Sciencemadness Discussion Board - Preparation of potassium hexafluoromanganate(IV) from household items

Preparation of potassium hexafluoromanganate(IV) from household items

MolecularWorld - 18-11-2015 at 19:16

In another thread, the preparation of fluorine was briefly discussed, particularly Karl O. Christe's preparation of fluorine without electrolysis:

Quote: Originally posted by j_sum1

There is one chemical method that involves antimony-fluoro-something-hazardous-and-unavailable.

Quote: Originally posted by MolecularWorld

2 KMnO₄ + 2 KF + 10 HF + 3 H₂O₂ → 2 K₂MnF₆ + 8 H₂O + 3 O₂↑
2 K₂MnF₆ + 4 SbF₅ → 4 KSbF₆ + 2 MnF₃ + F₂↑

Discussion of these reactions, particularly the first one, yielded the following comments:

Quote: Originally posted by j_sum1

That's the one. Actually, it is the 10HF that is scary there.

Quote: Originally posted by woelen

Both of Christe's reactions for making F2 without electrolysis require strictly anhydrous conditions. The first reaction hence requires anhydrous HF, anhydrous H2O2 and a very strong drying agent to absorb the water, formed in the reaction.

While I agreed that the second, fluorine-producing reaction would be difficult and dangerous to attempt, the first reaction seemed quite tame. Contrary to woelen's assertion, the original documentation suggests the reaction does proceed in aqueous solution (actually, concentrated hydrofluoric acid). Based on this, I attempted Christie's first reaction, using dilute hydrofluoric acid, in the form of a cleaning product, as my sole source of fluoride. All other reagents were also available as household products.

300ml of dilute hydrofluoric acid solution was added to a 500ml polypropylene beaker. The exact concentration is unknown; the MSDS states 1%-2.3%, and the quantities of other reactants were calculated assuming 1.2%, as an excess of hydrofluoric acid does not effect the reaction. To this was added 12g of 14% potassium hydroxide solution, 4.7g potassium permanganate dissolved in a minimum of water, and 52g 3% hydrogen peroxide.

<iframe sandbox width="280" height="160" src="//www.youtube.com/embed/PgXvecQw6yQ?rel=0" frameborder="0" allowfullscreen></iframe>

The reaction produces oxygen gas; avoid inhaling the fluoride-containing aerosol. The result is a light yellow solution/suspension.

K₂MnF₆ is a sparingly-soluble yellow compound, so the synthesis was successful. This liquid could be mistaken for a suspension of extremely small manganese dioxide particles, however, once enough peroxide has been added to remove all purple/pink color, addition of additional peroxide does not produce more gas, as would be expected from a suspension of manganese dioxide particles.

Though the above shows some fluorine chemistry can be performed with little danger, I do not take the real risks lightly. This post was intentionally made as my 50th post, when my rank transitions to "Hazard to Self". Due to the formation of a small amount of fluoride-containing mist, this was performed in a well-ventilated area, with googles and respirator, and the area was wiped down with limewater afterword. I will not be attempting the second of Christie's reactions for the production of fluorine.

This may or may not be part of my entry for j_sum1's competition.

[Edited on 19-11-2015 by MolecularWorld]

gdflp - 18-11-2015 at 19:46

Nice job. Did you isolate the potassium hexafluoromanganate(IV)? If so, what was your yield? I'm curious to see if using such a dilute solution of HF decreases yield significantly from the yield in the original paper. It's quite possible that the yellow solution arose from another manganese compound than K<sub>2</sub>MnF<sub>6</sub>, so I would recommend verifying that you actually synthesized what you thought you did.

[Edited on 11-19-2015 by gdflp]

blogfast25 - 18-11-2015 at 19:50

You're going to have to do a whole lot better than a 1 min UToob to convince me you've prepared K2MnF6. Like maybe a little elemental analysis, you know?

You do realise KMnO4 oxidises peroxide easily, right?

Quote:

K2MnF6 is a sparingly-soluble yellow compound, so the synthesis was successful.

Talk about an aversion to evidence...

[Edited on 19-11-2015 by blogfast25]

MolecularWorld - 18-11-2015 at 20:08

@gdflp: I did not isolate the compound, and destroyed the solution with calcium hydroxide. The only ways I know to isolate the compound from such a dilute solution would require very large amounts of acetone, or evaporating 500ml of hydrofluoric acid. I may repeat the procedure once I feel I can perform either isolation, and measure the purity of the product.

@blogfast: Good thing I wasn't trying to convince you! Just for fun, what do you think the reaction in the video was? I assure you, the reagents were exactly as I described. What else could account for the color change, and gas production, which ceased once the solution had turned light yellow (I added more peroxide after the end of the video)?
I currently have only the most limited means for analysis, the results of which surely wouldn't meet your standards. Though I may repeat the procedure in the future once my abilities have increased.

Texium - 18-11-2015 at 20:27

You could also try making more concentrated HF from a fluoride salt, though be very cautious if you do decide to do that, of course.

MolecularWorld - 18-11-2015 at 20:35

Quote: Originally posted by zts16

You could also try making more concentrated HF from a fluoride salt, though be very cautious if you do decide to do that, of course.

I do not plan to attempt any further fluorine chemistry, other than repeating the procedure above once I can better analyze the products. The purpose of this experiment was merely to see whether the desired complex could be produced from dilute solutions. I believe I provided enough evidence for this to be plausible, though blogfast is correct that my proof is far from absolute.

j_sum1 - 18-11-2015 at 20:42

I can see that I am going to have to look at this a whole lot more closely. I have always assumed that fluorine chemistry was almost completely out of reach for the normal person. Maybe you are not normal, MW. This all looks pretty exciting.

deltaH - 18-11-2015 at 21:52

It would appear that in the US nearly every reagent possible is OTC, as long as it contains 1% detergent

A very nice idea MolecularWorld, but I must admit that I too am somewhat sceptical.

