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Author: Subject: Preparation of potassium hexafluoromanganate(IV) from household items
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[*] posted on 21-11-2015 at 17:05


Based on woelen's post, I fully expected a 'control' test with 2% acetic acid would give nearly identical results to the procedure with HF.
However...

In the video below, the procedure I outlined above is performed with HF cleaning product on the left, and diluted cleaning vinegar on the right.

<iframe sandbox width="280" height="160" src="//www.youtube.com/embed/VDzFLuyDoqo?rel=0" frameborder="0" allowfullscreen></iframe>

Disclaimer for blogfast: These results have not been confirmed by independent laboratory testing. Restrictions apply, results may vary. Consult your psychiatrist before using.

The vinegar reaction produce some short-lived precipitate, so the solutions were stirred just after the end of the video; here is a picture of the fully-reacted solutions. The solution resulting from the acetic acid reaction is very pale pink, the solution from the HF is noticeably golden-yellow. I will withhold my interpretation of these differences for now.

acetic control.jpg - 25kB

[Edited on 22-11-2015 by MolecularWorld]




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[*] posted on 22-11-2015 at 04:37


I did the experiment with 48% HF and 50% H2O2. I proceeded as follows:

- Put appr. 1.5 ml of 48% HF in a PP transparent testtube
- Add a small spatula of finely divided KMnO4: The KMnO4 quickly dissolves. The liquid heats up noticeably (becomes luke-warm) and a colorless gas is produced. The resulting liquid is clear and has the usual intense color of a solution of KMnO4. The liquid, however slowly produces gas.
- Take a plastic stick and dip this in 50% H2O2, so that a small droplet remains attached to the stick.
- Dip the stick in the dark purple solution of KMnO4 in 48% HF: The liquid fizzles when the stick is immersed into it. The color of the solution changes from deep purple to reddish brown. The intensity of the color also is reduced considerably.
- Rinse the stick and dip it another time in 50% H2O2 and again immerse the top of the stick in the liquid with HF. A lot of tiny bubbles are produced, the liquid becomes turbid/white.
- Dilute with a lot of water: The turbidity does not disappear at once. The liquid remains somewhat opalescent.

Interesting observation is that KMnO4, when added to 48% HF produces a gas and heat. Apparently there is some reaction. I think that this reaction is formation of the MnF6(2-) anion. This is accompanied by a change of the oxidation state and this only is possible when oxygen is produced in the reaction. My speculation is that this reaction occurs:

4 MnO4(-) + 24 HF --> 4 MnF6(2-) + 3 O2 + 10 H2O + 4 H(+)

This reaction, however (if it occurs), is not complete. A lot of MnO4(-) remains in solution.
Another reaction which may occur simply is decomposition of MnO4(-) into colloidal hydrous MnO2 and oxygen, with consumption of HF, which is converted to F(-) and H2O. Maybe both reactions occur simultaneously.

On addition of H2O2 there is reduction of manganese to oxidation state +2, with formation of MnF2. MnF2 is only very sparingly soluble, just over half a gram in luke-warm water, probably much less when there is excess fluoride present. This explains the formation of the white solid and the turbidity and opalescence on dilution.

This experiment certainly does not prove that I made MnF6(2-) or K2MnF6. It does show some interesting behavior.






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[*] posted on 22-11-2015 at 07:28


Interesting results.
The published procedures use an ice bath, to slow decomposition of the complex (to Mn(III), then Mn(II)). I noticed no reaction when the permanganate was added to very dilute HF, even when left to sit for several minutes: no gas, no color change. My reactions did not produce noticeable heat, and gave practically identical results when conducted with ice-cold and room-temperature solutions. The published procedures also have additional potassium, from potassium fluorides (which I formed by using an excess of hydrofluoric acid and some potassium hydroxide). If the permanganate reacts with the hydrofluoric acid, perhaps stoichiometric amounts of hydrofluoric acid could be used, and the reaction conducted in a saturated solution of potassium fluoride to precipitate the product. I wonder, if you did the reaction in an ice bath, with KF, and with stoichiometric quantities combined slowly, you just might form the golden precipitate...

Any comments as to why the control with acetic acid gave different results from the experiment with hydrofluoric acid, but similar results to the control with sulfuric acid? My thoughts are that this indicates the hydrofluoric acid must be doing something the other acids aren't, as everything else is constant. Whether this something is the formation of an unstable complex, or mysterious colorless contaminants that turn yellow under these conditions, has yet to be shown.

