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Van der Waals radius of hydrogen deuteride

  1. Jun 28, 2013 #1
    Does anyone have a reference for experimental and/or estimated Van der Waals radius for the molecule hydrogen deuteride ? Symbolically, this would be a proton {(p)e-} + {(np)e-} deuterium.

    I would like to know how closely the two atoms can approach each other to maintain the stable state of a hydrogen deuteride molecule, and the Van der Waals radius should help provide the answer..?

    ==

    EDIT: OK, is it possible there is NOT any Van der Waals interaction for HD...that the two atoms within the molecule only have a covalent sharing of the two e- to close the 1s electronic orbital for each atom ?

    ==

    Also, it is my understanding that most of the deuterium (D) in the universe exists in the form of hydrogen deuteride (HD) and not D2 molecule, would this be correct ?

    Thanks for any help provided.
     
    Last edited: Jun 28, 2013
  2. jcsd
  3. Jul 8, 2013 #2

    hilbert2

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    This is correct and follows from simple statistics. As deuterium is a minority isotope of H, it's unlikely to have a molecule with two deuterium atoms, especially as a D atom binds no more strongly with another D atom than it does with regular hydrogen.
     
  4. Jul 9, 2013 #3
    Not sure about that.
    Deuteron is more massive than proton.

    It follows that an electron in deuterium possesses a bigger reduced mass and is more strongly bound to a deuteron than to a proton.

    Not sure how it effects the strength of molecular bond, but I would be surprised if the enthalpy difference were exactly zero.

    But there is another important effect.
    A diprotium molecule possesses no dipole momentum.

    It follows that if two protium atoms encounter each other, they cannot emit photons. And since they have no way of emitting their energy, they cannot form a molecule - they can only bounce off each other.

    The same reasoning - no dipole momentum - would apply to two deuterons, even though they have more energy to dispose of.

    But HD is different. Since deuteron is more massive, it is slightly more electronegative than proton. It follows that HD possesses a small dipole moment and the ability to emit photons, which H2 and D2 both lack.

    This would enable H and D, on their encounter, to dispose of their energy by emitting photons and form HD molecules, while the encounters between two H or two D merely lead to elastic bouncing.
     
  5. Jul 9, 2013 #4

    hilbert2

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    ^ Yeah, I noticed the mistake in my reasoning, there obviously is a difference in H-D and D-D bond strengths because of the mass difference affecting the moment of inertia of the molecule. However, the equilibrium constant for the reaction D + D -> D2 would have to be hugely larger than the one for D + H -> DH reaction if there were to be any significant fraction of D2 molecules in a sample of natural hydrogen at thermodynamic equilibrium.
     
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