Why is the boiling point of heavy water more than that of normal water?

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? 




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](http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hosc.html).


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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


 1

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.


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