# Why does ozone have a dipole moment different than 0 [duplicate]

First of all, I know that the reason for this is the bent structure of the molecule.

However, we were taught in my class that the dipole moment of a molecule is equal to the vector sum of the dipole moment of each bond in the molecule.

But ozone only has bonds between two oxygens, so why is the dipole moment of each bond not equal to zero?

• It can't really be understood without orbitals. – Ivan Neretin Oct 3 '20 at 17:13
• @IvanNeretin Please elaborate, we have learned about orbitals (the basic stuff), and I wonder if it's outside the scope of my course. – EL_9 Oct 3 '20 at 17:14
• Well, then you must know what is $sp^2$. Also, you surely have seen the structure which looks like $\ce{O=O+-O-}$, only bent. – Ivan Neretin Oct 3 '20 at 17:30
• Because it's not just oxygen atoms. The oxygen atom in the middle has two oxygen atoms bonded to it and the ones on the end have one. Because the environment around the atoms are different, they are different. Specifically, they have different electronic properties so there is a dipole moment. – Zhe Oct 3 '20 at 19:54

The electrons are not shared equally among the oxygen atoms. The central oxygen atom alone donates three electrons to the covalent bonding whereas the other two oxygen atoms combined donate the other three (there is a 3-center, 4-electron pi bond in which two of the rlectrons are shared between just the end atoms). Since the central oxygen atom is providing an excess of bonding electrons, it becomes positively charged. The other oxygen atoms, contributing fewer electrons to the bonds and retaining more "lone pairs", take the compensating negative charge.

Trisulfur is a similarly polar elemental species, with a bent structure and unequal electron sharing similar to ozone. Sulfur dioxide and disulfur monoxide, the latter found along with trisulfur on the Jovian moon Io, have a similar unequally-shared-electron structure with sulfur in the middle. But we take little notice in the latter two cases because the sulfur, being less electronegative than oxygen, would be positively charged anyway.

However, we were taught in my class that the dipole moment of a molecule is equal to the vector sum of the dipole moment of each bond in the molecule. But Ozone only has bonds between two Oxygens, so why is the dipole moment of each bond not equal to zero?

Chemical education is in such a sorry state. Why do they teach chemistry like chicken or eggs story? Did the chicken come first or the egg? I have yet to find a single general chemistry book which even mention how the dipole moment is measured (even in one line). If you want to pursue science, instead of thinking about imaginary fairy tales, ask the teacher how would you experimentally measure the dipole moment of a gas? How do we know that a given gas has a permanent dipole moment?

The dipole moment of gases is mostly determined by microwave spectroscopy. In chemistry, experiment usually comes first (unlike chicken and egg). So if you shine microwaves on ozone, ozone would show a rotational spectrum. This is a signature that yes ozone must have a dipole moment. Now you have to consider a consistent electron distribution which would lead to a permanent dipole moment.

Similarly, electron diffraction experiment will tell you that ozone molecule is not linear. When it is not linear and it is rather bent, and yet it has a dipole moment, now people have to think of the electronic distribution.

• Quite a tangent and not wholly relevant to the question asked. In fact, I’m arguing you’re not answering the question at all. – Jan Oct 5 '20 at 2:44
• In such cases, as I always suggest, write a separate answer if you feel there is room for improvement or have a better explanation, which you have done shortly. – M. Farooq Oct 5 '20 at 2:56

It is common practice at a high school or first year undergraduate level to determine whether bonds are in any way polar (and, by extension, whether a molecule can be a dipole) just by looking at the atoms on either side of the bond. If you do that for ozone, you will obviously come across two $$\ce{O-O}$$ bonds and would have to assume that these bonds are unpolar.

A more sophisticated on-paper analysis (and one that is rarely done in a lecture setting even at higher university levels) would not only take the atoms on either side of the bond but the entire fragments. It would recognise that the $$\ce{O-O}$$ bond is, in fact, an $$\ce{O-[O2]}$$ bond, where one side has a single oxygen atom but the other side a two-atom fragment. The mere presence of a third, distant atom means that the two sides of this bond are not equivalent and thus this bond must be ever so slightly polar.

Symmetry considerations then lead to understanding that the bond on the other side has exactly the same feature; thus, these two slightly polar bonds are even in strength but not antiparallel. The overall dipole moment loosely corresponds to vector addition of the two polar bonds as you would expect; naturally, it falls right into the axis and planes of symmetry as theory expects.

• Sorry to say but this is also hand waving and not much different from what the rest have stated. – M. Farooq Oct 5 '20 at 3:00
• @M.Farooq I don’t see any other answer stating anything about how to a priori derive at the two oxygens being different without invoking additional premises (such as a positively charged central oxygen). My method even works if you assume $\ce{^.O-O-O^.}$ as a structure for ozone. But then again, you do you. – Jan Oct 5 '20 at 3:10
• Jan, The OP is basically learning about dipole moments. I think Oscar Lanzi provided the classical explanation which the OP apparently understood. – M. Farooq Oct 5 '20 at 3:23

The hybridisation of the central oxygen atom is $$\mathrm{sp^2}$$. The major contribution in the dipole moment of ozone is due to the lone pair in the directional orbital of the central oxygen atom.
The orbitals containing valence electrons of other two oxygen atoms has more directional property. They unsuccessfully tend to cancel the dipole moment created due to lone pair on central atom .

• You act as if the central atom were the negative end of the dipole. It isn't. – Oscar Lanzi Oct 4 '20 at 20:54