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[*] posted on 6-4-2015 at 10:13
Remarkable reaction of antimony


Years ago I purchased 100 grams of this from a local pottery supplier, but I never did anything interesting with it. Last week, however, I stumbled upon this, while rearranging part of my lab, and I decided to read about antimony chemistry (which seems to be quite boring).

I read something about antimony(III) and antimony(V) and that in concentrated hydrochloric acid, these two compounds interact with formation of brown compounds. Normally, antimony chemistry is quite boring, all compounds are white or colorless, but I decided to give this a try.

I added some Sb2O3 to concentrated HCl (35% or so). The Sb2O3 dissolves very easily, giving the colorless SbCl4(-) ion in the concentrated acid. On dilution with water, the liquid becomes cloudy again, due to hydrolysis, with formation of basic antimony chloride and/or Sb2O3.

To the solution of Sb2O3 in conc. HCl I added a few drops of bleach. When this is done, no smell of chlorine is produced, but instead, the liquid becomes bright yellow. This probably is the brown compound. I kept adding drops of bleach and I managed to get a golden yellow/brown solution. Adding more bleach makes the liquid lighter again and at a certain point it becomes turbid as well. SbCl4(-) is oxidized by Cl2, formed from bleach and HCl. The colorless ion SbCl6(-) is formed, which like SbCl4(-) only is stable in concentrated HCl. On dilution it hydrolyses to Sb2O5.

Apparently, when both SbCl4(-) and SbCl6(-) are present, then indeed there is some interaction, leading to a yellow/brown solution. The book (a very old book, called "The chemical elements and their compounds" by Sidgwick) I read on mixed valency antimony(III/V) compounds is vague and talks about SbCl6(2-) and about precipitates of cesium and rubidium salts of this ion. I added a drop of a concentrated solution of CsCl to the yellow/brown solution and immediately, a deep indigo solid is produced :o This reaction really is remarkable. The solid, however, quickly fades and soon it is pale purple/violet.

I also tried adding a drop of a concentrated solution of CsCl to a solution of Sb2O3 in conc. HCl. This immediately leads to formation of a white solid. Probably this is CsSbCl4. Cesium is known to have many sparingly soluble salts of diverse cations (in this way in the past I already made CsICl4 and CsBr3, and also red Cs2CuCl4 and blue Cs2CoCl4).

If you have Sb2O3 and some cesium salt, then this experiment really is worth repeating. The dark indigo solid is really remarkable. Apparently, the yellow/brown liquid contains SbCl4(-) and SbCl6(-) and both ions most likely are precipitated in a single salt of Cs. I think that the intense color of this solid is due to charge transfer between the Sb(III) and Sb(V).

I tried to isolate the dark solid, but it is unstable. It quickly decomposes to pale purple and finally white material. I think that this is because of excess Sb(III) or Sb(V) which forms less soluble material than the mixed valency compound. More investigation is needed.

Pictures will follow soon.




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[*] posted on 6-4-2015 at 10:59


well I think that to post details of an interesting experiment
that I shall probably never do myself
then ask me to wait for photo's
is cruel.
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[*] posted on 6-4-2015 at 12:48


Certainly interesting. How did you try to isolate the c(a)esium compound? Does it decompose by itself or only when it comes into contact with air?




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[*] posted on 6-4-2015 at 13:31


Quote: Originally posted by Sulaiman  
... ask me to wait for photo's is cruel.

Patience patience.

He's already said further investigation is needed, and based on past form, you'll see photos.

Research Apostrophe Catastrophe while you wait.




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[*] posted on 6-4-2015 at 15:53


Very interesting, woelen.

Sb(III) is easily oxidised to Sn(V) by means of nitric acid (also hydrogen peroxide).

That would make it possible to prepare a solution of SbCl3 (in strong HCl), oxidise exactly half of it (by volume) to SbCl<sub>6</sub><sup>-</sup>, then mix both solutions to obtain a solution that’s equimolar in SbCl<sub>4</sub><sup>-</sup> (III) and SbCl<sub>6</sub><sup>-</sup> (V) (or SbCl<sub>6</sub><sup>2-</sup> (III, IV) ?).


[Edited on 7-4-2015 by blogfast25]




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[*] posted on 6-4-2015 at 18:09


Woelen can you also do some experiments with the metal? I have lots of it but no oxide.




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[*] posted on 6-4-2015 at 21:10


I'll have to add this to my to-do list. I'm excited to see some pictures.

@IrC - I once made the bright red SbI3 with I2 and Sb. If I remember correctly, I just refluxed the two in toluene for a while.


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[*] posted on 10-4-2015 at 06:19


Any further information?



