Question

The reaction of K₃[Fe(CN)₆] with freshly prepared FeSO₄ solution produces a dark blue precipitate called Turnbull’s blue. Reaction of K₄[Fe(CN)₆] with the FeSO₄. Solution in complete absence of air produces a white precipitate X, which turns blue in air. Mixing the FeSO₄ solution with NaNO₃, followed by a slow addition of concentrated H₂SO₄ through the sides of the test tube produces a brown ring.

Among the following the brown ring is due to the formation of:

पर्याय

• [Fe(NO)₂(SO₄)₂]²⁻

• [Fe(NO)₂(H₂O)₄]³⁺

• [Fe(NO)₄(SO₄)₂]

• [Fe(NO)(H₂O)₅]²⁺

Answer 1

[Fe(NO)(H₂O)₅]²⁺

Explanation:

In the brown ring test, FeSO₄ is mixed with NaNO₃, and concentrated H₂SO₄ is added slowly through the sides of the test tube. This test is used to detect Fe²⁺ ions, and it forms a brown ring at the interface between the two solutions.

The brown ring forms due to the reaction between Fe²⁺ and \[\ce{NO^{-}_3}\] (from NaNO₃) in the presence of concentrated sulfuric acid. The reaction leads to the formation of a nitrosyl complex.

• Fe²⁺ reacts with \[\ce{NO^{-}_3}\] (nitrate ions) to form the nitrosyl ion (NO):
  \[\ce{Fe^{2+} + NO^-_3 -> [Fe(NO)]^{2+}}\]
• This [Fe(NO)]²⁺ ion then complexes with water molecules from the H₂SO₄ (sulfuric acid solution), leading to the formation of the brown ring at the interface:
  \[\ce{[Fe(NO)]^{2+} + 5H2O -> [Fe(NO)(H2O)5]^{2+}}\]

The [Fe(NO)(H₂O)₅]²⁺ complex is responsible for the characteristic brown ring observed in the test.