[Edited on 22-11-2015 by MolecularWorld]




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[*] posted on 22-11-2015 at 07:49


I think that the difference with HF is the formation of insoluble MnF2, which forms a white precipitate. This white precipitate, contaminated with traces of hydrous MnO2 will have a yellow appearance. Only if the acid is strong and present in excess quantities, all MnO2 is reduced on addition of excess H2O2 and then you get a completely colorless solution, or in the presence of fluoride, you get a white precipitate.

Do you have a link to the puiblished procedures? What is in your original post only shows the first page, the rest is not accessible without payment.

[Edited on 22-11-15 by woelen]




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[*] posted on 22-11-2015 at 08:07


Quote: Originally posted by woelen  
I think that the difference with HF is the formation of insoluble MnF2, which forms a white precipitate. This white precipitate, contaminated with traces of hydrous MnO2 will have a yellow appearance. Only if the acid is strong and present in excess quantities, all MnO2 is reduced on addition of excess H2O2 and then you get a completely colorless solution, or in the presence of fluoride, you get a white precipitate.

I don't quite understand this. I did the experiment with hydrofluoric acid, sulfuric acid, and acetic acid. In all cases, the acid should have been present in [slight] excess. But, sulfuric acid is strong, acetic acid is weak, yet they gave results that were very similar to each other, and different from the results with hydrofluoric acid. I suppose there might be traces of manganese dioxide in the yellow stuff, though I would have thought at any visible concentration, manganese dioxide would noticeably decompose additional hydrogen peroxide; my yellow product did not. Plus, I'm working with such dilute solutions, I would expect even slightly-soluble manganese(II) fluoride to be mostly or entirely dissolved.

Quote: Originally posted by woelen  
Do you have a link to the puiblished procedures? What is in your original post only shows the first page, the rest is not accessible without payment.

I'm working from the freely-available parts of the references in my original post, and the excerpts posted by Pok and I in later posts (link, link, link).

[Edited on 22-11-2015 by MolecularWorld]




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[*] posted on 22-11-2015 at 11:42


Quote: Originally posted by woelen  

Do you have a link to the puiblished procedures? What is in your original post only shows the first page, the rest is not accessible without payment.


Here's the Christe paper, though it's only a paragraph over one page.

Chemical synthesis of elemental fluorine
Karl O. Christe
Inorganic Chemistry 1986 25 (21), 3721-3722
DOI: 10.1021/ic00241a001

Attachment: Chemical synthesis of elemental fluorine.pdf (288kB)
This file has been downloaded 496 times

Footnote 17 looks informative, but my online access to Agnew. Chem only goes back to 1962 :(

Here's footnote 18, described by Christe a an improvement upon the preparation of K2MnF6, though (18) actually appears to deal with K2MnF5*H2O (?!)

M.K. Chaudhuri, J.C. Das, H.S. Dasgupta, Reactions of KMnO4—A novel method of preparation of pentafluoromanganate(IV)[MnF5]−, Journal of Inorganic and Nuclear Chemistry, Volume 43, Issue 1, 1981, Pages 85-87, ISSN 0022-1902, http://dx.doi.org/10.1016/0022-1902(81)80440-X.
(http://www.sciencedirect.com/science/article/pii/00221902818...)

Attachment: A NOVEL METHOD OF PREPARATION OF PENTAFLUOROMANGANATE (176kB)
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[*] posted on 22-11-2015 at 12:23


Quote: Originally posted by mayko  
Footnote 17 looks informative

I quoted this (very short) article earlier and you can find it here. This source also emphasizes the need to cool (what woelen apparently didn't do). Otherwise Mn(III) will be formed.

The increased yield is already mentioned in the book I cited. Washing with acetone lead to 78.6 % yield instead of 73 % when washed with HF.

[Edited on 22-11-2015 by Pok]
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[*] posted on 22-11-2015 at 15:00


I indeed didn't cool. The red/brown color I observed indeed most likely is the color of manganese(III). I'll repeat the experiment with chilled chemicals. In my experiment, the most surprising thing was that addition of KMnO4 to 48% HF already produces a noticeable amount of heat and also a gas (must be oxygen). I expected it to simply dissolve without any further action. Probably I need to add the KMnO4 very slowly, while the PP-testtube is in ice cold water. I hope to find time for this on Tuesday evening.



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[*] posted on 22-11-2015 at 16:33


@mayko: Thanks for the references. I liked the end of the Christe article so much, I added to my signature.

. . . . .

