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

Question

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?

Answer 1

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

Answer 2

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 $\small Fe(II)$