Sciencemadness Discussion Board - Colours

Colours

nezza - 6-7-2014 at 03:00

As a keen photographer & chemist I have always been interested in pretty colours so have been synthesising/photographing interesting coloured materials. I made some bromine a long time ago and have recently done a small scale synthesis of chromyl chloride (scary stuff) to compare the colour with bromine in visible and infra red wavelengths. Both liquids look similar with chromyl chloride looking more reddish to me and bromine brownish. Both are also transparent to Infra Red. I have also uploaded a slow motion video of samarium dissolving in acid to give Sm2+ (blood red) as a transient colour.

Attachment: Samarium.mp4 (2.7MB)
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blogfast25 - 6-7-2014 at 05:29

Quote: Originally posted by nezza

I have also uploaded a slow motion video of samarium dissolving in acid to give Sm2+ (blood red) as a transient colour.

Are you sure that this is Sm (II)?

With regards to SmCl3 Wiki states that:

"it can also be prepared from samarium metal and hydrogen chloride.[3][4]

2 Sm + 6 HCl → 2 SmCl3 + 3 H2

Aqueous solutions of samarium(III) chloride can be prepared by dissolving metallic samarium or samarium carbonate in hydrochloric acid."

The Ln elements strike me as electropositive enough to be oxidised by H3O+(aq) (from hydrochloric acid) to their III oxidation states, including those that exhibit also the II oxidation state.

The Sm (III) salts are yellow, as solids, not red.

Nice video, BTW...

nezza - 6-7-2014 at 07:35

Yes Samarium III is virtually colourless, but in a reducing environment the Sm II ion can exist. It is blood red in colour. The video is of samarium dissolving in Hydrochloric acid and the solution goes colourless as soon as all of the metal is dissolved.

Eddygp - 6-7-2014 at 08:14

Last two weeks I was at a work experience program thingy and they were working with perylene-diimides and various complex asymmetric phthalocyanines. I was surprised because I didn't expect so many brilliant colours that changed depending on the concentration in the solvent and how they changed, visibly, during the reactions that they went through.

woelen - 6-7-2014 at 09:43

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?

----------------------------------------------------------------------------

How old is your ampoule of CrO2Cl2? I also made some of this stuff several years ago:

http://woelen.homescience.net/science/chem/compounds/chromyl...

I made a few drops of the chemical, made pictures and did some experiments, but did not keep it around for more than a day. I read in an old text book that CrO2Cl2 slowly decomposes, giving chromium(III) oxide/chloride and oxygen. If you have the liquid in a sealed ampoule, then (if the book is correct) it will explode due to overpressure at some bad day. Maybe I am wrong and it is safe to keep the compound around, but I just want to mention it to you.

blogfast25 - 6-7-2014 at 09:46

Quote: Originally posted by nezza

Yes Samarium III is virtually colourless, but in a reducing environment the Sm II ion can exist. It is blood red in colour. The video is of samarium dissolving in Hydrochloric acid and the solution goes colourless as soon as all of the metal is dissolved.

That's very interesting.

Try what woelen suggested, although the electrochemical series should provide the answer as to what will happen/not happen.

nezza - 6-7-2014 at 10:16

The vial I photographed is fresh. I always keep it double bottled in case of accident (It's really nasty stuff). I have an old vial with less than 1 ml in I prepared a year or so ago that is still intact. That one is a bit murky no so I will try snapping it observing adequate safety to see if there has been a build up of pressure. As for the Zinc and Samarium question I will have a go.

woelen - 9-7-2014 at 00:13

I tried the experiment with Sm2O3, dissolved in excess hydrochloric acid, as described by me above. When small granules of zinc are added to such a very pale yellow solution, then a lot of hydrogen is produced, but not even a trace of the red color of Sm(2+) can be observed.

blogfast25 - 9-7-2014 at 04:13

Woelen:

My CRC ’86 gives:

Sm3+ + e → Sm2+, with ERed = -1.55 V
Zn → Zn2+ + 2 e, with EOx = + 0.76 V

So this reduction of Sm (III) to Sm (II) with Zn is thermodynamically unfavourable: ERed + EOx < 0 (Delta G > 0)

But both Mg and Al should do it with oxidation potentials of resp. + 2.372 and + 1.662 V

But I still find it puzzling the 'reducing conditions' in nezza's experiment can lead to Sm (II)... It seems to suggest that H2/H+ can reduce Sm (III) to Sm (II)? How?

