Can He2 Molecules Form and Why Do They Dissociate?

[khanhhung2512]

This is a question in "Chemical Principles, 6th Edition, Steven Zumdahl":
Bond energy has been defined in the text as the amount of energy required to break a chemical bond, so we have come to think of the addition of energy as breaking bonds. However, in some cases the addition of energy can cause the formation of bonds. For example, in a sample of helium gas subjected to a high-energy source, some He2 molecules exist momentarily and then dissociate. Use MO theory (and diagrams) to explain why He2 molecules can come to exist and why they dissociate.
And this is from the solutions manual:
The ground state MO electron configuration for He2 is (σ1s)2(σ1s*)2 giving a bond order of 0.
Therefore, He2 molecules are not predicted to be stable (and are not stable) in the lowest energy
ground state. However, in a high-energy environment, electron(s) from the antibonding orbitals in
He2 can be promoted into higher-energy bonding orbitals, thus giving a nonzero bond order and
a “reason” to form. For example, a possible excited-state MO electron configuration for He2 would
be (σ1s)2(σ1s*)1(σ2s)1, giving a bond order of (3 – 1)/2 = 1. Thus excited He2 molecules can
form, but they spontaneously break apart as the electron(s) fall back to the ground state, where
the bond order equals zero.
What do you think about the solution? I think it's incorrect.

 

[Greg Bernhardt]

I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?

 

[Borek]

I think it's incorrect.

Why do you think so?

 

[khanhhung2512]

Why do you think so?

Because with this explanation, excited-state configuration (σ1s)2(σ1s*)1(σ2s)1 with bond order 1 seems stabler than ground-state configuration (σ1s)2(σ1s*)2 with bond order 0, while it's not the case.

 

[Borek]

with this explanation, excited-state configuration (...) seems stabler than ground-state configuration (...) while it's not the case.

Question starts telling you it IS the experimental fact, doesn't it?

 

[khanhhung2512]

Question starts telling you it IS the experimental fact, doesn't it?

I think the question didn't tell us anything about the configuration of the He2 molecule in high-energy environment.

 

[Borek]

Well, it doesn't say anything about configuration, but it tells us the molecule exist. And the explanation given is quite plausible to me.

You have stated

while it's not the case

but you never explained WHY it is not the case.

 

[khanhhung2512]

Well, it doesn't say anything about configuration, but it tells us the molecule exist. And the explanation given is quite plausible to me.

You have stated
but you never explained WHY it is not the case.

Because the ground state is lower in energy, and thus is stabler than the excited state, whatever bond orders they have.
My explanation for this He2 question is as followed: In high-energy environments, some He atoms will exist in the excited state. Two He atoms might then form a He2 molecule with configuration (σ1s)2(σ1s*)2 which is stabler than the two excited He atoms. After that, this He2 molecule will decompose to ground-state He atoms.

 

[Borek]

Because the ground state is lower in energy, and thus is stabler than the excited state, whatever bond orders they have.

That's where you are wrong. For a "standard" molecule - one that doesn't have filled antibonding orbital - that would be true. But we are talking about molecule that is not a standard one - it has a filled antibonding oribtal, which makes is completely unstable in the ground state. However, antibonding effect of the excited antibonding orbital is weaker than the antibonding effect of the antibonding orbital in the ground state - so once the antibonding orbital gets excited, its overall effect on the molecule stability gets lower and the molecule becomes stable.

Two He atoms might then form a He2 molecule with configuration (σ1s)2(σ1s*)2 which is stabler than the two excited He atoms.

Two excited atoms would not create a molecule in the ground state.

 

[khanhhung2512]

That's where you are wrong. For a "standard" molecule - one that doesn't have filled antibonding orbital - that would be true. But we are talking about molecule that is not a standard one - it has a filled antibonding oribtal, which makes is completely unstable in the ground state. However, antibonding effect of the excited antibonding orbital is weaker than the antibonding effect of the antibonding orbital in the ground state - so once the antibonding orbital gets excited, its overall effect on the molecule stability gets lower and the molecule becomes stable.



Two excited atoms would not create a molecule in the ground state.

Suppose you're correct, then why would stabler excited He2 molecules spontaneously break apart as the electron(s) fall back to the ground state?

 

[Borek]

Suppose you're correct, then why would stabler excited He2 molecules spontaneously break apart as the electron(s) fall back to the ground state?