I suggest you perform this reaction side-by-side with an acidified control of similar volume and concentration but minus the fluoride cleaner. Take care to match the two as closely as possibly, the one simply lacking the small amount of HF.

I think the principle doubt here is whether highly diluted hydrofluoric acid is capable of dissolving the formed MnO2 suspension at such dilute concentrations and so a control would be most convincing.

Finally, you might also want to add an excess of your dilute hydrofluoric acid and see if you can't dissolve the suspension completely as a solution of MnF6(2-)(aq).

[Edited on 19-11-2015 by deltaH]

Pok - 19-11-2015 at 04:36

This is not K2MnF6. The salt is decomposed by water and you are working in nearly pure water with neglectible concentrations of reagents. The whole procedure is different from your experiment. And the "original documentation" is a different one. This:

http://onlinelibrary.wiley.com/doi/10.1002/ange.19530651108/...

Quote:

K2MnF6 is a sparingly-soluble yellow compound, so the synthesis was successful.

You did't get a precipitate, so your conclusion is wrong.

Quote:

This liquid could be mistaken for a suspension of extremely small manganese dioxide particles, however, once enough peroxide has been added to remove all purple/pink color, addition of additional peroxide does not produce more gas, as would be expected from a suspension of manganese dioxide particles.

Only MnO2 and K2MnF6 can be produced? You are working in a diluted acidic medium! Your neither get one of these products but simply MnF2 which explains your observation that additional H2O2 isn't decomposed. The yellow colour could be anything, probably side products.

[Edited on 19-11-2015 by Pok]

MolecularWorld - 19-11-2015 at 07:16

@Pok: Thanks for your input. I can't translate your German text above, but I assume from your comment it stipulates anhydrous conditions are required for this reaction. This contradicts with the paper I cited above, which states:

Quote:

The salt selected for this study was K₂MnF₆. It has been known since 1899 and is best prepared from aqueous HF solution.

As for the lack of precipitate, you know I'm working with very dilute reagents, and not nearly enough HF concentration to force the complex to precipitate, and the lack of precipitate is still surprising? I'll see what can be done to extract the solid compound from this solution.
And finally, MnO₂ and K₂MnF₆ are the only compounds/complexes that could be formed in this mixture that are yellow, that I know of. You say the yellow must be a side product, but don't state what it could be.

Pok - 19-11-2015 at 09:46

Take a look at the references of your source. Your source quotes my source. And your source doesn't specify the reaction conditions. Anhydrous conditions aren't necessary, but highly concentrated solutions - e.g. 30 % H2O2 instead of 3 %, 40-50 % HF instead of 1-2 %. H2O2 has to be added drop by drop. The reaction has to be cooled with ice.

Here is a similar instruction from Christie's description:

http://onlinelibrary.wiley.com/doi/10.1002/9781118409466.ch1...

The lack of a precipitate isn't surprising. But your observation is only based on a colour and nothing else! There is not the slightest evidence for the existence of the desired product.

What the yellow stuff could be? I stated it: impurities. You used household chemicals. These aren't pure reagents but can contain all sorts of contamination. It's impossible that it is K2MnF6, because that hydrolyses in water and your whole procedure lacks in some essential conditions.

fluorescence - 19-11-2015 at 09:50

Oh hey POK

Nice to have at least one other VC member here.

There is a short passage on that compound in "The Chemistry of Manganese, Technetium and Rhenium".

MolecularWorld - 19-11-2015 at 09:59

@Pok: Interesting. Based on your reference, I will attempt the procedure again, in an ice bath and with ice-cold solutions, to see if a precipitate forms. I'm also considering forcing the compound to precipitate with a large quantity of acetone, although I'm slightly concerned with the potential formation of acetone peroxide.
The yellow could be from impurities in the reagents, but I doubt it. Three of my four reactants were perfectly colorless solutions, and the permanganate is sold for use in iron-removing drinking water purification devices, so it's unlikely to contain other transition metal impurities.

[Edited on 19-11-2015 by MolecularWorld]

softbeard - 19-11-2015 at 10:45

MolecularWorld, I think you should be careful before making statements like "K2MnF6 is a sparingly-soluble yellow compound, so the synthesis was successful".
Anyone with a chemistry background would know that it's impossible to synthesize K₂MnF₆ under the conditions you're describing. But even leaving that aside, you can't confirm a compound's identity by wishful thinking and because it has a colour similar to your desired compound.

Have fun & take care with the fluoride experiments. I would suggest giving up on producing elemental fluorine for now. I think copper chemistry is a lot more fun and colourful!

MolecularWorld - 19-11-2015 at 11:00

Quote: Originally posted by softbeard

MolecularWorld, I think you should be careful before making statements like "K2MnF6 is a sparingly-soluble yellow compound, so the synthesis was successful".
Anyone with a chemistry background would know that it's impossible to synthesize K₂MnF₆ under the conditions you're describing. But even leaving that aside, you can't confirm a compound's identity by wishful thinking and because it has a colour similar to your desired compound.

