woelen
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Molybdenum and nitric acid, what is the reaction product???
When molybdenum is dissolved in concentrated nitric acid, then a red powder remains. This really puzzles me, I don't understand what is happening. I
expected to get a colorless solution of a molybdate, or a white suspension of hydrous MoO3. Instead I get a strange red/brown precipitate, which also
is fairly inert. Hydrogen peroxide, however, quickly dissolves this precipitate.
http://woelen.homescience.net/science/chem/riddles/Mo+HNO3/i...
Any ideas what this could be?
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[Edited on 7.1.14 by bfesser]
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DJF90
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Could it be possible that you have changed from one allotrope of molydenum to another? A quick google search yeilds nothing but a red allotrope is
known for several other elements (selenium comes to mind...). This would also allow for an explanation of why it still produces coloured peroxy
complexes (the H2O2 would oxidise it from 0 to +6).
I just remembered allotropes are because of a different arrangement of the atoms, not something that happens in metallic elements... my bad.
[Edited on 25-5-2008 by DJF90]
[Edited on 25-5-2008 by DJF90]
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The_Davster
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From Handbook of Organic Chemicals, Patniak, P
| Quote: |
the metal is attacked by nitric acid and aqua regia. The reaction is
rapid with dilute nitric acid but slow with concentrated nitric acid due to the
formation of a protective oxide film over the metal. With excess nitric acid, the
solution becomes colorless due to formation of molybdic acid, H2MoO4. Excess
molybdenum turns the solution red |
MoO2, Mo4O11 are described as brown-purple powders in one reference, and in Elements of Chemistry (1871)
is described as below
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woelen
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| Quote: | Originally posted by The_Davster
From Handbook of Organic Chemicals, Patniak, P
| Quote: |
the metal is attacked by nitric acid and aqua regia. The reaction is
rapid with dilute nitric acid but slow with concentrated nitric acid due to the
formation of a protective oxide film over the metal. With excess nitric acid, the
solution becomes colorless due to formation of molybdic acid, H2MoO4. Excess
molybdenum turns the solution red |
|
My experience is quite different. I use very fine Mo-powder, and reacts vigorously with nitric acid. I only use a small spatula of this metal powder
(100 mg or so, at most) and I use a few ml of concentrated nitric acid, so there is a large excess of nitric acid. Still, I don't get a colorless
liquid, it becomes red, as shown in the pictures in the webpage. Also, this quote only describes observations, but does not give an explanation.
I like the old 1871 much better, and this could be closer to what really is happening. The copper/MoO3 test is something, which I can try. I'll come
back on that when I have done the experiment.
[Edited on 26-5-08 by woelen]
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The_Davster
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I just tried it as well, -200mesh Mo and 70% nitric. No reaction until I added a few drops of distilled water at which point the usual metal-nitric
reaction occurred leaving me with exactly the results you describe. Adding many more mL of nitric did not cause a disappearance of the strange red
color.
Molybdenum is pretty good at coordination, I wonder if it is some sort of nitrogen containing complex?
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12AX7
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Nitroso- or nitrito- something? Ah, but I recall platinum dissolved in aqua regia forms a Pt(IV) nitroso- compound, not to mention all the
other transition metal nitrite complexes that are known. Could be plenty of those sorts of things formed by the NO2 gas.
Filter a sample and heat it, see if brown gas comes off maybe?
Tim
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Jor
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I think a nitrito/nitroso-complex can't be that stable/inert, to withstand a concentrated nitric attack.
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woelen
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I also have severe doubts on the theory of nitrito/nitroso complexes, but I'll try to isolate some of the powder and heat the dry stuff.
==============================================================
I also did the experiment, described in the 1871 text, posted by The_Davster.
I took 1 spatula full of coarse copper powder (small granules, 0.1 mm or so, like table salt)
I took 1 big spatula full of finely powdered MoO3 (excess MoO3)
I added a few ml of 35% HCl, lab reagent grade
The mix fairly quickly dissolves, especially if it is heated somewhat. The solution becomes totally black and opaque. No solid particles remain, all
copper and MoO3 are in solution now. Apparently the MoO3 oxidizes the copper, itself forming an intensely dark compound, it seems to have a blue hue.
Next, I diluted somewhat and added an excess amount of ammonia, 25%. The dark liquid slowly changes color. It becomes dark blue and a red precipitate
is formed. The dark blue color, however, slowly fades and after some time, it is only light blue. The red precipitate is very fine and only settles
very slowly. For this reason, the liquid with the precipitate was heated till boiling. This makes the precipitate much coarser and now it quickly
settles. The liquid is only pale blue at this point, but it absorbs oxygen and near the surface it is much darker blue.
The ammonia solution is decanted from the red precipitate. The decanted liquid very quickly becomes dark blue. So, the red precipitate is the reducing
agent, which keeps the solution nearly colorless. When the solution is separated from the red precipitate, then it soon becomes very dark blue. The
colorless [Cu(NH3)2](+) is oxidized by oxygen from air, resulting in formation of hydroxide and blue [Cu(NH3)4](2+).
The red precipitate is rinsed a few times with distilled water.
AND NOW THE DECEPTION: The red precipitate is nothing else than fine copper powder! When the red powder is added to dilute sulphuric acid or dilute
nitric acid, then it does not dissolve, but on addition of some H2O2 it slowly dissolves. After a few minutes, a pale blue solution is obtained. The
color of aqueous copper(II) ions.
The red precipitate does not dissolve in conc. HCl, but when some H2O2 is added, it quickly dissolves and a green/yellow solution is obtained. The
color of [CuCl4](2-).
Adding some conc. nitric acid to the red powder gives a vigorous reaction. The liquid becomes sky blue. Color of copper(II) ions at somewhat higher
concentration.
So, I think that the author of the 1871 text was totally wrong. His observations were right, indeed a red powder is obtained, but this powder
definitely is not a molybdenum compound, but it is plain copper metal, finely divided. This explains the observations, described above, and it also
explains why the ammoniacal solution first was blue, but became much lighter. The [Cu(NH3)4](2+) complex is quite good of oxidizing copper metal,
giving rise to formation of a copper(I) complex:
[Cu(NH3)4](2+) + Cu --> 2[Cu(NH3)2](+)
The author of the 1871 text probably was mislead by the blue color of the liquid, assuming that all copper was in that, and that the red powder was
the Mo-compound. I think that this author did not do any further tests with the red powder. The red powder does not contain any Mo in measurable
quantities, otherwise there would be a red color on addition of H2O2 in acidic environment, due to formation of a peroxo complex. The powder, made
from Mo and HNO3 is totally different, as shown by my webpage.
[Edited on 27-5-08 by woelen]
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