Talk:Spin isomers of hydrogen

Needs explanation

Needs explanation as to why the release of heat is undesirable. -- Leonard G. 03:55, 19 Sep 2004 (UTC)

I added some content about this aspect a while ago. DMacks (talk) 11:35, 10 February 2018 (UTC)[reply]

It would also be very helpful if someone could explain why the electron spins can be ignored. I think this is because the electrons are already tightly bond in a singlet state and Pauli exclusion prevents them from leaving that at a reasonable energy. Jess_Riedel (talk) 16:15, 8 February 2018 (UTC)[reply]

Yes, your reason is correct. To expand a little, the two electrons interact strongly in a bonding molecular orbital, so promoting one to a higher orbital requires a large energy and does not happen at reasonable temperatures. This is in contrast to the nuclear spins which interact very weakly with each other, so that the energy difference between spins parallel and spins opposed is very small and both states are found at normal temperatures. Dirac66 (talk) 23:53, 9 February 2018 (UTC)[reply]

Triplet singlet states

Why is orthohydrogen refered to as the triplet state, and parahydrogen is the singlet state?

If this page is correct (i.e. ortho- and para- hydrogen refer to alignment of the spins of the two protons in a H₂ molecule, not alignment of electron and proton spins in a H atom), then there's a problem with http://en.wikipedia.org/wiki/SHGb02_14a . The problem is that in that article they are referring to alignment of electron and proton spins. I am assuming this article is correct, and removing the reference from the SHGb02_14a article to ortho- and para-hydrogen. 21:22, 30 March 2006 (UTC)

Rename?

Move to a more general and descriptive name like nuclear spin modifications of hydrogen or spin isomers of hydrogen? Femto 13:25, 10 August 2006 (UTC)[reply]

Question

Is there a significant difference in the energy released when para or ortho hydrogen is combined with oxygen to form water? I got the impression from a patent that one is less explosive than the other....??

Another question

How much heat? in other words how many kJ/mol is released when turning ortho into para hydrogen? —Preceding unsigned comment added by 87.10.127.103 (talk) 12:22, August 29, 2007 (UTC)

New title?

Since virtually all the ambiguitiies mentioned in earlier questions have been addressed and the theoretical basis of ortho- and parahydrogen is now clear, I ve changed this article to Start class. I don't think I've seen the name "spin isomers" appled anywhere else, so we should consider a different title to this article - I admit that an alternative doesn't occur to me at the moment. Gadolinist (talk) 20:51, 25 November 2009 (UTC)[reply]

(I have moved your comment to the bottom of the talk page as a new section.) I suggest the title "Orthohydrogen and parahydrogen". These words are redirects now, and I think they would be the words most readers would search for. Dirac66 (talk) 22:16, 25 November 2009 (UTC)[reply]

Image cleanup

The figures in this article need improvement. The ortho vs. para figure would be better with a white background, while the graphs on this page are overcompressed in jpeg format. SVG would be better! 129.128.157.245 (talk) 21:38, 9 December 2009 (UTC)[reply]

I'm not willing to re-do the image again just to satisfy the artistic sensibilities of one user. This is Wikipedia - as soon as I make you happy, someone else will just come along and say the original was better and revert it. Xaa (talk) 02:19, 12 December 2009 (UTC)[reply]

para/ortho ratio at 20K

The article claims that at 20K the ratio of para-H2 in thermal equilibrium is 99.95%. I've added a "citation needed" for that number. If I do the math with the partition functions given later in the article (using E_J = B*J*(J+1) with B=7.35meV) I arrive at Z_para=1.0000 and Z_ortho=0.0018, so the given ratio should be ~99.82%. I've cross-checked my T-dependent curves for Z_ortho/Z_para with Fig.1 from ApJ 516 (1999) 371, and they perfectly match, so I'm pretty sure my math is correct. —Preceding unsigned comment added by 129.206.21.102 (talk) 15:58, 5 March 2010 (UTC)[reply]

I inserted a reference to Rock's text which has a table showing No/Np = 0.002 at 20 K, corresponding to 99.8% para which value I inserted. And then I noticed your comment, and I see that you and Rock are in agreement. Dirac66 (talk) 02:02, 6 March 2010 (UTC)[reply]

Bogus article

At the very least, an equation for the position of an electron requires three variables: one for each physically perpendicular plane in three-space. The basis for this article implies that there is only a single plane in which the electrons spin.