[Edited on 9-7-2014 by blogfast25]

HgDinis25 - 9-7-2014 at 04:39

Samarium isn't like the other lanthanides in this aspect.

When the reaction kicks in, there are a lot of hydronium ions, comparing to the surface area of the samarium lump and the solutions isn't too hot, so the following reaction happens (mind the water in the equation):
Sm(s) + 2H3O+(aq) = Sm2+(aq) + H2(g) + 2H2O(l)

Sm2+ is red in colour, that's why you see the solution go red. However Sm2+ is unstable in the presence of hydronium ions and, as the reaction goes on, the solution gets hotter catalyzing the following reaction:
2Sm2+(aq) + 2H3O+(aq) = 2Sm3+(aq) + H2(g) + 2H2O(l)

woelen - 9-7-2014 at 04:44

Exactly for that reason I found it still interesting to try the experiment with Zn in acid. Sometimes the use of Zn as reductor gives surprising results. Not with Sm(3+) though.

Use of Al or Mg I can test, but I do not expect much of that, because the pure metals give very fast reactions with acids. Maybe the use of Devarda's alloy gives better results. I have some of that as well.

---------------------------------------------------------

@nezza: What IR-filter did you use for the IR-pictures. I have seen modified cameras (e.g. Canon SX150 or Sony CyberShot), sold by a UK-based company, where you must specify the filter to be built into the camera. Choices are
1) Monochrome Infrared: below 700nm opaque, above 800nm transparent, inbetween smooth transition from opaque to transparent.
2) Faux Colour Infrared: below 600nm opaque, above 695nm transparent, inbetween smooth transition from opaque to transparent.
3) Goldie: Below 550 nm opaque, above 550 nm transparent. Sharp transition from opaque to transparent at 550 nm.
4) Super Enhanced Colour: opaque below 650 nm, transparent above 750 nm, inbetween smooth transition from opaque to transparent. Also a /\-shaped area with max. transparency around 460 nm.
5) Deep Contrast: below 850nm opaque, above 900nm transparent, inbetween smooth transition from opaque to transparent.
6) No filter, full spectrum (UV/VIS/IR)

What filter would be necessary for the kind of pictures you made of the Br2 and CrO2Cl2?

[Edited on 9-7-14 by woelen]

blogfast25 - 9-7-2014 at 06:13

Quote: Originally posted by woelen

Use of Al or Mg I can test, but I do not expect much of that, because the pure metals give very fast reactions with acids. Maybe the use of Devarda's alloy gives better results. I have some of that as well.

Why not part neutralise with ammonia, thus decreasing [H3O+] substantially? Then test the metals with this quasi-neutral solution.

@HgDinis:

Yes, I can see that too but the question is WHY?

This has to be related to exactly how Sm is oxidised by hydronium ions.

[Edited on 9-7-2014 by blogfast25]

IrC - 9-7-2014 at 09:34

Quote: Originally posted by woelen

I have seen modified cameras (e.g. Canon SX150 or Sony CyberShot), sold by a UK-based company, where you must specify the filter to be built into the camera. Choices are
1) Monochrome Infrared: below 700nm opaque, above 800nm transparent, inbetween smooth transition from opaque to transparent.
2) Faux Colour Infrared: below 600nm opaque, above 695nm transparent, inbetween smooth transition from opaque to transparent.
3) Goldie: Below 550 nm opaque, above 550 nm transparent. Sharp transition from opaque to transparent at 550 nm.
4) Super Enhanced Colour: opaque below 650 nm, transparent above 750 nm, inbetween smooth transition from opaque to transparent. Also a /\-shaped area with max. transparency around 460 nm.
5) Deep Contrast: below 850nm opaque, above 900nm transparent, inbetween smooth transition from opaque to transparent.
6) No filter, full spectrum (UV/VIS/IR)

