Sciencemadness Discussion Board - Weird Mn compound

Weird Mn compound

guy - 16-7-2006 at 18:04

I prepared Mn2+ compound by oxidizing an acidified (sulfamic acid) aqueous solution of glycerol with KMnO4. This resulted in a clear solution which was expected. Then I wanted to precipitate the Mn2+ with NaOH. It formed a gelatinous precipitate. Then 3% was added to it. First it formed orange-brown ppt which was MnO2, but after adding more H2O2, it formed a clear red-brown solution with no ppt. This solution will turn clear when the pH is lowered. What is this compound? I know this doesn't happen with regular Mn(OH)2 and peroxide, but this could have formed a complex with the oxidation products of glycerol.

[Edited on 7/17/2006 by guy]

woelen - 16-7-2006 at 22:35

What you obtained could be a manganese (III) solution. Under highly acidic conditions, this ion forms a nice rose/red/brown aqueous ion, and at raising the pH it hydrolyses.

This can be made by careful reduction of permanganate or MnO2 in acidic solution. I made it with malonic acid as reductor, but it also works with some more difficulty with citric acid, and according to your post, also with glycerol.

Could you compare your solution with the one I obtained and which is manganese (III)? Here follows the page with pictures. Look at the section for oxidation state +3:

http://woelen.scheikunde.net/science/chem/solutions/mn.html

[Edited on 17-7-06 by woelen]

guy - 16-7-2006 at 23:39

Mine is an orange-red solution that only forms in BASIC solution, when concentrated, it looks like a nice cherry red color. It acid with a reducing agent(H2O2 in my case), it quickly turns to Mn2+.

I wonder what are the ligands that stabilize this stage. Oxidation of glycerol with KMnO4 can form many things since it has 3 hydroxyl groups. So either I formed mesoxalic acid (COOH)2CO, or glyceric acid, or (COOH)2CHOH (what would that be called?). I think in basic solution, these will form the conjugate bases which would allow complexing.

I will take pictures tommorow.

[Edited on 7/17/2006 by guy]

woelen - 17-7-2006 at 01:18

Sorry that I missed that point of formation in basic solution. I was mislead by the phrase "This solution turns clear when the pH is lowered". I should have read "turns colorless". With "clear" I mean "no solid particles", but it still can have color. Native language-speaking people, am I correct with this?

But guy, that experiment looks really interesting. I definitely will try this experiment this evening. What is remarkable is that on addition of a small amount of H2O2 to the basic solution with gelatinous Mn(OH)2 precipitate, you get a brown precipitate (which can be a manganese (III) compound) and that on addition of more H2O2 you get the red/brown clear solution. This is an indication of formation of a complex of higher than +3 oxidation state, but if the initial brown precipitate already is hydrous MnO2, then this even is an indication of a higher than +4 oxidation state! That would be really interesting.

Time to take the test tubes out of their container again

YT2095 - 17-7-2006 at 01:45

yes something can be clear but have a color, like copper sulphate soln for example.

woelen - 17-7-2006 at 10:29

As promised, I have taken the test tubes from the cabinet, despite the VERY warm weather (35 C in my lab

, 30+ C outside) I did the following experiment:

Dissolve some MnSO4.H2O in water.
Add a small amount of glycerol and dissolve that as well.
In a separate test tube dissolve quite some NaOH
Add both solutions together.

Result is a slimy precipitate, light brown. This is impure Mn(OH)2 with traces of brown oxidized material (there always is some oxygen dissolved in the solutions).

Add a small amount of H2O2 (3%). This results in formation of a dark brown precipitate, and marginal bubbling.

Add much more H2O2 (3%). The solution becomes clear again, but it has a very intense red color, much more intense than the rose/red/brown color of manganese (III) in acidic solution. This deep red color most likely is due to some peroxo complex of manganese (IV), together with glycerol. It is remarkable how intense the color of the complex is.

On dilution with water, the complex decomposes and brown MnO2 is precipitated.

On dilution, however, with a moderately concentrated solution of NaOH, the complex does not decompose. The resulting solution then has a beautiful red wine like color. Very neat. On addition of a small amount of HCl, the complex decomposes again, forming brown MnO2.

So, one can conclude that this is a complex of glycerol, possibly together with peroxo, and this complex only is stable at very high pH. When things have cooled down a little in my lab (< 30 C), then I'll put some more effort in researching this interesting phenomenon.

guy - 17-7-2006 at 11:03

Yes that is exactly what my results were. So now I now that glycerol is the ligand. That is interesting, I wonder what the formula could be and why it only forms in basic conditions.

guy - 17-7-2006 at 11:12

A paper I found on Metal-gylcerol complexes.Some Complexes of Glycerol and Their Applications

Does anyone know what the pH of glycerol is or how easy it is to deprotonate? Its should be more acidic than ordinary alcohols.

