Van der Waals radius of hydrogen deuteride

In summary: 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
  • #1
Salman2
95
0
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.
 
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  • #2
Salman2 said:
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 ?

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.
 
  • #3
hilbert2 said:
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.

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.
 
  • #4
^ 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.
 

1. What is the Van der Waals radius of hydrogen deuteride?

The Van der Waals radius of hydrogen deuteride is approximately 120 picometers (pm). This is the distance between the centers of two atoms in a hydrogen deuteride molecule when they are not bonded to each other.

2. How is the Van der Waals radius of hydrogen deuteride calculated?

The Van der Waals radius of hydrogen deuteride is calculated using the Lennard-Jones potential equation, which takes into account the attractive and repulsive forces between atoms at a certain distance. The equation is dependent on the size and mass of the atoms involved.

3. What factors can affect the Van der Waals radius of hydrogen deuteride?

The Van der Waals radius of hydrogen deuteride can be affected by the size and mass of the atoms, as well as the temperature and pressure of the environment. It can also be influenced by the presence of other molecules nearby, as their interactions can affect the distance between the atoms in a hydrogen deuteride molecule.

4. How does the Van der Waals radius of hydrogen deuteride compare to other molecules?

The Van der Waals radius of hydrogen deuteride is smaller than most other molecules, as the hydrogen and deuterium atoms are both very small. It is also smaller than the covalent radius, which is the distance between the nuclei of the hydrogen and deuterium atoms when they are bonded together.

5. Why is the Van der Waals radius of hydrogen deuteride important?

The Van der Waals radius of hydrogen deuteride is important in understanding the behavior and interactions of molecules in a gas or liquid state. It is also used in various calculations and simulations in chemistry and physics, such as in determining the strength of intermolecular forces and predicting the properties of substances.

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