Woelen is it possible to buy just the filters from this company? Below list is the dimensions of the IR block filters that come factory in these two cameras. I have replaced them many times doing various experiments. Having no way to alter the built in software I have had to alter the thickness to get the camera back into focus. A tedious, delicate (and intricate) job replacing them but easy after you do it a few times. Main trouble is keeping dust out while working. I had toyed with building a uV version but no way can I afford to replace that big Zeis front lens with one made from a uV transparent material. I have however found it does pass uV well enough to be useful if lighting is not too low. On your #6 in the list they must be installing a transparent plate of exactly the correct thickness and index of refraction to maintain focus. Or are they able to alter the internal software in the camera CPU? When you switch between night shot (or night framing) and normal a solenoid physically rotates the filter in or out of the light path thus requiring focus to be readjusted (done automatically controlled by software to one of two focus alterations).

Sony DSC 707/717
1.24 mm thick, 12.95 mm X 10.95 mm IR block filter for CCD Assy.

Sony DSC 505
2.52 mm thick, 11 mm X 9.5 mm IR block filter for CCD Assy.

woelen - 9-7-2014 at 12:29

I will not attempt to do this rebuilding myself. I have two new and good digital cameras (Panasonic DMC FZ 50 for more advanced photography, having a very good lens and a Panasonic DMC FZ 200 which is somewhat less advanced, but has a very fast CCD and allows me to make movies at 200 frames per second), and do not want to take these apart. I have another camera (Canon PowerShot 610), but that I have given to one of my children, who uses it for decent photography. So, I have to purchase a camera anyway and then I feel more comfortable if I purchase a modified one for a decent second hand price plus a few tens of euros for the mod. In this way I can get such a Canon SX150 for EUR 180 or so, completely converted and tested OK. Many sellers sell this same camera for EUR 180, even without the conversion.

Brain&Force - 9-7-2014 at 12:49

I just happened upon this thread, and I have found a video that shows the exact same thing occurring with both samarium and praseodymium (skip to 0:46)

<iframe sandbox width="420" height="315" src="//www.youtube.com/embed/IFmAhhiam9g" frameborder="0" allowfullscreen></iframe>

The potential for the reaction Sm₂₊ + 2e_- ⇌ Sm is -2.68 V. This is greater than the reaction Sm₃₊ + 3e_- ⇌ Sm whose potential is -2.304 V. The oxidation to samarium(II) is favored as well as the oxidation to samarium(III), but the +2 state is too reducing to be stable in water, so it oxidizes again in the presence of hydronium. A similar thing likely happens with praseodymium, but it's likely moving up to the +4 state and being reduced to the +3 state.

Samarium(II) iodide, used in Barbier reactions and one-electron reductions, is blood red. It's normally prepared in THF, which is not oxidizing enough to produce the +3 state by itself (but probably will with air).

Bezaleel - 9-7-2014 at 13:16

I just tried to perform an electrolysis of a Sm(OAc)3 solution, 1 spatula tip full of small crystals that dissolved to an opalescent solution.

The reaction conditions were: 15V between electrodes (stainless steel), current: 230mA, electrode separation 6 mm (smallest distance). On the cathode some precipitate formed, stronlgy resembling Sm2O3. No red colouring at all. I do know it is mentioned in literature that Sm2+(aq) has a strong red colour, but it did not show up in this experiment.

Pictures of setup:

Some samarium acetate

Electrolysis at 12V, 230mA

Residue after electrolysis (about 1 minute)

blogfast25 - 9-7-2014 at 13:31

@B&F:

Great find, that video. You're giving me reasons to finally dissolve my SmCo magnet!

@Bezaleel:

You are electrolysing water but with a rather expensive electrolyte! The hydronium ions in the solution are much easier to reduce than the Sm(III) ions and are reduced preferentially. The samarium acetate acts only as a charge carrier.

To reduce Sm(III) by electrolysis would require an aprotic solvent, NOT water. Or an anhydrous molten salt of Sm.

[Edited on 9-7-2014 by blogfast25]

[Edited on 9-7-2014 by blogfast25]

Bezaleel - 9-7-2014 at 14:07

Yes, was afraid of that, and the experiment proved it.

Fortunately, Sm isn't that expensive, if you are content with a 99.5% RE purity level, and take the oxide as your starting point.

