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Author: Subject: The trouble with neodymium...
elementcollector1
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[*] posted on 2-12-2016 at 20:28


Also, quick question about reducing NdF3: Wasn't calcium metal suggested at one point as a suitable alternative?

HoF (CaF2) = -1228 kJ/mol (Wolfram Alpha)

2 NdF3 + 3 Ca -> 2 Nd + 3 CaF2: -370 kJ/mol, more than the equivalent with lithium. Is there some reason this is not preferred that I missed? I read over the entire thread, but couldn't seem to find it. Is it due to calcium fluoride's high melting point?




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[*] posted on 6-12-2016 at 12:40


Update: Just got my sample of pure neodymium oxide from Snaucke Elements on EBay. It arrived as a white/gray powder, with distinct pink tinges under incandescent light.



Most of the sample was then mixed with diluted acetic acid and boiled, in the hopes that this would dissolve the oxide. To my surprise, this worked!



Once the bubblegum-pink neodymium acetate had crashed out, I then decided to test if it dissolved in ethanol. The results gave some important clues:

-The filtrate upon first coming out was the same dark pink as the concentrated aqueous solution. This is presumably due to residual water in the filtered acetate.

-As time went on, however, the filtered liquid became more dilute in coloration, and the drops coming out of the bottom of the filter paper were clear in coloration. The solid acetate on top retained mostly the same volume, though it had largely disintegrated into a paste. This indicates that the solid neodymium acetate is not soluble in ethanol, making separation of iron by selective dissolution in ethanol a viable procedure, provided the iron/neodymium acetates are boiled to dryness.



EDIT: Well, what do you know! The ethanol and water ended up separating just a few minutes after I made this post, and something cloudy has formed at the interface.



[Edited on 12-6-2016 by elementcollector1]




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[*] posted on 6-12-2016 at 12:49


Yay!





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[*] posted on 23-5-2017 at 10:42


Quote: Originally posted by elementcollector1  
Also, quick question about reducing NdF3: Wasn't calcium metal suggested at one point as a suitable alternative?

HoF (CaF2) = -1228 kJ/mol (Wolfram Alpha)

2 NdF3 + 3 Ca -> 2 Nd + 3 CaF2: -370 kJ/mol, more than the equivalent with lithium. Is there some reason this is not preferred that I missed? I read over the entire thread, but couldn't seem to find it. Is it due to calcium fluoride's high melting point?


I'm confident that you're absolutely right and that it's superior to commercial lithium.





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[*] posted on 4-10-2017 at 10:44


The reason lithium is preferred is its fluoride's melting point. Ideally the reaction is run similar to a thermite, where all products are in the liquid phase. This affords good metal-slag separation.
Melting points:
Nd - 1021 C
LiF - 845 C
CaF<sub>2</sub> - 1418 C

The reaction's target minimum running temperature is one where everything is molten, thus using LiF reduces this requirement to 1021C.

The other consideration is reaction enthalpy. Careful choice of reducing agent yields extra heat as part of the reaction, helping keep your external heating requirements low. Coincidentally I just yesterday found my calculations for reaction enthalpies for various reducing agents (the more negative the value, the greater the heat released):

Zn - deltaH = 1020.8
Ga - deltaH = 494
Mg - deltaH = -58.6
Na - deltaH = -72.8
Li - deltaH = -193.79
Ca - deltaH = -363.73



These were calculated by using the reaction equation for each:
A NdF<sub>3</sub> + B M == C Nd + D MFx
where ABCD are the coefficients and M is the chosen metal. deltaH for elements is 0, so the calculation boils down to
dH = D*deltaH(MFx) - A*deltaH(NdF<sub>3</sub>;)

In retrospect, I used the values for the solids but the products will be liquid. This changes the numbers a little, but preserves their order in the list.

Calcium produces the most heat but requires the highest temperature to melt, and I'm not sure if the tradeoff is favorable or not. If you have calcium metal, it might be worth a shot!

