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Brain&Force
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I think we need to move this discussion to "Obtaining rare earths around the house." Or maybe a new thread, "The trouble with samarium..." 
At the end of the day, simulating atoms doesn't beat working with the real things...
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blogfast25
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Ok, so far my Sm/Co separation (on an alleged SmCo5 magnet) can be summarised in two words: ‘iron’ and ‘loads’.
I dissolved the 6.8 g mini-magnets in 50 ml HCl 37 % with low heat, over the course of several hours. Immediately on addition of the acid I saw some
swirls of dark green emanating from the magnets, alongside the bubbles of hydrogen. This emerald green then gave way for the cobalt blue of
CoCl<sub>4</sub><sup>2-</sup>.
After dissolution was complete I diluted to 100 ml and the colour shifted to a wine red: this is ligand exchange from the cobalt (II) chloro complex
to the aqua complex.
While still hot, 10 g of solid K2SO4 was added which dissolved promptly but no precipitate formed (with Nd the double sulphate starts precipitating at
that point). Cooling and chilling overnight yielded a few crystals (of K2SO4, I think) but no Sm/K double sulphate hydrate. So it appears the Sm/K
double sulphate doesn’t exist or is far too soluble to be used here. The Sm sulphate solubility limit clearly wasn’t exceeded either.
Time for plan B. 150 ml of 33 % NH3 was added and the slurry was kept warm for about ½ hour. The supernatant liquid looked like dilute cobalt (II)
hexammine complex.
But the precipitate (expected to be mainly Sm(OH)3) raised suspicions: it peptised through the Buchner (most of it anyway) and looked much like
Fe(OH)3. A careful test on a few drops of the suspension with Fe free H2SO4 tested ludicrously positive for iron: first the suspension cleared up to a
light yellow solution with addition of a small amount of conc. H2SO4, then on adding 1 ml of 2 M KSCN the solution turned almost black with
FeSCN<sup>2+</sup>! The iron seems also to tie in with the initial green swirls.
I’m not sure it's worth pursuing this three tier separation for a few gram of Sm: without using the double sulphate method, I can only see
separation of the Fe and Sm as trisoxalatoferrate complex and Sm oxalate.
Time to have a look on eBay for Sm or its compounds…
[Edited on 11-7-2014 by blogfast25]
Just bought a 10 g sample of Sm metal for £10. Sorted.
[Edited on 11-7-2014 by blogfast25]
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Brain&Force
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blogfast, this is suspiciously similar to my separation of terbium.
This is probably a really dumb question, but have you checked the HCl for Fe contamination? When I dissolved the terbium in the Fe-loaded HCl I had,
it turned slightly yellowish before turning clear. I'm not sure what caused this.
Second, I've always had serious issues with filtering terbium hydroxide: it keeps on peptizing as well, just as you said. Terbium sulfate is also
soluble enough to run through a filter when rinsed - the filtrate turned cloudy white with terbium oxalate when sodium oxalate-bearing solution was
filtered and mixed with the terbium-containing filtrate.
Third, I've had the exact same issue with a lack of precipitation from a neodymium magnet solution with potassium sulfate and sulfuric acid. Samarium
sulfate is less soluble than neodymium sulfate, as evidenced here. It's probably the second-least soluble sulfate of all.
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blogfast25
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B&F:
My HCl is very, very slightly contaminated with iron, but that is not the cause here: there is far, far, far too much of it. This is coming from the
sample, NOT the reagents or cutting tool (none was used).
The precipitate that didn't dissolve in the ammonia (strong!) looked effectively like Fe(OH)3 and behaved like it. I'm sure you've noticed this
slightly iridescent layer floating on water when there's a lot of ferric hydroxide about? Well, I could see that here too!
With Nd the double K sulphate method has always worked splendidly for me, as it has for MrHomeScientist. It's well referenced too.
What I did here was the same as what I do to neomagchloride soup.
I will investigate the double sulphate (Sm/K) when I take delivery of the metal.
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Going back a bit to the dissolution of the SmCo5 magnet:
Here’s the solution just a few minutes into the process:
The ‘surprising’ green is likely caused by Fe (II) ions being formed.
Another 5 minutes later and the blue cobalt tetrachloride complex is now masking:
I think the Fe contamination of the magnet need not be very high to produce my results: at 5 % (e.g.) it would already wreak havoc, produce Fe(OH)3,
produce a strong thiocyanate response etc.
[Edited on 12-7-2014 by blogfast25]
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nezza
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I have an update on Samarium and chromyl chloride. I have tried reducing samarium (III) with various metals and reducing agents and none of the ones
available to me are capable of reducing Sm 3+ to Sm 2+. THat includes zinc & magnesium powder and sodium dithionite.
Woelen - The old chromyl chloride did not seem to be under pressure when I broke the ampoule and reacted as it should. Both my old sample and the
fresh one gave a brownish opaque deposit inside the ampoule after a few days. I wonder if this is some impurity or a reaction with residual water
vapour inside the vial.