[Edited on 7/17/2006 by guy]

12AX7 - 17-7-2006 at 13:20

Hmm interesting, that's the same color I would describe Mn(III) acid solutions. I have some sulfuric solution downstairs, obtained by reacting pottery grade MnO2 with pretty strong H2SO4.

Seems suprising to me that another ligand would have such a similar color at such a different pH!

Tim

chemoleo - 17-7-2006 at 15:48

I'd like to see what'd happen if glycerol was substituted with other organic solvents, alcohols etc.
How about ethylene glycol, or methanol, or ethanol? Acetone, Isopropanol? What if glycerol is left out altogether?
How about erythritol, or higher sugars? mannitol?

I am not entirely convinced that this is a complex with an organic ligand ... what is the evidence for this so far?

guy - 17-7-2006 at 16:00

Well I've seen a picture of Mn(III) without ligands and the color of mine is much more red and darker. And this will only work in the presence of glycerol. And this will only work in a basic solution. If glycerol is let out, only MnO2 will be formed.

mericad193724 - 17-7-2006 at 16:42

Woelen,

AWSOME site. I love it! So much good information. I really like and admire it.

Is there anyone else with a site like that besides you and Bromic acid (awesome site too).

Mericad

12AX7 - 17-7-2006 at 17:25

I have some stuff, mostly how to make some basic metal salts...

guy - 17-7-2006 at 19:36

Sodium manganese triglycerolate?

Concentrated vs. Dilute

2Mn(OH)2 + H2O2 + 6OH- + 6C3H8O3 -----> 2[Mn(C3H6O3

3]3- + 12H2O

[Edited on 7/18/2006 by guy]

woelen - 17-7-2006 at 22:34

Quote:

Originally posted by chemoleo
I'd like to see what'd happen if glycerol was substituted with other organic solvents, alcohols etc.
How about ethylene glycol, or methanol, or ethanol? Acetone, Isopropanol? What if glycerol is left out altogether?
How about erythritol, or higher sugars? mannitol?

I am not entirely convinced that this is a complex with an organic ligand ... what is the evidence for this so far?

Best evidence is that without glycerol, addition of H2O2 to a suspension of Mn(OH)2 in a solution of NaOH only results in formation of a brown flocculent precipitate and a lot of bubbles of oxygen. With the glycerol present in the alkaline solution, a deep wine-red solution is formed, and no precipitate is formed. That is enough evidence for me that there is a complex with an organic ligand.

Quote:

Sodium manganese triglycerolate?

Now the second step. What is this complex? I'm absolutely not sure about this being a glycerolato manganate (III) complex, as suggested by guy. It might as well be a mixed ligand complex, e.g. with glycerol (or a deprotonated version of that) and with peroxo ligands. This is something which also can be tested easily. Take another oxidizer than H2O2. I'll try tonight with Na2S2O8 as oxidizer. One could also try with oxygen from air as oxidizer by prolonged shaking of the liquid with contact with air.

It might well be true that also complexes can be formed with other organic compounds. This evening I can try the same experiment with glucose, with pentaerythritol and with plain sugar. All of them have plenty of -OH groups attached. I'll come back on this later.

I once did the experiment with ethanol, and that does not give such a complex. Most likely because this can be at most a mono-dentate ligand. I do think that ligands with multiple bites are needed for this type of complex.

Quote:

Is there anyone else with a site like that besides you and Bromic acid (awesome site too).

Thanks for the compliment

. There is an index of websites of many members over here on sciencemadness. It is a sticky topic in the forum matters subsection. Many websites are mentioned there, and some are quite interesting. So, yes, there are more good websites around, fortunately!

[Edited on 18-7-06 by woelen]

guy - 17-7-2006 at 23:02

Well this is kind of Off-Topic, but I tried this with Cu2+ and glycerol in NaOH forms a dark blue solution, much like ammonia does. (See the link I provided before for reference). This leads me to think that glycerol will only form complexes in basic conditions because, it will deprotonate with the aid of a metal ion. Glycerol is more acidic than normal alcohols because there are less electron donating groups. The metal ligand will form a bond between the ion and the OH, which will cause the hydrogen to become more positive and more easily deprotonated.

