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Chemical Complex Responsible for the Brown Ring Test

Among the following options, the formation of the brown ring is due to which coordination species?

Options

A

[Fe(NO)2(SO4)2]2−[\text{Fe}(\text{NO})_2(\text{SO}_4)_2]^{2-}

B

[Fe(NO)2(H2O)4]3+[\text{Fe}(\text{NO})_2(\text{H}_2\text{O})_4]^{3+}

C

[Fe(NO)4(SO4)2][\text{Fe}(\text{NO})_4(\text{SO}_4)_2]

D

[Fe(NO)(H2O)5]2+[\text{Fe}(\text{NO})(\text{H}_2\text{O})_5]^{2+}

Correct

Step-by-Step Solution

The brown ring test is a classic qualitative inorganic analysis test used to detect the presence of nitrate (NO3−\text{NO}_3^-) ions in a solution.

  1. Chemical Reactions Involved: When freshly prepared ferrous sulfate (FeSO4\text{FeSO}_4) solution is added to an aqueous solution containing nitrate ions in the presence of concentrated sulfuric acid (H2SO4\text{H}_2\text{SO}_4), nitrate ions are reduced to nitric oxide (NO\text{NO}) by Fe2+\text{Fe}^{2+} ions: NO3−+3Fe2++4H+⟶NO+3Fe3++2H2O\text{NO}_3^- + 3\text{Fe}^{2+} + 4\text{H}^+ \longrightarrow \text{NO} + 3\text{Fe}^{3+} + 2\text{H}_2\text{O}

  2. Formation of the Brown Ring Complex: The generated nitric oxide (NO\text{NO}) ligand coordinates with the remaining unoxidized aqueous iron(II) complex, [Fe(H2O)6]2+[\text{Fe}(\text{H}_2\text{O})_6]^{2+}, displacing one water molecule to form a brown-colored coordination complex at the junction of the two liquid layers: [Fe(H2O)6]2++NO⟶[Fe(H2O)5(NO)]2++H2O[\text{Fe}(\text{H}_2\text{O})_6]^{2+} + \text{NO} \longrightarrow [\text{Fe}(\text{H}_2\text{O})_5(\text{NO})]^{2+} + \text{H}_2\text{O}

  3. Oxidation State and Bonding: In the complex ion [Fe(H2O)5(NO)]2+[\text{Fe}(\text{H}_2\text{O})_5(\text{NO})]^{2+}, an electron transfers from the NO\text{NO} molecule to Fe2+\text{Fe}^{2+}, resulting in iron in the +1+1 oxidation state (Fe+\text{Fe}^+) and the nitric oxide ligand present as a nitrosonium ion (NO+\text{NO}^+).

Therefore, the brown ring is due to the formation of the complex ion [Fe(NO)(H2O)5]2+[\text{Fe}(\text{NO})(\text{H}_2\text{O})_5]^{2+}.

Correct Option: (D)