Clearly. The more experienced members here will settle for nothing less than mass spectrometry and x-ray crystallography to confirm the identity of a compound. I drew my conclusion from the fact that there's a very limited number of yellow products that could be produced from these reactants, and the fact that my synthesis is similar (though not identical) to that which was reported in the literature. As Pok pointed out, and I agreed, this evidence is far from absolute. But, it's not just "wishful thinking". Compare, for example, all the preparations of tetraamminecopper(II) complexes reported on this board, where color is used as the primary, and sometimes only, evidence of reaction.

Quote:

Have fun & take care with the fluoride experiments. I would suggest giving up on producing elemental fluorine for now. I think copper chemistry is a lot more fun and colourful!

In the very first post:

Quote: Originally posted by MolecularWorld

I will not be attempting the second of Christie's reactions for the production of fluorine.

Followed by:

Quote: Originally posted by MolecularWorld

Pok - 19-11-2015 at 11:50

It indeed is wishfull thinking. You see what you want to see. Nothing else. Thousands of colourless chemicals can form yellow/brown products by reacting with KMnO4. Well, one thousand still is a "limited number".

There is no need for spectroscopy to check the yellow stuff. K2MnF6 hydrolyzes in water. It's simply impossible to make a solution of this material in 1-2 % HF!

MolecularWorld - 19-11-2015 at 12:10

Name three yellow products formed from the reaction of a colorless solution with potassium permanganate, not including potassium hexafluoromanganate(IV) and manganese dioxide. Be sure to take my comment above about likely impurities (ie, there wouldn't be other metals in the permanganate, or heavy metals in medical-grade peroxide) into consideration.

The comments here, particularly yours (Pok), have convinced me that my product might not be potassium hexafluoromanganate(IV). But this has yet to be proven definitively. Rather, my baseless assertions have been countered by... baseless assertions. My hypothesis is that the main reason concentrated hydrofluoric acid is used in the literature preparations is to precipitate the complex before it can hydrolyze, but that small amounts of unstable potassium hexafluoromanganate(IV) could form in dilute solution/suspension. At least I conducted an actual reaction; if someone repeats my procedure, and proves the yellow stuff produced is not potassium hexafluoromanganate(IV), then I'll be convinced.

[Edited on 19-11-2015 by MolecularWorld]

MolecularWorld - 19-11-2015 at 12:45

Here's an excerpt from the reference fluorescence mentioned. My wishful thinking has me focused on the keyword "slow".

[Edited on 19-11-2015 by MolecularWorld]

deltaH - 19-11-2015 at 13:01

Why don't you carry out the simple control experiment I suggested?

MolecularWorld - 19-11-2015 at 13:08

This one?

Quote: Originally posted by deltaH

I suggest you perform this reaction side-by-side with an acidified control of similar volume and concentration but minus the fluoride cleaner. Take care to match the two as closely as possibly, the one simply lacking the small amount of HF.

I'll do it, although I don't fully understand what it's intended to prove.
Which acid would be suitable as a substitute for the HF? HCl is liable to react and release chlorine, would sulfuric acid do?

[Edited on 19-11-2015 by MolecularWorld]

deltaH - 19-11-2015 at 13:22

H2SO4 should be fine.

[Edited on 19-11-2015 by deltaH]

Pok - 19-11-2015 at 13:37

Quote: Originally posted by MolecularWorld

My wishful thinking has me focused on the keyword "slow".

Not "slow" is the keyword but "even in HF solution"! Again the original literature:

https://books.google.de/books?hl=de&lr=&id=DbACnFyLS...

So you see that your source simplified the main statement from the original source!

In "aqueous solution" can mean anything but it doesn't necessarily mean "in water". 50 % HF is an aqueous solution, too. If you look at the decomposition products: Mn(III) ions will form. These are yellow/brown in solution as far as I know.

deltaH - 19-11-2015 at 13:46

Speaking of which, I suspect the precipitate is a suspension of hydrated Mn2O3 or manganese oxyhydroxide, exactly why the control is needed. You should see a similar result.

Here's why:

Attachment: Pourbaix_diagram_for_Manganese.svg (52kB)
This file has been downloaded 619 times

[Edited on 19-11-2015 by deltaH]

MrHomeScientist - 19-11-2015 at 14:28

Suspensions of very fine, freshly precipitated MnO<sub>2</sub> appear yellow. See: the final state of the 'chemical chameleon' reaction (see my blog here). The final picture is a bit on the orange side, but most times I've done this reaction it ends at a nice pale yellow.

People's reaction to your experiment has been a bit harsh, in my opinion, but they do make a valid point. Color alone isn't definitive when looking at a reaction product, though it certainly can be one point of evidence. You keep challenging others to come up with a possible yellow product, but it's nearly impossible for anyone to say what it could be because you're using household chemicals. Who knows what impurities are present? From my experiment above, only very small amounts of chemicals led to very visible colors. Maybe some detergent or inactive ingredient in your cleaners reacted in some way. There's no way to tell without knowing exactly what's in there (and the MSDS's aren't always all-inclusive, either).

Clearly no one's expecting someone at home to utilize state-of-the-art analysis equipment, but one or two more pieces of evidence are necessary before we can claim with certainty that you got what you think you got. If you were able to isolate the colored compound and run some more tests on it, that would help greatly. For example, when I made Chevreul's Salt, the color was pretty indicative but I went ahead and ran some extra tests to show it contains both copper(I) and copper(II).
I thought this thread was all pretty interesting, for what it's worth.

Edit: Speaking of "tetraamminecopper(II) complexes," I've made hexaaminenickel(II) chloride before (it's on my YouTube Channel). Near the end of the video, you can see that some leftover on the filter paper turned from purple to green upon drying in air. I wasn't sure at the time, but I've since learned that if it's left out, the ammonia leaves the complex and it reverts to normal hydrated nickel(II) chloride. So an ammonia smell of the purple compound vs. no smell of the green compound is another indication I made what I think I did.

