I read that Ne has $S=0$. How can this be found, knowing the electron configuration?

Electrons, protons and neutrons all have 1/2 spin. The Pauli exclusion principle implies that the even number of electrons are half up-spin and half down-spin, so that they sum to 0. Is that argument valid, and if so, can the same be applied to show that the total spin of neutrons and protons, which occupy the same space, is 0?

I suppose that would imply that $S=0$ for all molecules with even numbers of electrons and protons+ńeutrons.

  • 1
    $\begingroup$ This is a very confusing question. As Boris says, the nuclear spins don't measurably affect chemical processes so they're ignored. Moreover, neon isn't a molecule: it is an atom. Spins of molecules are subtle things because the nuclei would have to be counted. One may talk about the "rotational spectrum" of the molecules in which the angular momentum is changing by the discrete amounts (very thinly spaced) but the spin of a molecule itself may be considered to be zero or macroscopic. $\endgroup$ – Luboš Motl Apr 10 '12 at 15:15
  • $\begingroup$ If your question is really about atoms and not molecules, it's far from true that $S=0$ for all atoms with an even number of electrons. Hund rules, see en.wikipedia.org/wiki/Hund_Rules , on the contrary state that $S$ tries to be maximized. And it possible to do so because your Pauli principle argument is just wrong: electrons' states are not given just by spin up or down but many other additional quantum numbers so there may be lots of electrons with spin up, and no electrons with spin down. Your reasoning just seems to be flawed at every level... $\endgroup$ – Luboš Motl Apr 10 '12 at 15:17

If you stick to atoms, and use the atomic orbital approximation, it's easy to work out the electronic spin by working out what atomic orbitals are populated. The atomic orbitals are populated in a slightly strange order that's given in http://en.wikipedia.org/wiki/Electron_configuration#Aufbau_principle_and_Madelung_rule. You have to remember that the p orbitals are really 3 orbitals of equal energy, the d orbitals are 5 orbitals of equal energy and if you get as high as the f orbitals there are 7 of these.

So hydrogen with 1 electron is 1s$^1$ and has spin half. helium is 1s$^2$ and the spins pair so you get spin zero. Likewise Lithium is 1s$^2$2s$^1$ with spin half and Beryllium is 1s$^2$2s$^2$ with spin zero. But take Carbon, which is 1s$^2$2s$^2$2p$^2$. It looks as if the spin should be zero because you have an even number of electrons. However the two p electrons will not be in the same p orbital because there are three p orbitals with equal energy. Because the p electrons are in different orbitals their spins are not necessarily paired, so the spin could be zero or it could be one. As a general guide Hund's rule tells you the higher spin is favoured, so the electronic spin of the carbon would be one (actually I can't remember if this is true or not).

Neon is 1s$^2$2s$^2$2p$^6$ and has spin zero because there are two electrons in each of the 1s, 2s and all three 2p orbitals, so the spins are all paired.

For molecules you proceed in a similar manner but you have to fill the molecular orbitals. It's rarer to get degenerate molecular orbitals since molecules don't have the spherical symmetry of atoms.

I'm not sure why Boris is telling you to ignore nuclear spin. True, it doesn't play any role in chemistry, but it is important in the spectra of atoms. In fact interactions between the electronic and nuclear spin are responsible for the Hydrogen 21cm radiation, and there's a lot of that in the universe!

| cite | improve this answer | |

In chemistry, you are only interested in electronic spin. So forget neutrons and protons.

| cite | improve this answer | |
  • $\begingroup$ Ok, I'm trying to find the allowed electron transitions in Na (Ne and $e^-$). Is my reasoning about the zero total electron spin sound? $\endgroup$ – Per Apr 10 '12 at 14:24
  • $\begingroup$ Pauli exclusion priciple implies only that only 2 electrons can occupy 1 orbital. Energetic degeneracy of more distinct orbitals can cause some orbitals to be only singly occupied. There is also a Hund principle of maximum multiplicity (electrons having same spin repel less each other resulting in lower energy). So not all atoms/ions/compounds with even number of electrons have zero spin. $\endgroup$ – Boris Apr 10 '12 at 14:29

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.