The question is in the title itself.

My guess: Higher for Normal water.

My Reasoning: Internet told me Deuterium is more electronegative than Protium. So, there should be less difference between the electronegativity values of $\ce{D}$ and $\ce{O}$. Consequently, the dipole moment for $\ce{D2O}$ should be lower.

Is this a valid reasoning or should I take into account factors like lesser vibration amplitude of $\ce{D}$ than $\ce{H}$ that forms more stable bonds.

  • $\begingroup$ Dipole moment= Delta × distance So the more the distance the more dipole moment. H2o have more dipole moment due to greater distance, so more vibration. $\endgroup$ Oct 18, 2022 at 12:39

1 Answer 1


There is no measurable difference.

HOH 1.8546 ± 0.0006 D

DOD 1.8558 ± 0.0021 D

Dipole moment of water from Stark measurements of H2O, HDO, and D2O J. Chem. Phys. 59, 2254 (1973)

  • $\begingroup$ they are indeed very close to each other. But if I am to predict theoretically, how would I predict that? My reasoning seems to not match witg the data $\endgroup$ Oct 23, 2016 at 12:17
  • 5
    $\begingroup$ @AmritanshSinghal you could think about bond length differences and bond angle differences, in addition to what you were already thinking about. But I think it's completely unreasonable to decide which factor predominates when there is no measurable difference. $\endgroup$
    – DavePhD
    Oct 23, 2016 at 12:24

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