[Edited on 11-19-2015 by MrHomeScientist]

MolecularWorld - 19-11-2015 at 18:46

@MrHomeScientist: I was well aware that the solution could be a fine suspension of manganese dioxide, which is why I tested it with additional peroxide, as was already noted in my original post and later posts. Additional peroxide did not decompose, so it's (probably) not manganese dioxide.
I don't mind harsh, though I do mind harsh-and-useless, like blogfast's comment above. I very much appreciate Pok providing detailed explanations and references, in addition to voicing doubt. You and Pok are technically correct in that the yellow product 'could be anything' as the contaminants 'could be anything', though I've also explained above how unlikely it is that these products are contaminated with anything that could give a colored product. I've been researching it, and the only possibility I've found is if my drain-opener potassium hydroxide was contaminated with lead(II) hydroxide, it could form yellowish lead(II) oxide, but really, how likely is that?

@deltaH: I believe Mn₂O₃ would also decompose hydrogen peroxide. If so, per my above testing, it's not that.

@Pok: I'm probably misunderstanding things, but that would seem to confirm that the desired complex could form, then hydrolyze. Just because I couldn't (yet) extract and purify the compound, doesn't mean it didn't form. Also, you almost had me convinced that the resulting solution contains a fluoride of manganese, but both MnF₂ and MnF₃ are pink. Since it's not manganese dioxide, and not a fluoride of manganese, where did the manganese go?

Despite the apparent futility, I'll be attempting to purify a solid compound from [a fresh batch of] the solution prepared by my above procedure. I'll also do deltaH's 'control' reaction.

[Edited on 20-11-2015 by MolecularWorld]

deltaH - 19-11-2015 at 20:08

Quote: Originally posted by MolecularWorld

@MrHomeScientist: I was well aware that the solution could be a fine suspension of manganese dioxide, which is why I tested it with additional peroxide, as was already noted in my original post and later posts. Additional peroxide did not decompose, so it's (probably) not manganese dioxide. I don't mind harsh, though I do mind harsh-and-useless, like blogfast's comment above. I very much appreciate Pok providing detailed explanations and references, in addition to voicing doubt. You and Pok are technically right that the yellow product 'could be anything' as the contaminants 'could be anything', though I've also explained above how unlikely it is that these products are contaminated with anything that could give a colored product. I've been researching it, and the only possibility I've found is if my drain-opener potassium hydroxide was contaminated with lead(II) hydroxide, it could form yellowish lead(II) oxide, but really, how likely is that?

@deltaH: I believe Mn₂O₃ would also decompose hydrogen peroxide. If so, per my above testing, it's not that.

Not necessarily, let's say that it's some kind of manganese oxide. The way it would decompose H2O2 is by acting as a heterogeneous catalyst. All reactions that happen on heterogeneous catalysts are surface reactions and the surfaces of these catalysts are subject to something called 'poisoning', usually by having some other species adsorb strongly/preferentially onto the surface and so make it inert towards the reagents it's meant to adsorb initially in a catalytic mechanism. It's a kind of passivation if you will.

Now these surfaces can be very sensitive to poisoning and you do have fluoride in there. I would imagine that the fluoride ions would adsorb onto the surfaces of manganese oxides very strongly.

So what you might have made could be a highly porous manganese oxide nanoparticles whose surfaces have been fluoridated and so passivated.

HF is also added at a couple percent strength to inhibit red fuming nitric acid (RFNA) in rockets so as to passivate the metal fuel tanks and prevent corrosion.

MolecularWorld - 19-11-2015 at 20:43

For what it's worth, I attempted a very crude experiment which may or may not be relevant to the above theory. A small quantity of manganese(II) sulfate solution was added to dilute ammonia, resulting in a reddish-brown suspension. I'm not going to state what this suspension is, but I will think it loudly. The precipitate was allowed to settle, the supernatant decanted, and the solids divided among two beakers. To one was added 3% hydrogen peroxide. This resulted in much oxygen gas production, and the liquid turning black. To the other was added my dilute HF cleaning solution, which dissolved the brown particles to a faintly pink solution.. When hydrogen peroxide was added to this solution, no gas was evolved and the color did not change.

I'll be performing the reactions more closely relating to the original post over the next day or two.

deltaH - 19-11-2015 at 21:18

That's a nice experiment MW. It is interesting that even when dissolved (homogeneous mode), the fluoride binds the manganese ions so strongly that they lose their catalytic action.

It would also be interesting to see if you can passivate the black MnO2 with fluoride against H2O2 decomposition without dissolving it. Bear this in mind for future experiments if you can.

If, however, you can dissolve a fine suspension of MnO2 in excess fluoride cleaner, then I would say it is likely that you indeed have prepared a solution of MnF6(-2).

If that happens, you might want to try to saturate with potassium sulfate or chloride to see if you can precipitate something (exploiting the common ion effect to precipitate a dilute solution). Adding potassium hydroxide solution would not be convincing because one might argue that you simply precipitated an oxide again by raising the pH.

Anyway, lots of strategies and suggestions for investigation.

MolecularWorld - 20-11-2015 at 00:01

A black suspension of freshly-precipitated manganese dioxide was prepared. Addition of hydrogen peroxide to this suspension produced much oxygen gas, with no color change. Addition of dilute hydrofluoric acid yielded no obvious reaction, even after sitting for several minutes. Addition of hydrogen peroxide to the manganese dioxide and hydrofluoric acid mixture produced no gas (!), but dissolved the manganese dioxide to a nearly colorless solution. Saturating this solution with potassium chloride gave no obvious reaction.