I've been thinking about this experiment again recently. God, I need to just try the last step already.
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[*] posted on 4-10-2017 at 11:35


Interesting.
"A binary eutectic composition of Formula and Formula was observed to melt at 769°C."
From
http://jes.ecsdl.org/content/104/11/661
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[*] posted on 5-10-2017 at 07:32


Formula and Formula you say? :P

From the link (in case it breaks): "A binary eutectic composition of 80.5 mole%LiF and 19.5 mole%CaF<sub>2</sub> was observed to melt at 769°C."

Somewhat less than LiF alone, but the reaction still needs to achieve 1021C to melt the Nd.
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[*] posted on 10-4-2018 at 15:50


This thread's been quiet for some time now... I hope my intrusion isn't too unwelcome, but I've been reading and re-reading the earlier bits of this thread a lot in the past few weeks, and now I'm finally throwing my hat in the ring here with some questions!
I'm a simple man: all I really want is to grow some pretty pink crystals of Nd2(SO4)3*8H2O (yeah, there are other, more accessible compounds that fill this role, but honestly I also just want the coolness of them being a Nd compound :D). Based on earlier discussions in this thread, I processed a half-inch cube Nd magnet using the oxalate separation method: demagnetize, remove plating, dissolve in sulfuric acid, filter, oxidize all Fe(II) to Fe(III), precipitate Nd2(Ox)3 with oxalic acid. That worked fairly well, I think, as it has for other people in this thread. Later, I was planning on converting the oxalate to the oxide, then redissolving in H2SO4 for the final sulfate.
...Buuut, then I read Blogfast's posts on page 4, noting his discovery that Nd2(Ox)3 from this method contains significant Fe(III) contamination. :(
...BUUUT, then I read Blogfast's OTHER posts on page 4 about separating Nd+3 from Fe+3 using fresh iron hydroxide/oxide-hydroxide/whatever you want to call it! It seems perfect: FeO(OH)*nH2O is easy to prepare, and apparently, when added in excess to Nd/Fe sulfate solution, produces almost perfect precipitation of iron as... something, the old posts are actually very unclear about what the iron becomes, but it appears that an oxide of iron is involved.

And that's basically my question. I've literally been thinking about this iron separation problem all day, and I can't work it out. How does adding solid Fe(OH)3 (or FeO(OH), or whatever) to a solution of a ferric salt cause all the iron cations to precipitate out? What do they turn into? What does the Fe(OH)3 turn into? What's the reaction/mechanism? And why does it leave the Nd virtually untouched?

[Edited on 4/10/2018 by Lab Rat]
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[*] posted on 16-4-2018 at 06:02


Doesn't look like anyone knows! Unfortunately blogfast isn't around currently to answer. Perhaps some sort of common ion effect?

I had great results with separation via the potassium sulfate double salt route. If your goal is to make crystals, though, iron removal may be less important. The very act of growing crystals will purify your desired product! You just might have to brush the iron crud off the surface.
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[*] posted on 16-4-2018 at 12:44


Hey, thanks for replying! Eh, that's ok, I'll try the iron thing out on a small batch anyway just in case. And yeah, even just separating Nd salt by crystallization is probably sufficient for me, but... I have just as much doing the chemistry as I do looking at the final product!

As for the double salt method, GAAH, it seems like everyone has had luck with that except for me! Recently, I dissolved a 45.3 gram "de-shelled" magnet in 80 mL of 9M H2SO4, diluted it to 600 mL (to let all salts stay dissolved), filtered it, and added 7.3 g dissolved K2SO4 (which I believe is stoichiometric)... And nothing happened. I added a bunch more, and I DID finally get precipitate, but the reaction was very slow. So okay, whatever, chemistry isn't as clean as I dream of it being, I need to add more than one equivalent.

But the other problem is that I can tell that the reaction was incomplete, because the supernatant solution was still different colors under fluorescent and incandescent light, indicating remaining Nd.
I figured, hey, maybe I just needed to add even more K2SO4. To test that theory before doing it, I instead made a small amount of saturated K2SO4 solution and added a few drops of my magnet solution into that. Weirdly, though, no additional precipitate!

