JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Redox Reactions and Electrochemistry

The brown ring formed in the test for $\mathrm{N}{\mathrm{O}}_3^{-}$ is the complex $[\mathrm{Fe}({\mathrm{H}}_2\mathrm{O}{)}_5(\mathrm{NO}){]}^{2+}$, formed when freshly generated $\mathrm{F}{\mathrm{e}}^{2+}$ reduces $\mathrm{N}{\mathrm{O}}_3^{-}$ to NO, which then coordinates to iron.

In this complex the nitrosyl ligand is bonded as $\mathrm{N}{\mathrm{O}}^{+}$ (nitrosonium), and the five water ligands are neutral. Balancing the overall +2 charge on the complex:


The oxidation number of iron in the brown ring complex is +1.