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woelen
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Very interesting what you write about the CrO2Cl2. I might make a new sample then, of high purity and ampoule that in a nice glass ampoule, like I did
with Br2, SOCl2, SO2Cl2 and CH3COCl. It is a nice compound to have around as a "safe" sample.
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blogfast25
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Quote: Originally posted by nezza  | I have an update on Samarium and chromyl chloride. I have tried reducing samarium (III) with various metals and reducing agents and none of the ones
available to me are capable of reducing Sm 3+ to Sm 2+. THat includes zinc & magnesium powder and sodium dithionite.
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What conditions were these reductions attempted in?
I now have some Sm too and will have a stab at this...
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Brain&Force
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Try using an aprotic solvent like THF, if possible. Commerically, solutions of samarium(II) iodide are available in this form. Samarium(II) iodide
itself appears to be green, or somewhat dark bluish in THF, and oxidizes extremely rapidly in air:
<iframe sandbox width="420" height="315" src="//www.youtube.com/embed/eMBPNGTCqRE" frameborder="0" allowfullscreen></iframe>
At the end of the day, simulating atoms doesn't beat working with the real things...
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blogfast25
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The only aprotic solvent I have worth mentioning is ether. I'd have to start from an anhydrous Sm salt too. Or maybe react Sm with
I<sub>2</sub> in an aprotic solvent? That could work...
[Edited on 15-7-2014 by blogfast25]
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Brain&Force
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Well, I never had success with iodine and methanol, but samarium may be more willing to react than terbium. An oxygen-free environment will definitely
help, if available.
At the end of the day, simulating atoms doesn't beat working with the real things...
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nezza
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Blogfast25. All of my reduction attempt were made in dilute (approx 2M) hydrochloric acid solution. I too suspect an aprotic solvent and an oxygen
free atmosphere may be required to keep samarium in the 2+ oxidation state.
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blogfast25
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Another possible way to oxidise Sm to Sm(II) might be to react clean Sm with dry HCl at 'high' temperature. Like in the case of Fe, dry HCl may only
be able to oxidise it to SmCl2, not SmCl3.
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Brain&Force
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I'd think you'd have to limit the amount of HCl to prevent oxidiation to samarium(III). Or you could try comproportionating samarium metal with a
samarium(III) salt. If all else fails, try making a samarium chalcogenide.
At the end of the day, simulating atoms doesn't beat working with the real things...
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blogfast25
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With a Standard ΔH of Formation for SmCl3 of – 1026 kJ/mol and – 92 kJ/mol for HCl(g) (both CRC ‘86), the chlorination of Sm with HCl(g) will
run all the way to SmCl3, not SmCl2.
But I’m not sure such a highly ionic compound would dissolve in anything aprotic.
Tomorrow I’ll see if I can get Sm to react with liquid I<sub>2</sub> (at about 120 C). Suppose SmI3 is formed, that stands a better
chance of being soluble in e.g. ether, as the difference in ionic radii is larger for the iodide than the chloride. Reduction to
SmI<sub>2</sub> could then be attempted. It’s a long shot…
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Brain&Force
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I think diethyl ether may not cut it if you want it to dissolve in the solvent. But I was thinking of something you said:
| Quote: Originally posted by blogfast25 | | I like DA's idea a lot. For want of a better term, 'reactivities' of Tb and Zn are probably quite comparable. This has the potential of being highly
exothermic, especially if the TbI<sub>3</sub> turns out to be insoluble in the solvent. |
This may be enough to drive the reaction along, but it may not proceed to completion without stirring to dislodge any SmI3 from the
samarium pieces/powder you use.
If you can make and stabilize SmI2 in the solvent immediately, I'd be surprised. If not, I would crystallize it out and react it with more
samarium powder to make solid samarium(II) iodide.
At the end of the day, simulating atoms doesn't beat working with the real things...
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blogfast25
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Today I attempted to react iodine with samarium metal: about 0.5 g of Sm (small pieces) with few round pellets of resublimed
I<sub>2</sub>. These were placed in a large test tube, immersed in an oil bath from about 95 C onwards. Temperature of the oil then
gradually increased to about 200 C. The iodine could clearly be seen melting, then boiling and as luck would have it the colder upper part of the tube
provided more or less 100 % reflux. No attempt to exclude air oxygen was made in this first run. Set up:
Nothing I saw pointed to any reaction taking place, considering the formation of SmI2 or SmI3 should be highly exothermic. So I retrieved the tube and
let it cool. Then in an attempt to find unreacted metal I clamped up the tube at 45 degrees and started heating to bottom with a medium Bunsen flame
to sublime off the iodine. At some point reaction did take place: the pieces of samarium metal lit up to a bright red heat!
Of course I cannot claim this is a samarium iodide, as oxygen was present (particularly because of the cooling of the tube). I proceeded with fuming
off the unreacted iodine as completely as possible.
On cooling the following material was left at the bottom of the tube:
A yellow material, a little darker around the edges.
To this was added a few ml of water and the tube put on steam bath. Almost immediately some interesting things happened. Firstly the water turned
brown, which I first attributed to residual iodine. But that faded very quickly and gave way to a clear solution with partial white precipitation.