Brain&Force - 9-7-2014 at 14:13

blogfast, removing samarium from a solution of samarium and cobalt should be trivial - add concentrated ammonia and let the cobalt complex away as the samarium precipitates. Metallic Sm is only US$12 for 5 g, or $35 for 40g (at Metallium) so it's not expensive in that form either.

Does the samarium oxide or acetate fluoresce under shortwave UV?

nezza - 9-7-2014 at 22:35

Sorry about the delay in replying. The ir filter I use is 720nm.

woelen - 9-7-2014 at 22:41

Thanks for the info, then I should take option 1 for a converted camera. Now I first need to get some budget for this, probably after holidays.

blogfast25 - 10-7-2014 at 04:53

Quote: Originally posted by Brain&Force

blogfast, removing samarium from a solution of samarium and cobalt should be trivial - add concentrated ammonia and let the cobalt complex away as the samarium precipitates. UV?

I'll still try the double sulphate method first, as it stinks less.

Some of these ammine complexes don't form so easily: the Ni one is quite hard. Dunno about the Co one...

Bezaleel - 11-7-2014 at 08:34

Quote:

Nickel on Atomistry
Preparation of Nickel Salts free from Cobalt

Another method consists in adding ammonia to the impure bromide in aqueous solution, whereby nickel bromide hexammoniate, NiBr2.6NH3, separates out in beautiful violet crystals. Since cobalt does not yield a similar derivative under like conditions, a very pure salt of nickel may be obtained in this way. The hexammoniate of nickel chloride, namely, NiCl2.6NH3, may similarly be used.

So separating cobalt from samarium with ammonia may not be successful.

Brain&Force - 11-7-2014 at 09:28

The separation above relies on the fact that nickel is far, far more likely to complex under slightly alkaline conditions. Cobalt will just precipitate with dilute ammonia - you need concentrated ammonia to get it to complex, then you need to oxidize it with hydrogen peroxide to stabilize the complex in solution as hexaaminecobalt(III). Samarium doesn't complex at all, and will just precipitate as a white to slightly yellowish hydroxide.

If I had a lab I would attempt this myself.

DraconicAcetate - 11-7-2014 at 09:29

Cobalt does form a complex with ammonia. The nickel one simply has a low solubility with chloride or bromide counterions. The cobalt one will stay in solution.

Brain&Force - 11-7-2014 at 09:48

I think we need to move this discussion to "Obtaining rare earths around the house." Or maybe a new thread, "The trouble with samarium..."

blogfast25 - 11-7-2014 at 12:44

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 CoCl42-.

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 FeSCN2+! 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]

Brain&Force - 11-7-2014 at 13:22

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.

blogfast25 - 11-7-2014 at 13:37

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.

--------------------

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]

nezza - 14-7-2014 at 10:41

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.

woelen - 14-7-2014 at 22:54

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.

blogfast25 - 15-7-2014 at 08:24

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.

What conditions were these reductions attempted in?

I now have some Sm too and will have a stab at this...

Brain&Force - 15-7-2014 at 09:09

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>

blogfast25 - 15-7-2014 at 11:38

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 I2 in an aprotic solvent? That could work...

[Edited on 15-7-2014 by blogfast25]

Brain&Force - 15-7-2014 at 12:56

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.

nezza - 16-7-2014 at 23:40

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.

blogfast25 - 17-7-2014 at 04:09

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.

Brain&Force - 17-7-2014 at 12:02

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.

blogfast25 - 18-7-2014 at 07:52

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 I2 (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 SmI2 could then be attempted. It’s a long shot…

Brain&Force - 18-7-2014 at 17:28

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 TbI3 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.

blogfast25 - 19-7-2014 at 08:26

Today I attempted to react iodine with samarium metal: about 0.5 g of Sm (small pieces) with few round pellets of resublimed I2. 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)3:

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]

Brain&Force - 19-7-2014 at 09:46

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?

blogfast25 - 19-7-2014 at 11:51

Well, it appears that the direct union of Sm and I2 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.

Brain&Force - 15-8-2014 at 10:32

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)?

woelen - 3-4-2016 at 03:53

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.

-------------------------------------------------------------------------

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]

DraconicAcid - 8-1-2023 at 14:43

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.