Also, I will try the Mn experiment with a different oxidizer and see the result too.

guy - 18-7-2006 at 12:25

I have just conformed that this is compound is at least Mn4+ or higher. I started this time with pure MnO2. I added a basified solution of glycerol to it. Then H2O2 was added. The same red solution was formed.

woelen - 18-7-2006 at 12:41

I also just did some experiments.

I did the original experiment, starting off with manganese (II) hydroxide, suspended in a solution of NaOH.

I did the experiments with Na2S2O8 as oxidizer.

The procedure was as follows:

Step 1: Prepare a solution of MnSO4
Step 2: Add some ligand** to the solution of MnSO4 and dissolve that as well.
Step 3: In a separate test tube, prepare a solution of NaOH, fairly concentrated.
Step 4: Add the solution of MnSO4, ligand** and NaOH to each other.
Step 5: Add some solid Na2S2O8 as oxidizer to the result of step 4.
**: I repeated the experiment for the following four ligands:
A) Glycerol
B) Pentaerythritol
C) Glucose
d) Saccharose (plain sugar)

With glycerol the result is the same as with H2O2 as oxidizer. From this I conclude that the complex is not a peroxo complex, it is sufficient to have glucose as ligand.
With pentaerythritol as ligand, no complex is formed. On addition of Na2S2O8 the suspension of Mn(OH)2 turns dark brown and more flocculent.
With glucose also no complex is formed. Same brown precipitate.

Finally, with plain sugar a really interesting effect is created. The solution becomes VERY eager for oxygen. No suspension of Mn(OH)2 is formed when the solution of NaOH is added to the MnSO4/sugar solution. At the surface of the liquid a red/brown complex is formed, which slowly moves into the liquid. A little later, the liquid becomes covered by a yellow/brown crust and the walls of the test tube, which are exposed to air, also first become brown, and lateron scales of a yellow/brown compound slowly move downwards along the glass. When the liquid is shaken, then all brown scales redissolve again, and the liquid becomes clear and red/brown. On standing, all places, which are in contact with air soon become covered by the yellow/brown solid again. After some time, the entire liquid becomes dark brown and turbid and I think, finally MnO2 is formed.

It looks as with sugar, that a complex is formed with manganese in the +3 oxidation state and that on further oxidation this is destroyed again.

===================================================

Your experiment with H2O2, added to MnO2 also makes me think more and more that it is a complex with manganese in the +3 oxidation state. You start off with a suspension of MnO2 in NaOH/glycerol solution and on addition of H2O2 it becomes red and clear? If that is the case, then I would say that the H2O2 in this case acts as reductor and reduces the MnO2 to Mn(III), itself being converted to water and oxygen. It is well known that H2O2 can act as reductor. That even is applied in cleaning glassware from MnO2 stains, by rinsing it with a dilute H2O2/acid mix. So, why could H2O2 not act as reductor in the alkaline medium with the ligand as stabilizer for the +3 species of manganese?

This is very interesting stuff and I'll perform a few more experiments, trying to see whether this is a +4 (or higher) oxidation state complex, or +3.

guy - 18-7-2006 at 12:56

Well I tried the experiment this time with MnO2, NaOH, glycerol, and NaOCl(household bleach). The results were almost the same. It formed a yellow solution first, but on heating, the same red solution was formed.

What does this mean? Can hypochlorite reduce Mn4+ to Mn3+?

Can you try all your experiments this time starting only with MnO2? Then use substances that can ONLY oxidize, such as persulfate or hypochlorite. This will be more conclusive.

woelen - 18-7-2006 at 13:32

Your experiment with NaClO indeed is a strong argument against my +3 oxidation state hypothesis. NaClO indeed cannot reduce as far as I know.

Did you use commercially obtained MnO2, or a hydrous suspension, made from Mn(OH)2. I have MnO2 from a pharmacy, but this is a dry crystalline very fine and glittering powder. It seems to me that such MnO2 would be very slow in its reactions.

Tomorrow evening I again have the time to experiment further. If you could give some info on your MnO2 before that, that would be very helpful. Right now, I'm determined to solve this "riddle", at least the oxidation state part of the riddle.

guy - 18-7-2006 at 13:41

It is formed from the decomposition of the red compound. It is the hydrous form. I cleaned and washed it of course, so it would be fairly pure.

[Edited on 7/18/2006 by guy]

woelen - 20-7-2006 at 13:55

I did some further experimenting on this and now I come to the conclusion that the complex either is a manganese (V) or higher compound with glycerol ligands, or it is a manganese (IV) compound with an oxidized species of glycerol as ligand.

I did the following:

Prepare some hydrous MnO2 from MnSO4, NaOH and Na2S2O8. Filter and rinse the precipitate well with water.