To it I added a few ml of HCl 33 % and the solution turned wine red again, while gas evolved. I believe there may have been some unreacted Sm metal,
dissolving in the HCl. Here it is after the addition of the HCl:
A few drops of strong sodium thiosulphate removed the wine red colour, proving it was caused by free iodine.
Strong ammonia was then added and what is presumed to be off-white Sm(OH)<sub>3</sub>:
I’m not sure whether any samarium iodide was formed here but I did make the metal oxidise in anhydrous conditions, at least that's something!
This experiment will now be repeated using argon fluxing.
[Edited on 19-7-2014 by blogfast25]
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Brain&Force
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I would guess the yellowish color is due to formation of samarium(III) triiodide (Sm(I3)3). When I attempted the same reaction
with terbium I got the yellow crust. I would suggest adding a drop of water to the metal to start the reaction (as you would with aluminum) but it
wouldn't be anhydrous samarium(III) iodide, would it?
At the end of the day, simulating atoms doesn't beat working with the real things...
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blogfast25
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Well, it appears that the direct union of Sm and I<sub>2</sub> is indeed possible.
About the same amounts of both reagents as previous were loaded into a large test tube, clamped at 45 degrees. Argon was the used to flush out the
oxygen, using a glass tube that went all the way down to the bottom. High Ar rate was maintained for a few minutes, then reduced and the tube
retracted until only about 2 cm protruded into the test tube, thereby maintaining the argon blanket.
The bottom of the test tube was then fired up and a couple of minutes later two pieces of samarium ‘lit up’ to a bright red glow for a few
seconds. It looked almost identical to the first test.
The excess iodine was then fumed off, this time taking advantage of the argon flow.
Here’s what it looked like:
The two black bits are unreacted samarium. Due to the coarseness of the metal and the volatility of the iodine making this reaction run to complete
oxidation of the Sm metal will not be easy.
Two side experiments were conducted too. The Sm(OH)3 from the previous post was redissolved in HCl 33 % and solid K2SO4 added. This dissolved easily
and on cooling a sandy precipitate of (presumed) samarium potassium sulphate hydrate formed.
Secondly, about 0.13 g of the Sm metal was dissolved in HCl 33 %. I did NOT observe any red swirls of Sm(II) were observed. Dissolution in the cold
proceeded only slowly and some white precipitate formed. This test will be repeated.
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Brain&Force
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I am way too late on this, but if you're going to or have continued the experiment:
- Have you attempted to dissolve any samarium in dilute hydrochloric acid? I think the excess hydronium may be ruining the effect.
- I don't think you have enough samarium for this, but have you tried igniting the samarium and then dropping it into the argon atmosphere?
Also, do you mind if I use the picture above on the Sciencemadness Wiki (http://http://sciencemadness.wikia.com/wiki/Sciencemadness_Wiki)?
At the end of the day, simulating atoms doesn't beat working with the real things...
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woelen
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| Quote: Originally posted by woelen | | The samarium is quite interesting. Do you also get the red color if you add powdered zinc to a nearly colorless solution of Sm(3+) in dilute
hydrochloric acid? |
I made an ampoule of yttrium for my element collection and had some left-over yttrium granules. I added these to dilute hydrochloric acid and when
this is done, you get a bright yellow solution and a lot of hydrogen is evolved. As soon as all metal is dissolved, the yellow color quickly fades and
the liquid becomes clear and colorless. Yttrium(III) ions are colorless.
So, apparently, yttrium also can exist in lower oxidation state in water, but only for a short time, just like samarium.
---------------------------------------------------------------------------------------
I also have some 99.95% samarium, intended as sample for my element collection, but half of it is oxidized and covered in a thick layer of Sm2O3 (pale
yellow powder) 
I decided to salvage the metal from this (ugly looking pieces, covered in dust, soft long needles of metal) and added some of this to dilute HCl. You
indeed get a deep red solution, but it seems not to be completely clear. Once all metal is gone, the liquid almost instantly loses its red color and
it also becomes clear and a pale yellow color remains.
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Yttrium is more promising for experimenting, its transient color remains for a longer time than samarium's transient color. The price of yttrium also
is acceptable, albeit not really cheap.
A good affordable and pure source (free of iron and other first row transition metals) is the following:
http://www.ebay.nl/itm/252032408773?_trksid=p2060353.m2749.l...
Ask for smaller granules, otherwise you get a single lump of well over 10 grams, but yttrium is quite hard and not easy to break up in smaller parts).
If you want to experiment, then I think you'd better receive precisely 10 grams in smaller particles, than e.g. 12 grams as a single lump.
[Edited on 4-4-16 by woelen]
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DraconicAcid
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I was looking for this thread! I also got red Sm(II) ions when dissolving samarium metal in hydrochloric acid. However, the samarium I was using was
leftover from an attempt to make samarium(III) sulphate, so there as some sulphate leftover, and I got red precipitate which seemed to be SmSO4. I
could probably isolate it if I was set up for working with an inert atmosphere.
Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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