Oxidation state of Fe(Iron) in complex formed in Brown ring test.

The chemical test called a brown ring test is conducted to find out the presence of Nitrate ions in any solution, in which the solutions form a brown ring in the test tube.

The above test is also called the brown ring test, in which the brown ring complex compound is formulated as$\left[\mathrm{Fe}\right({\mathrm{H}}_2\mathrm{O}{)}_5\mathrm{NO}]{\mathrm{SO}}_4.$ 

Here the oxidation state of Iron in the complex, can be found as:

We can assume that the oxidation state of Iron metal is x.

Here the oxidation state of the Sulphate ion is$-2$, the oxidation state of water is zero and the oxidation state of Nitroso $\mathrm{NO}$ is $+1$.Hence, the oxidation state of Iron can be calculated as:
$$\begin{gathered} \Rightarrow \mathrm{x}+(+1)+5\left(0\right)=+2 \\ \Rightarrow \mathrm{x}=+1 \end{gathered}$$