I noticed that, on sitting, my yellow fluoride solution turns pink. I had assumed that this was due to the complex decomposing. But my acetic acid control, as suggested by woelen, showed that in the presence of a dilute weak acid, the reaction I've been performing first forms manganese dioxide, which quickly dissolves to give a pink colored Mn(II) solution. I had thought that woelen's theory, that the golden liquid was a very pale pink liquid with suspended manganese dioxide particles, was unlikely, based on the fact that the yellow stuff didn't decompose peroxide. But his comment was enough for me to doubt this test, so...

I prepared a suspension of extremely fine manganese dioxide particles via a well-known procedure (as pointed out upthread by MrHomeScientist), and diluted it to approximate the color of my fluoride solutions. To this I added hydrogen peroxide. The peroxide did not decompose, or rather, decomposed very slowly, almost imperceptibly. Hence, my peroxide test is invalid for dilute suspensions of very fine manganese dioxide particles.

Though the hydrofluoric version gave no noticeable precipitate, I'm now inclined to agree with woelen's theory: my reaction probably produced a suspension of extremely fine particles of manganese dioxide, in a solution of manganese(II) fluoride. These particles then gradually dissolve in the excess HF. Though I'll note an above test where freshly precipitated manganese dioxide wouldn't dissolve in dilute HF without hydrogen peroxide.

But this begs the question: If potassium hexafluoromanganate(IV) can be produced in solution, what is the lowest possible concentration of hydrofluoric acid in which the complex will form?
@woelen: If you can successfully prepare the complex by following the literature procedures, attempting the synthesis with lower concentrations of HF could be an interesting followup.

[Edited on 23-11-2015 by MolecularWorld]




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[*] posted on 23-11-2015 at 02:53


Quote:
what is the lowest possible concentration of hydrofluoric acid in which the complex will form?

The more relevant question is: what concentration is needed to produce the salt in an isolable amount. It's useless to have a dilute K2MnF6 solution as long as it isn't possible to isolate the compound. There are no signs that addition of e.g. acetone could help precipitate low concentrations of this stuff.

I think the critical concentration of HF is in the range of about 30 %, but certainly much higher than only a few %. The paper from 1953 used 40 % HF with 32 % yield. Christe used 50 % HF with 73 - 78.6 % yield. The other reaction conditions are nearly identical except for the 1 liter setup (Christe) compared to 100 ml (in the 1953 article). Usually, higher volumes give slightly higher yields due to technical reasons. But this can't explain the large difference between the yields alone. I could imagine that a higher solubility is the main reason for the lower yield if the HF concentration is slightly decreased (e.g. from 50 to 40 %) and in much lower concentrated HF the salt is completely hydrolized.

Quote:
I liked the end of the Christe article so much, I added to my signature.

Well, Christe was not the first one who prepared fluorine in a strictly chemical way. So there wasn't any "dogma", except in the brains of illiterate people. The only real achievement is, that he managed it in a more convenient way.

[Edited on 23-11-2015 by Pok]
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[*] posted on 23-11-2015 at 08:24


The goal was never to produce the pure complex in quantity, otherwise I would have concentrated my HF and used a procedure more closely following the literature. Here, again, is the purpose of my experiment:
Quote: Originally posted by MolecularWorld  
The purpose of this experiment was merely to see whether the desired complex could be produced from dilute solutions.

We could quibble on the phraseology (I should have said "formed" rather than "produced"), but I think the intention was clear enough. Yes, it would be better to be able to isolate the yellow stuff as a solid, for better analysis. But, as you say, there may be no way to isolate the yellow stuff before it turns pink, and as I noted immediately above, I'm inclined to agree with woolen's theory that my yellow stuff is just manganese(II) fluoride with traces of manganese dioxide. I'm sure there are methods and machines capable of analyzing such a solution quickly enough to prove definitively whether or not the complex was present in visible quantities, but as I don't have access to such things, I've done all I can.

Though I see no reason why my acetone procedure doesn't work in theory. Acetone is used in the published procedures to wash the solid product, leading to reduced loss compared to washing with HF, so I conclude the complex doesn't (quickly) react with acetone, and is even less soluble in acetone than in HF. As for how quickly the complex might react with acetone, I can't say, but I can say that permanganate doesn't appear to react with or dissolve appreciably in acetone, even when some of my dilute HF is added. I'll probably try the acetone precipitation again, once I've arranged for vacuum filtration, but I'll refrain from posting my results and conclusions until I've tested the solids in such a way as to prove they do or don't contain Mn(IV). Considering this complex readily decomposes to Mn(III) and Mn(II), and I've already stated I'm not going to produce elemental fluorine, what is a qualitative test I could do to prove the yellow stuff is definitely not Mn(IV)?

