The ice formed by heavy water sinks in normal water or not? Why?

assuming pure D2O ice and ignoring that natural water has a minor proportion of DHO and D2O (the proportions are small enough to ignore for a first order evaluation, besides which the density values for "water" are based on the existence of those natural proportions). the mass of D2O is roughly 20, relative the roughly 18 for natural ("normal") water (H2O). That is, D2O is about 11 % denser than H2O (the presence of the neutrons would not be expected to affect the atomic radius, which is defined primarily by the electron shell, although perhaps some shielding could occur and allow the electrons to expand outward slightly).

When water ice forms, there is an expansion which results in a lowering of density by about 8-9%. Thus, the increase in density of 11% because of the replacement of two H by two deuteriums would be expected to offset the 9% decrease resulting from the volumetric expansion occurring because of the distortion of the H-O bonds during crystallization. D2O ice would be expected to be slightly denser than H2O liquid at the same temperature.

It turns out that, in fact, D2O solid is even more dense than that simplistic examination would predict. Apparently there is also an impact on how the O-H bonds distort, so the increase in volume upon freezing is less than that expected 8-9%, and the density of D2O is actually about 10% more than that of water at about 0 degrees C. D2O ice sinks in regular water that is at about 0 degrees C (that is, about 1.1 g/cm3 versus about 1g/cm3).

Due to the difference in density, an ice cube of deuterium oxide (D2O or heavy water) will sink in regular water instead of floating.7 Aug 2013