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Chapter 17: Problem 53

Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in presence of air gives a Prussian blue precipitate. The blue colour is due to the formation of (a) \(\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}\) (b) \(\mathrm{Fe}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{2}\) (c) \(\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{2}\) (d) \(\mathrm{Fe}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}\)

Short Answer

Expert verified
The blue color is due to \(\text{Fe}_4\left[\text{Fe(CN)}_6\right]_3\) (option a).

Step by step solution

01

Understanding the Reaction

Aniline is treated with sodium to create sodium cyanide in the sodium fusion test. The product is a cyanide compound.
02

Reaction with Iron (II) Sulphate

When the sodium cyanide is treated with iron (II) sulfate (\(\text{FeSO}_4\)), the cyanide ions (\(\text{CN}^-\)) from the sodium fusion extract form a complex with iron ions in the presence of air, creating iron cyanide complexes.
03

Prussian Blue Formation

In the presence of air, which oxidizes \(\text{Fe}^{2+}\) to \(\text{Fe}^{3+}\), an insoluble Prussian blue precipitate forms. This complex is known chemically as \(\text{Fe}_4\left[\text{Fe(CN)}_6\right]_3\).
04

Identify the Correct Option

Compare the Prussian blue chemical structure \(\text{Fe}_4\left[\text{Fe(CN)}_6\right]_3\) with the options given: Option (a) matches the composition of Prussian blue.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sodium Fusion Test
The Sodium Fusion Test, also known as Lassaigne's test, is a fundamental procedure in chemistry used to detect elements such as nitrogen, sulfur, and halogens within an organic compound. When performing this test on aniline, a primary aromatic amine, it is heated with metallic sodium. This reaction can decompose the organic compounds and convert them into more easily detectable inorganic salts. Specifically for nitrogen detection, it converts nitrogen to sodium cyanide.Here's what essentially happens:
  • The sodium reacts with the organic matrix, breaking the bonds and releasing atoms of nitrogen.
  • These nitrogen atoms combine with sodium to form sodium cyanide ( \( ext{NaCN} \)).
This test plays a crucial role in identifying nitrogen-containing compounds due to its effectiveness in forming distinct and identifiable compounds such as sodium cyanide.
Aniline Reaction
Aniline, an aromatic amine characterized by its amino group attached to a phenyl group, undergoes interesting reactions during the sodium fusion test. When fused with sodium, aniline's nitrogen is converted to cyanide. This conversion is essential in further reactions where it serves as a precursor to complex formations.The key steps during this reaction involve:
  • The breakdown of aniline in the sodium fusion test, releasing cyanide ions.
  • The cyanide ions ( \( ext{CN}^- \)) become pivotal in subsequent reaction steps with other compounds.
Understanding this fundamental reaction helps in grasping how complex cyanide structures are later formed, as they play a significant role in producing Prussian Blue.
Iron (II) Sulphate Reaction
The next phase in this series of reactions involves the interaction between sodium cyanide and iron (II) sulfate, \( ext{FeSO}_4 \). This step is crucial for the development of complex cyanide networks. When these cyanide ions meet iron (II) ions, they start forming a series of iron cyanide complexes, which are pivotal in producing the famed Prussian Blue.Some key details of this reaction include:
  • The initial bond between the cyanide ions and iron (II) ions forms weak iron-cyanide complexes.
  • The presence of air oxidizes iron (II) ions (\( ext{Fe}^{2+} \)) to iron (III) ions (\( ext{Fe}^{3+} \)).
This oxidation step is vital as it sets the stage for the dense blue precipitate, indicating the conversion to Prussian Blue.
Cyanide Complex Formation
The formation of a cyanide complex is an intricate process essential in the production of Prussian Blue. Initially, cyanide ions produced from the sodium fusion extract react with iron ions in the presence of \( ext{H}_2 ext{SO}_4 \) and air. This reaction sees the delicate interplay between cyanide ions and multiple iron ions, particularly once oxidized.The process can be detailed by:
  • The generation of intermediate iron-cyanide complexes as \( ext{Fe(CN)}_6^{4-} \).
  • A subsequent rearrangement where oxidized iron establishes strong, stable complexes with cyanide ions.
  • The resulting complex, \( ext{Fe}_4 ext{[Fe(CN)]}_3 \), forms a dense blue precipitate, known as Prussian Blue.
This final complexation provides the identifying deep blue color, facilitating identification within the reaction scheme.

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Most popular questions from this chapter

A current of \(1.70 \mathrm{~A}\) is passed through \(300.0 \mathrm{~mL}\) of \(0.160 \mathrm{M}\) solution of a \(\mathrm{ZnSO}_{4}\) for \(230 \mathrm{sec}\). with a current efficiency of \(90 \%\). Find out the molarity of \(\mathrm{Zn}^{2+}\) after the deposition of \(\mathrm{Zn}\). Assume the volume of the solution to remain constant during the electrolysis.

The standard reduction potential of the \(\mathrm{Ag}^{+} / \mathrm{Ag}\) electrode at \(298 \mathrm{~K}\) is \(0.799 \mathrm{~V}\). Given that for AgI, \(K_{s p}=8.7 \times 10^{-17}\), evaluate the potential of the \(\mathrm{Ag}^{+} / \mathrm{Ag}\) electrode in a saturated solution of AgI. Also calculate the standard reduction potential of the \(\mathrm{I}^{-} / \mathrm{AgI} / \mathrm{Ag}\) electrode.

Electrolysis of dilute aqueous \(\mathrm{NaCl}\) solution was carried out by passing 10 milli ampere current. The time required to liberate \(0.01\) mol of \(\mathrm{H}_{2}\) gas at the cathode is \(\left(1\right.\) Faraday \(\left.=96500 \mathrm{C} \mathrm{mol}^{-1}\right)\) [2008S] (a) \(9.65 \times 10^{4} \mathrm{sec}\) (b) \(19.3 \times 10^{4} \mathrm{sec}\) (c) \(28.95 \times 10^{4} \mathrm{sec}\) (d) \(38.6 \times 10^{4} \mathrm{sec}\)

For the electrochemical cell, [Adv. 2018] \(\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\) the standard emf of the cell is \(2.70 \mathrm{~V}\) at \(300 \mathrm{~K}\). When the concentration of \(\mathrm{Mg}^{2+}\) is changed to \(x \mathrm{M}\), the cell potential changes to \(2.67 \mathrm{~V}\) at \(300 \mathrm{~K}\). The value of \(x\) is (given, \(\frac{F}{R}=11500 \mathrm{~K} \mathrm{~V}^{-1}\), where \(F\) is the Faraday constant and \(R\) is the gas constant, \(\ln 10=2.30\) )

The standard electrode potentials \(\left(\mathrm{E}^{\circ} \mathrm{M}^{+} / \mathrm{M}\right)\) of four metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) are \(-1.2 \mathrm{~V}, 0.6 \mathrm{~V}, 0.85 \mathrm{~V}\) and \(-0.76 \mathrm{~V}\), respectively. The sequence of deposition of metals on applying potential is: [Main Online April 9, 2014] (a) \(A, C, B, D\) (b) \(\mathrm{B}, \mathrm{D}, \mathrm{C}, \mathrm{A}\) (c) \(\mathrm{C}, \mathrm{B}, \mathrm{D}, \mathrm{A}\) (d) \(\mathrm{D}, \mathrm{A}, \mathrm{B}, \mathrm{C}\)
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