The still wet, but cleaned, precipitate is added to a solution of NaOH, to which also some glycerol is added. If this is done, then there is no visible reaction. The solid does not dissolve, the liquid becomes turbid and very dark brown.
When some Na2S2O8 is added, then the solid MnO2 dissolves and a deep red/brown complex is formed. So, indeed the oxidizer is needed. Hydrous MnO2, NaOH and glycerol alone do not form the complex.

I also tried the reaction with some commercial MnO2 from a pharmacy. This does not react at all under these conditions. This is something I expected. The MnO2 I have consists of many small very hard refractive glittering crystals, which only can be dissolved with difficulty in conc. HCl.

guy - 20-7-2006 at 19:27

Is there anyway to isolate this compound? It is soluble in alcohol and it wont decompose if you attempt to evaporate it. Maybe it will be easier to analyze once it has been isolated, such as how much MnO2 will be formed with X grams of it.

OR

If you have the time, you can try to see how many grams of a certain reactant is needed to fully react, assuming that it goes to completion. You could find out the number of glycerol ligands there are at least.

This is a way to determine oxidation state: How much of a certain oxidizing agent is required to form the compound.

I would do this but I don't have any measuring tools.

[Edited on 7/22/2006 by guy]

guy - 25-7-2006 at 21:09

What is the reason why MnO4(2-) has to be in basic solutions? My guess is that it is unstable enough to react with H+ formed by water. What do you think?

woelen - 25-7-2006 at 23:24

The ion MnO4(2-) is resonance stabilized, while the molecule H2MnO4 is not.

H2MnO4 has structure MnO2(OH)2, with the O-atoms without hydrogen double bonded to the Mn, and the O-atoms with the H-atom single bonded to Mn. Hence, the O-atoms are not equivalent in this compound. When the two H-atoms are removed, then all O-atoms become equivalent, each having "on average" 1.5 bonds. This structure is much more stable than the structure of HMnO4(-) and the structure of H2MnO4.

Many ions are much more stable than their corresponding acids. E.g. NO3(-) is very stable in water, HNO3 is a strong oxidizer. The same is true for ClO4(-), which is very inert in water, while molecular HClO4 is an extremely dangerous oxidizer (which only exists in 72+% solutions, at lower concentration all is ionized).

Resonance stabilization is present in MANY compounds, and also in MANY ions. The classical example is benzene, but lateron it became clear that it is not special at all.

guy - 25-7-2006 at 23:35

Thanks that was a good explanation. So since this weird compound of manganese is stable in basic conditions, is there reason to believe that it has oxygens around it?

not_important - 26-7-2006 at 01:11

OK

1) Remember that you can run the haloform reaction on alcohols and get a positive test for ketones, becase

RR`CHOH + OCl- => RR`CO + H2O + Cl-

Read the history of sugar chemistry, the oxidation of crabohydrates with halogen and base was often used.

So rather than the reduction of Mn4+, you most likely had the oxidation of the alcohol.

2) The oxidation by permanganat of alkenes to diols is thought to have a transition compound like

C-O O-
| \ /
| Mn
| / \
C-O O-

As does oxidations wih OsO4

So a complex of Mn and diols (or higher polyols) whould not be unexpected.

3) H2O2 is a strong oxidiser, about as strong as permanganate. At the same time, it can reduce some substances while oxidising itself to O2 and water. So mixtures with H2O2 can give unexpected oxidation levels.

4) Mn2+ and Mn3+ appear to form stable aqueous complexes. While many of these include amino groups, some may only have H-O- and -O-C as the ligands

http://stratingh.eldoc.ub.rug.nl/FILES/root/FeringaBL/2004/M...

5) The colours seen are quite similar to those I made when attempting a good, strong cone 10 red. I was trying for Mn3+, but I know that sometimes I was getting some Mn4+ About the only 'ligand' in these is oxygen and phosphorus, and as there's no water Mn3+ is a bit more stable than with aqueous chemistry; plus going to Mn4+ doesn't mean its going to clump up as MnO2 (at least at low concentrations).

Try adding some borax to the red coloured stuff. Borates form fairly strong complexes with polyols, so it might affect the colour.

guy - 27-7-2006 at 17:04

Can anyone with measuring equiptment try finding how much glycerol per MnO2 or Mn2+ is needed? That would really help.

chemoleo - 27-7-2006 at 17:24

If there are truly ketones or aldehydes present, there are numerous great analytical tests that determine the rpesence of these. For instance dinitro phenylhydrazine. I doubt this is what it is because these oxidations don't happen this rapidly, do they? Add some H2O2 to small amounts of EtOH in NaOH, and see how long it takes till you smell Acetaldehyde. By the way, what happens if you heat the complex? Is it stable? What happens if you add a surplus of oxidiser, to then heat it? Any change? This might tell whether oxidation of the ligand is detrimental to complex formation.

