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As far as I understand, magnetism comes from the 'unpaired electrons' in the subshells of atoms. Atoms with paired electrons are diamagnetic ('not magnetic') while atoms with unpaired electrons are paramagnetic.

However, Calcium is said to be paramagnetic, even though it has no free electrons. How come?

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    $\begingroup$ One of us is confused, because I thought calcium was metallic and diagmagnetic. And wikipedia agrees. Clarify? $\endgroup$ Commented Jul 9, 2013 at 15:41
  • $\begingroup$ I got it from here: periodictable.com/Properties/A/MagneticType.html but now that you say so, I can't find any other references that Calcium would be paramagnetic. Sooo, I think an error on periodictable.com and a reason to send them a mail :-) Thanks! $\endgroup$ Commented Jul 10, 2013 at 9:52
  • $\begingroup$ Oh, and Wolfram Alpha is also saying that it is paramagnetic... wolframalpha.com/input/?i=is+calcium+paramagnetic $\endgroup$ Commented Jul 10, 2013 at 9:56
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    $\begingroup$ We should probably a find a reliable original source, since I also find contradictory reports $\endgroup$ Commented Jul 10, 2013 at 16:49
  • $\begingroup$ Also vis-a-vis your original question, its not really clear what your first two sentences mean. Are you refering to unpaired electrons in the sense of Hund's rules? Or are you talking about free electrons as in a conduction band? $\endgroup$ Commented Jul 10, 2013 at 16:53

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Let's just agree that this official looking document

http://iopscience.iop.org/0953-8984/12/45/304/pdf/0953-8984_12_45_304.pdf

from the D0 experiments website is correct, and that calcium is paramagnetic

Now, the statement that everything with "paired electrons" is diamagnetic is problematic.

We would like to say that isolated atoms, with no net electron spin are diamagnetic, because the leading effect is the coupling of the orbital angular momentum which is diamagnetic. (To even distinguish between spin and orbital momentum we need weak spin-orbit coupling). The usual estimate of the net electron spin comes from Hund's rule, which says that any unfilled shell has net electronic spin. Basically the electrons don't like being stuck together in the same atom, so they minimize their contact by making their spins parallel.

But the above applies only to isolated atoms. We can apply it to insulators as long as we keep in mind that interaction between the atomic spins can lead to all sorts of fun.

Metallic systems are a whole different ballgame. Because the electrons are delocalized, we don't have this picture of isolated atoms with little electron spins attached. The electrons are no longer trapped together in an atom, leading to net spin. Instead we have extended electrons with equal number spin up and spin down, and when we apply a magnetic field we get Pauli-paramagnetism. This is quite weak in a good metal and it competes with the also weak diamagnetism of the core electrons and such. The diamagnetism can dominate: gold and zinc are diamagnetic. Calcium by the way is metallic even though the $s$ shell is full - the $s$ and $p$ bands mix and cross so that their is a Fermi surface.

You can complicate the above paragraph in about a hundred ways, with spin-orbit coupling, by combining unlocalized electrons with localized spins, by adding disorder and interaction which try to localize the electrons or order their spins. Its a real mess. I certainly don't know enough to explain the magnetic susceptibility of the whole periodic table.

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    $\begingroup$ An answer exactly at my knowledge level; great, thanks! :-) $\endgroup$ Commented Jul 13, 2013 at 20:47
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Ca2+ is believed to be paramagnetic due to the excitation of one electron from the s-orbital to the emptied d-orbital (s and d orbital are closer in energy, thereby causing transition to occur between both orbitals) which renders the s orbital unpaired in its excited state and attracted to the magnetic field (PAULI PARAMAGNETISM). It is worthy of note that calcium in its ground state is diamagnetic... C.U.EBONG

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  • $\begingroup$ U.EBONG, The answer is really interesting. Do you have a reference paper to support this statement? Thanks. $\endgroup$ Commented Oct 11, 2022 at 12:03

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