I On the necessary conditions to form the hydrogen molecule

[DaTario]

TL;DR Summary

Is it true that it is needed a third body in order to form the $H_2$ molecule from two hydrogen atoms?

Hi All,

Is it true that in order to form the $H_2$ molecule from two hydrogen atoms one needs a third body so that the energy difference between initial and final states can be released?

Best Regards,

DaTario

 

[Drakkith]

I don't think so. The energy should be able to be radiated away.

 

[DaTario]

I don't think so. The energy should be able to be radiated away.

I have heard that this impossibility (two hydrogen atoms, both in 1s state, colliding in isolation and forming $H_2$) has to do with the symmetry of these states. It sounded to me as it was forbiden by some selection rule. A quadrupole transition could be a way but it is slow compared with the small interval of time the elastic collision takes.

 

[Drakkith]

Ah, well, the details are above my knowledge level, so I cannot say anything with certainty.

 

[Vanadium 50]

I have heard that

Where?

You also seem to be making two different arguments, perhaps three: the excess energy cannot be radiated away (or cannot be radiated away quickly enough) or that it is blocked by some symmetry principle. Please show us where you are getting this so we can untangle it.

 

[Baluncore]

2⋅H⁺ + 2⋅e^- → H₂
Where do the two electrons come from, that are then shared by the two hydrogens, to form their 1S² covalent bond?
Is the reaction catalysed by a metal?

 

[DaTario]

Where?

You also seem to be making two different arguments, perhaps three: the excess energy cannot be radiated away (or cannot be radiated away quickly enough) or that it is blocked by some symmetry principle. Please show us where you are getting this so we can untangle it.

Hi, Vanadium 50, the source is the webinar



having the title: "The gymnastics of two hydrogens to form H2 in rarefied media".
The specific discussion I am addressing here you will find from time 4:25 to 5:37. The webinar is in Portuguese, but the translation seems to be available to English. Professor Eduardo Montenegro refers to some papers in the slide that corresponds to this discussion. I had no access to them.

 

[DaTario]

2⋅H⁺ + 2⋅e^- → H₂
Where do the two electrons come from, that are then shared by the two hydrogens, to form their 1S² covalent bond?
Is the reaction catalysed by a metal?

There seems to exist two alternative routes to the formation of $H_2$, as shown below (slide of prof. Montenegro)

 

[Baluncore]

"The gymnastics of two hydrogens to form H2 in rarefied media".

I did not see any mention of "rarefied media" in the OP.

 

[Vanadium 50]

It appears that the OP's point (and I am not going to sit through a lecture in a language I don't understand in order to figure it out) is that if I have an initial state of two free atoms, and a final state of one bound molecule and nothing else, the reaction does not go.

Of course.

If you don't conserve energy, the reaction doesn't happen.

This is the same as saying I can't drop a baseball if it can't gain kinetic energy. True. But seldom helpful.

The system will radiate (I believe as an M1).

 

[DaTario]

I did not see any mention of "rarefied media" in the OP.

Hi, Baluncore,

when I mentioned the need of a third body I was implicitly saying that the two hydrogen atoms were isolated.

 

[DaTario]

It appears that the OP's point (and I am not going to sit through a lecture in a language I don't understand in order to figure it out) is that if I have an initial state of two free atoms, and a final state of one bound molecule and nothing else, the reaction does not go.

Of course.

If you don't conserve energy, the reaction doesn't happen.

This is the same as saying I can't drop a baseball if it can't gain kinetic energy. True. But seldom helpful.

The system will radiate (I believe as an M1).

But the role played by the symmetry, specially related to the condition of 'homonuclear diatomic molecule', is stressed by the lecturer as essential in forbiding the process of molecule formation. Do you agree with this?

 

[DaTario]

I used to believe that when two $H$ atoms approach in a collision such that their wave functions interfere constructively, a sigma bonding orbital is created and the $H_2$ molecule is likely to be formed. Now the symmetry seems to appear as an obstacle to the formation of the molecule in such an isolated scenario.

