In class, we learned that London forces become stronger as relative molecular mass increases. Not just in organic chemistry but in things like the halogens.

However, as I understand, the London forces are just temporary dipoles forming. This indicates that this would have nothing to do with the molecular mass but instead the number of electrons. My reasoning is that there will be more electrons to form the dipole. This would mean that the trend would be stronger London Forces with more electrons instead of increasing molecular mass.

How can that explain the slightly higher boiling points of heavy water (deuterium oxide ~101˚C) than normal water?

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    $\begingroup$ possible duplicate of Why do molecules having a higher $M_r$ have stronger inter-molecular forces? $\endgroup$
    – Philipp
    Dec 7, 2014 at 20:46
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    $\begingroup$ The linked question only provides an answer with respect to changing number of electrons available to the molecules. That is not the case here, where water is compared to heavy water. $\endgroup$
    – tschoppi
    Dec 8, 2014 at 0:15
  • $\begingroup$ @tschoppi Good point, I had remembered that we already had a question about the relation between mass and inter-molecular forces some time ago but I didn't read through the answer carefully again. I'll retract my close vote. $\endgroup$
    – Philipp
    Dec 8, 2014 at 13:19
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    $\begingroup$ At atmospheric pressure D2O has the higher boiling point; however, at higher pressure H2O has the higher boiling point. When the pressure is high enough that the boiling point is 220 degrees C, the boiling points are equal. $\endgroup$
    – DavePhD
    Dec 9, 2014 at 19:15
  • $\begingroup$ The fact that D2O is 10% heavier surely counts for something. $\endgroup$
    – matt_black
    Sep 15, 2020 at 14:44

3 Answers 3


The hydrogen bonds in deuterium oxide are slightly stronger than those in water. This is due to a quantum mechanical effect; the bonding interaction has a lower zero point energy due to the greater mass of the deuterium atom. It, therefore, requires more energy to excite the bonding electrons from the ground level to the dissociation point, and a higher boiling point is observed.

You can find a good explanation of the physics behind these effects on this page.

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    $\begingroup$ Due to this effect, heavy water is toxic beyond certain concentrations in the human body. $\endgroup$
    – TAR86
    Apr 18, 2017 at 18:56

I think in water and D2O the main factor is not Van Der Waals but rather hydrogen bonding. So the boiling point of D2O is higher than H2O not because of the London forces, but because of the difference in electronegativity between the hydrogen and the water, creating an electrostatic attraction.

This begs the more interesting question: why is deuterium's electronegativity different?

Just to clarify, electronegativity is the relative attraction an atom has for an electron.

Deuterium has a lower electronegativity than hydrogen, i.e it wants to give away its electron more. This is because the extra neutron increases the size of the nucleus and I think partially reduces the effect of the positive charge. Since deuterium has a lower electronegativity, there is a greater electronegativity difference between the deuterium and the oxygen, resulting in a stronger hydrogen bond.

Stronger hydrogen bond = stronger intermolecular forces = greater boiling point.

I don't think van der Waals is relevant here.

TL;DR: In this case, hydrogen bonding is more important than London forces


The above explanation is dodgy, I am still not sure to what extent it is true. There are many conflicting sources on the internet.


Heavy water is heavier, right? So it requires more kinetic energy (i.e more heat) to escape the liquid and vapourise. Hence it has a higher boiling point

  • $\begingroup$ You make a good point about the main force being hydrogen bonds. However, I don't see a reason why the neutron will affect electronegativity as the attraction of the electron to the nucleus depends on charge. Since the neutron has a neutral charge, the strength of the pull would not have increase/decrease. The neutron would have no way of reducing the charge, which means no change in electronegativity. $\endgroup$ Dec 8, 2014 at 12:04
  • $\begingroup$ @Jonathanjaya fair point, but this article ought to clear things up. The point is, deuterium's electronegativity IS different,and as you have diligently pointed out though this explanation may not be intuitive (or could be wrong), it was the best out of the others.I will edit my response. en.wikipedia.org/wiki/Kinetic_isotope_effect $\endgroup$ Dec 9, 2014 at 11:38

I think due to the greater size of deuterium atom, positive charge would be more dispersed, thus more stable. So deuterium's electron can come out with comparatively more ease, thereby reducing deuterium's electronegativity.

  • $\begingroup$ The description of the behavior you are trying to describe is not clear/wrong, I think you can afford to use some more technical language to describe the situation. $\endgroup$
    – J. Ari
    Apr 18, 2017 at 18:42
  • $\begingroup$ In D2O ,there would be more extensive hydrogen bonding due to greater electronegativity difference between deuterium and oxygen. Greater the difference more stronger will be the bonding. I just explained why electronegativity difference would be greater in my above comment @J.Ari $\endgroup$
    – Nehal Garg
    Apr 18, 2017 at 19:22
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    $\begingroup$ I think you should edit your answer to be clearer and include some of this newer information. $\endgroup$
    – J. Ari
    Apr 18, 2017 at 20:28

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