I must say, I was quite surprised that the peroxide produced no gas when added to the manganese dioxide suspension in dilute hydrofluoric acid, even though it agrees with the assertions I've been making in this thread. It seems fluorine chemistry is more interesting than even I expected.

deltaH - 20-11-2015 at 00:49

I'm thoroughly confused by these observations. Since the MnO2 only dissolved upon addition of H2O2 AND no gas was evolved, we must conclude that the manganese's oxidation state was raised and this formed a colourless ion?

If it was a fluoride, could it be MnF6(-) ??

Does anyone have literature on such an ion, is it colourless? Usually Mn(V) complexes are bright blue AFAIK.

If I were you, I'd really focus my efforts to try to crystallise the potassium salt of this mystery colourless ion. Sounds amazing.

[Edited on 20-11-2015 by deltaH]

MolecularWorld - 20-11-2015 at 00:59

I couldn't find anything in the literature, though I didn't search very much.
I can tell you what didn't happen:

<strike>MnO₂ + 6 HF + H₂O₂ >>> H₂MnF₆ + 4 H₂O</strike>

...because, if such a simple reaction was possible, under such mild conditions, it would surely have been reported in the literature by now, right? Internet searches for "hexafluoromanganic acid" only turn up a few obscure patents, with little supporting documentation.

Edit: The above was written at 3:00am. I see the errors now.

[Edited on 21-11-2015 by MolecularWorld]

deltaH - 20-11-2015 at 01:10

Yes I think we can safely conclude from your observations that you are NOT dealing with MnF6(2-) ions. So far the only possibility I can think of is MnF6(-), which is rather crazy I must admit.

Your equation isn't possible because manganese isn't changing its oxidation state, yet the peroxide is getting reduced??? Not possible and it's not balanced.

You really should try to isolate a crystalline salt of whatever is there. It might be weakly coloured in concentrated form which might give us a clue as to what it is.

[Edited on 20-11-2015 by deltaH]

MolecularWorld - 20-11-2015 at 01:19

Quote: Originally posted by deltaH

]Your equation isn't possible because manganese isn't changing its oxidation state, yet the peroxide is getting reduced??? Not possible and it's not balanced.

It's not balanced? I agree the kinetics don't make sense.

Quote:

You really should try to isolate a crystalline salt of whatever is there. It might be weakly colored in concentrated form which might give us a clue as to what it is.

I added potassium chloride to the solution until some solids remained undissolved. I assumed this meant the solution was saturated with potassium chloride with no reaction, but I suppose it might have been a white precipitate...

Boffis - 20-11-2015 at 06:39

Quote: Originally posted by MolecularWorld

I couldn't find anything in the literature, though I didn't search very much.
I can tell you what didn't happen:

MnO₂ + 6 HF + H₂O₂ >>> H₂MnF₆ + 4 H₂O

...because, if such a simple reaction was possible, under such mild conditions, it would surely have been reported in the literature by now, right? Internet searches for "hexafluoromanganic acid" only turn up a few obscure patents, with little supporting documentation.

What's the H2O2 for? MnO2 is already tetravalent. You are more likely to reduce the Mn4+ to Mn2+.

MnO<sub>2</sub> + 6HF → H<sub>2</sub>MnF<sub>6</sub> + 2H<sub>2</sub>O

Looks more likely

Texium - 20-11-2015 at 07:05

Quote: Originally posted by deltaH

Yes I think we can safely conclude from your observations that you are NOT dealing with MnF6(2-) ions. So far the only possibility I can think of is MnF6(-), which is rather crazy I must admit.

Your equation isn't possible because manganese isn't changing its oxidation state, yet the peroxide is getting reduced??? Not possible and it's not balanced.

You really should try to isolate a crystalline salt of whatever is there. It might be weakly coloured in concentrated form which might give us a clue as to what it is.

So crazy it might just be some "unreferenced speculation" that doesn't hold water.

Manganese(V) is the trickiest oxidation state to obtain out of all of them 2-7. I highly doubt that it can be achieved simply by adding a tiny bit of HF to the manganese dioxide/peroxide reaction. And, as you pointed out, Mn(V) solutions are known to be an intense blue color, which should be obvious if this was actually a Mn(V) compound by some miracle.

AJKOER - 20-11-2015 at 08:10

To expand the thinking on some of the possible underlying chemistry, first note that the action of dilute HF on dilute H2O2 may form hypofluorous acid, HOF, in a manner noted by Watts as to the action of dilute H2O2 on dilute HCl forming HOCl. Per Wikipedia ( https://en.m.wikipedia.org/wiki/Hypofluorous_acid ) on pure
HOF says that it is a very unstable at RT and a yellow liquid above -170 C. Its salts, hypofluorites, exist and my speculation is that there could also be yellow (?). We may also consider, akin to basic chlorides and hypochlorites (for example, dibasic magnesium hypochlorite), such related fluorine salts.

Note, if attempting this reaction with concentrated reagents, like in the case of concentrated HCl and H2O2 (see, for example, discussion at https://sites.google.com/site/unusualchemistry/halogen-hydro... ), and assuming parallel products, then in place of chlorine in water, one would expect fluorine water (F2 and H2O, where the intermediate HOF quickly decomposes, releasing O2 and HF fumes). However, there is some discussion, for H2O2 that HF actually acts as a stabilizer (see, for example, discussion by KMnO4 in this old SM thread with a cited source at:
http://www.sciencemadness.org/talk/viewthread.php?tid=14490 and also https://books.google.com/books?id=rpfsCAAAQBAJ&pg=PA156&... ).