 

[Vanadium 50]

I'm afraid that learning Portuguese to critique a video is a bit more than I am willing to do.

 

[dextercioby]

I want to understand, as well. Perhaps what they tell people in high-school chemistry is wrong. One tells that at regular temperature and pressure, there is no atomic Hydrogen. Any atom will search another one to form a steady molecule. So which part of that is wrong? I didn't study inorganic or quantum chemistry in college and my QM curriculum had no hydrogen molecule (4-body problem for SE) and neither the hydrogen molecule with a missing electron (3-body problem for SE).

 

[DaTario]

May be this is a question to those who study Comp Chem. Thank you all anyway and sorry for not having brought suitable references.

 

[DaTario]

I'm afraid that learning Portuguese to critique a video is a bit more than I am willing to do.

No problem, Vanadium 50. Thank you anyway.

 

[biffus22]

I want to understand, as well. Perhaps what they tell people in high-school chemistry is wrong. One tells that at regular temperature and pressure, there is no atomic Hydrogen. Any atom will search another one to form a steady molecule. So which part of that is wrong? I didn't study inorganic or quantum chemistry in college and my QM curriculum had no hydrogen molecule (4-body problem for SE) and neither the hydrogen molecule with a missing electron (3-body problem for SE).

"Any atom will search another one to form a steady molecule." A rather misleading analogy, I think. It implies that somehow the most primitive, still ionized elements in the universe are forever actively "seeking out" what they need to transform themselves into pre-selected more stable configurations. In fact such "hook-ups" are the result of random encounters.

One should also notice that, at least in the case where more complex molecules are already around, the original "naked" hydrogen atom might simply first "collide" with something else, like, say, a carbon, oxygen, silicon, fluorite, chlorine, or sodium atom isotope, etc., and bond with it instead of waiting for another naked hydrogen molecule to show up.

In short, although they can sometimes be wary and dismissive of the many new "suitors" pursuing them, they generally tend to accept the "offer" of the first coutier they encounter.

 

[Vanadium 50]

Any atom will search another one to form a steady molecule. So which part of that is wrong?

I don't like this analogy either. Chemistry is not Tinder.

bond with it instead of waiting for another naked hydrogen molecule

Naked? And we're back to Tinder.

The auto-translate, as has been said before, says that binding requires losing energy. This is not news, and quibbling about that and words like "isolation" seems unhelpful. That's why I think we need to go back to the original Portuguese.

There seems to be a suggestion that this reaction is "too slow". That begs the question "too slow for what"? Certainly H2 molecules form all the time at STP-like conditions, and while there is discussion of super-dilute mixtures, I don't think this is helpful either. If I have two atoms a zillion light years apart, of course they won't form a molecule.

I'm afraid we have no clear statement of the problem unless one speaks Portuguese.

 

[DaTario]

I respectfully desagree with you Vanadium 50. I said in the OP that a collision betweem two atoms of hydrogen, in isolation (far from anything else -- vacuum), is an almost impossible scenario to produce $H_2$
. If one agrees with this I asked for an explanation, if one doesn't, I asked for a refutation. I see no unclear statement here.
If you don't know the subject, I am just not expecting you to provide and answer here.

best regards

P.S. with respect to what you said in the last post : "There seems to be a suggestion that this reaction is "too slow". That begs the question "too slow for what"?"

I said that a quadrupole effect could be (is said to be) a way to produce $H_2$, but the quadrupole process is slow and the collision happens during a relatively very short time interval.

 

[Baluncore]

I said in the OP that a collision betweem two atoms of hydrogen, in isolation (far from anything else -- vacuum), is an almost impossible scenario to produce H₂.

That will depend on the initial ionisation state of the hydrogen atoms. Where do the two electrons come from, that are needed to form the covalent bond?
If the two separate atoms start out neutral, a collision is all it will take to form the bond and release the radiant energy.

 

[DaTario]

If the two separate atoms start out neutral, a collision is all it will take to form the bond and release the energy.