This article and the article about "Spin Isomers" should be deleted.

See the Wikipedia article on "Isomers" which, by that definition, is applicable to a compound though may also be applied to unbound, elemental hydrogen. (Two atoms of hydrogen). A "Nuclear Isomer" refers to physical differences at a sub-atomic level and to other elements known to radiate in the EM spectrum, so is not applicable.

(Incidentally, if this were remotely possible it would also apply to Deuterium. In such case, the basis for a "nuclear isomer" is more likely to apply.) Kernel.package (talk) 23:10, 4 June 2010 (UTC)[reply]

The article is not bogus. The physics described is real and supported by references to reliable scientific books and articles.

It is true that the word "isomers" is confusing here because it does not have the same meaning as for chemical isomers which are of course much better known and more important. But "spin isomers" is the term used in the scientific literature so it is not is up to Wikipedia to change it. Dirac66 (talk) 00:38, 9 June 2010 (UTC)[reply]

I have now added a recent (2011) reference which uses the term spin isomers, and added a sentence to the intro to specify that these isomers do not differ in chemical structure as the term isomers implies elsewhere. Dirac66 (talk) 02:08, 30 June 2016 (UTC)[reply]

NMR section

Although this section is factually correct, it is poorly written and cited. First off, the reference for PASADENA is not the actual paper where the term is first published. There is also no mention of the related ALTADENA effect. Second the oxford-instruments page is no longer valid. 3rd, the Duckett group site from York and one of their papers are referenced but oddly there is no mention the the SABRE phenomenon in the article. I think this section needs to be rewritten and expanded. The PHIP review article by C.R. Bowers would be an excellent source to add. — Preceding unsigned comment added by Danny31292 (talk • contribs) 22:01, 27 May 2014 (UTC)[reply]

Relevance tags for magnetic field and non-degeneracy

There is a relevance tag for the statement that the 3-fold degeneracy of orthohydrogen can be broken by a magnetic field. I think that the 3-fold degeneracy is important because it is responsible for the 3:1 population ratio. However if we just call it a degeneracy without referring to a magnetic field, some readers will say yes, but the levels are not degenerate in a magnetic field. So we need to say in the absence of a magnetic field in order to be completely correct. I think that phrasing is a bit more to the point than can be broken by a magnetic field.

As for the other relevance tag two lines further down, the non-degeneracy of parahydrogen is also necessary to explain the population ratios. The only flaw I see with that sentence is that it repeats the point 3 times in slightly different words, but in a general encyclopedia this may be necessary for some readers. I am going to remove these 2 relevance tags. Dirac66 (talk) 02:31, 30 June 2016 (UTC)[reply]

Energy difference wrong?

Is the statement "The para form whose lowest level is J = 0 is more stable by 1.06 kJ/mol" really true? Isn't it the energy difference between 75% ortho + 25 % para and 100 % para?`-- Wassermaus (talk) 21:48, 6 November 2020 (UTC)[reply]

The statement does agree with Ref.5, which implies that the ortho-para energy difference is 5.27 kJ/kg. Multiply by (1 kg / 1000 g) x (2.016 g H₂ / 1 mol H₂) to obtain 1.06 kJ/mol. Have you a source for your version? Dirac66 (talk) 00:39, 7 November 2020 (UTC)[reply]

My first-class source is Die Entdeckung des para-Wasserstoffs. (Max Planck Institute), unfortunately in German. There you find 1455 kJ/mol; 3/4 of this is 1091 kJ/mol. Another source is Ortho and Parahydrogen in Interstellar Material where they have 0,34 kcal [per mol, I assume], which means 1,423 kJ. 3/4 of this is 1068 kJ. -- Wassermaus (talk) 13:09, 8 November 2020 (UTC)[reply]

In the first ref, the other number in the same phrase is in units of J/mol not kL/mol. In "0,34", does it really make sense for the comma to be a thousands-separator, or is this the standard German convention of using comma a decimal-separator (and period as thousands-separator) in reverse of English? Those two as mis-readings would therefore support the ballpark of ca. 1 kJ/mol. DMacks (talk) 20:12, 8 November 2020 (UTC)[reply]

