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I came across a question asking me the temperature at which $\ce{D2O}$ will have maximum density. I didn't really know, so I checked the answer, and I found that the answer is $\pu{11.6^oC}$. But why ? Why is the density of $\ce{D2O}$ maximum at $\pu{11.6^oC}$? I can't find the reason anywhere...

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The maximum density of $\ce{H2O}$ occurs when cooled to about $\pu{4^oC}$. At this point, the thermal motion of the molecules has slowed down enough such that water molecules can start to orient themselves in a manner more resembling that found in the expanded crystalline ice structure, and density thus starts to decrease with decreasing temperature. This pre-freezing expansion happens at a somewhat higher temperature for $\ce{D2O}$ due to the greater hydrogen bonding strength of $\ce{D2O}$ at any given temperature.

Also note that $\ce{D2O}$ freezes at about $\pu{4^oC}$, so clearly the maximum density of $\ce{D2O}$ must occur at a greater temperature than that of $\ce{H2O}$. In the case of $\ce{H2O}$, the temperature of maximum density occurs at about $\pu{4^oC}$ above it's freezing point, where for $\ce{D2O}$ the temperature of maximum density occurs at about $\pu{7^oC}$ above it's freezing point.

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    $\begingroup$ I understand, what you are trying to explain when you say 'molecular vibrations have slowed down' but this is not correct; the frequency is the same irrespective of temperature. What would be more accurate is to explain that as the freezing temperature is approached regions of structured H bonds have a chance to form without thermal motions disrupting them by too much. On balance (H bonds expansion vs. thermal motion causing contraction) the density is increased. On freezing the H bonds are now 'fixed' into a larger lattice and then occupy a larger volume than in the liquid. $\endgroup$ – porphyrin Apr 27 '17 at 8:15
  • $\begingroup$ @porphyrin , yea, I didn't have that all right. I made some changes, thx for the heads up. $\endgroup$ – airhuff Apr 27 '17 at 17:36

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