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[*] posted on 10-4-2015 at 12:12


The weekend just started, I planned to continue with this in the weekend. During the week I am very busy now. I have to do exams soon and my evenings are filled with teaching myself all kinds of stuff: Oracle's OCA-JP 8 and OCP-JP 8 for the people who know this (and probably understand the amount of work needed for this). This must be done besides my work at daytime, so unfortuntely I have very limited time for experimenting.


[Edited on 10-4-15 by woelen]




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[*] posted on 10-4-2015 at 14:33


SELECT MAX(time) as MyTime from life;



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[*] posted on 11-4-2015 at 08:28


I found something new while experimenting. It is important to have excess cesium in the solution in order to keep the black solid.

I tried the suggestion of blogfast25. I did as follows:

- Take 5 ml of concentrated HCl (appr. 35% by weight)
- Add a spatula of Sb2O3. Dissolve all of it.
- Divide the liquid in two equal parts (at appr. 0.1 ml accuracy).
- To one part add appr. 0.25 ml of 50% H2O2 and heat gently. When this is done, then the liquid quickly turns pale green and a strong smell of Cl2 appears. Heat this liquid to nearly boiling. It foams/bubbles quite strongly, probably giving a mix of O2 and Cl2. At a certain point in time, however, the liquid becomes quiet and no bubbles appear anymore. At this point all H2O2 is used up and destroyed. I continued heating until the liquid started boiling to drive off any Cl2. The final liquid is very pale yellow/green.
- Add both parts to each other. When this is done, then the liquid obtains a deep golden color.
- In a separate test tube dissolve a fairly large amount of CsCl in conc. HCl. The solid dissolves quite well.
- Add appr. 2/3 of the golden yellow liquid to the solution of CsCl in conc. HCl. This results in formation of a very dark purple, nearly black solid. The solid slowly settles at the bottom and the liquid above the nearly black solid is very pale purple, nearly colorless. So, the black solid must be nearly insoluble.
- To the remaining 1/3 of the golden yellow liquid at a few crystals of solid CsCl and swirl. Immediately the liquid turns dark purple, but the color quickly fades again and half a minute later, the solid matter is white and the liquid again is golden yellow.

I now have set aside the test tube with the nearly black solid. I let it stand for a day in order to allow it to settle at the bottom. I hope to isolate a small amount of the black solid so that I can do further experiments with it. Hopefully it does not decompose on standing.




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[*] posted on 11-4-2015 at 09:12


I can confirm the pale yellow/green colour of Sb(V) in strong HCl, as per my own past experiments.

My initial theory was that, as woelen suspects, that Sb(III) and Sb(V) in strong HCl are in equilibrium as follows (complex formula is tentative in this scheme):

[Edited]

SbCl<sub>6</sub><sup>3-</sup> + SbCl<sub>6</sub><sup>-</sup> < === > 2 SbCl<sub>6</sub><sup>2-</sup> (III/V)

… with a low K (equilibrium mainly to the left), explaining the light colour of the combined Sb(III)/Sb(V) solution.

On adding Cs<sup>+</sup>, insoluble Cs<sub>2</sub>SbCl<sub>6</sub> would form, pulling the first equilibrium to the right (by removal of one of the reaction products, Le Chatelier).

But the slight purple colour of the suopernatant liquid after addition of the CsCl solution doesn’t really support that. I would have expected the colour of that supernatant to have been unaltered after the CsCl addition.

And the last experiment also doesn’t really support what I thought either.

It’ll be really interesting to see how the dark solid behaves. Oxygen could well slowly oxidise the Sb(III) part, thereby destroying it.

There seems to me to be some interaction between the caesium and the suspected Sb(III/V) complex (other than forming an insoluble compound with it). And perhaps even HCl plays a role, other than simply preventing hydrolysis? For instance HSbCl<sub>6</sub> (V), acc. Holleman, can be isolated as a 4.5 hydrate.

This is all VERY interesting and unusual for antimony.

Very nice work, can’t wait for any photos.

[Edited on 11-4-2015 by blogfast25]

[Edited on 12-4-2015 by blogfast25]




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[*] posted on 11-4-2015 at 15:58


It seems to be Cs2SbCl6: http://books.google.de/books?id=_1gFM51qpAMC&pg=PA410#v=...

There is also an explanation for the dark colour, which confirms woelens assumption about charge transfer between the two ions in the crystal lattice.

At least (NH4)2SbBr6 can be isolated (and prepared in a similar way) according to other books.

[Edited on 12-4-2015 by Pok]
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[*] posted on 12-4-2015 at 04:33


Quote: Originally posted by Pok  
It seems to be Cs2SbCl6: http://books.google.de/books?id=_1gFM51qpAMC&pg=PA410#v=...

There is also an explanation for the dark colour, which confirms woelens assumption about charge transfer between the two ions in the crystal lattice.

