Why is sulfur dioxide more stable to UV radiation than ozone?
I'm going to attempt to answer this from three points of view.
Qualitative considerations and MO theory:
First, why isn't ozone linear? You can imagine $\ce{O3}$ like a $\ce{CO2}$ molecule that lacks two electrons. This lack of electrons will result in less $s$-character, and less bonding &quot;attitude&quot;. That's why ozone is bent.
Still, the repulsion in its electrons ($\ce{O}$ is a small central atom, very small) will want it to be free. Oxygen's van der Waals radius is 1.5 $\mathrm{\overset{\circ}A}$. Thus, $\ce{O3}$ has a bond length of $\approx 1.2\ \mathrm{\overset{\circ}A}$. That's really cramped. Electrons aren't happy with this and will try to get free as soon as $E_a$ and other conditions are provided.
Now compare that with sulfur's 1.8 $\mathrm{\overset{\circ}A}$ radius; which, results in a $\approx 1.43\ \mathrm{\overset{\circ}A}$ bond length. Less cramped, the electrons aren't as much complaining as they do in $\ce{O3}$'s case. We can't get really further as these two molecules aren't isoelectronic.
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Approach using classic high school knowledge:
It's because oxygen has an arbitrarily positive formal charge. Take a look at $\ce{O3}$'s electron-dot diagram:
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The oxygen atom, which has an electronegativity of $3.6$ in Allen scale, has formed $sp^2$ orbitals. This, a very electronegative atom with a positive formal charge, is highly unfavorable for the molecule, and acts very &quot;non-bonding&quot;.
The sulfur, on the other hand, has an electronegativity of $\approx 2.5$. It will also get a positive formal charge. But since it's less electronegative, it bears it.
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Thermodynamics and kinetics:
Ozone decays into diatomic oxygen under certain conditions:
$$\ce{2O3 -&gt; 3O2}$$
Three moles of oxygen will be produced per two moles of ozone. The measured enthalpy of formation for ozone is $142~ \mathrm{kJ/mol}$, while it's 0 for diatomic oxygen. So
$$\mathsf{\Delta H_f~of ~the~ products - \Delta H_f ~of ~the ~reactants} = \mathsf{reaction~ enthalpy}$$
$$ 0 - 142 = \fbox{-142} ~\mathrm{kJ/mol}$$
Furthermore, this reaction will increase entropy, as two moles of a gas produce another three moles. Thus, it's thermodynamically highly favorable.
Above that, its energy of activation is a measly $\approx 32 \mathrm{~kJ/mol}$, which is a very small kinetic barrier.
Thus, you can also conclude that $\ce{O3}$ is exceptionally &quot;inclined to decompose&quot;. Also, note that ozone is a very strong oxidant; while $\ce{SO2}$ is a weaker one.