I agree the first part of the Christe line is a little too self-aggrandizing, which is why I'm only using the end. A lot of people have said similar things, but this is the first time I've seen it published by a chemist.

[Edited on 23-11-2015 by MolecularWorld]




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[*] posted on 23-11-2015 at 09:07


Even K3MnCl6 (III) is stable in aqueous solution, but when acidified, releases chlorine. Apparently K2MnCl6 (IV) also exists. If you are unable to produce any K2MnF6 (IV) maybe there is not enough potassium in your mixture?

Qualitatively, if the Mn(IV) salt oxidizes water, then if the Mn(III) salt did not, at least at a certain pH, that is a possible test. I would not recommend tasting for piquancy, but that could, conceivably, prove insightful

[Edited on 23-11-2015 by halogen]




F. de Lalande and M. Prud'homme showed that a mixture of boric oxide and sodium chloride is decomposed in a stream of dry air or oxygen at a red heat with the evolution of chlorine.
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[*] posted on 23-11-2015 at 11:07


MW, I know what your goal is. But I don't think that this is the relevant question. A K2MnF6 solution this is totally useless. That's the point.

I never said that your yellow stuff may not be isolable. I was talking about K2MnF6 which may not be isolable if the procedure is done correctly in with slightly (!) decreased HF concentrations. You didn't even cool the reaction. This is a knock-out criterion according to literature!

Quote:
Though I see no reason why my acetone procedure doesn't work in theory.

Both liquids could form two phases. Who knows? Acetone can form two phases with highly concentrated salt solutions. I didn't say that there are definite reasons for this procedure not to work but that there are no good reasons to believe that it does work.

Quote:
what is a qualitative test I could do to prove the yellow stuff is definitely not Mn(IV)?

I see no need to do this. You first would have to isolate the stuff. Filtering with a HF resistant material (some synthetic fiber or so) should work. K2MnF6 has some behaviours that can be used to identify it. I translate from a 1899 paper (see below):

"If you heat the salt on a platinum sheet or in a test tube it gets red-brown, but becomes yellow again if heating wasn't to intense. If you heat stronger (in air) it decomposes under evolution of HF fumes and yields a purple residue. Under very strong heating in a platinum crucible it gets gray-black with constant evolution of HF fumes. The residue consists of a mixture of Mn2O3 and KF. [...] The salt is slowly decomposed by cold water under precipitation of brown hydrated MnO2, faster in boiling water. Alkali hydroxides and carbonates act likewise while alkali fluorides go into solution. Cold HCl dissolves it yielding a dark brown coloured solution. Even at moderate heating the solution evolves chlorine. Concentrated cold H2SO4 dissolves it under evolution of HF yielding a dark brown solution. On heating HF, O2 and ozone are evolved (KI paper gets blue) and you get a violet solution. [...] Treating the salt with HNO3 precipitates MnO2 and releases HF. The solution is free from dissolved manganese. H3PO4 dissolves the salt with brown-red colour [...]. The salt is insoluble in glacial acetic acid but dilute acetic acid precipitates hydrated MnO2. Oxalic acid in aqueous solution gets oxidized to CO2 immediately. Indigo solution gets decolourized. H2O2 gets turbid and soon evolves O2 even without addition of HCl. The residue is reddish."

Original source:
Unbenannt.png - 123kB
http://onlinelibrary.wiley.com/doi/10.1002/zaac.620200106/ab...

I think you now have enough information to identify your product as "no K2MnF6".

[Edited on 23-11-2015 by Pok]
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[*] posted on 23-11-2015 at 11:42


@Pok: Thanks for the reference and translation.

@halogen: Do you have a reference for the preparation of those chloride analogues? I only found this (in English).

[Edited on 23-11-2015 by MolecularWorld]




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[*] posted on 24-11-2015 at 08:17


Mm! Mellor, now that I have been reminded, is where I read about the compound! The full volume is available courtesy of sciencemadness.org/library and you would be able to track down those original references it gives.

But the day I posted, I did a very quick search for "K2MnCl2" on google, and an unreferenced note on it's preparation was "from HCl and KMnO4". So it must be common knowledge.




F. de Lalande and M. Prud'homme showed that a mixture of boric oxide and sodium chloride is decomposed in a stream of dry air or oxygen at a red heat with the evolution of chlorine.
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