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Is it possible for two hydrogen atoms with parallel electron spins (i.e. a pair of spin-up electrons, or a pair of spin-down electrons) to form a covalent bond? I've seen some say there is no sigma bond formation in that situation due to the Pauli exclusion principle, whereas others say one of the two electrons undergoes a spin flip, with bond formation occurring naturally after that.

I would expect 50% of collisions in a gas of free hydrogen atoms to have up spins, and 50% to have down spins. So, on average, any given pair of hydrogen atoms are equally likely to either have parallel or antiparallel spins. Does this imply that only 50% of H–H collisions in that situation result in a bond, i.e. would this impact the yield of the reaction?

How would this look with other elements?

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  • $\begingroup$ The mixed answers you get are actually describing the same picture. Initially there is no sigma bond in that situation, and then there occurs a spin flip. $\endgroup$ Jan 11, 2022 at 23:28
  • $\begingroup$ @IvanNeretin After the flip has passed, however, a sigma orbital is created. Where does the energy for the flip come from? Shouldn't it come from the fact that some is released by the bond? $\endgroup$
    – jemeto7849
    Jan 11, 2022 at 23:43
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    $\begingroup$ The flip does not need any energy. $\endgroup$ Jan 11, 2022 at 23:45
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    $\begingroup$ @IvanNeretin So one electron reorients itself by the magnetic field of the other before the sigma bond happens so that both then have an antiparallel spin and can then transition into the sigma orbital? $\endgroup$
    – jemeto7849
    Jan 11, 2022 at 23:59
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    $\begingroup$ When you say "parallel" and "antiparallel", what do you mean? As in, parallel to what? Normally when people use these words it's in the context of an external magnetic field being applied, is this the case here? If not, then you may want to just use "up" and "down" spin (and in this case, the spin states are degenerate, so indeed no energy is required to go from up to down spin or vice versa). Only when you have an external magnetic field do spin flips require energy (and even then, only in one direction; the reverse flip releases energy). $\endgroup$ Jan 12, 2022 at 0:45

3 Answers 3

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I'll specifically answer this part of your question:

I would expect 50% of collisions in a gas of free hydrogen atoms to have up spins, and 50% to have down spins. So, on average, any given pair of hydrogen atoms are equally likely to either have parallel or antiparallel spins. Does this imply that only 50% of H–H collisions in that situation result in a bond, i.e. would this impact the yield of the reaction?

TL;DR: No, only 25% of collisions will be able to result in a bond.

In general, when you "measure" something in quantum mechanics, you force the system into the eigenvalues and eigenfunctions to some operator. In your case, you are looking at the electron spins, and once you measure them, your overall system is collapsed into an eigenstate of the spin operator. For a hydrogen molecule with its two electrons, those eigenstates are a singlet state $\alpha \beta - \beta \alpha$, and a triplet state with three "microstates" $\alpha \alpha$, $\beta \beta$ and $\alpha \beta + \beta \alpha$.

In the non-interacting case, we assume that each spin result ($\alpha$ or $\beta$) is measured with 50% probability on each of the free atoms.1 Then, for any pair of atoms, we measure each binary spin combination $\alpha \alpha$, $\alpha \beta$, $\beta \alpha$ and $\beta \beta$ with the same 25% probability. If those atom pairs now draw closer together and start to interact, then the $\alpha \beta$ and $\beta \alpha$ combinations must combine to form either the triplet or singlet microstates with equal probability; in other words, isolated atoms with opposite spins have equal chances of forming a singlet or a triplet state when they interact. However, the $\alpha \alpha$ and $\beta \beta$ combinations are also still possible, and since they belong to the triplet state, the latter still wins over the competing singlet state with a 3:1 ratio. And since we know that only the singlet state leads to formation of a bond, we expect 25% of atom collisions to be able to form a hydrogen molecule.

NOTE: As @Paul pointed out in the comments, the explanation here is solely regarding the probabilities of bond formation that come from the spin statistics. There are further effects from collision cross-sections and energy dissipation that I am not taking into account here. In reality, less than the 25% of atom-atom collisions stated above would actually result in the formation of a bonded molecule.


You may wonder: "If we measure eigenstates to the spin operator, how can the isolated atoms yield the specific spins combinations $\alpha \beta$ and $\beta \alpha$? The spin operator is still the same, so why wouldn't we expect to see the singlet and triplet states again? What happened to them?"

At this point, it's vital to remember one important rule: Any linear combination of eigenfunctions to the same eigenvalue is again an eigenfunction. 2 When two atoms are non-interacting, measuring the singlet or triplet microstates of $\alpha \beta \pm \beta \alpha$ is still perfectly possible; but the same is true for their linear combinations that form the "distinct" $\alpha \beta$ and $\beta \alpha$ states.3 What you measure now depends on what answer you seek, and each set of answers is as valid as the other. (But again, this is only true if there is no energy difference between the triplet and singlet states.)


