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$[ Fe ( {H}_{2}O{)}_{5}NO ] {SO}_{4} .$ 

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 $NO$ is $+ 1$.Hence, the oxidation state of Iron can be calculated as:
$⇒ x + ( + 1 ) + 5 ( 0 ) = + 2 ⇒ x = + 1$