Because in the ground state the antibonding effect of the antibonding orbitals gets higher than in the excited molecule - and it gets high enough for the molecule to become unstable

Think about is this way: at the end of the excited states of the molecule there is an ion, He₂⁺ - and it is perfectly stable even in the ground state, as single electron on the antibonding orbital is not able to counteract two electrons on the bonding orbital. When you ad an electron and neutralize He₂⁺ to He₂ - but leave the molecule excited - it is still stable. It will get unstable once the electron drops to the ground state.

 

[khanhhung2512]

Because in the ground state the antibonding effect of the antibonding orbitals gets higher than in the excited molecule - and it gets high enough for the molecule to become unstable

Think about is this way: at the end of the excited states of the molecule there is an ion, He₂⁺ - and it is perfectly stable even in the ground state, as single electron on the antibonding orbital is not able to counteract two electrons on the bonding orbital. When you ad an electron and neutralize He₂⁺ to He₂ - but leave the molecule excited - it is still stable. It will get unstable once the electron drops to the ground state.

I mean, what's the reason the stable, low-energy excited state would spontaneously drop electron to transform to the unstable, high-energy ground state?

 

[Borek]

I mean, what's the reason the stable, low-energy excited state would spontaneously drop electron to transform to the unstable, high-energy ground state?

Excited state doesn't have lower energy! It has a higher energy, which is what makes the antibonding effect weaker.

When the bonding orbital gets excited, bonding effect is lower.

When the antibonding orbital gets excited, antibonding effect is lower.

Molecules stability reflects the difference between bonding effect and antibonding effect.

 

[khanhhung2512]

Excited state doesn't have lower energy! It has a higher energy, which is what makes the antibonding effect weaker.

When the bonding orbital gets excited, bonding effect is lower.

When the antibonding orbital gets excited, antibonding effect is lower.

Molecules stability reflects the difference between bonding effect and antibonding effect.

I still haven't understood all of what you're saying, but it's making sense to me now. Thank you very much. One more question, please. What kind of book do I need to read or study to have such a good knowledge of physical chemistry as yours, or at least to be able to fully comprehend this He2 problem?
Thanks again.

 

[Borek]

I am afraid I won't be of much help when it comes to books - the ones I can suggest will be in Polish, so of no use for you.

 

[snorkack]

And this is from the solutions manual:
The ground state MO electron configuration for He2 is (σ1s)2(σ1s*)2 giving a bond order of 0.
Therefore, He2 molecules are not predicted to be stable (and are not stable) in the lowest energy
ground state.

(He-3)2 molecules are indeed not stable, and neither are He-3-He-4 molecules. But (He-4)2 molecules ARE stable.

 

[Borek]

But (He-4)2 molecules ARE stable.

Never heard about it. Do you have any source for that?

 

[snorkack]

Never heard about it. Do you have any source for that?

Are you more interested in calculations or in experiments?

Predicted but questioned theoretically for some time. Experimentally proven in 1994. Main properties measured by 2000: bond length 52 angströms (sic! over 17 times the potential well minimum inside of 3 angströms), binding energy 95 neV (sic! about 5000 times less than the binding energy of helium 4 atom to bulk liquid helium 4).

Oh, and a lot of practical importance, too. The bound nature of helium 4 dimer AND all higher clusters to bulk (as well as of all clusters containing two or more helium 4 and any number of helium 3) means that helium 4 dimer is easily produced by cooling helium 4 (if you know what to look for) and is essential for condensation of helium containing appreciable amounts of helium 4.

An example source:

http://pubs.acs.org/doi/abs/10.1021/ed084p860

 

[Borek]

Please don't post links to copyrighted material, I have replaced it with a link to acs abstract.

[STRIKE]While it technically fits IUPAC definition of a molecule ("An electrically neutral entity consisting of more than one atom, with at least one vibrational state." - I assume there are vibrational states, I see no reason why there would be none),[/STRIKE] calling something with a binding energy of 95 neV "stable" doesn't make much sense to me. Typical bond energies are in the eV range, so they are tens of millions times stronger. Calling these dimers "stable" is a misnomer IMHO.

Edit: after reading the paper - no, it isn't a molecule. JChemEd paper clearly states that there are no excited states:

The He₂ potential well can support just a single bound rotational–vibrational state (v = 0, J = 0) with an energy of approximately 1.176 mK. The dissociation energy of the molecule in its only rotational–vibrational state is smaller than the potential well depth by a factor of about 10^-4. The bound state is just barely below the dissociation continuum threshold. The binding between two He atoms is so weak that even rotational excitation will result in dissociation.