# Why is iron in the brown ring compound in a +1 oxidation state?

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 at 0:46

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
@F'x Where exactly did you get that info from. Some citations might be helpful. –  Bolt64 Aug 19 '13 at 13:57
This answer is wrong! @ManishEarth what you read was right. –  Ayush Pateria Jan 7 at 20:06
Wrong........... –  Awesome Mar 10 at 13:40
$Your$ basic assumption is incorrect. –  Awesome Mar 10 at 15:55

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

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

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

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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

"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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ALL ARE WRONG... You read it right.

Draw Molecular orbital diagram of NO... You will see...

See? That electron in antibonding orbital... So easy to lose it... Hence it easily goes to $NO^+$ and $Fe(II)$ to $Fe(I)$.

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

Wikipedia also says Iron is in $+1$

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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.