Do you have a dessicator? You could try crystallising something that way. But I think first of all these experiments have to be done stoichiometrically, so far they are qualitative at best. Lots of work I know. Work with larger amounts, and it is easier.

guy - 27-7-2006 at 17:47

Quote:

Originally posted by chemoleo
If there are truly ketones or aldehydes present, there are numerous great analytical tests that determine the rpesence of these. For instance dinitro phenylhydrazine. I doubt this is what it is because these oxidations don't happen this rapidly, do they? Add some H2O2 to small amounts of EtOH in NaOH, and see how long it takes till you smell Acetaldehyde. By the way, what happens if you heat the complex? Is it stable? What happens if you add a surplus of oxidiser, to then heat it? Any change? This might tell whether oxidation of the ligand is detrimental to complex formation.

Do you have a dessicator? You could try crystallising something that way. But I think first of all these experiments have to be done stoichiometrically, so far they are qualitative at best. Lots of work I know. Work with larger amounts, and it is easier.

I tried boiling it, nothing happens, you can actually crystallize it. I tried boiling it in excess NaOCl and nothing happens. The problem is I have no accurate measuring tools so I am asking anyone who has the time ad equiptment and actually cares about this

.

[Edited on 7/28/2006 by guy]

guy - 31-7-2006 at 15:23

Some New Information

I just crystallized some of this product and washed it with alcohol. Its a bright orange solid.

Ok so I put some of this in a crucible and heated it. It decomposes quickly to MnO2 (black solid) and some smoke (water vapor + ??). The solid I thought must have some Na2CO3, and I tested this by adding vinegar to it. It bubbled indicating CO3(2-).

What could I do with this? Measure the amount of Na+ present by precipitating some insoluble CO3(2-).

Measure the change in mass and calculate what constituted the missing mass through guess and check?

not_important - 31-7-2006 at 19:00

Can you dissolve a bit of it in plain water, get the original color, and get a bit of it to crystalize back from that solution? This might help determine if it is stable on its own, or only in the presence of an excess of one of the reagents.

Going to MnO2 sure suggests it is Mn(IV) or higher, but see below.

(oops, see that you don't have measuring gear. But I'm leaving this anyway)
If you have a sensitive scale you could weigh a crucible, weight some of the orange crystals into that, heat, cool away from air/moisture, and weight again to get an idea of how much organics and water are in it. The leach it out with distilled water, dry and weigh again for how much is Mn. Evaporate the water solution to get the weight of the soluble stuff, then test that for Mn(II) - if the orange stuff is a complex of Mn(III) it could got to II + IV when it decomposes.

Wish all my gear wasn't in storage.

guy - 31-7-2006 at 20:05

You can only dissolve it in basic solution or it will decompose into MnO2. Its like manganate where it is only stable in alkaline solution. I crystallized before by evaporating most of the water but not all (made sure to not get any NaOH precipitating out).

Another thing I wanted to do if I had equiptment was to titrate the decomposition product to see how much Na2CO3 was in there.

guy - 1-8-2006 at 17:52

Actually I did some more experiments and I was wrong. THe solid is SPARINGLY soluble in water and almost insoluble in basic. In acid it will disproportionate to dissolve to form a brown-cinnamon solution and MnO2. The solid is DIFFERENT than the solution it came out of.

The brown solution will react with H2O2 to form Mn2+. It will react with NaOH to very slowly form MnO2. It does not react with NaOCl or H2O2 in basic solution.

not_important - 2-8-2006 at 01:29

Good job, guy.

Here's some related information, from Sidwick's Chemical Elements and Their Compounds.

Mn(III) oxide has two forms. The hydrated one, MnO(OH) is formed when wet MnO is exposed to air, espeically if bases are present, when MnCl2 is oxidised by air in the presence of excess NH4Cl, and several other ways. It can be grey, brown, or black depending on how it was made. It will generate chlorine if treated with HCl.

Mn(III) forms complexes with H3PO4, oxalic acid, malonic acid, and other oxyo compounds. These generally are a deep red, red-violet, or red-brown in solution, the pure solid compounds are usdually red or green, some are yellow or yellow-brown ,and others red-brown.

