Hydrogen Versus Deuterium Bonds
Name: A. K.
Date: Spring 2012
If you have a compound, say HCl, why does the bond length decrease when you substitute the hydrogen atom with a deuterium atom?
Hi A. K.
Think of the bond in a diatomic molecule (H2 versus D2) as a spring between two spheres (the atoms). Let's say the available energy to make the spring vibrate is constant (you supply a specific amount of energy). The H2 spring will vibrate more because the masses being held by the spring is lighter than that of the D2 spring (remember that D is twice the mass of H). In effect the H2 system can extend farther away from the center than the D2 system. Since the closest approach of the H2 and D2 systems are the same (the same sized spheres touching each other) and the H2 system can get farther out than the D2, we can see that the average distance between the two H's is larger than that of the two D's.
Greg (Roberto Gregorius)
The bond lengths of HCL and DCL are the same to a high degree of accuracy and precision (1.27455 and 1.27458 Angstroms, respectively) -- taken from the CRC Handbook of Chemistry and Physics. When determining the molecular constants of a molecule it is not only important to make careful experimental measurements with high precision instrumentation. It is also important to analyze the experimental data with a highly accurate theoretical analysis. In many cases the data do not warrant that kind of analysis. But that is not the case in HCL and DCL, where the data are very accurate and precise.
The “correct” analysis, seldom seen in elementary texts, was derived by J. L. Dunham, Physical Review, Vol. 41, pg. 721 ff., (1932). What you will see in that paper is that the theoretical analysis is rather complicated.
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Update: June 2012