# Why is neodymium the most paramagnetic lanthanide?

My textbook says that paramagnetism rises to a maximum in neodymium.

I don't understand how gadolinium has 8 unpaired electrons, whereas neodymium has only 4. Shouldn't paramagnetism be higher in gadolinium?

Has this got anything to do with the magnetic moment associated with orbital angular momentum?

The magnetic moment of many d-metal ions can be calculated by using the spin-only approximation because the strong ligand field quenches the orbital contribution. But, for the lanthanoids, where the spin orbital coupling is strong, the orbital angular momentum contributes to the magnetic moment, and the ions behave like almost free atoms. Therefore, the magnetic moment must be expressed in terms of the total angular momentum quantum number J: $$\mu = g_J{\{J(J+1)\}}^{1/2}\mu_B$$ where the Landé g-factor is $$g_J=1+\frac{S(S+1)-L(L+1)+J(J+1)}{2J(J+1)}$$ and $\mu_B$ is the Bohr Magneton.
Old question, but what is your textbook? Gadolinium is ferromagnetic just below room temperature and the most paramagnetic just above. Looking at this table we extract $$\begin{array}{rl}\text{Metal}&\chi_M/10^{-6}\text{cm}^3\text{mol}^{-1}\\\hline \text{La}&95.4\\ \text{Ce}&2500\\ \text{Pr}&5530\\ \text{Nd}&5930\\ \text{Pm}&\cdots\\ \text{Sm}&1278\\ \text{Eu}&30900\\ \text{Gd}&185000\\ \text{Tb}&170000\\ \text{Dy}&98000\\ \text{Ho}&72900\\ \text{Er}&48000\\ \text{Tm}&24700\\ \text{Yb}&67\\ \text{Lu}&182.9\\ \end{array}$$ The Gadolinium result is quoted at $$350 \text{ K}$$ so as not to be too close to its Curie point. So Gd is in fact the most paramagnetic at temperatures where it's not actually ferromagnetic. You should not think of lanthanide metals as atoms because they are most typically $$+3$$ ions with the $$6s$$ electrons at least stripped from the atoms. Europium and Ytterbium maybe as $$+2$$ ions and Ce and Tb perhaps as $$+4$$ ions to achieve half- or fully-filled $$4f$$ orbitals.
So your count should be $$2$$ unpaired electrons for Nd and $$7$$ for Gd.