Is it possible that my magnet solution was too acidic, causing the double salt precipitation to become unfavorable or something? Or maybe the reaction DID go to completion, and I'm wrong about the different colors thing. When I get the chance to work on it again, I'll try oxidizing to Fe(III) with H2O2 and testing for Nd with oxalic acid.

[Edited on 4/16/2018 by Lab Rat]
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[*] posted on 17-4-2018 at 05:35


I ran into a similar problem. We found that just adding the solid K2SO4 directly to the magnet solution is far better than adding a solution. I guess the extra water interferes with precipitation for some reason (doesn't make a lot of sense to me, but it does work. It could be a pH thing like you mentioned). Just add the solid salt, stir vigorously for something like 30 minutes, allow to settle, then check the supernatant for color. Changing solution colors definitely indicates Nd still in solution.
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[*] posted on 17-4-2018 at 15:11


Ooooooh, VERY good info, thanks so much! And now that I think about it, I'd be too afraid to try and modify the pH with base, because introducing alkali or ammonium ions in the presence of sulfate might trigger early precipitation of the double salt before I want it.
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[*] posted on 3-5-2018 at 10:42


Ok, here's a little update from me, if anyone's interested. In my second post from a couple weeks ago, I mentioned that my double salt Nd precipitation didn't appear to go to completion, as evidenced by different solution color under different lighting. When I got to work on it again, I took a small amount of my solution (containing mostly Fe(ii) with suspected leftover Nd(iii) sulfates, as well as a lot of excess potassium sulfate), filtered it, oxidized the iron, and added plenty of oxalic acid to test for neodymium. To my pleasant surprise, the test failed! No precipitate.

Either I was completely wrong before and the reaction DID go to completion and I literally just observed the color wrong, or leaving it alone for two weeks allowed it to move towards completion. Honestly, I'm thinking it was probably the first thing lol

Filtered and washed the double salt, then reacted it with excess KOH, and filtered and washed the Nd hydroxide (pale green under fluorescent, pink under incandescent). I'm gonna wait to continue until I finish my vacuum filter setup--all I need is the right water pump to connect to my aspirator, and the tubing--so that the filtering and washing steps are faster.
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[*] posted on 7-5-2018 at 08:33


Thanks for posting all this, Lab Rat! Great work!

I've been working on Nd as well. Separating it with oxalic acid worked well for me, and rinsing with ~5% H2O2 removed most of my Fe contamination. I then thermally decomposed the Nd-oxalate, to be left with a brown oxide, the brow colour being due to contamination with Pr. A few percent of Pr is enough to make it dark brown. When dissolving this "didymium"oxide in H2SO4 soln., the sulphate crystals were a bit more reddish than those of completely pure Nd-sulphate.

If you decide to thermally decompose some of your Nd-oxalate, please do not strongly calcine it, or your oxide may become hard to dissolve in sulphuric or any other acid.

Besides, I've been working on the dissolution of industrially calcined Nd2O3 (which is pure with respect to Pr-content), which is a very difficult task. So far I failed in coming up with a good method to do this.
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[*] posted on 10-5-2018 at 12:21


Hey man, sounds good! Yeah, to be honest, there's no way I can expect my sample to be free of praseodymium, but the ionic radii of Pr and Nd are so similar that I would expect Pr contamination to simply fill the same holes as Nd in the crystal structure.

As for the calcination, I've seen that issue arise once or twice in the depths of this thread, but I don't really understand it... If you heat Nd2(Ox)3, you get Nd2O3. And if you heat it REALLY hot, you still get Nd2O3, right? How can it be that strongly calcining it makes it harder to dissolve? Does it transform into an incredibly stable solid state structure at high temperatures or something?


As for dissolving Nd2O3, I wonder if you've seen this paper, it might be userful: https://www.degruyter.com/downloadpdf/j/ncrs.2002.217.issue-...