Sorry, my mistake, I used the German way of writing numbers where the comma is indeed the decimal-separator.
So for the first reference: At the bottom it says: "Die Umwandlungsenthalpie von ortho- in para-Wasserstoff bei tiefen Temperaturen ist [..] um 50 % höher als die Verdampfungsenthalpie des Wasserstoffs (1455 gegenüber 904 J/mol)." translation: "The transition enthalpy from o-H2 to p-H2 at low temperatures is by 50 % higher than the evaporation enthalpy of hydrogen (1455 vs. 904 J/mol)."
The second reference says 0.34 kcal (Page 164 top left), which corresponds to 1423 J/mol.
I added a third reference (see below) - this time with correct decimal points. -- Wassermaus (talk) 21:22, 8 November 2020 (UTC)[reply]

My mistake too then, I was assuming the comma was in the original and you were 1) directly quoting it and 2) mis-interpretting it. DMacks (talk) 22:38, 8 November 2020 (UTC)[reply]

[1] at 5:00 min. 170.5 kelvins - multiplied by boltzmann constant corresponds to 1.42 kJ/mol -- Wassermaus (talk) 19:49, 8 November 2020 (UTC)[reply]

I think now that Wassermaus is correct in questioning the statement based on the reference now cited in the article = anonymous lecture notes from University of Florida (as the URL includes ufl.edu). The fifth slide of these notes includes the sentence "Ortho to Para-hydrogen conversion releases a significant amount of heat (527 kJ/kg)". This is ambiguous as it is not clear if the initial state is ALL ortho, or 75% ortho corresponding to equilibrium at room (or high) temperature. The article now assumes all, but as Wassermaus has shown, it must be 75% to agree with the value in the other references.

So I suggest we delete the ufl.edu reference, correct the heat value, and use the first two references of Wassermaus: the English first as this is English Wikipedia, and then the German. But not the YouTube please: I don't think readers should have to watch YouTube films to verify facts. Dirac66 (talk) 21:59, 8 November 2020 (UTC)[reply]

Please do it. I'm not a native speaker therefore I'm reluctant to do it myself. -- Wassermaus (talk) 17:27, 9 November 2020 (UTC)[reply]

OK, done. I decided to put the German source first because the value is more precise with 4 significant figures. The English article only gives the value in kcal to 2 figures, so I rounded off the value in kJ to 2 figures also. Dirac66 (talk) 01:55, 10 November 2020 (UTC)[reply]

Doublet?

Are there any known molecules with two spin-paired Hydrogen atoms that form a doublet instead of a triplet? If not, is it worth explaining in the article why a doublet is prohibited? Ronnotel (talk) 00:16, 11 January 2021 (UTC)[reply]

A doublet state requires a spin of 1/2 and therefore an odd number of electrons. For a molecule with two hydrogens to have a total spin of 1/2, there would have to be a third atom with an odd number of electrons. An example would be the amino radical ^•
NH
₂. Dirac66 (talk) 01:56, 11 January 2021 (UTC)[reply]

If all we need is an odd number of electrons (well, an odd number of unpaired electrons), then H₂⁺ (the monocation of the H₂ molecule) would suffice. However, isn't the issue here the nuclear spin (not electronic)? What would ²H¹H (deuterium being spin +1) allow? DMacks (talk) 03:04, 11 January 2021 (UTC)[reply]

Ah, now the question is clearer, thank you. I will consider nuclear (only) spin, and try to answer first for H₂ and then for HD.

For H₂, Ronnotel asked why a doublet is prohibited. This is an example of the rules of quantum mechanics for the coupling of two angular momenta J₁ and J₂. The total angular momentum has the only possible values J₁ + J₂, J₁ + J₂ - 1, ... |J₁ - J₂|. The successive possibilities differ by 1. For H₂, both nuclear spins are 1/2, so the total can only be 1 or 0, corresponding to a triplet and a singlet.

For DH as suggested by DMacks, J₁ = 1 and J₂ = 1/2 so J can be 3/2 and 1/2, corresponding to a nuclear spin quartet and yes, a doublet. However for DH the two nuclei are not identical, so the Pauli exclusion principle here does not require the wavefunction to be either symmetric or antisymmetric with respect to the exchange of the two nuclei: DH → HD. This means that there is no constraint on the symmetry of the rotational levels, and for DH all rotational levels are allowed. This isotopomer is therefore less interesting and usually not mentioned in textbooks. Dirac66 (talk) 23:38, 11 January 2021 (UTC)[reply]