Footnotes

1 Note that I'm treating this as the premise of your question. If your ensemble of atoms does not conform to this assumption, e.g., due to the presence of external fields, then the answer will also be different.

2 This is very easy to show. Assume $\hat{O} \Psi_1 = E_1 \Psi_1$ and $\hat{O} \Psi_2 = E_2 \Psi_2$, with $E_1 \ne E_2$. Then, $\hat{O} (a\Psi_1 + b\Psi_2) \ne (E_1 + E_2) (a\Psi_1 + b\Psi_2)$. However, if $E_1 = E_2 = E$, then $\hat{O} (a\Psi_1 + b\Psi_2) = E (a\Psi_1 + b\Psi_2)$.

3 Let $\alpha \beta + \beta \alpha = \Psi_1$ and $\alpha \beta - \beta \alpha = \Psi_2$, then $\frac{1}{2} (\Psi_1 + \Psi_2) = \alpha \beta$ and $\frac{1}{2} (\Psi_1 - \Psi_2) = \beta \alpha$.

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    $\begingroup$ your analysis is only based on spin statistics and ignores reaction cross sections. In addition, it is good to note that a collision between two atoms in the gas phase can only produce a bound molecule if it dissipates the bonding energy by release of a photon or by collision with a third body. Since radiative transitions in molecular hydrogen are dipole forbidden, three-body collisions are required to form molecules. $\endgroup$
    – Paul
    Jan 12, 2022 at 20:36
  • $\begingroup$ True. I'll add a disclaimer. $\endgroup$
    – Antimon
    Jan 12, 2022 at 20:45
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    $\begingroup$ This is a good answer to the core question (whereas I've only challenged the premise), but I object to the statement "In the non-interacting case, we assume that 50% of our free hydrogen atoms have α and 50% β spin." If you haven't measured them, the correct statement is "we assume a 50% probability of measuring $\alpha$ and 50% probability of measureing $\beta$ spin." $\endgroup$
    – Andrew
    Jan 12, 2022 at 23:28
  • $\begingroup$ Okay, I see what you mean. I'll incorporate that. $\endgroup$
    – Antimon
    Jan 12, 2022 at 23:30
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I think the premise of your question is flawed, which isn't your fault. It's the way students are introduced to quantum mechanics at the intro level. But if you could somehow generate a gas phase of free hydrogen atoms, it isn't correct to say that 50% would have "up" electron spin and 50% "down." What is correct is that if you measured the individual spins you would get 50% up and 50% down.

What's the difference? That's the weirdness of quantum mechanical systems. Individual electrons do not have defined spin until you measure it. Prior to that, they exist in a superposition of both possible spin states with a 50% chance of each result if you take a measurement.

So in your hypothetical, when two hydrogen atoms interact, there are no different cases to consider. Every pair of hydrogen atoms are the same. You have two electrons of undefined spin. When the atoms bond, the electrons become entangled so that the now diatomic system has net zero spin, but still neither electron has defined up or down spin. Only the net spin is defined.

If this doesn't make much sense, you're not alone. Quantum mechanics in many ways does not make sense. All we can say is that the empirical description we've come up so far does a very good job of predicting experimental outcomes. Why that description works is utterly unknown.

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  • $\begingroup$ "when two hydrogen atoms interact, there are no different cases to consider. Every pair of hydrogen atoms are the same" - This seems wrong to me. If we measure their spins, we must get an eigenvalue to the total spin operator, and those eigenvalues will correspond to overall triplet or singlet states with their corresponding 3:1 weights. $\endgroup$
    – Antimon
    Jan 12, 2022 at 18:23
  • $\begingroup$ @Antimon - Yes, if we measure their spins, they must collapse to a state represented by an eigenvalue. But prior to measurement, they have no defined spin. So if you somehow picked out a pair of atoms and measured their spins, you could create a pairing of up/up or down/down. But in a bulk mixture that hasn't been measured, all atoms are the same. $\endgroup$
    – Andrew
    Jan 12, 2022 at 18:50
  • $\begingroup$ I'm still missing a connection between the bulk state of isolated atoms and the formation of a bond. I'll address this in an answer of my own, see if you can comment on that. $\endgroup$
    – Antimon
    Jan 12, 2022 at 18:54
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Yes -- if one or both electrons are in an excited state and they thereby occupy different orbitals. Of course, the excited state electron(s) would decay rapidly, so either the atoms blow apart from the resulting energy release or an electron spin-flips to preserve/strengthen the bond.

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