Sciencemadness Discussion Board - Remarkable reaction of antimony

Remarkable reaction of antimony

woelen - 6-4-2015 at 10:13

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

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.

Sulaiman - 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.

Eddygp - 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?

aga - 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.

blogfast25 - 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 SbCl6-, then mix both solutions to obtain a solution that’s equimolar in SbCl4- (III) and SbCl6- (V) (or SbCl62- (III, IV) ?).

[Edited on 7-4-2015 by blogfast25]

IrC - 6-4-2015 at 18:09

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

Justin Blaise - 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.

Eddygp - 10-4-2015 at 06:19

Any further information?

woelen - 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]

aga - 10-4-2015 at 14:33

SELECT MAX(time) as MyTime from life;

woelen - 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.

blogfast25 - 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]

SbCl63- + SbCl6- < === > 2 SbCl62- (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+, insoluble Cs2SbCl6 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 HSbCl6 (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]

Pok - 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]

Eddygp - 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?

woelen - 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):

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

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]

woelen - 12-4-2015 at 11:49

Quote: Originally posted 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.

blogfast25 - 13-4-2015 at 05:50

Thanks for the photos, wonderful.

Revisiting my thread on KSbCl6 (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 KSbCl6 I found strong evidence pyrolysis generates SbCl5 (much like pyrolysis of K2SnCl6 produces SnCl4

. Would perhaps a mixture of SbCl3 (BP 224 C) and SbCl5 (BP 140 C) be generated?

[Edited on 13-4-2015 by blogfast25]

deltaH - 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]

woelen - 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]

blogfast25 - 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)3 or 5 (assuming they even exist at STP?), do you?

deltaH - 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?

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]

DraconicAcid - 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.

deltaH - 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.

DraconicAcid - 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]

blogfast25 - 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]

DraconicAcid - 13-4-2015 at 15:12

Quote: Originally posted by blogfast25

DeltaH:

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

[Edited on 13-4-2015 by blogfast25]

I don't say that they don't exist (the CRC says that lead(II) cyanide will dissolve in KCN(aq), so presumably there is a cyano complex of lead(II) in existence), but they won't show the great stability of transition metal complexes. The CRC doesn't list lead(IV) cyanide, which leads me to think that the cyanide would be oxidized by it.

deltaH - 13-4-2015 at 22:33

It's possible that Sb(CN)5 is not stable, yet that a mixed insoluble Sb(III) and Sb(V) cyanides might, nevertheless, be stable (if it forms at all).

That said, I do agree it's unlikely based on what we know, but woelen has said he'll try it out, so let's wait and see. I also agree that formation of cyanogen is likely and that cyanide will not bond strongly to the metal as with transition metals.

[Edited on 14-4-2015 by deltaH]

woelen - 14-4-2015 at 00:06

I can give it a try on a small scale (tens of mg, in a test tube) and I am not scared of HCN or (CN)2 in such small amounts. If I have enough of the dark blue compound with the chloride, then I could try adding a solution of KCN to the solid. I expect just formation of a white solid, but it is an easy try, so why not?

Other experiments I want to try are
- adding a solution of N(CH3)4Cl to the golden yellow solution with antimony(III) and antimony(V)
- dissolving Sb2O5 in 40% HBr and dissolving Sb2O3 in HBr and mixing these solutions and then adding a solution of CsBr to it. I can imagine though that the Sb2O5 could oxidize the HBr to Br2. I also need to make Sb2O5, I only have Sb2O3. I expect, however, that making Sb2O5 should not be too difficult.

My first goal, however, is to obtain more of the dark blue solid. This evening I'll filter it and put it aside for drying.

blogfast25 - 14-4-2015 at 06:19

Quote: Originally posted by woelen

I can imagine though that the Sb2O5 could oxidize the HBr to Br2. I also need to make Sb2O5, I only have Sb2O3. I expect, however, that making Sb2O5 should not be too difficult.

Acc. these here reduction potentials (http://web.archive.org/web/20070518092613/http://www.northla...):

Br- to Br2(aq) = - 1.09 V

Sb2O5 to Sb2O3 (acid conditions) = + 0.65 V

So, no chance of Sb(V) oxidising bromide to bromine, at least acc. these values. (Sb(V) does oxidise iodide to iodine, as per my experience and SRP values).

