Ozone has three oxygen molecules. Since all three are the same atom, there is no difference in electronegativity. As per theory, formation of a polar covalent bond is not possible but there are two resonance equivalent structures with one single and one covalent bond.
As I searched in the net, those are the only two structures repeated often. Also whenever there are more than two Lewis structures, the structure which is valid as per VSEPR theory will be taken are my understanding. So, technically even though the three interconnected single bonds are valid as in this diagram,, why doesn't the structure hold good? Is it because there are more number of close lone pair of electrons which will cause repulsion as per VSEPR theory? But, still how does the polar bond form when there is no difference in electronegativity between the oxygen atoms? I am asking this because, I am not able to get validate this question with third structure in the google search.
I am adding one more detail which I found regarding this, called 'formal charge':
The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.
Thus, we calculate formal charge as follows: formal charge = # valence shell electrons (free atom) − lone pair electrons −1/2*( bonding electrons)
I calculated the formal charge for O1, O2, O3 and all three resulted in +1. Sum of the formal charges didn't give net charge of zero. I guess, that could be one of the explanations. If anyone can confirm, it will be helpful. I will meanwhile also search. Thanks for the help!