Now, if no HOF is created, then a modified Fenton reaction between H2O2 and whatever Manganese ion is first created (in place of the usual Fe(ll) ) could take place with favorable pH. However with HOF, then a related Fenton-type reaction may still proceed, where any created HOF supplants HOCl, for example, as seen in biological systems. Source, see "Fenton chemistry in biology and medicine*" by Josef Prousek available at https://www.google.com/url?sa=t&source=web&rct=j&...

698ab943000000.pdf&ved=0CB0QFjAAahUKEwjCjvjHqrDHAhVKz4AKHZrKDuc&usg=AFQjCNERAlvJfhkQp1Z7LrFk1zdFyLaPMQ&sig2=QUzbx2NQWIZHre_kitaEbQ . To quote reaction 15 on page 2330:

"For Fe(II) and Cu(I), this situation can be generally depicted as follows [20,39],

Fe2+/Cu+ + HOX → Fe3+/Cu2+ + HO• + X- (15)

where X = Cl, ONO, and SCN. "

In the current context, without or with any HOF, the transition metals Fe and Cu are replaced by Mn and HOX is either HOOH or possibly HOF. Note, this reaction shows how the valence state of the transition metal can increase.

[Edit] I should also note on the possible issue of impurities, I recall reading there may be some Fe in the MnO2.

Also, in the presence of any formed hydroxide radical and the flouride ion, to quote a source:

"Well known examples of this are the reactions of .OH with halide and pseudo-halide ions (X−) where the first product is HOX.− although the measured product is .X2−.

.OH + X− → HOX.−

HOX.− → X. + OH- "

Now, in the case of the flourine radical, F., one reaction I would expect is the formation of F2, but not a gaseous release in dilute conditions, with its subsequent reaction with water:

F. + F.→ F2

F2 + H2O = HF + HOF

And then, various other reactions including the possible Fenton reaction noted above, reforming .OH, or a reaction with H2O per a source above:

HOF + H2O → HF + H2O2

where the hydrogen peroxide could feed a Fenton reaction, or its self-decomposition:

2 HOF → 2 HF + O2

[Edited on 20-11-2015 by AJKOER]

[Edited on 21-11-2015 by AJKOER]

MolecularWorld - 20-11-2015 at 09:33

Quote: Originally posted by Boffis

What's the H2O2 for? MnO2 is already tetravalent. You are more likely to reduce the Mn4+ to Mn2+.

MnO<sub>2</sub> + 6HF → H<sub>2</sub>MnF<sub>6</sub> + 2H<sub>2</sub>O

Looks more likely

I added H₂O₂ because the MnO₂ didn't dissolve in HF alone. When H₂O₂ was added to the suspension of manganese dioxide in ~1% HF, it dissolved without producing gas. For comparison, Mn₂O₃ dissolved in HF without H₂O₂, also without producing gas. (All this is in my posts above.)

I haven't found any references to the reaction of manganese dioxide, hydrofluoric acid, and hydrogen peroxide, but I did find one obscure reference to the existence of the acid I mentioned above. Note that in my tests, MnO₂ didn't dissolve in dilute HF alone.

Chemical News and Journal of Physical Science, Volume 4, 1869, William Crookes.png - 84kB

These additional experiments are causing my HF supply to run low, so I'll be postponing study of this line of experiments until after I do the 'control' test, and attempt to precipitate the originally desired complex, as stated on the first page.

MolecularWorld - 20-11-2015 at 11:56

I attempted deltaH's control experiment. The procedure in the original post was performed, with half the reagents, side-by-side with a similar reaction, with ~1.5% sulfuric acid substituting for the hydrofluoric acid. Both reactions appeared similar: no obvious reaction until the peroxide was added, then much gas, followed by lightening of the color of the solution. I noted two differences: the sulfuric acid mix gave a small amount of dark precipitate, which quickly redissolved to give a nearly colorless solution, while the hydrofluoric acid mix gave no precipitate and left a noticeably yellow-orange solution. A picture of the result is below, a video clip of the reactions is available, but filming had to be aborted due to the overproduction of fluoride foam.

I also prepared a new batch of !hexafluoromanganate, using the same procedure and half the quantities of reactants specified in my original post. I made one modification to the procedure: all of the reactants were chilled to near 0*C before mixing, to slow any hydrolysis that might take place. The peroxide was also added slowly, in increments and with stirring, to ensure no excess was used.

<iframe sandbox width="280" height="160" src="//www.youtube.com/embed/FjGKL_UTqA0?rel=0" frameborder="0" allowfullscreen></iframe>

This resulted in a yellow solution identical to that produced in my first post. This was quickly added to 500ml acetone, which resulted in a voluminous, light-pink-orange precipitate and a small amount of gas evolution. The pink may not be the true color of the precipitate: there were a few droplets of permanganate on the top of the beaker walls, which mixed in as it was poured, coloring the otherwise yellow stuff pink.

There may be a risk of forming acetone peroxide with this procedure. Unfortunately, I couldn't think of any other way to extract the complex before it could hydrolyze. Both the fluoride solution and the acetone were ice-cold. A small quantity of the pink stuff was heated in a flame and did not explode.

The supernatant was decanted and the ~100ml of thin pink slurry placed into two improvised desiccators with calcium chloride, one of which is being kept near 0*C while drying.

