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

However, as I understand, the London forces are just temporary dipoles forming indicating that this would have nothing to do with the molecular mass but instead the number of electrons because 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 '14 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 '14 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 '14 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 '14 at 19:15
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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 '17 at 18:56
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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

EDIT

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

MUCH MORE INTUITIVE EXPLANATION

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

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  • $\begingroup$ 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. $\endgroup$ – Jonathanjaya Dec 8 '14 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 $\endgroup$ – surelyourejoking Dec 9 '14 at 11:38
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I think due to greater size of deuterium atom...+ve charge would be more dispersed...thus more stable ...so deuterium's electron can come out with comparatively more ease,thereby reducing deuterium's electronegativity.

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  • $\begingroup$ 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 '17 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 '17 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 '17 at 20:28

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