It's relatively well-known that neodymium forms extremely strong magnets. However, it seems rather strange that this element in particular makes such strong magnets. What is it about neodymium that allows it to form such strong magnets?
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$\begingroup$ have you tried wikipedia? $\endgroup$– niels nielsenCommented May 4, 2018 at 4:11
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1$\begingroup$ It is its ability to form an alloy with iron and boron with a special property that all atoms are aligned while changing this alignment is very hard: quora.com/Why-are-neodymium-magnets-so-strong $\endgroup$– safesphereCommented May 4, 2018 at 6:00
1 Answer
The element neodymium is not magnetic. What people refer to when talking about neodymium magnets is the compound with approximate composition Nd$_2$Fe$_{14}$B. This is a ferromagnet containing mostly iron. The saturation magnetization is about the same as that of iron, approximately 2 tesla.
The difference with pure iron is that it is a permanent magnet with a strong coercive field. And that is due to fact that it is difficult to move the narrow walls of the magnetic domains. The domain walls are narrow because of the large magnetocrystalline anisotropy.
Anisotropy is larger than in iron for a number of reasons. Iron has a cubic crystal structure, the structure of Nd$_2$Fe$_{14}$B has lower symmetry.
The role of neodymium is to increases the magnetocrystalline anisotropy because of spin-orbit coupling mediated by the Nd $4f$ orbitals. In this respect, it is related to samarium cobalt magnets (SmCo$_5$ etc) but with iron which is much cheaper.
The coercivity is also strongly influenced by metallurgy, the microstructure of crystal grains, impurities, and inclusions.
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1$\begingroup$ This isn't an answer, I should think. Your (correct) observation simply pushes the question to "what is it about neodymium that makes its alloys with iron so much more magnetic than ordinary magnets?". $\endgroup$ Commented May 4, 2018 at 17:51
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4$\begingroup$ @EmilioPisanty That was not what the poster asked, but I have now added stuff like spin-orbit coupling and a bunch of other words that may not be very enlightening to the poster. $\endgroup$– user137289Commented May 4, 2018 at 17:56