At least (NH4)2SbBr6 can be isolated (and prepared in a similar way) according to other books.

[Edited on 12-4-2015 by Pok]


It would be interesting to know more about the stability of this compound, since it might be stable as a solid under an argon atmosphere or it disproportionates into some sort of antimony halide and caesium halide too.

If it is possible to isolate as a solid, while this is from a hypothetical perspective, crystallography can be useful to learn more about the nature of the bonds.
Does anyone have an X-ray analyser for crystals at home? :D




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[*] posted on 12-4-2015 at 11:21


I now have pictures of the solution and of the precipitate (according to my previous post):


Here follows a picture of the golden/yellow solution (in sunshine, with a grey background):

Sb_III_V_yellow.jpg - 202kB


And here follows a picture of the precipitate (also in sunshine, so that nice glittering crystals can be observed):

Sb_III_V.jpg - 221kB


I filtered the dark precipitate (which was a rather unpleasant thing to due, due to the fumes of the concentrated HCl) and pressed the solid material between the filter paper and a lot of paper tissues so that I get almost dry material and I put it away for further drying. After half a day of drying it is almost dry and lost its pungent smell of hydrochloric acid. Tomorrow I expect it to be completely dry so that I can transfer it to a vial. The material seems to be air-stable, as long as it is kept dry. It cannot be rinsed with water. Doing so will decompose the material, it turns white in a minute or so when water is added. It is stable in conc. HCl (appr. 35% by weight).

Please click the pictures for more detail.

Some remarks:
- The dark precipitate consists of very fine glittering crystals. It is not slimy nor flocculent.
- The precipitate really has a very low solubility. When a small amount is added to concentrated HCl, then after a few minutes the liquid is perfectly colorless and the solid settled at the bottom.
- The black precipitate is decomposed by water quickly (I made a movie of its decomposition in water, that will follow lateron when I made a webpage of this experiment).
- Formation of the black precipitate occurs with Cs-salts, not with K-salts (I tried that). When only a little amount of Cs is present, then the precipitate quickly turns white (but still it remains a precipitate, probably Cs-stibnite or Cs-stibnate). When there is an excess amount of Cs, then the precipitate is stable.

Tomorrow I expect to have the dry black solid (a few hundreds of mg) and then I expect to be able to put some in a vial so that I can make good pictures of that. It looks like a very dark blue, nearly black powder. It will be hard to capture this color, most likely it will look just black on a picture.

[Edited on 12-4-15 by woelen]




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[*] posted on 12-4-2015 at 11:49


Quote: Originally posted by Pok  
It seems to be Cs2SbCl6: http://books.google.de/books?id=_1gFM51qpAMC&pg=PA410#v=...

There is also an explanation for the dark colour, which confirms woelens assumption about charge transfer between the two ions in the crystal lattice.

At least (NH4)2SbBr6 can be isolated (and prepared in a similar way) according to other books.

[Edited on 12-4-2015 by Pok]

Small, but interesting read. If this information is correct, then the compound can best be formulated as

Cs3[SbCl6].Cs[SbCl6],

with one Sb in oxidation state +3 and the other in oxidation state +5.




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[*] posted on 13-4-2015 at 05:50


Thanks for the photos, wonderful.

Revisiting my thread on KSbCl<sub>6</sub> (Sb(V)) I was reminded that this salt also hydrolyses with pure water but does dissolve easily in strong HCl:

http://www.sciencemadness.org/talk/viewthread.php?tid=15022

It could be interesting to submit the caesium 'double salt' (III/V) to strong heat and observe. With KSbCl<sub>6</sub> I found strong evidence pyrolysis generates SbCl<sub>5</sub> (much like pyrolysis of K<sub>2</sub>SnCl<sub>6</sub> produces SnCl<sub>4</sub>;). Would perhaps a mixture of SbCl<sub>3</sub> (BP 224 C) and SbCl<sub>5</sub> (BP 140 C) be generated?

[Edited on 13-4-2015 by blogfast25]




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[*] posted on 13-4-2015 at 10:36


Wonderful work woelen and stunning photos. On a side note, this makes me think of Prussian blues and so I find myself wondering if similar complexes can be prepared with antimony and cyanide as ligand. Have you tried this by any chance?

[Edited on 13-4-2015 by deltaH]




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[*] posted on 13-4-2015 at 13:10


I did not try the experiment with cyanide. I just discovered this remarkable property of antimony with chloride (and according to the link provided, it should also be possible with bromide). I, however, am willing to investigate this property of antimony with other anions and also with other large cations (e.g. Rb(+), NH4(+), N(CH3)4(+)). I'll do these experiments on a test tube scale.

I also can confirm for sure now that this compound is air-stable. I left a small amount on a small plastic spoon, just lying on my workbench for one and a half day. It still is a black powder.