That is one of the hot points of OP.
According to what I have studied, in a collision between two atoms of $H$, the wave functions of both atoms are put to interfere either constructively or destructively. In the first case this collision is likely to give birth to a molecule ($\sigma_{1s1s}$ case), in the second case ($\sigma^*_{1s1s}$ case) this collision will follow as a simple elastic one. In the intermediate case, randomness enters giving probabilities for this molecule formation to occur.

Recently I watched this webinar in Portuguese (which is my mother language by the way), the lecturer said that a process in which two neutral H atoms collide and form the molecule $H_2$ is forbiden by symmetry reasons. This is what I am trying to understand here. May the symmetry in this collision forbid the radiation to go out as expected by the principle of conservation of energy?

 

[Baluncore]

... in which the lecturer said that two neutral H atoms colliding and forming the molecule is forbiden for symmetry reasons.

Symmetry will not survive the collision.

 

[DaTario]

Symmetry will not survive the collision.

How so? Spontaneous symmetry breaking?

 

[DaTario]

Symmetry will not survive the collision.

Maybe this was a really good joke and I didn't know how to make the most of it.

 

[Vanadium 50]

What do we have:

1. It is a fact that atomic hydrogen (under laboratory conditions) forms molecular hydrogen.
2. We have a YouTube video in Portuguese that says, at least under some circumstances, that it doesn't. The English translation at the selected time talks about the system needing to lose energy (which nobody seems to dispute) and a cryptic comment about this being too slow.
3. The OP seems to think there is some sort of unspecified symmetry argument preventing molecules from forming. This argument is not clearly presented in the English translation in the identified segment. Maybe its in the Portuguese, I don't know.

It's not clear what you want us to do - or even what we can do.

As far as SSB, is there some argument for that, or are you guessing?

 

[DaTario]

1. It is a fact that atomic hydrogen (under laboratory conditions) forms molecular hydrogen.

No doubt, but there are different routes to this formation. I posted a figure where two of these routes are shown (#8). I am addressing one specific route. Two atoms of $H$ colliding in vaccum (rarefied media)

1. We have a YouTube video in Portuguese that says, at least under some circumstances, that it doesn't. The English translation at the selected time talks about the system needing to lose energy (which nobody seems to dispute) and a cryptic comment about this being too slow.

What professor Montenegro says / states in short is:
a) $H_2$ is the most abundant molecule in the universe.
b) most people believe two atoms of hydrogen when colliding may form $H_2$.
c) two atoms of hydrogen colliding are basically an elastic collision.
d) symmetry of the states 1s in these two $H$ atoms are a fundamental factor in forbiding production of $H_2$ from the collision of $H$ atoms in vaccum.
e) he points that a third physical system is needed for the above process to occur.
f) he discusses others work in which hypothesis are considered, like for example dust, as it could act as an intermediator in the production of $H_2$ from two $H$ atoms.

1. The OP seems to think there is some sort of unspecified symmetry argument preventing molecules from forming. This argument is not clearly presented in the English translation in the identified segment. Maybe its in the Portuguese, I don't know.

It's not clear what you want us to do - or even what we can do.

The aspect of symmetry is not fully explained in his webinar. He seems to suggest that it prevents dipole transitions. The next alternative, he says, would be the quadrupole transitions, but he said these processes demand long time intervals, compared to the typical values of the duration of this collision.

Dear Vanadium 50, I wonder:
1) whether this analysis is valid
2) if the impossibility $H + H \rightarrow H_2$ in vaccum is a well known fact
3) whether symmetry really plays a determinant role.
4) how (details) symmetry acts as a prohibiting agent in this route of formation of the $H_2$ molecule.

Thank you once more for the attention (Thanks also to all that contributed)

P.S. I know very little about SSB. I was just trying to understand in which sense symmetry would not survive after the collision, according to Baluncore's contribution (#23). Perhaps it was a (good) joke.

