[kkjc2]

What is the spceies formed in the brown-ring test for nitrate (acidified ferrous sulfate and nitrate/ conc sulfuric acid boundary)

and how is it formed?

[madscientist]

Show your working so far and the genius's on this site may be able to help you.

cheers,

madscientist :laughbounce:

[kkjc2]

My thought is the formation of some iron(III) nitrosyl complex.
The nitrate oxidise iron (II) to iron (III) and is reduced to nitrite.
Nitrite reacts with conc sulfuric acid, which condensate to give some complex??

I couldn't find any book where the mechanism is explained. even in thick inorganic analysis text.

It's not an example that I need to hand in or any set work, it's just a curiousity to understand a bit more about this commonly done test. It have been a question in my mind for one year or so now.

If some genius could enlighten me, that will be great!

[AWK]

NO3- + 3Fe2+ + 4H+ = NO + 3Fe3+ +2H2O

Fe2+ + NO = [Fe(NO)2+] brown solution

More precise:
Fe(H₂O)₆²⁺ + NO = [Fe(H₂O)₅(NO)²⁺ ] + H₂O

[Mitch]

Awk is the smartest person on the planet.

[kkjc2]

Thank you AWK.

So NO replaces one of the water ligands in Fe2+ (aq) and form a brown complex ion.
Am I correct to think that it only occur at the conc sulfuric acid end because the H+ conecntration is high enough to shift the equilibrium to favour the formation of NO?

and does the concentrated acid stablise the brown complex ion? it seemed that it is destoryed when diluted with water or is it just diluted out?

[AWK]

Concentrated sulfuric acid decomposes this complex and brown ring disappers in a few tens seconds
when H2SO4 diiffuse to the upper layer.
 And just this disappering indicate nitrate. Some other anions form also brown ring that do not disappear and diffuse to the whole upper layer And just this behaviour indicate nitrate when masking anions are missing

[woelen]

Thank you AWK.

So NO replaces one of the water ligands in Fe2+ (aq) and form a brown complex ion.
Am I correct to think that it only occur at the conc sulfuric acid end because the H+ conecntration is high enough to shift the equilibrium to favour the formation of NO?

and does the concentrated acid stablise the brown complex ion? it seemed that it is destoryed when diluted with water or is it just diluted out?

This answer is close to it, but not precisely so.

NO indeed replaces a water ligand, but by doing so, it is oxidized to a nitrosyl ligand, NO⁺. So, you get the compound [Fe(H2O)₅NO]²⁺, but in this compound there is no NO ligand, but a NO⁺ ligand. The iron has oxidation state +1 in this complex, which is quite remarkable! Normally, iron has oxidation state +2 or +3.

On dilution, the nitrosyl complex is destroyed. It hydrolyses. The NO⁺ entity is not stable in water and is quickly converted to NO₂^- and/or HNO₂. Iron in oxidation state +1 is not stable in water and so, the nitrite is reduced to gaseous NO.

[kkjc2]

Thanks for all the replies, leant a lot of this reaction.

From AWK: conc sulfuric acid decompose the nitrosyl complex and from woelen: dilution hydrolyse the complex ion.

Am I right in thinking that the brown complex occur as a band at a medium concentration acid level (in the boundary of conc acid and dilute acid) where it is at equilibrium of being decomposed when diffused to both the upper layer and the lower layer and being formed at the same time?

Is it correct that the iron in [Fe(H₂O)₅NO]²⁺ is at oxidation state I? It seemed wrong that nitrate to NO⁺ is a reduction : N(V) --> N(III), so how does this couple for another reduction to occur: Fe(II) --> Fe(I) ??

It is more convincing as from AWK that iron is oxidised to Fe(III) rather than reduced to iron (I); but then it must be a NO^- ligand rather than NO⁺ for the ion to have a charge of +2.  

Then, is the oxidation state of N in NO^- +1 ??

[woelen]

Thanks for all the replies, leant a lot of this reaction.

From AWK: conc sulfuric acid decompose the nitrosyl complex and from woelen: dilution hydrolyse the complex ion.

Am I right in thinking that the brown complex occur as a band at a medium concentration acid level (in the boundary of conc acid and dilute acid) where it is at equilibrium of being decomposed when diffused to both the upper layer and the lower layer and being formed at the same time?

I'm not sure about that. I have the impression that the reason for the appearance of a band is that below the band, there is no ferrous ion (it is mainly concentrated H2SO4) and above the band, there is too much water, such that the complex hydrolyses. But in order to test that, you could take some solid KNO3 and some solid FeSO4 and mix this thorougly and add a small amount of this to H2SO4. If the liquid becomes dark brown, then the complex can also exist at high acid concentration.

Is it correct that the iron in [Fe(H₂O)₅NO]²⁺ is at oxidation state I? It seemed wrong that nitrate to NO⁺ is a reduction : N(V) --> N(III), so how does this couple for another reduction to occur: Fe(II) --> Fe(I) ??

A small part of the nitrate is decomposed by the strong acid. Part oxidizes ferrous ions to ferric ions, forming iron (III). What remains is NO, which in turn reacts to form the iron (I) complex:

HNO₃ + 3Fe²⁺ + 3H⁺  -->  NO + 3Fe³⁺ + 2H₂O  (simplified without water ligands)

NO + Fe(H₂O)₆²⁺  --> Fe(NO)(H₂O)₅²⁺  + H₂O   (this causes the brown color)

Look at page 1094 of the book "Chemistry of the elements", written by Greenwood and Earnshaw, which mentions this complex as a complex with iron in oxidation state +1.

It is more convincing as from AWK that iron is oxidised to Fe(III) rather than reduced to iron (I); but then it must be a NO^- ligand rather than NO⁺ for the ion to have a charge of +2.  

Then, is the oxidation state of N in NO^- +1 ??

See explanation above.

[AWK]

My conclusion came from an analogous reaction of nitrite (NO2-). We use diluted H2SO4 in this reaction and the colour is much more intensive in the whole volume of solution.

[woelen]

My conclusion came from an analogous reaction of nitrite (NO2-). We use diluted H2SO4 in this reaction and the colour is much more intensive in the whole volume of solution.

This observation can be perfectly explained.

In the acidic medium, the nitrite forms HNO2, which in turn disproportions to HNO3 and NO (and some side reactions, forming N2O3 and NO2, which are reduced immediately by the ferrous ions). The NO in turn forms the brown complex, but now in the whole liquid. On strong dilution, however, the color fades and the liquid starts bubbling, giving off NO. In fact, adding nitrite to an acidified solution of a ferrous salt is a good way to produce quite pure NO gas. It goes through the nitrosyl complex, which in turn decomposes and gives off NO. Any NO2 and N2O3 from side reactions is effectively neutralized by the ferrous ions.

[dtjohnnyb]

http://pubs.acs.org/cgi-bin/article.cgi/inocaj/2002/41/i01/html/ic010628q.html
Apparently its not actually NO+, its NO- thats coordinated to Fe(III)

[sunshines_rain]

am i right to draw the figure of the brown ring test like this?

thank you.