It's actually about dissolving praseodymium oxide to get the sulfate, but I'm pretty confident the method would work with Nd, too. Basically, they wet the oxide powder with water first, then added conc. H2SO4 dropwise to dissolve the oxide and form the sulfate. I guess then for additional purification/clean crystal growth, they dissolved it in 1:1 methanol:water (v/v), and added a tiny bit of 2,2'-bipyridine (presumably to form some kind of bipy coordination complex). Evaporation yielded green needle crystals.
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[*] posted on 14-5-2018 at 03:47


Strong heating of oxides makes the oxide more compact and more crystalline. The reactivity of the sample then strongly decreases and this can go so far that it becomes nearly impossible to dissolve the oxide in anything else than molten glass or molten alkalies.

I have encountered many oxidis which have become like this:
- Nd2O3
- Er2O3
- Pr2O3
- SnO2
- TiO2
- Co3O4
- Cr2O3
- Al2O3
- Fe2O3
- Sb2O5
- Nb2O5

Once you have an oxide in this inert state it has become nearly inaccessible for aqueous chemistry. In this way I lost well over 100 euros worth of money by buying oxides of metals and not being capable of using them in any meaningful way.




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[*] posted on 8-6-2018 at 06:07


Quote: Originally posted by Lab Rat  
(...)

As for dissolving Nd2O3, I wonder if you've seen this paper, it might be userful: https://www.degruyter.com/downloadpdf/j/ncrs.2002.217.issue-...

It's actually about dissolving praseodymium oxide to get the sulfate, but I'm pretty confident the method would work with Nd, too. Basically, they wet the oxide powder with water first, then added conc. H2SO4 dropwise to dissolve the oxide and form the sulfate. I guess then for additional purification/clean crystal growth, they dissolved it in 1:1 methanol:water (v/v), and added a tiny bit of 2,2'-bipyridine (presumably to form some kind of bipy coordination complex). Evaporation yielded green needle crystals.


Well, it is evident from their description, that they use a weakly calcined Pr-oxide. Otherwise the dropwise addition of sulphuric acid would not make any sense. Wetting a hard-calcined oxide does nothing at all to such oxide. (I had a bit of success by partially dissolving hard-calcined Nd2O3 in sulphuric acid by boiling it for over an hour in a distilation set-up.)

Adding the sulphuric acid dropwise seems to me like a way of constraining the excess of sulphuric acid to a minimum. I assume that an excess of the acid is needed in order to solve the oxide (even if it is only weakly calcined). They are still left with at least a little excess acid of though, because they heat it to only near-dryness, instead of evaporating the excess H2SO4.

It seems to me that the addition of the CH3OH/H2O mixture is their way to neutralise the excess acid. And indeed a precipitate forms, presumably of Pr(OH)3.nH2O. It could be that here the role of the pyridine comes in, as an agent to dissolve the hydroxide again, so that the hydroxide will not serve as a nucleation agent for the Pr-sulphate they are after. I only suppose this; if anyone knows a better explanation, please post!

So I think that the mixture of methanol, water and the excess H2SO4 yields them a more or less neutral solution, which would prevent H2SO4 from entering the crystal lattice. (When you crystallise Pr-sulphate from a strongly acid solution, H2SO4 *will* enter into the lattice - I know from experience. Successive steps of calcining such results and recrystallising will reduce the amount of acid in the Pr-sulphate by a factor of ~100 in each step.)

For me the big question about the article you mention is the role of the 2,2'-bipyridine. It is used in a large amount. The calculated molar ratio is

Pr2O3 : bipy = 1 : 1.9

or

Pr : bipy = 2 : 1.9 (almost one bipy ligand for each Pr atom)

That their solution turns orange shows that there must be a strong interaction between the bipy ligands and the f-orbitals, since the green colour of Pr is due to transitions involving the f-electrons.
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[*] posted on 19-6-2018 at 08:10


Sorry, not related to what we were just talking about, but two quick things about Nd magnet separation:

1. In the sulfate double salt precipitation method, don't heat the mixture above 50°C. Honestly there's no reason to heat it at all (I don't really know why I did it anyway), but one batch I warmed and stirred just to quicken the precipitation (which, in my experience, is a very slow reaction, taking upwards of 24 hours to complete), and another batch I warmed upwards of 70°C and stirred, and that precipitate appears noticeably orange (probs iron 3) under fluorescent light. I'll probably be able to separate it out somehow later.