Also, based on the Cu(II) + cyanide == > copper(I) cyanide + cyanogen reaction, I think Sb(V) should also by strong enough to oxidise cyanide to cyanogen.

For preparation of Sb2O5, I suggest preparing a HSbCl6 solution from Sb2O3, HCl and hydrogen peroxide (as per above). Then dilute a lot and neutralise with ammonia solution. Sb2O5 should precipitate.

[Edited on 14-4-2015 by blogfast25]

kmno4 - 14-4-2015 at 09:03

Some info about colour of Cs chloroantimonates:
cssb2.bmp - 380kB

Some info about spectra Sb(III), Sb(V) and their mixture (in HCl(aq)):

abs.bmp - 150kB

Deeply coloured compounds can be also prepared with rubidium salts (Wienlan, Berichte, 1905). I tried [(n-Bu)4N]Cl, but only white precipitate formed.
Potassium salt is golden-yellow.

woelen - 15-4-2015 at 09:23

@kmno4: Interesting info. Also interesting to see that there are other stoichiometries than the one I have seen in Pok's reference.

The color I have indeed can be described as black, but if you observe very carefully, then you can see that the material has a very dark purple/blue color. You can best see this, if you keep the solid near a really black powder, such as carbon powder.

I tried to repeat the experiment with N(CH3)4Cl instead of CsCl, but this does not lead to a colored precipitate. I do get a precipitate, but it is white. Probably it is either N(CH3)4SbCl4 or N(CH3)4SbCl6, with only antimony in oxidation state +3 or oxidation state +5, but not both.

I also did an experiment with the black solid by adding it to 60% HNO3. When this is done, then it quickly becomes white and a small amount of a brown/orange gas can be observed (either NO2 or ONCl). The compound probably is oxidized by the nitric acid and it loses its chlorine, leaving Sb2O5 behind, which is known to be insoluble in nitric acid.

blogfast25 - 15-4-2015 at 10:34

Quote: Originally posted by woelen

Probably it is either N(CH3)4SbCl4 or N(CH3)4SbCl6, with only antimony in oxidation state +3 or oxidation state +5, but not both.

That can be verified by isolating and carefully washing the precipitate. Then prepare a little slurry of it in weak HCl and add strong KI solution to it.

SbCl6-/ SbCl4- => Ered = + 0.75 V
I-/I2 => Eox = - 0.54 V

So if the precipitate is the Sb(V) salt, iodine should form. I've used this test to differentiate between tin and antimony (Sn(IV) can't oxidise iodide).

[Edited on 15-4-2015 by blogfast25]

woelen - 15-4-2015 at 12:38

The final result of my synthesis of this compound is quite well. According to theory I should obtain just over 4.1 grams. I measured 3.9 grams of dark material, which is a yield of better than 95%. I think that the reaction has nearly 100% yield (measured, based on amount of Sb2O3 used), due to the very low solubility of the compound. The losses are mechanical.

I tested the remaining liquid after the material settled at the bottom. It does not show any opalescence on dilution with water. This means that it must be nearly free of antimony. Even a tiny amount of Sb2O3, dissolved in HCl, gives a white opalescence on dilution with water. I kept the colorless liquid, it is appr. 30 ml of concentrated HCl with a few tenths of grams of CsCl dissolved in it.

Below follow pictures of my 3.9 grams of dark powder:

dark_powder1.jpg - 95kB

Click the pictures for more detail. (The left picture is somewhat too blue, it was made in late afternoon daylight with blue sky, wityh the vial in the shadow).

I'll make a webpage about this interesting experiment (probably next weekend).

[Edited on 15-4-15 by woelen]

kmno4 - 16-4-2015 at 11:41

I decided to check preparation (given by Wienland) of "(Sb, Sn)CI6(NH4)2".
Equal volumes of solutions of Sn(IV) and Sb(III/V) were mixed and nothing happened. Then some of solid NH4Cl was added, with continuous mixing : it caused precipitation of black-violet ammonium complex salt. It is stable in strongly acidic media, water decomposes it to colourless products.
(both solutions of Sn and Sb were prepared from metals, HCl(aq) and calculated amounts of H2O2, to obtain ~0,1 mol Sn(IV) and ~(1+1) Sb(III)/Sb(V) concentrations)

[Edited on 16-4-2015 by kmno4]

woelen - 23-4-2015 at 11:06

I did the experiment with RbCl instead of CsCl, but this was not successful. When a solution of RbCl in conc. HCl is added to the golden yellow liquid, mentioned above, then nothing seems to happen. No dark color is produced, no solid material appears. However, after several days of standing, I have a small amount of a crystalline solid at the bottom, which is colorless/white.