 

[Vanadium 50]

While elastic collisions surely can and do occur, so do inelastic collisions: H + H → H2 + energy. That energy is in the form of EM radiation, and I believe it is dominated by the M1 (magnetic dipiole). Each hydrogen atom has a magnetic dipole equal to the electron magnetic moment, and the molecule has zero, so this transition can occur. (I am ignoring nuclear magnetic moments which are three orders of magnitude smaller)

Does this dominate over a three-body reaction? Depends on the temperature and pressure.

This is not SSB. As far as what the YouTube speaker means, I believe for more detail you will need to ask him (or read his papers, which is probably a better start)

 

[DaTario]

Ok, you say that this route of formation is not impossible. Now I believe the papers he used as references may provide the complete answer. It seems to have become a question of comparing rates (# transitions / time) for each process (route of formation). Professor Montenegro was discussing basically in this seminar, held at the Federal University of Rio de Janeiro, which processes are plausible to be considered as relevants, in order to account for the number of $H_2$ molecules in the universe.

I inform you that I consider myself satisfied with the debates we had and the answers given to me and I am grateful to everyone who contributed to this discussion.

Best wishes

 

[DrClaude]

1) whether this analysis is valid
2) if the impossibility $H + H \rightarrow H_2$ in vaccum is a well known fact
3) whether symmetry really plays a determinant role.
4) how (details) symmetry acts as a prohibiting agent in this route of formation of the $H_2$ molecule.

1) Yes.

2) Yes, any (respectable ) molecular physicist knows this.

3) Yes.

4) For two atoms in the same electronic state, the transition matrix element for a dipole transition boils down to something proportional to
$$
\propto \int \psi^* \hat{d}_\mathrm{N} \psi \, \mathrm{d}\tau_\mathrm{N}
$$
where $\psi$ is the nuclear wave function, $\hat{d}_\mathrm{N}$ the dipole moment operator, and $\tau_\mathrm{N}$ the nuclear degrees of freedom (this is in the Born-Oppenheimer approximation, and the electronic degrees of freedom have already been integrated over).

For a homonuclear diatomic molecule, there can be no dipole moment by symmetry (for the two atoms initially in the same electronic state), so $\hat{d}_\mathrm{N} = 0$ and there is no dipole transition.

As @Vanadium 50 pointed out, M1 magnetic dipole transitions are dominant in this case. So while two colliding identical atoms can temporarily form a diatomic molecule, M1 transitions are slow with respect to the timescale of dissociation, so the two atoms just fly apart.

The only way to form a stable molecule is to have a 3-body collision, with the 3rd body carrying away the extra energy. The rate at which this happens depends on density and temperature.

 

[DaTario]

Thank you very much, DrClaude. I also apologize to you for not having organized the text of the question as it should have, with the appropriate references in English.

This phenomenon reminds me of another one, which was studied by Serge Haroche (Nobel Prize winner), namely, the suppression of a given electronic transition in an atom whose electron is in an excited state, when it is placed in front of a mirror, at a given distance.

 

[DrClaude]

This phenomenon reminds me of another one, which was studied by Serge Haroche (Nobel Prize winner), namely, the suppression of a given electronic transition in an atom whose electron is in an excited state, when it is placed in front of a mirror, at a given distance.

That is slightly different in that it is due to the electromagnetic field inside the cavity not having a mode corresponding to the energy of the transition in the atom (so it is a property of the field, not the atom). But it all stems from the same physical framework, namely quantum electrodynamics.

 

[DaTario]

Yes, I understand there are differences, but I 'always' found interesting the metaphorical idea that particles are like 1 and fields are like 0,99999...

In a certain sense, although they propagate differently, the wave function of the atom and the EM field have some degree of correspondence, haven't they?

OBS.: Perhaps it is not appropriate here to conduct a discussion that departs from the original.

 

[snorkack]

For an English source including some introduction, see for example:
https://articles.adsabs.harvard.edu//full/1991ApJ...372..161L/0000161.000.html
Note the link to van Dishoek, 1989.

That symmetry suppresses radiation is something I had in bachelor level chemistry... bonds with symmetry around the bond are inactive in IR, and weak when it is near symmetry.