Chemistry - Is iron in the brown ring compound in a +1 oxidation state?

Solution 1:

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 $\ce{Fe^{I}}$ and $\ce{NO}$ oxidized to $\ce{NO+}$, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin $\ce{Fe^{III}}$ ($S=5/2$) antiferromagnetically couples with $\ce{NO-}$ ($S=1$) for an observed spin of $S=3/2$.

Solution 2:

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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Solution 3:

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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Solution 4:

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 $+3$.