron from the shared pair.
• The C–Cl bond breaks preferentially because its bond enthalpy (~339 kJ mol$^{-1}$) is lower than that of the C–F bond (~485 kJ mol$^{-1}$).
• Chlorine radicals catalyse ozone destruction via a chain reaction, regenerating $Cl\bullet$ after each cycle — one radical can destroy thousands of ozone molecules.
• The overall effect is conversion of $O_3$ to $O_2$, reducing the stratosphere's capacity to absorb UV radiation.

The ozone layer and CFCs topic is a focused but high-value area of the halogenoalkanes unit. Mastering the distinction between homolytic and heterolytic fission, using correct radical notation, and linking bond enthalpy to selectivity will position you to answer any AQA question on this topic with precision and confidence.