Sciencemadness Discussion Board

Extracting Boron from a neo-magnet

symboom - 3-4-2011 at 22:23

a neodymium magnet is composed of iron neodymium and boron if i add hydrochloric acid will it cause it to beak apart the compound leaving only insoluble elemental Boron. then dry the mix and add acetone to the mix of iron and neodymium chloride neodymium insoluble in acetone
then mix NdCl3 with either lithium or calcium forming neodymium metal under argon or a polar solvent that allows for the reaction to occur and that does not boil at that temperature
tell me thoughts improvements if this can work not sure if the magnet is affected by acids

[Edited on 4-4-2011 by symboom]

blogfast25 - 4-4-2011 at 05:49

There's a lengthy thread on extracting Nd from neomagnets here:

http://www.sciencemadness.org/talk/viewthread.php?tid=14145#...

You need to remove the nickel coating first, as this will not dissolve in anything but nitric acid.

HCl is a good solvent for neomagnets (if you've removed the Ni cover!)

Acid insoluble boron is left behind as an amorphous brown powder (I've seen it).

Your idea of leaching out the FeCl3 (don’t forget to oxidise first!) with acetone is worth a try: you’ll need to grind your mixture of NdCl3 + FeCl3 quite finely and extract with several small aliquots of acetone, until the last batch of acetone no longer tests positive for Fe3+.

One potential problem is that FeCl3 has a strong tendency to hydrolyse. To prevent this, slowly evaporate the chloride mix, keeping the HCl content high, otherwise you end up with all sort of ferric oxychlorides which aren’t soluble in acetone…

But synthing neodymium metal by reduction in solvent? You can more or less kiss goodbye to that. The way to go is the reduction of NdF3 (neodymium trifluoride) with Mg in a thermite-like set up. NdF3 is fortunately water insoluble so can be obtained by adding a water soluble fluoride (NH4F, really - caution: very toxic!) to NdCl3 solution. Filter, wash and dry. React this with Mg powder acc. NdF3(s) + 3/2 Mg(s) === > Nd(l) + 3/2 MgF2(l) + reaction heat. The reaction enthalpy will be high enough so that the reaction products Nd + MgF2 will be obtained in the molten state (T > 1300 C), they will separate with the Nd sinking to the bottom, then cool and solidify with the slag of MgF2 nicely protecting the Nd against air… No argon needed! Light the mix as you would most thermite mixtures.

It's tempting to try this with the chloride: don't because there is not enough reaction enthalpy for it to work... Reduction of the chloride would probably work by heating anyhdrous NdCl3 with either Ca or Mg (possibly Li or Na) in a closed (steel) reactor. Additional problem: NdCl3 is very hygroscopic, so making anhydrous NdCl3 (and keeping it that way!) isn't child's play either...


[Edited on 4-4-2011 by blogfast25]

symboom - 4-4-2011 at 08:52

thank you very informative