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In the standard brown ring test for the nitrate ion, the brown ring is $$\ce{[Fe(H2O)5 (NO)]^{2+}}$$

In this compound, the nitrosyl ligand is positively charged, and iron is in a $+1$ oxidation state.

Now, Iron has stable oxidation states $+2,+3$. Nitrosyl, as a ligand, comes in many flavours, of which a negatively charged nitrosyl is one.

I see no reason why the iron doesn't spontaneously oxidise to +3 and reduce the NO to -1 to gain stability. But I don't know how to analyse this situation anyway. I think that there may be some nifty backbonding increasing the stability, but I'm not sure.

So, why is iron in +1 here when we can have a seemingly stable situation with iron in +3?

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Textbooks say it is FeII and NO . FeII is rather obvious, because it is used in big excess in the reaction. And NO- is unknown to textbooks. –  Georg Sep 17 '12 at 18:50
It's a matter of assignment. I guess it comes from stablility argument, that NO$^+$ is isoelectron to N$_2$, which is a stable triple bond ion. On the other hand, Fe$^{2+}$ or Fe$^{3+}$ will produce spin polarization around NO molecule, which may be detected from EPR spectroscopy(?) –  user26143 Feb 11 '14 at 0:46

8 Answers 8

up vote 3 down vote accepted

Your basic assumption is incorrect: the iron in [Fe(H2O)5(NO)]2+ is Fe(III), and the ligand is NO.

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That is strange, I'm quite sure I've read the contrary. I'll look for the exact book and reread the section :/ –  ManishEarth May 14 '12 at 0:23
The link in the answer contradicts the answer itself. However pubs.acs.org/doi/pdf/10.1021/ic010628q supports the answer. –  DavePhD Jun 11 '14 at 14:42

According Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of Nitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction Revisited

The correct structure is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$

For many years it was thought that iron was reduced to Fe(I) and NO oxidized to NO+, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin Fe(III) (S=5/2) antiferromagnetically couples with NO- (S=1) for an observed spin of S=3/2.

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The other answers are all wrong. Draw Molecular orbital diagram of $\ce{NO}$ and you will see.

That electron in antibonding orbital, it is so easy to lose, hence it easily goes to $\ce{NO^+}$ and $\ce{Fe^{(II)}}$ to $\ce{Fe^{(I)}}$.

Remember if you read something extraordinary which claims something out of daily routine (especially in chem) then it must be true.

Link : http://en.wikipedia.org/wiki/Nitrate_test

Wikipedia also says Iron is in $+1$

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"The overall reaction is the reduction of the nitrate ion by iron(II) which is reduced to iron(I) and formation of a nitrosonium complex where nitric oxide is oxidised to NO1+." - http://en.wikipedia.org/wiki/Nitrate_test#Brown_ring_test

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The $\ce{Fe(NO)2}$ fragment has strongly delocalized electrons, it is considered as covalently bound. The semiempirical calculations show that the electron density is polarized in a manner that allows to ascribe the delta+ charge to $\ce{NO}$ groups. Therefore formally Fe is considered +1. Remember this formal assignment is somewhat simplified... If interested, read e.g. http://pubs.rsc.org/en/content/articlehtml/2011/dt/c0dt01244k and references.

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Welcome to Chemistry.SE! Could you summarize the contents of that article you linked. It is behind a paywall for some folks. –  Ben Norris Feb 10 '14 at 16:31

Oxidation State of "Fe" in brown ring complex depends on the binding mode of NO ligand to Iron (whether is it "bent" or "linear"). Unless we specify the IR frequencies of NO ligand for different modes, we can not say the oxidation state of "Fe". Depending on the IR stretching frequencies of NO ligand which is bound to the Iron, it can be NO+, NO- or simply NO.

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Here iron becomes +1 due to presence of the ligand $\ce{NO}$ +1. The compound $\ce{[Fe(H2O)6]^{2+}}$ is formed when $\ce{FeSO4}$ dissolves in water during the brown ring test.

Now, when we add an aqueous solution of nitrate ions in $\ce{FeSO4}$ solution $\ce{Fe^{(II)}}$ are converted to $\ce{Fe^{(III)}}$ and $\ce{NO3-}$ are converted to $\ce{NO}$.
These $\ce{NO}$ being an odd electron compound can use three of its electrons among which it donates two electrons to $\ce{Fe^{(II)}}$ while one it shares with incompletely filled $\ce{d}$ orbital and displace one water ligand bonded with secondary valency to $\ce{Fe}$.

This way $\ce{Fe^{(II)}}$ is converted to $\ce{Fe^{(I)}}$ and not $\ce{Fe^{(III)}}$ In these regard we get the colour brown instead of yellow for $\ce{Fe^{(III)}}$ or green for $\ce{Fe^{(II)}}$

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Are you sure? Wikipedia says it's Fe+3, see above answer. Also, I find it rather hard to follow your arguments, please be a little clearer :) –  ManishEarth Sep 16 '12 at 4:56

In the brown ring complex, $\ce{[Fe(H2O)5(NO)]^{2+}}$ five water molecules are present, they are harder to stabilise the $+3$ oxidation state of $\ce{Fe}$ and hence here $\ce{NO}$ present as $-1$.
If we consider that $\ce{NO}$ is $+1$, then $\ce{Fe}$ is $+1$, which is very unstable in the co-ligand region like $\ce{H2O}$.
i.e, in brown ring complex oxidation state of Iron is $+3$.

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Welcome to Chemistry.SE! For formatting help visit the help center. Your answer could likely benefit from a bit of cleanup to clarify what you're trying to say. –  Geoff Hutchison Oct 21 '14 at 15:48
Also, since another answer also gives the oxidation states, can you be specific about what you want to add to the discussion? Thanks! –  Geoff Hutchison Oct 21 '14 at 15:50

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