9
$\begingroup$

Is there any atom or molecule that has spin 1 in its ground state?

Do Hund's rules keep this from happening for an atom?

The reason I'm curious is that it would be nice to have a spin-1 example for use in pedagogical discussions of the Stern-Gerlach experiment.

[EDIT] Clarification: when I say "spin," I mean the total angular momentum, not just the sum of the spin-1/2's. (The total angular momentum is what you are seeing in the Stern-Gerlach experiment.) I deleted the part of the question about ions, because, as pointed out by Orthocresol, they won't be usable in a normal Stern-Gerlach spectrometer.

$\endgroup$
10
$\begingroup$

Atoms

Atomic carbon with its $\mathrm{1s^2 2s^2 2p^2}$ configuration has a triplet ground state ($S = 1$), precisely because of Hund's first rule.

However, in the context of the Stern–Gerlach experiment, you might run into a problem with orbital angular momentum, as carbon's ground state also has nonzero orbital angular momentum ($^3\mathrm{P}$ ground state, $L = 1$). The behaviour in a magnetic field will be rather more complex and you probably need to take into account spin-orbit coupling.

At the moment I can't think of any atoms with a ground state term symbol of $\mathrm{^3S}$. I actually suspect that it's impossible, but I'm not really up to proving it right now.


Molecules

Triplet dioxygen (as Zhe mentioned) has no orbital angular momentum ($^3\Sigma_\mathrm{g}^-$ ground state, $\Lambda = 0$), but I'm not sure if the inhomogeneity of the electron density would have any impact. (as in, I'm genuinely not sure.)


Ions

The $\ce{^2H}$ nucleus (which us chemists usually refer to as $\ce{D+}$) is lightest stable nucleus that has a spin of 1.

The only issue is that you then have a moving charge in a magnetic field. If you could ignore it, or take it out of the equation, somehow...

$\endgroup$
  • $\begingroup$ The behaviour in a magnetic field will be rather more complex and you probably need to take into account spin-orbit coupling. I don't think the behavior of a neutral particle can be any more complicated than what it normally is in a Stern-Gerlach spectrometer. The Hamiltonian is $z\mu_z=zgJ_z$, and spin versus orbital angular momentum would just have an effect on $g$. I'm not sure if the inhomogeneity of the electron density would have any impact. For similar reasons, I don't think this matters. $\endgroup$ – Ben Crowell Jul 11 '17 at 17:21
  • 1
    $\begingroup$ In an S state (L=0) you only have room for 2 electrons, one with spin up, one down. $\endgroup$ – Magicsowon Jul 11 '17 at 17:38
  • $\begingroup$ @BenCrowell It is slightly complicated (imo, at least!). The coupling of the angular momenta $S$ and $L$ gives rise to the overall angular momentum $J$, which takes values $0, 1, 2$ (Clebsch-Gordan series). Of the three resulting terms, Hund's third rule dictates that $\mathrm{^3P_0}$ is the ground state (which unfortunately isn't magnetic). $\mathrm{^3P_1}$ (the one we're probably interested in) and $\mathrm{^3P_2}$ are then $16$ and $43~\mathrm{cm^{-1}}$ above the ground state. $\endgroup$ – orthocresol Jul 11 '17 at 17:41
  • $\begingroup$ @orthocresol: You're talking about carbon, right? All I'm saying is that a Stern-Gerlach experiment only measures the component of the total magnetic moment along the field axis. There is no other, independent parameter that is available to measure. The angular momentum isn't independent of the magnetic moment. Since the first two excited states are so close in energy to the ground state, they would presumably be equally populated in an atomic beam. Therefore you would probably see a superposition of results corresponding to the three energy states, each with its own quantized magnetic moment. $\endgroup$ – Ben Crowell Jul 11 '17 at 17:48
  • $\begingroup$ @BenCrowell Yes, I agree entirely. By "complicated", what I meant was that if you performed a S–G experiment on a beam of carbon atoms, the result wouldn't be a clean separation into three different beams. If you consider carbon to be a good enough example, though, then I guess we're done here? I was hoping to find an example that doesn't have a mixture of multiple states, but I don't think it's forthcoming. $\endgroup$ – orthocresol Jul 11 '17 at 17:53
8
$\begingroup$

Triplet oxygen has two unpaired electrons with the same spin, and a total spin value of 1.

In fact, by Hund's rule, the triplet states are preferred over the singlet states which have two electrons with opposite spins.

Further reading:

  1. https://en.wikipedia.org/wiki/Triplet_oxygen
  2. https://en.wikipedia.org/wiki/Singlet_oxygen
$\endgroup$
1
$\begingroup$

Rubidium 87 is one candidate, if you take into account hyperfine splitting, the ground state (which is part of hyperfine manifold of 5$S_{1/2}$) has total angular momentum $\vec{F}=\vec{J}+\vec{I}$ momentum of $F=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.