Are these reaction equations for the formation of the brown ring complex correct?

Question

\begin{align} \ce{NO3- + 3Fe^2+ + 4H+ &-> NO + 3Fe^3+ + 2H2O}\tag{1}\\ \ce{[Fe(H2O)6]^2+ + NO &-> \underset{\text{(brown)}}{[Fe(H2O)5(NO)]^2+} + H2O}\tag{2} \end{align}

The above are the reactions given in my book [1] for the qualitative analysis of nitrate ion (formation of the brown ring complex). I think that the reactions above are incorrect. The first one shows formation of $\ce{Fe^3+}$, which should be attacked by excess of $\ce{H2O}$ to give ferrum(III) hexaaquasulphate, not ferrum(II) hexaaquasulphate. Now that should be attacked by $\ce{NO}$ as a ligand and give $\ce{[Fe(H2O)6NO]SO4}$, where the oxidation number of $\ce{Fe}$ is being $3+$.

Here the reactions show that the $\ce{Fe}$ already present gives $\ce{[Fe(H2O)6]^2+}$ and then $\ce{NO}$ acts as a neutral ligand to form the complex. But by the reactions shown in the book, even after formation of the brown ring complex $\ce{Fe^3+}$ is still present in the solution, which according to me shouldn't be because if $\ce{Fe^3+}$ still remains in the solution then water should also have attacked it as a ligand to give $\ce{[Fe(H2O)6NO]SO4}$.

I need some clarity on this. Are the reactions given in the book correct?

Edit: After reading the answers, I do get that the hexaaqua iron(II) will be formed and the $\ce{Fe^3+}$ will be attacked by water to give $\ce{[Fe(H_2O)_6]^3+}$ but why won't $\ce{NO}$ take electrons from the oxidation of iron ($\ce{Fe^2+ -> Fe^3+ + e- }$), make $\ce{NO-}$ and attack the above and substitute one water molecule and give more of the brown ring complex?

1. NCERT. Chemistry: Textbook for class XII; National Council of Educational Research and Training: New Delhi, 2007. ISBN 978-81-7450-648-1.

Answer 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 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.

Other than the above, I agree with Dhanajay Gupta that the equations in the book are correct, that some of the Fe2+ is oxidized to Fe3+ in the first equation forming NO, and NO coordinates to remaining Fe2+ according to the second equation, but the second equation doesn't specify the charge on Fe or NO, just the overall charge of the complex.

In the first equation in the book, it should be understood that the ions are aqueous. For example H+ doesn't mean there are bare protons, it is understood to mean $\ce{H_3O+}$ and Fe2+ is $\ce{[Fe(H2O)6]^{2+}}$ and Fe3+ is $\ce{[Fe(H2O)6]^{3+}}$.

In the second equation in the book, the inner coordination sphere is explicitly shown.

In the article cited above the mechanism of the second equation is explained. Neutral NO attacks hexaaquairon(II), forming a 7-coordinate transition state, a water leaves, and "This rate-determining displacement of coordinated water is followed by a rapid intramolecular charge-redistribution process to lead to the final $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+}}$ product".

but why won't NO take electrons from the oxidation of Iron ($Fe^{2+} \rightarrow Fe^{3+} + e^-$), make $NO^-$ and attack the above and substitute one water molecule and give more of the brown ring complex?

To get the electron from Fe2+, NO would enter the inner coordination sphere, and when the electron transfer occurs, the brown ring complex is formed at that Fe. If the NO- then left that Fe, it would destroy one instance of the brown ring complex. Hypothetically, the Fe2+ could reduce NO via an outer sphere electron transfer, and then NO- coordinate Fe3+, but this is not what the article reports occurring.

Answer 2

Since this reaction takes place in aqueous solution, is is implied, that the reaction equations are involving solvated ions. In general both of the given equations are correct. However, I would prefer the first equation to include the highly charged metal ions as aqua complexes: $$\ce{NO3- + 3[Fe(H2O)_6]^{2+} + 4H+ -> NO + 3[Fe(H2O)_6]^{3+} + 2H2O}$$

Also note that this is a test for nitrate, and nitrite ions will interfere accordingly. $$\ce{NO2- + [Fe(H2O)_6]^{2+} + 2H+ -> NO + [Fe(H2O)_6]^{3+} + H2O}$$

It is important to realise, that the Iron (II) solution is given in massive excess for this test to work. Hence the newly reduced nitrogen monoxide will react with the yet unreacted excess of $\ce{Fe^{2+}}$: $$\ce{NO + [Fe(H2O)_6]^{2+} -> [Fe(NO)(H2O)_5]^{2+} + H2O}$$

Since the iron hexaaqua complexes are nearly colourless, they do not interfere with this visibility of this reaction.

Answer 3

The reactions given are correct. The nitric acid oxidizes the ferrous ions into ferric ions, and itself get's reduced to NO. But this reaction doesn't happen all at once. It's not as if all the ferrous ions are oxidized to ferric in one go. The remaining ferrous ions then form the brown ring complex with NO. Furthermore, the formation of the brown ring complex itself is a redox reaction. Iron is in the rare +1 oxidation state in the complex, whereas NO is actually NO+.

So yes, ferric ions are present in the solution. They were produced as a result of the first reaction, but are not required in the second reaction.