[Edited on 20-11-2015 by MolecularWorld]

Pok - 20-11-2015 at 15:49

Quote: Originally posted by AJKOER

HOF...flourine radical...speculation

Speculation is a very friendly word. This is totally senseless. Fluorine radicals? This would mean that this simple reaction could produce elemental fluorine. HOF is extremely unstable and could never be formed by mixing these chemicals together.

@MolecularWorld: as your MnO2 didn't dissolve in (extremely diluted) HF, you can say that the reference does NOT describe what you have done. So there is no reason to quote this reference. You should find something else that deals with your reaction and isn't 150 years old. And in your video you didn't "isolate" a compound. Isolation means that you make the pure compound and not a slurry of it.

MolecularWorld - 20-11-2015 at 16:01

Quote: Originally posted by Pok

...make the pure compound and not a slurry of it.

I'm working on it. I don't have a centrifuge or vacuum-filtration apparatus, so all I can do is let it settle, and try to remove the liquid by evaporation.

I may also attempt another batch, careful to ensure there is no excess permanganate (tricky, as I also don't want an excess of peroxide) to get a better idea of the color of the precipitate.

Any comments on why the version with sulfuric acid instead of hydrofluoric gave a much lighter colored (almost colorless) product?

AJKOER - 20-11-2015 at 20:48

Quote: Originally posted by Pok

Quote: Originally posted by AJKOER

HOF...flourine radical...speculation

OK, anytime you are combining three reagents, I would suggest one considers all reaction pairs to explain products.

Also, more from Wikipedia on HOF, to quote:

"Hypofluorous acid in acetonitrile (generated in situ by passing gaseous fluorine through "wet" acetonitrile) serves as a highly electrophilic oxygen-transfer agent.[2][3] Treating phenanthroline with this reagent yielded the previously elusive 1,10-phenanthroline dioxide,[4] more than 50 years after the first unsuccessful attempt.[5]"

Yes, so the presence of HOF is speculative as it is certainly unstable in H2O, reacting to form HF and H2O2. But under dilute conditions, perhaps the equilibrium shift more back to HOF and H2O, as has been reported for the action of dilute HCl on H2O2 forming hypochlorous acid and water, or not as fluorine is a little more unique.

Interestingly, the quote from Wikipedia suggests to me, that perhaps the HOF created in situ (possibly concentrated?) is not so unstable so as not to be able to effect any reaction. [Edit] Or, is it the highly concentrated and unstable H2O2 it forms with water?

Also, if you are forming hydroxyl radicals as transition metals tend do in Fenton-type reactions, in the presence of a halogen ion (including the flouride ion), you can get halogen radicals, which are generally more detectable upon further reacting with another halogen ion. For example:

.Cl + Cl- → .Cl2-

I did edit my post to make it clear that under these dilute conditions I am not expecting gaseous F2 or HF.

[Edited on 21-11-2015 by AJKOER]

deltaH - 20-11-2015 at 21:00

I'd venture that the version with sulfuric was reduced to MnSO4 (colourless).

You might have recovered crude MnF2 (see wiki) from the fluoride media. Permanganate is not particularly stable under acidic conditions, so I doubt that's causing your light pink colour. MnF2 is described as being a light pink (looks almost white on photos).

I think what happened in both beakers was roughly similar, except that the counter ion varied, i.e. manganese dioxide oxidised hydrogen peroxide to oxygen, itself being reduced to a manganese(II) salt.

MolecularWorld - 20-11-2015 at 21:27

I agree the sulfuric acid mix probably resulted in a solution of manganese(II) sulfate. But, I'd thought manganese(II) sulfate and manganese(II) fluoride were effectively the same color, very pale pink, the color being due to the manganese(II), not the anion. The concentration of manganese is also nearly identical in the two beakers in the above photo, so manganese(II) fluoride would have to be a much stronger and more yellow color than manganese(II) sulfate for that to explain the difference in coloration.

I just looked at my precipitate again. It appears to be turning a darker pink. It's still wet with acetone/water mix, so this suggests to me that is may have actually been (partly or mostly) potassium hexafluoromanganate(IV), but is slowly hydrolyzing to MnF₃, which is dark purple-pink. I'm currently trying to arrange for some sort of vacuum filtration, so that I can attempt my acetone precipitation procedure again (ensuring neither permanganate or peroxide are in excess), and then pump the precipitate dry, to see if it's yellow and stays yellow (the complex is supposed to be quite stable when dry).

deltaH - 21-11-2015 at 00:59

Quote:

... (partly or mostly) potassium hexafluoromanganate(IV), but is slowly hydrolyzing to MnF3, which is dark purple-pink.

What you describe is not hydrolysis, but a reduction, manganese (IV) being reduced to manganese (III). What was the reducing agent?

Also, your crude product might be a mix of many things, you would need to recrystallise it first before doing anything else.

MolecularWorld - 21-11-2015 at 06:14

That was based on the reference cited by Pok here. The pink is darker than any picture of manganese(II) fluoride I can find, and while I agree Mn(III) and Mn(IV) are unlikely to be formed, and permanganate wouldn't last long in the presence of acid and acetone, these are the only explanations I can think of. That or color-changing mystery contaminants. Of course, it's not pure MnF₃: Wikipedia states that it too is prone to "hydrolysis", to MnF₂, and there would also be some colorless potassium salts.

If my product contains an unstable complex prone to hydrolysis/reduction/decomposition, recrystallization could be problematic. I'll retain the pink stuff for future study, but I'm now focusing my efforts on preparing a new batch of precipitate, and drying it as quickly as possible.

