# What is the direction of the net electric dipole moment in hydronium H3O+?

Assuming electric dipole moment points towards the negative charge, what would its direction be in hydronium?

Although the bonding electron density is distorted towards the more electronegative oxygen, it already has a formal charge of +1. Assuming the partial charge on oxygen due to distorted shared e.density is less than O.5, wouldn’t the actual electric dipole moment points away from the oxygen(opposite to distortion)? I can’t find any source to check this

• This is why describing molecules with Lewis structures isn't adequate. The formal charge overestimate the sharing of electrons. In addition, you can even see the oxygen pulling even more right because its formal charge. The formal charge does not reflect the overall electron density, and the latter is more near the oxygen, as you probably already know. – Alchimista Mar 16 at 10:27
• @Alchimista I know the distortion of shared electron density isnt really affected by formal charges. However, I thought the formal charge on each atom in a perfectly covalent case supplemented with the distortion of shared e density provided by electronegativity difference would provide an accurate picture of the actual charges on each atom, no? – OVERWOOTCH Mar 18 at 5:33
• If you know that, that your second logical passage is not on ground. Note that the example you use can not exist. Basically formal charges are not poles of whatever component or sum dipole moment. – Alchimista Mar 18 at 10:08

In aqueous medium, the proton ($$\ce{H+}$$) is thought to exist as hydronium, $$\ce{H3O+}$$. However, in reality, probably bigger complexes of water may also be associated with one proton. In liquid water, everything is interconnected with strong hydrogen bonds anyways, and it is difficult to distinguish the free species. All of this means that it would be difficult to measure the dipole moment of $$\ce{H3O+}$$ experimentally. I haven't found any reference that did this, and I suspect it is impossible.

So we have to turn to computational chemistry to calculate the dipole moment of $$\ce{H3O+}$$. For one isolated $$\ce{H3O+}$$ (i.e. in gas phase):

The z-axis is set along the $$C_3$$ axis.

$$\text{M06-2X/aug-pcseg-2}$$ gives a dipole moment of $$\text{1.515 D}$$ in z-direction. In x and y direction, the dipole values are negligible:

         DX          DY          DZ         /D/  (DEBYE)
-0.007834    0.000072    1.515546    1.515566


This means the dipole points along +z-axis in the picture. Partial charges(ESP) fitted to the dipole moment are:

 NET CHARGES:
-------------------------------------
ATOM                CHARGE    E.S.D.
-------------------------------------
O                  -0.3668    0.0000
H                   0.4561   -0.0000
H                   0.4553   -0.0000
H                   0.4554    0.0000
-------------------------------------


As you can see, the oxygen atoms has a partial negative charge and hydrogen atoms have partial positive charges. So, the formal charges that we assign to the Lewis structure does not have much significance in this case.

(IUPAC convention of dipole moment is that the arrow points from the negative charge towards the direction of the positive charge. This is exactly opposite of what you wrote in the question)

Another method, $$\text{MP2/aug-pcseg-2}$$ gives $$\text{1.507 D}$$ along z-axis, which is close to what we got from the DFT calculation.

(Note that I have not enforced symmetry for the calculation, which is why the partial charges are slightly different for each hydrogen. This is a numerical error, because in reality all the hydrogens are equivalent)

• Careful! The calculation of the dipole moment in ions is dependent on the choice of coordinate system. Please check that the center of mass is also the center of the coordinate system. The direction should be the same as in ammonia. – Martin - マーチン Mar 16 at 23:51
• @Martin-マーチン I have symmetrized the molecule to ensure the C3 axis falls along z. Does the centre of mass actually matter? All the dipole moment is along the z-axis anyways, so moving it up and down along z would not change anything. – S R Maiti Mar 17 at 10:34
• If you shift the whole molecule along an axis, you will shift a positive charge away from the origin. I would not be certain whether the program actually does account for that, corrects that, is even able to do these kinds of translations. Also, even if you have symmetrised the molecule, it is not a given that the program will use this kind of orientation. You would have to have insight into the algorithms to know about that. It should be relatively easy to check though, but it is important to do that. || Another note: Please use MathJax only for Maths or Chemistry expressions. – Martin - マーチン Mar 17 at 22:21
• Abstractly speaking, the whole molecule is a point charge. If you calculate all vectors from the origin, those vectors will change depending on where you put the molecule. I have not used GAMESS in a long time, but I would be surprised if it didn't do coordinate transformation internally. Your output shows that you have no symmetry, your output shows that the dipole moment is not aligned with the z-axis. While there still are numerical inaccuracies, this is not one of those. You can easily check: run a single point on a shifted coordinate system. (Add 10 to all x/y/z coordinates.) – Martin - マーチン Mar 17 at 23:08
• I guess that's good news then. The topic recently came up at CCL: ccl.net/cgi-bin/ccl/message-new?2021+03+12+002 – Martin - マーチン Mar 20 at 22:53