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?