2. I don't recommend washing/drying the sulfate double salt with acetone. I didn't have a vacuum filter setup until just now, so to dry my product, I'd often suspend it in acetone and re-gravity-filter, because the acetone evaporates quicker than water. But in my experience, the double salt appears slightly soluble or at least colloidal in acetone. Either way, I don't really recommend it.

Honestly, neither of these things have been suggested anywhere above, but like... just so ya know, I guess!

[Edited on 6/19/2018 by Lab Rat]
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[*] posted on 19-8-2018 at 14:55


Quote: Originally posted by Lab Rat  


And that's basically my question. I've literally been thinking about this iron separation problem all day, and I can't work it out. How does adding solid Fe(OH)3 (or FeO(OH), or whatever) to a solution of a ferric salt cause all the iron cations to precipitate out? What do they turn into? What does the Fe(OH)3 turn into? What's the reaction/mechanism? And why does it leave the Nd virtually untouched?

[Edited on 4/10/2018 by Lab Rat]



This isn't a strictly chemical process. It has a lot to do with adsorption. Ferric oxyhydroxide (from ferric chloride plus a hydroxide) has long been used to wash heavy metal ions out of ground-water.

It's unlikely that this method would necessarily leave Nd totally untouched, but this physisorption process is very pH sensitive and operating near the notch of the typical V-shaped solubility curve for the targeted metal hydroxide is the aim. Normally, for ferric iron that is about pH 7.8. This is close to the pH 7 that blogfast25 mentioned on p. 4. Presumably, the solubility minima for Nd lies at a somewhat higher pH.





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[*] posted on 20-8-2018 at 02:44


Is there any solvents that can dissolve amorphous B? CS2?



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[*] posted on 21-8-2018 at 17:57


Similar dissolves simular. Most like the way silicon oil cleans dust better than soap does clean it. I remind carbon is more soluble in water than alcohol, and > than organic solvents. Thats because pi bounds and some e pairs make their fashion to surpass the dynamics of the solvents requirements to carry the C particles along. Also, not less important you have to first of all undo the clumps of amorphic B then they will inevitably solubilize. Then you go 2 steps: create a silly putty of B + surfactant (anything that matches the maximum dipole moment achievable for B bonds), heat and stir, then dilute in suitable solvent for both, same way you would first rub grease with soap and they dilute both in water. For the surfactant I would suggest sulphonic acid.

[Edited on 8-22-2018 by Poppy]
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[*] posted on 8-9-2018 at 13:50


when I use oxalic acid to precipitate Nd3+ from the magnet chloride solution, should I add excess oxalic to complex all Fe3+ or slight excess relative to Nd3+ is enough?



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[*] posted on 26-9-2018 at 09:04


Quote: Originally posted by Dan Vizine  

This isn't a strictly chemical process. It has a lot to do with adsorption. Ferric oxyhydroxide (from ferric chloride plus a hydroxide) has long been used to wash heavy metal ions out of ground-water.


Ahh, very interesting, thanks a ton! I've ultimately decided that I don't trust my abilities/equipment enough to rely on this adsorbtion process, and will instead probably just recrystallize my Nd2(SO4)3 at the end to purify. Also, I'm highly suspicious that the double salt precipitation step is pH sensitive, and its efficiency is low in the presence of acid. I tried to use Fe(OH)3 to neutralize excess acid, but making Fe(OH)3 is a pain, and so is filtering the excess back out (small particles). I'll try using bicarbonate in the future instead.
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