Atoms or molecules with spin 1 in the ground state?

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](https://chemistry.stackexchange.com/a/78639/), they won't be usable in a normal Stern-Gerlach spectrometer.




**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.




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**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.)




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**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...


 9

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>


 7

[Rubidium 87](http://steck.us/alkalidata/rubidium87numbers.pdf) 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 $F=1$.


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