These all seem to contain the structure

-O
\ /
Mn
/ \
-O

A test for Mn(III) might be as follows. If you have very concentrated H3PO4, 98%, or fairly concentrated H3PO4 and 98% H2SO4 (mix a cc of each), try dissolving a bit of the mystery compound in that. In strong acid the colour will be violet, if the solution is diluted then grey-gren MnPO$ will precipitate.

guy - 2-8-2006 at 22:14

How light is the color of Mn(III) phosphate complex? I am trying to test for Mn(III) after I acidified the red solution. When the red solution is acidified with acetic acid, it forms MnO2 and a brown solution (manganese(III) acetate?). I added some phosphoric acid (from shower cleaner) and it formed a very light-green precipitate.

I suspect this compound might oxidize the glycerol ligands in acid to form an aldehyde or ketone (glycerol can do either or both), but I don't have the neccesary chemicals to do it!

[Edited on 8/3/2006 by guy]

not_important - 2-8-2006 at 23:48

I'm not sure that the precipitate is MnO2, it could be MnO(OH); it's not simple to tell.

The several descriptions of MnPO4 just say grey or grey-green, as 'dark grey' is used for other things I would assume it is fairly light in shade. The Mn(II) phospates are all pink to red.

Mn(III) acetate isn't stable in water, the brown may be yet another complex or colloidal manganese hydroxide/oxides. I suspect that MnO(OH) would give the phosphate, MnPO4 is very insoluable which is mainly why it can be made in water.

I think you could get the oxidation even in somewhat basic solution, and getting carboxylic acids isn't out of the question.

Off the top of my head, the only way that comes to mind to investigate the organics would be to make 10 or 20 grams of the complex, hit it with trisodium phosphate and sodium carbonate to drop out all the manganese while still in an alkaline solution. Then filter and distill, simple ketones and aldehydes would come over with the steam; I don't think any of the hydroxy-(acids,ketone,aldehydes) would. Evaporate the remainder of the carbonate/phosphate solution, check for smell and then try extracting with acetone to see if there is relatively non-volatile organics.

If you have strong sulphuric acid, try dropping a small crystal of the orange stuff into 1cc H2SO4 + icc tile cleaner. Mn(III) forms sulfate complexes as well as phosphate, if the acid is strong enough.

guy - 3-8-2006 at 21:46

Here's my guess on the compound based on my observations so far.

Edit:
Deleted pic.

[Edited on 8/4/2006 by guy]

Nicodem - 4-8-2006 at 00:46

Just a few comments on mechanism drawing:
The net electron transfer is of two electrons so the Mn(V) should go to Mn(III). So your mechanism is wrong in showing MnO2 as the product. However it is true that fast disproportionations lead to MnO2 as the end Mn species in neutral of weakly acidic media, but you omited that. There also lack an electron pair transfer arrow from H-C bond to C-O bond. Also, you should not depict the "glycerolmanganic acid" in the dissociated form or else the protonation step makes no sense.

guy - 4-8-2006 at 11:32

Thanks for pointing that out. New drawing

[Edited on 8/5/2006 by guy]

not_important - 4-8-2006 at 22:37

Nice drawings, what are you using to do them?

I've done more researching, and it still looks as if you have a complex of Mn(III), Mn(III) and Mn(II), or just maybe Mn(III) and Mn(IV). Interestingly complexes with Mn(III) often have several Mn atoms in each complex, similar with the mixed oxidation state cases. I've seen refernces, full article tucked away behind 'for fee' walls, of complexes based around 4 Mn(III) ions

guy - 4-8-2006 at 22:51

Quote:

Originally posted by not_important
Nice drawings, what are you using to do them?

I've done more researching, and it still looks as if you have a complex of Mn(III), Mn(III) and Mn(II), or just maybe Mn(III) and Mn(IV). Interestingly complexes with Mn(III) often have several Mn atoms in each complex, similar with the mixed oxidation state cases. I've seen refernces, full article tucked away behind 'for fee' walls, of complexes based around 4 Mn(III) ions

I used ChemDraw, unfortunetely its a trial version.

Did you find the information about the Mn(III) complex from Google scholar? I doubt this is a +3 because Woelen tried MnO2, Na2S2O8, NaOH, glycerol and it had the same results. Therefore it has to be at least +5. It will work with OCl- but OCl- oxidizes the glycerol faster than the MnO2 so it wont work as well.

not_important - 4-8-2006 at 23:51

Ah, but Mn(IV) and higher get reduced by polyols. Given the very low solubility of MnO2 and hydrated kin, I wonder if the oxidiser isn't converting a bit of it to a higher state that gets the Mn into solution; a bit like a trace of Cr(II) solublizing CrCl3. And the complex gives enough stability to Mn(III) that it can stay around. We need ESR gear.

