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The above are the reactions given in my book 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 hexaaqua (III) sulphate, not ferrum hexaaqua (II) sulphate. 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?

Source: "Chemistry Part I Textbook for Class XII" published by National Council Of Education Research And Training (India). http://www.flipkart.com/chemistry-class-xii-part-i-english-01/p/itmdyu9bmzghhwn5

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 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?

\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.

The above are the reactions given in my book 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 hexaaqua (III) sulphate, not ferrum hexaaqua (II) sulphate. 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 $Fe^{3+}$ 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?

Source: "Chemistry Part I Textbook for Class XII" published by National Council Of Education Research And Training (India). http://www.flipkart.com/chemistry-class-xii-part-i-english-01/p/itmdyu9bmzghhwn5

EDIT:

After reading the answers, I do get that the hexaaqua iron(II) will be formed and the $Fe^{3+}$ will be attacked by water to give $[Fe(H_2O)_6]^{3+}$ 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?

The above are the reactions given in my book 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 hexaaqua (III) sulphate, not ferrum hexaaqua (II) sulphate. 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?

Source: "Chemistry Part I Textbook for Class XII" published by National Council Of Education Research And Training (India). http://www.flipkart.com/chemistry-class-xii-part-i-english-01/p/itmdyu9bmzghhwn5

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 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?