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

For convenience I also ordered some Sb2O5, with the idea to do easier experimenting. But unfortunately, the Sb2O5 does not (or only very slowly) dissolve in conc. HCl

It is a pale yellow powder, which is remarkably inert. It does not dissolve in a hot solution of KOH, nor in hot conc. HCl.

I did an experiment by adding some Sb2O5 to conc. HCl and then heating it. It does not dissolve. Then I added some Sb2O3 to the still hot liquid. When this touches the liquid there is a brief hissing noise and it dissolves at once. The color of the liquid is very pale yellow. The Sb2O5 still does not dissolve.
Next, I added a few drops of a solution of CsCl in conc. HCl. This leads to formation of a white precipitate with a faint purplish/grey hue. So, only a very small amount of the Sb2O5 dissolved in HCl.

blogfast25 - 23-4-2015 at 14:17

Fusing the stubborn pentoxide with KOH or NaOH, followed by neutralisation of the antimonate should yield a more soluble form of the oxide. But you already knew that, I'm sure.

I'd also suggest to dissolve elemental Sb in aqua regia. It dissolves remarkably easily to HSbCl6 (+5).

[Edited on 23-4-2015 by blogfast25]

kmno4 - 24-4-2015 at 12:34

Quote:

I did the experiment with RbCl instead of CsCl, but this was not successful. When a solution of RbCl in conc. HCl is added to the golden yellow liquid, mentioned above, then nothing seems to happen. No dark color is produced, no solid material appears...

It is just a matter of concentrations. When Sb(III)/(V) concentration in solution is high (it is then coloured brown), both CsCl and RbCl give dark violet precipitates. Cesium salt is easier to prepare. In paper by Wienland, rubidium salt is prepared via slightly different (and complicated) route (the interested may bother to find and read the article)

woelen - 25-4-2015 at 12:54

@blogfast25: I tried dissolving in hot conc. KOH-solution, but this also does not work. Use of NaOH, according to literature, does not work, NaSbO3 is not soluble in water, it is one of the very few insoluble sodium salts. I did not fuse it with KOH.

@kmno4: What concentration is needed for the dark brown color? With the reagents I have I tried making the liquid as concentrated as possible, but the best I can reach is a golden yellow. If I add more Sb2O3 and H2O2, then the liquid becomes turbid. I have the impression that the dark brown solution, which you mention, only can be obtained with SbCl3 in HCl and SbCl5 in HCl. Both SbCl3 and SbCl5, however, are not something you will find outside of a lab in a home/amateur setting.

blogfast25 - 25-4-2015 at 13:19

Fusing with KOH should work, I think (Holleman says so

). Having said that, I have some Cr2O3 that completely resisted fusing with KOH, not even the slightest bit of dissolution. Some oxide grades are very stubborn, as we know.

Yes, my Holleman confirms sodium antimonate to be insoluble. Used for sodium determinations.

woelen - 27-4-2015 at 12:15

I made videos of the component, while it is still on the filterpaper. These filter papers were used for preparing the powdered compound and after scraping the solid from the paper, some material was left. I did an experiment with one of the filter papers, which was still wet with conc. HCl and another experiment with the dried filterpaper, which was perfectly dry.

These videos nicely show that the dark compound quickly decomposes in the presence of water:

http://www.homescience.net/chem/exps/Sb_III_V/filter_still_a...
http://www.homescience.net/chem/exps/Sb_III_V/filter_dried.a...

The still acidic one decomposes slower, but it finally also completely decomposes.

woelen - 28-6-2015 at 10:05

It took some time, but I finally found the time to make a real webpage of the experiments with the mixed valency antimony compound:

http://woelen.homescience.net/science/chem/exps/Sb_III_V/ind...

aga - 28-6-2015 at 14:25

Fabulous !

It is very hard to make any comment at all, due the sheer Awe inspired.

j_sum1 - 28-6-2015 at 15:16

Very seriously cool.
That caesium complex is very vivid in colour. Just amazing. Not something that I would have ever guessed.
And the disappearing black trick is amazing to watch.