Both online and books on hand research. None of the mentioned classes of Mn(IV) and higher complexes are the proper colur, most are greenish. And most of them require nitrogen containing ligands as well, while there is mention of purely oxo complxes of Mn II and III. When making Mn(OAc)3, Mn(OAc)2 and acetic acid stablize the III state.

Wish I had easy access to my gear. Be interesting to try to make some Mn(OAc)3 and add gylcerol.

guy - 5-8-2006 at 00:36

Are you thinking that a higher oxidation state of Mn oxidized some of the glycerol and itself being reduced to 3+? Is that possible in basic conditions?

I guess I could try making MnO(OH) and adding glycerol (in basic conditions). MnOOH is made by Mn(OH)2 and shaking with oxygen? I guess I could try that tommorow.

[Edited on 8/5/2006 by guy]

guy - 5-8-2006 at 12:13

EVIDENCE for +3 state

Mn(OH)2 was precipitated from a solution of Mn(OAc)2. This was slightly brown due to oxidation from air (MnOOH). More NaOH was added along with glycerol. No apparent reaction until the air was bubbled into the mixture forming the same red solution!

My guess for why this reaction worked with MnO2 as the starting reagent was becaused MnO2 was oxidized to a higher state, which then oxidized the glycerol, reducing itself to MnOOH and reacting with glycerol.

My hypothetical structure for this is:

When pH is lowered the Glycerolate get re-protonated leaving MnOOH again (which can be confused for MnO2)

Nicodem - 5-8-2006 at 12:30

And why not like in the attachment?

The Mn(III) salts are not oxy salts. For example, Mn(OAc)3 and not MnO(OAc) so I don't see why such a complex would have to have any Mn=O bonds.

[Edited on 5-8-2006 by Nicodem]

Mn(III)glycerol.gif - 3kB

guy - 5-8-2006 at 12:34

Quote:

Originally posted by Nicodem
And why not like in the attachment?

The Mn(III) salts are not oxy salts. For example, Mn(OAc)3 and not MnO(OAc) so I don't see why such a complex would have to have any Mn=O bonds.

[Edited on 5-8-2006 by Nicodem]

Because how can it get rid of the oxy groups in a really high pH? The usual Mn(III) salts are formed in low pH's.

not_important - 5-8-2006 at 19:31

Ok, thanks guy! I suspected the possibility of Mn(III) simply because none of the complexes of higher oxidation states had colours in the right range.

This isn't going to be easy, Mn forms polynuclear complexes. As an example, even though it's done at 100 K, here is one with two Mn(III) and four Mn(II)

http://scripts.iucr.org/cgi-bin/paper?gk2005&buy=1

Note that the ligands ketones and carboxylic acids, previous to finding this and your experiment with Mn(OH)2, I wondered if oxyidised glycerol is at least part of the complex, even though Mn does complex with polyols. This doesn't discourage me from thinking that.

edit -

I suspect Mn(III) or a mixed state complex. But it is possible that there are higher states, complexing may lower the cost of getting there enough that even air oxidation can reach them; think of the Co(III) complexes.

[Edited on 6-8-2006 by not_important]

Nicodem - 5-8-2006 at 22:06

Quote:

Originally posted by guy
Because how can it get rid of the oxy groups in a really high pH? The usual Mn(III) salts are formed in low pH's.

But you can't say acetic acid makes for a low pH.
Also, the complex in the reference Not_important provided, the Mn(III) is not in the form of an oxo salt. Furthermore, I can't imagine how can an oxo salt (having covalent bonds besides the ionic charge) be suitable to interact with complexing ligands at all. Can, for example BiO(+) cation, form a complex with some ligands? I don't know much about inorganic chemistry, so I could be wrong.

Edit: A Mn(III) complex would not behave like a Mn(III) salt at high pH. For example, CuCl2 will precipitate Cu(OH)2 at high pH in water solution, but its complex [Cu(NH3)4]Cl2 will not.

[Edited on 6-8-2006 by Nicodem]

guy - 5-8-2006 at 22:25

Ah ok I see they you are right Nicodem. I was thinking they were covalent like in higher oxidation states like MnO4-. So the complex makes more sense without the oxo-ligands.

guy - 5-8-2006 at 23:07

Quote:

Originally posted by not_important
Ok, thanks guy! I suspected the possibility of Mn(III) simply because none of the complexes of higher oxidation states had colours in the right range.

