Ozone readily decomposes when impacted by UV radiation.

Sulfur dioxide, having the same bond structure and valence electrons in each atom as ozone, is much more stable.

Would someone please help me understand what makes sulfur dioxide a more stable compound?


2 Answers 2


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 "attitude". 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.


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:

$\hspace{45ex}$ $\small{Source}$

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 "non-bonding".

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.


Thermodynamics and kinetics:

Ozone decays into diatomic oxygen under certain conditions: $$\ce{2O3 -> 3O2}$$ 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 "inclined to decompose". Also, note that ozone is a very strong oxidant; while $\ce{SO2}$ is a weaker one.

  • $\begingroup$ Just wanted to say thanks for the quick response and thorough work on this! $\endgroup$ Nov 30, 2015 at 20:03

The bonding of ozone is explained in other questions. The important point is that its bonds are weaker than SO$_2$ even in the ground state. The dissociation energy of SO$_2$ ground state is approx 5 times larger than O$_3$. The effect of 3 vs 4 bonds to put it casually.

However, this question asks about the stability of SO$_2$ vs. O$_3$ to ultra-violet radiation. This means that the thermodynamic stability and bonding in the ground state is not important as dissociation occurs from one of the excited states. Excited states are generally less stable than the ground state as a photon promotes an electron from a bonding to an anti-bonding orbital

As it happens ozone can be dissociated from its Chappius bands (visible absorption approx 500 to 800nm) into an oxygen atom (in term symbol notation $^3P$) and an oxygen molecule ($^3\Sigma ^-_g$) with a yield of >90 %. (The products also form for absorption up to 320 nm. At shorter wavelengths O($^1D$) and O$_2$ ($^1\Delta_g$)are produced). Dissociation happens from the lowest excited state because this excite state is repulsive, i.e. has no minimum as the dissociating bond stretches. The ground state does have such a minimum which is why it is stable.

Dissociation for other excited states either occurs by pre-dissociation, where one excited state crosses another, or when the uv light is of sufficient energy that it exceeds the dissociation energy of the excite state. Clearly this is a complex subject but I hope you get the idea.

In SO$_2$ the first excite state absorbs in the 340-400 nm region (far blue) and is the result of interaction between two excited state. It has a large dissociation energy compared to ozone and fluorescence can be observed from this state. However, SO$_2$ will dissociate from this state if excited at a short enough wavelength.

The picture shows a schematic only of several potential energy profiles, energy vs bond extension (e.g. O$_2$ - O or SO-O). The units are arbitrary but typical if in nanometres and energy in wavenumbers but only for molecule with low energy excited states. The shapes of the curves are typical (Morse potential) but do not represent any particular molecule) PES

The black curve is the ground state, the grey a repulsive excited state (no minimum on energy) that causes the red curve to pre-dissociate where it crosses it. The blue curve is another higher energy excited state. You can see that excited states generally have a smaller dissociation energy than the ground state as an electron is promoted into an anti-bonding orbital


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