[Edited on 21-11-2015 by MolecularWorld]

blogfast25 - 21-11-2015 at 10:24

Quote: Originally posted by MolecularWorld

Rather, my baseless assertions have been countered by... baseless assertions. My hypothesis is that the main reason concentrated hydrofluoric acid is used in the literature preparations is to precipitate the complex before it can hydrolyze, but that small amounts of unstable potassium hexafluoromanganate(IV) could form in dilute solution/suspension.

You really haven't got the foggiest idea with regards to what is 'evidence' and 'burden of proof', do you? Nor do you seem to care: too busy putting out triumphant and misleading vids on the UToobs, the preferred medium of the brain dead.

You are beneath contempt, Sir.

Make a note to self: 'must not give up the day job yet'.

[Edited on 21-11-2015 by blogfast25]

ave369 - 21-11-2015 at 10:33

Quote: Originally posted by AJKOER

Its salts, hypofluorites, exist and my speculation is that there could also be yellow (?).
[Edited on 21-11-2015 by AJKOER]

Do hypofluorite salts really exist? All my google-fu on hypofluorites yields some organic, definitely non-salty compounds. And one source claims that HOF does not exhibit any acidity whatsoever, and its "acid" name is a misnomer.

Who is right? Summoning Woelen here, our resident fluorine expert.

(mumbles to herself: Crazy, crazy people. Experimenting with H *cking F like it's nothing...)

[Edited on 21-11-2015 by ave369]

MolecularWorld - 21-11-2015 at 10:37

Another borderline-trolling comment from blogfast. Past the one sentence in my first post, I never claimed the videos proved anything. Rather, like the images and text descriptions, they simply serve to provide information for consideration. And since nobody's attempting to replicate my procedure, I thought being able to actually see the reaction take place would be helpful.

As for proof, I'm working on it. There's only so much I can do in a limited amount of time with an improvised setup.

I was wrong to state the products of the reaction without further testing, and I've agreed that my products may not be what I've originally stated. Further berating me over that is pure trolling, and truly beneath contempt.

Edit: blogfast: I'm also confused as to why you quoted me stating my hypothesis, then went on to complain about 'evidence' and 'proof'. By definition, a hypothesis is "made on the basis of limited evidence". Yes, I should have titled this thread "Attempted preparation of potassium hexafluoromanganate(IV) from household items", and shouldn't have declared the identity of my product based mostly on color: get over it.

[Edited on 21-11-2015 by MolecularWorld]

woelen - 21-11-2015 at 14:46

I also am not convinced at all by the results, posted in this thread, but it is interesting to see that no strictly anhydrous conditions are needed.

Tomorrow I'll try with 48% HF, 50% H2O2 and KMnO4. Right now it is near midnight, time to go asleep, but this is a nice Sunday-afternoon project, especially with the bad weather we will have tomorrow

And yes, I will be careful, VERY careful! I'll work in PP test tubes and use micro quantities.

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

I also am not impressed by the results with the use of acetone. You really think that a beast like MnF6(2-) can coexist with acetone? Acetone is too strongly reducing. I think you just made hydrous MnO2 or maybe hydrous Mn2O3.

I bet that if you replace your HF with acetic acid, that your results will be very similar. HF is a weak acid and hence the formation of colorless (or very pale pink) Mn(2+) is going very slowly or does not complete at all. Reduction of permanganate results in formation of hydrous MnO2 or Mn2O3 (the latter is yellow/brown when very finely suspended and not too concentrated) when there is insufficient free acid. Only at low pH (in the presence of excess strong acid) you get the nearly colorless Mn(2+) ion. So, repeat your experiment with dilute acetic acid and then you'll get quite similar results as with dilute HF.

I even do not expect too much of my experiment I will try tomorrow, but it is interesting enough to try it.

[Edited on 21-11-15 by woelen]

MolecularWorld - 21-11-2015 at 15:17

Quote: Originally posted by woelen

Tomorrow I'll try with 48% HF, 50% H2O2 and KMnO4.

Thanks. I'll be very interested to see your results.

Quote:

I also am not impressed by the results with the use of acetone. You really think that a beast like MnF6(2-) can coexist with acetone? Acetone is too strongly reducing. I think you just made hydrous MnO2 or maybe hydrous Mn2O3.

Acetone is used in the published procedures to wash the product.
My acetone precipitate does not decompose peroxide.
As noted above, I do think my product was reduced (to Mn(III)) by prolonged contact with the acetone/water mix.

Quote:

I bet that if you replace your HF with acetic acid, that your results will be very similar. HF is a weak acid and hence the formation of colorless (or very pale pink) Mn(2+) is going very slowly or does not complete at all. Reduction of permanganate results in formation of hydrous MnO2 or Mn2O3 (the latter is yellow/brown when very finely suspended and not too concentrated) when there is insufficient free acid. Only at low pH (in the presence of excess strong acid) you get the nearly colorless Mn(2+) ion. So, repeat your experiment with dilute acetic acid and then you'll get quite similar results as with dilute HF.

At the suggestion of deltaH, I did a 'control' experiment with sulfuric acid, and produced a differently colored product.
I agree a test with a different weak acid would be a better control, and I will try that.

[Edited on 21-11-2015 by MolecularWorld]

MolecularWorld - 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.

[Edited on 22-11-2015 by MolecularWorld]

woelen - 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.

MolecularWorld - 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]

woelen - 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]

MolecularWorld - 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]

mayko - 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 498 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)
This file has been downloaded 382 times

Pok - 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]

woelen - 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.

MolecularWorld - 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]

Pok - 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]

MolecularWorld - 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]

halogen - 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]

Pok - 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:

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]

MolecularWorld - 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]

halogen - 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.