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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 the formation of \( \mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3} \).

Step by step solution

01

Understand the Scenario

Sodium fusion extract is obtained from aniline. This extract is then treated with iron (II) sulphate (FeSO鈧) and sulfuric acid (H鈧係O鈧) in the presence of air, leading to the formation of a Prussian blue precipitate.
02

Identify the Iron and Cyanide Reaction

In this reaction, cyanide ions (CN鈦) formed in the fusion extract react with ferrous ions (Fe虏鈦 from FeSO鈧) and participate in a redox reaction in the presence of air (oxygen). This process forms hexacyanoferrate (II) ions ([Fe(CN)鈧哴鈦粹伝).
03

Formation of Prussian Blue

The hexacyanoferrate (II) ions ([Fe(CN)鈧哴鈦粹伝) react with ferric ions (Fe鲁鈦, formed by the oxidation of Fe虏鈦 in air), leading to the formation of a precipitate. The stoichiometry of this reaction results in a complex known as Prussian blue.
04

Determine the Correct Formula

The correct formula for Prussian blue, the characteristic blue precipitate, is known to be \( ext{Fe}_4[ ext{Fe(CN)}_6]_3 \). This corresponds to option (a). It arises from the combination of [Fe(CN)鈧哴鲁鈦 with Fe鲁鈦 ions, resulting in the solid complex.

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Key Concepts

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

Sodium Fusion Extract
Sodium fusion extract is an essential technique used in organic chemistry to detect certain elements within an organic compound. This method is particularly useful for identifying elements like nitrogen, sulfur, and halogens in unknown substances. In this procedure, an organic compound is fused with sodium metal, creating sodium compounds that are water-soluble. These water-soluble compounds form after the organic compound is broken down in the presence of molten sodium, resulting in simpler, easily detectable fragments. Once this extract is obtained, it can be tested using various reagents to reveal the presence or absence of specific elements, leading to important insights into the composition of the original compound.
Iron and Cyanide Reaction
When cyanide ions (-\[ \text{CN}^- -\]) from the sodium fusion extract come into contact with iron ions in the solution, an intriguing redox reaction takes place. Initially, the cyanide ions react with ferrous ions (-\[ \text{Fe}^{2+} -\]) provided by iron (II) sulphate (FeSO鈧). This reaction occurs in the presence of oxygen from the air, enabling a redox process that results in the formation of hexacyanoferrate (II) ions (-\[ [\text{Fe(CN)}_6]^{4-} -\]). The involvement of air is crucial here because oxygen acts as an oxidizing agent, converting the ferrous ions (-\[ \text{Fe}^{2+} -\]) into ferric ions (-\[ \text{Fe}^{3+} -\]). This transformation is necessary for the subsequent formation of the Prussian blue complex.
Hexacyanoferrate Formation
Hexacyanoferrate formation is a fascinating process that occurs during the reaction sequence for Prussian blue synthesis. In the presence of ferric ions (-\[ \text{Fe}^{3+} -\]), the hexacyanoferrate (II) ions (-\[ [\text{Fe(CN)}_6]^{4-} -\]) interact to generate the characteristic Prussian blue precipitate. The ferric ions, having been oxidized from ferrous ions by the action of atmospheric oxygen, combine with the hexacyanoferrate ions to form a solid complex. This complex has the formula -\[ \text{Fe}_4[\text{Fe(CN)}_6]_3 -\]. The intense blue color of this compound makes it easily recognizable and serves as an indicator of the reaction's successful completion. Such complexes are not just chemically interesting but also historically significant as pigments and indicators in analytical chemistry.

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Most popular questions from this chapter

How many moles of electrons, are transferred in the following reduction- oxidation reaction? \(2 \mathrm{MnO}_{4}^{-}(\mathrm{aq})+16 \mathrm{H}^{+}(\mathrm{aq})+10 \mathrm{Cl}^{-}(\mathrm{aq}) \rightarrow\) $$ 2 \mathrm{Mn}^{2+}(\mathrm{aq})+5 \mathrm{Cl}_{2}(\mathrm{~g})+8 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ? $$ a. 2 b. 5 c. 10 d. 12

\(\mathrm{aK}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+\mathrm{bKCl}+\mathrm{cH}_{2} \mathrm{SO}_{4} \rightarrow\) \(\mathrm{xCrO}_{2} \mathrm{Cl}_{2}+\mathrm{yKHSO}_{4}+\mathrm{zH}_{2} \mathrm{O}\) The above equation balances when a. \(a=1, b=6, c=4\) and \(x=6, y=2\), \(\mathrm{z}=3\) b. \(a=2, b=4, c=6\) and \(x=2, y=6\), \(z=3\) c. \(a=1, b=4, c=6\) and \(x=2, y=6, z=3\) d. \(a=4, b=2, c=6\) and \(x=6, y=2, z=3\) e. \(a=6, b=4, c=2\) and \(x=6, y=3, z=2\)

The emf of the cell \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.01 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.001 \mathrm{M})\right| \mathrm{Fe}\) at 298 \(\mathrm{K}\) is \(0.2905\) volt. Then the value of equilibrium constant for the cell reaction is a. \(\mathrm{e}^{0.32 / 0.0295}\) b. \(10^{0.32 / 00295}\) c. \(10^{0.26 / 0.0295}\) d. \(10^{0.32 / 00591}\)

The reduction potential of a half cell consisting of a Pt electrode immersed in \(1.2 \mathrm{M} \mathrm{Fe}^{2+}\) and \(0.012 \mathrm{M}\) \(\mathrm{Fe}^{3+}\) solution at \(25^{\circ} \mathrm{C}\) is \(\left(\mathrm{E}^{\circ} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.770\right)\) a. \(-0.326 \mathrm{~V}\) b. \(0.652 \mathrm{~V}\) c. \(0.326 \mathrm{~V}\) d. \(-0.652 \mathrm{~V}\)

Which of the following are disproportionation reactions? (1) \(\mathrm{UO}^{2+}+\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{H}^{+} \rightarrow \mathrm{UO}_{2}^{2+}+\mathrm{Cr}^{3+}+\mathrm{H}_{2} \mathrm{O}\) (2) \(\mathrm{X}_{2}+\mathrm{OH}^{-} \rightarrow \mathrm{XO}^{-}+\mathrm{X}^{-}+\mathrm{H}_{2} \mathrm{O}\) (3) \(\mathrm{HO}_{2}^{-}+\mathrm{Br}^{-}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{OH}^{-}+\mathrm{Br}_{2}\) (4) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}+\mathrm{OH}^{-} \rightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}+\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{OH}\) a. 1 and 2 b. 2 and 3 c. 3 and 4 d. 2 and 4

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