This isn't going to be easy, Mn forms polynuclear complexes. As an example, even though it's done at 100 K, here is one with two Mn(III) and four Mn(II)

http://scripts.iucr.org/cgi-bin/paper?gk2005&buy=1

Note that the ligands ketones and carboxylic acids, previous to finding this and your experiment with Mn(OH)2, I wondered if oxyidised glycerol is at least part of the complex, even though Mn does complex with polyols. This doesn't discourage me from thinking that.

edit -

I suspect Mn(III) or a mixed state complex. But it is possible that there are higher states, complexing may lower the cost of getting there enough that even air oxidation can reach them; think of the Co(III) complexes.

[Edited on 6-8-2006 by not_important]

Damn thats a huge molecule. Maybe I should try oxidizing all the Mn(OH)2 to MnOOH then add glycerol to see, then that could narrow it down.

not_important - 6-8-2006 at 02:46

I did pick one of the larger complexes I'd encountered; it was interesting for being purely oxygen links. Many of the complexes seem to be 2 to 4 Mn per complex; all the Mn may be in the same oxidation state or it may be a mixed state complex.

The MnO(OH) might be able to oxidise glycerol, which would give you Mn(II) again. However the test of using as close to 100% MnO(OH) as possible does sound worthwhile.

I'm thinking on some ways to figure what organics are in the complex, without resorting to equipment or supplies that aren't easy to be had.

guy - 6-8-2006 at 12:46

I tried it again first by oxidizing the Mn(OH)2 in base to MnOOH, some were converted to MnO2 since Mn(OH)2 + 2OH- -----> MnO2 + 2H2O + 2e E=0.05. Then I added glycerol and stirred it and same red solution.

I think that it could be just as simple as a glycerolate complex. Do you think it is possibe to deprotonate glycerol when it complexes? Glycerol has only weaker electron donating groups than regular alcohols. The link I have on the first page had Cu glycerol complex and a boric acid gylcerol complex.

How can we test for a presence of glycerol? Or if it was oxidized to an aldehyde you could use Tollen's reagent (which I don't have).

[Edited on 8/6/2006 by guy]

not_important - 6-8-2006 at 16:09

Did the red colour form quickly - complex formation vs oxidation then complex?

How about making some MnO(OH) in as little base as possible. Try to determine how much base is needed for the complex to form.

It could be a simple glycerolate complex, I'll agree. However manganese appears to like to form polynuclear complxes, so while this may be a glycerolate complex, it may not be simple although not s bad as the monster I posted the link to.

Add a liitle borax to a bit, see if borate displaces Mn; boric complexes of polyols are pretty stable.

Add a bit of bisulfite, glucose or invert sugar, or other mild reductant. What doeas it take to get back to Mn(II). I think you've shown that Mn(II) alone isn't making any interesting complex.

Charge transfer complexes, with an element in two different oxidation states, tend to have strong, deep colours. Another reason to suspect mixed valency in this, but hard to tell without fancy gear we don't have.

guy - 6-8-2006 at 16:16

Yes it formed quickly almost the same speed. It takes about 20-30 seconds to fully form. I will try it with as little base as possible.

-------------
Edit:

It takes quite a lot of base to form this. I couldn't measure the exact amount since I have no scale though. To get it back to Mn2+ you need acid and a reducing agent like H2O2.

[Edited on 8/7/2006 by guy]

[Edited on 8/7/2006 by guy]

not_important - 6-8-2006 at 20:44

Even little vs alot is useful, as it suggests the hydroxide is being used in a reaction. Many of the glycolates don't require a base or acid to form. If the managese is already Mn(III) the base must be serving another purpose, I think. The energetics of complex formation are often such that you don't need to deprotonate the diol/glycl so much as to soak up the protons levered off the OHs during complex format. Even bicarb works in some cases.

Problem with H2O2 is that it can potentially also be oxidising the glycerol. That's why I suggested bisulfite or other reducing agents.

I am going to cook some dry cell guts and ammonium sulfate together this evening, to get some MnSO4. Hopefully I will be able to replicate you experiment and join in the fun.

woelen - 12-8-2006 at 10:03

I'm back from vacation and I'm pleased to read that this research continued! Guy, could you please describe in detail how you dried the material and obtained crystals. Although you don't have weight measuring gear, try to give estimates of amounts.

I have weight measuring gear and quite a lot of chemicals, such as diverse phosphate salts, phosphoric acid, concentrated H2SO4, so I could try many of the suggestions. If I can obtain a gram or so of the orange/brown compound in pure state, then I can do a lot of experiments and help solve this riddle

guy - 12-8-2006 at 10:11

I obtained the orange crystals by evaporating about 90% of the solution. Then I filtered it and washed with denatured alcohol. The solid is bright orange and quite stable in air. I hope you have good results when you try it.