The structure of ozone is traditionally depicted using two resonance forms:

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However, what I am wondering is instead of using such an idea to explain the structure why can't we show the structure of ozone like this?

Triangular ozone

Some people have told me that such a structure is not possible but they fail to mention the reason. What I have in mind is maybe electron repulsion. But I would like to know exactly what mechanism would prevent this from happening?


3 Answers 3


Here is the Walsh diagram depicting all the valence molecular orbitals (a diagram showing how individual molecular orbitals change in energy due to bending around the central atom). Oxygen has 6 valence electrons, so ozone has 18 electrons in total. If we start on the right where ozone would be linear, we can see that all the orbitals up to the $2\pi_\mathrm u$ orbitals (don't worry about why they are named this) are doubly occupied and the two $2\pi_\mathrm u$ orbitals are both singly occupied.

Walsh Diagram

Note: 6a$_1$ and 2b$_1$ have the incorrect sign on the outer p-orbitals.

If we bend the molecule slightly (moving to the left on the diagram), we can see that there is a favorable interaction between the p-type orbitals on the end as well as between the black of the central p-type orbital and the black of the outer p-type orbitals (Someone drew the $\mathrm{6a_1}$ and $\mathrm{2b_1}$ orbitals wrong, flip the central orbital). This lowers the energy of the molecule.

So why doesn't it keep bending? As a good first-order approximation, we can estimate the relative energy of configurations by the highest energy orbital (provided the other orbitals don't change too much). We can see that if we keep bending, the $\mathrm{1a_2}$ orbital and the $\mathrm{4b_2}$ orbital start rising in energy, eventually rising above the $\mathrm{6a_1}$ orbital. Thus too much bending will be unfavorable, and thus ozone prefers a bond angle of around $117^\circ$.

This same diagram can be used for other molecules, such as $\ce{CO2}$. Try using it to figure out why $\ce{CO2}$ is linear.


Your guess about electronic repulsion is right: the repulsion between the lone pairs of electrons on the Oxygen atoms leads to intense repulsion at such close ranges leading to instability. In comparison to this resonance is a far far more stable configuration.

Strain at the joints in the cyclic structure does its part as well because experimentally it is found to be abnormal. The strain causes instability in bond and eventually the molecule will break into constituents. The ideal angle for stability is 109 degrees and 28 minutes (I don't know why, but experiments show this is true). Angular strain can be found in water as well: to decrease that strain water collapses to an angle of 104 degrees.

Ring strain or angular strain is a type of instability that exists when bonds in a molecule form angles that are abnormal (i.e., not generally found in nature).

  • $\begingroup$ 109 is for carbon in sp3, is the o here in sp3 too? $\endgroup$
    – RE60K
    Commented Jan 2, 2015 at 16:36
  • $\begingroup$ the hybridisation of O in the cyclic structure is sp3 because it (each O) has two lone pairs and each O is bonded to two other O's which gives us Ab2e2 = 4 =sp3, Also there will be bond pair bond pair repulsions which are instability provocative $\endgroup$
    – user10153
    Commented Jan 2, 2015 at 16:38
  • $\begingroup$ You are correct but the other answer was more appropriate. $\endgroup$
    – Yashbhatt
    Commented Jan 3, 2015 at 13:27
  • $\begingroup$ @Yashbhatt, yes the other answer explains the details well $\endgroup$
    – user10153
    Commented Jan 3, 2015 at 14:18

Possible Reasons:

  • Electronic Repulsion: In order to form a 3-member ring your electron pairs would have to bend at 60 degrees; the energy required to push naturally repulsive electrons so close together would make the molecule very unstable.
  • Unstability of three membered rings (angular strain):

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