I decided to make much more of this chemical so that I can do more experiments with it. I did the following:
- Dissolve 0.500 grams of Sb2O3 in appr. 15 ml of conc. HCl. The resulting liquid is colorless and slightly opalescent.
- Dissolve another portion of 0.500 grams of Sb2O3 in another portion of appr. 15 ml of conc. HCl.
- To the second portion, add appr. 0.5 ml of 50% H2O2. This is a large excess of H2O2. Boil off the second solution, so that all H2O2 is destroyed and the smell of chlorine only is weak. This second portion of antimony now certainly is in oxidation state +5, while no free oxidizer is left. This liquid is appr. 12 ml and it has a pale green/yellow color. It also is weakly opalescent.
- Mix both portions, this yields well over 25 ml of golden yellow liquid. This liquid is less opalescent than the two original liquids. The faint opalescence probably is due to slight hydrolysis of Sb(III) and/or Sb(V). It can be removed, but then a lot of conc. HCl is needed. I decided to accept the weak opalescence, it can never be more than mg-quantities in total and it hardly will contaminate the end-product.
- Dissolve well over 2.5 grams of CsCl in 20 ml of conc. HCl. This is an excess amount of Cs(+), assuming formation of Cs3[SbCl6].Cs[SbCl6].
- Slowly pour the golden yellow liquid in the solution of CsCl, NOT the other way around. Do this while swirling the beaker. In this way, the solution never is exposed to low concentrations of freely available Cs(+) ions. This is to assure that no white precipitate is formed. Previous experiments show that excess Cs(+) is important.
- A lot of very dark blue/purple precipitate is produced.

Tomorrow I expect that most will have settled at the bottom. Then I will filter and dry the solid. The yield should be around 4 grams.

My experiment is according to the following stoichiometry:

Sb2O3 + 4CsCl + 8HCl + H2O2 --> Cs3[SbCl6].Cs[SbCl6] + 5H2O

I used large excess H2O2 and very large excess HCl and slight excess CsCl. For 1.00 gram of Sb2O3, 2.31 grams of CsCl are needed and 4.12 grams of Cs3[SbCl6].Cs[SbCl6] are produced.

Tomorrow more results will follow.

[Edited on 13-4-15 by woelen]




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[*] posted on 13-4-2015 at 13:49


Quote: Originally posted by deltaH  
On a side note, this makes me think of Prussian blues and so I find myself wondering if similar complexes can be prepared with antimony and cyanide as ligand. Have you tried this by any chance?


Interesting thought but iron forms very strong coordination complexes with cyanide. By contrast, I don't even see an 'easy' way to prepare Sb(CN)<sub>3 or 5</sub> (assuming they even exist at STP?), do you?




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[*] posted on 13-4-2015 at 14:17


Quote: Originally posted by blogfast25  
Quote: Originally posted by deltaH  
On a side note, this makes me think of Prussian blues and so I find myself wondering if similar complexes can be prepared with antimony and cyanide as ligand. Have you tried this by any chance?


Interesting thought but iron forms very strong coordination complexes with cyanide. By contrast, I don't even see an 'easy' way to prepare Sb(CN)<sub>3 or 5</sub> (assuming they even exist at STP?), do you?


I wouldn't prepare Sb(CN)3 or 5, rather I'd use a basic soluble solution of mixed valence chlorides described and add a cyanide salt to it, relying on a ligand substitution reaction. Cyanide is a strong ligand and chloride is labile, so it might/should proceed to produce a mixed valence cyanide precipitate. Maybe it has a nice colour :)

Another variation is adding a thiocyanate salt to produce maybe a different hue.

[Edited on 13-4-2015 by deltaH]




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[*] posted on 13-4-2015 at 14:24


I would not even try to make a cyanide complex with antimony(V) or any other strong oxidizing agent, considering how copper(II) will precipitate copper(I) cyanide and give off cyanogen.



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[*] posted on 13-4-2015 at 14:32


That is a distinct possibility and one I believe woelen is well familiar with, hence the precautions he takes.



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[*] posted on 13-4-2015 at 14:47


Quote:
Cyanide is a strong ligand and chloride is labile, so it might/should proceed to produce a mixed valence cyanide precipitate.


Cyanide binds very nicely to transition metals, but I don't think it's much of a ligand for main-group metals (apart from zinc/cadmium/mercury, which are arguable).

[Edited on 13-4-2015 by DraconicAcid]




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[*] posted on 13-4-2015 at 15:01


DeltaH:

If you are going to rely on ligand substitution and start from an antimony chloride, those Sb cyanide complexes had better be extremely stable because antimony chlorides require very low pH and HCN is a weak acid.

Like DA, I don't believe main group cyanide complexes exist.

[Edited on 13-4-2015 by blogfast25]




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