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

The Brown Ring Test is used to detect nitrate ions ($$NO_3^-$$). In this test, ferrous sulfate ($$FeSO_4$$) solution is added to the sample, followed by carefully layering concentrated sulfuric acid, resulting in the formation of the brown ring complex $$[Fe(H_2O)_5(NO)]^{2+}$$.

This nitrosyl complex of iron involves NO acting as a neutral ligand: $$NO^+$$ donates to the metal and becomes $$NO$$ in the coordination sphere.

To determine the oxidation state of Fe, we use charge balance. The overall charge on the complex is +2, each water ligand is neutral (charge = 0), and NO is treated as nitrosonium ($$+1$$). Writing the charge equation gives $$\text{Fe} + 5(0) + 1 = +2$$, so $$\text{Fe} = +1$$.

Therefore, the oxidation state of Fe in the brown ring complex is +1. The iron starts as $$Fe^{2+}$$ and is reduced to $$Fe^{+1}$$ upon coordination with NO, an unusual oxidation state stabilized by strong back-bonding with the NO ligand.

The final answer is 1.