It’s both. The electron $g$-factor within an atom contains a term proportional to the electron spin, and also a term proportional to the electron’s orbital angular momentum quantum number. In an atom with an unpaired $s$-wave electron, there isn’t any orbital angular momentum, so only the spin contributes. But if a valence electron has nonzero $L$, the orbital state contributes to the atom’s magnetic dipole moment as well.

For example, in iron, there are two unpaired d-shell electrons which contribute to ferromagnetism. However, beware that chemists label the relevant states $d_{z^2}$, which has $m=0$, and $d_{x^2-y^2}$, which is a superposition of $m=\pm2$. The orbital magnetic moment is proportional to $m$, and vanishes if you use the non-complex wavefunctions in which projections like $m=+2$ are unavailable.

Note that ferromagnetism is a condensed-matter phenomenon, which depends on a material’s crystal structure and whose presence or absence is the subject of a phase transition. You ask in a comment whether ferromagnetism is spin or orbital in nature. I’m pretty sure the answer is “both.” The iron ground state has angular momentum $^{2S+1}L_J = {}^5D_4$, which corresponds to spin $2\hbar$, orbital $2\hbar$, aligned for a total of $4\hbar$. The spin contributes more, since the spin g-factor is twice the orbital g-factor.

However, I’m not a condensed-matter expert, and there are a couple of ways I might be wrong. It could be the iron ions which polarize, since one conduction electron is delocalized to the metal, in which case the angular momentum increases to $^6D_{9/2}$. But it could also be that the orbital degrees of freedom are eaten up by the interatomic binds which form the lattice, so that only the spin degrees of freedom are available to polarize. If the orbital degrees of freedom are used to define the crystal lattice, that would also explain why chemists tend to use the real-valued wavefunctions instead of the complex-valued wavefunctions with definite $m$.

I’m about 90% certain that, of the two magnetic materials forming the torsion pendulum in this nifty experiment, one was magnetized entirely by electron spin, and one with zero electron spin. I don’t recall whether the other material was magnetized by electron orbital momentum or (more likely) by nuclear spin, and I don’t currently have access to the paper or it’s references. However, nuclei are absolutely magnetizable. Furthermore, in nuclei, the spins and orbital angular momenta of the nucleons are not separate observables, so their magnetizability always involves both terms.

It’s both. The electron $g$-factor within an atom contains a term proportional to the electron spin, and also a term proportional to the electron’s orbital angular momentum quantum number. In an atom with an unpaired $s$-wave electron, there isn’t any orbital angular momentum, so only the spin contributes. But if a valence electron has nonzero $L$, the orbital state contributes to the atom’s magnetic dipole moment as well.

For example, in iron, there are four unpaired d-shell electrons which contribute to ferromagnetism. However, beware that chemists label the relevant states $d_{z^2}$, which has $m=0$, and $d_{x^2-y^2}$, which is a superposition of $m=\pm2$. The orbital magnetic moment is proportional to $m$, and vanishes if you use the non-complex wavefunctions in which projections like $m=+2$ are unavailable.

Note that ferromagnetism is a condensed-matter phenomenon, which depends on a material’s crystal structure and whose presence or absence is the subject of a phase transition. You ask in a comment whether ferromagnetism is spin or orbital in nature. I’m pretty sure the answer is “both.” The iron ground state has angular momentum $^{2S+1}L_J = {}^5D_4$, which corresponds to spin $2\hbar$, orbital $2\hbar$, aligned for a total of $4\hbar$. The spin contributes more, since the spin g-factor is twice the orbital g-factor.

However, I’m not a condensed-matter expert, and there are a couple of ways I might be wrong. It could be the iron ions which polarize, since one conduction electron is delocalized to the metal, in which case the angular momentum increases to $^6D_{9/2}$. But it could also be that the orbital degrees of freedom are eaten up by the interatomic binds which form the lattice, so that only the spin degrees of freedom are available to polarize. If the orbital degrees of freedom are used to define the crystal lattice, that would also explain why chemists tend to use the real-valued wavefunctions instead of the complex-valued wavefunctions with definite $m$.

I’m about 90% certain that, of the two magnetic materials forming the torsion pendulum in this nifty experiment, one was magnetized entirely by electron spin, and one with zero electron spin. I don’t recall whether the other material was magnetized by electron orbital momentum or (more likely) by nuclear spin, and I don’t currently have access to the paper or it’s references. However, nuclei are absolutely magnetizable. Furthermore, in nuclei, the spins and orbital angular momenta of the nucleons are not separate observables, so their magnetizability always involves both terms.

It’s both. The electron $g$-factor within an atom contains a term proportional to the electron spin, and also a term proportional to the electron’s orbital angular momentum quantum number. In an atom with an unpaired $s$-wave electron, there isn’t any orbital angular momentum, so only the spin contributes. But if a valence electron has nonzero $L$, the orbital state contributes to the atom’s magnetic dipole moment as well.

For example, in iron, there are two unpaired d-shell electrons which contribute to ferromagnetism. However, beware that chemists label the relevant states $d_{z^2}$, which has $m=0$, and $d_{x^2-y^2}$, which is a superposition of $m=\pm2$. The orbital magnetic moment is proportional to $m$, and vanishes if you use the non-complex wavefunctions in which projections like $m=+2$ are unavailable.

It’s both. The electron $g$-factor within an atom contains a term proportional to the electron spin, and also a term proportional to the electron’s orbital angular momentum quantum number. In an atom with an unpaired $s$-wave electron, there isn’t any orbital angular momentum, so only the spin contributes. But if a valence electron has nonzero $L$, the orbital state contributes to the atom’s magnetic dipole moment as well.

For example, in iron, there are two unpaired d-shell electrons which contribute to ferromagnetism. However, beware that chemists label the relevant states $d_{z^2}$, which has $m=0$, and $d_{x^2-y^2}$, which is a superposition of $m=\pm2$. The orbital magnetic moment is proportional to $m$, and vanishes if you use the non-complex wavefunctions in which projections like $m=+2$ are unavailable.

Note that ferromagnetism is a condensed-matter phenomenon, which depends on a material’s crystal structure and whose presence or absence is the subject of a phase transition. You ask in a comment whether ferromagnetism is spin or orbital in nature. I’m pretty sure the answer is “both.” The iron ground state has angular momentum $^{2S+1}L_J = {}^5D_4$, which corresponds to spin $2\hbar$, orbital $2\hbar$, aligned for a total of $4\hbar$. The spin contributes more, since the spin g-factor is twice the orbital g-factor.

However, I’m not a condensed-matter expert, and there are a couple of ways I might be wrong. It could be the iron ions which polarize, since one conduction electron is delocalized to the metal, in which case the angular momentum increases to $^6D_{9/2}$. But it could also be that the orbital degrees of freedom are eaten up by the interatomic binds which form the lattice, so that only the spin degrees of freedom are available to polarize. If the orbital degrees of freedom are used to define the crystal lattice, that would also explain why chemists tend to use the real-valued wavefunctions instead of the complex-valued wavefunctions with definite $m$.

I’m about 90% certain that, of the two magnetic materials forming the torsion pendulum in this nifty experiment, one was magnetized entirely by electron spin, and one with zero electron spin. I don’t recall whether the other material was magnetized by electron orbital momentum or (more likely) by nuclear spin, and I don’t currently have access to the paper or it’s references. However, nuclei are absolutely magnetizable. Furthermore, in nuclei, the spins and orbital angular momenta of the nucleons are not separate observables, so their magnetizability always involves both terms.