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Hint: Lassaigne’s test is carried out for the detection of sulphur, nitrogen and halogens in an organic compound. The blue colour is due to the formation of a compound named Prussian blue.
Complete answer:
We use Lassaigne’s test or the sodium fusion test for the detection of presence of nitrogen, sulphur and other halogens in an organic compound.
We know that sulphur, nitrogen and the halogens are linked to the organic compound through a covalent bond. To detect their presence, we need to convert the compounds in their ionic forms.
To convert these organic compounds to their ionic forms, we fuse them with sodium metal. There are different methods through which we can do the fusing. But generally, we heat the organic compound strongly with sodium metal and then we mix the heated mixture in water.
In the aqueous solution, the ionic compounds formed during the fusion are present and we can identify them through various chemical tests.
The nitrogen, sulphur and the halogens are converted into inorganic sodium salts like sodium nitride, sodium sulphide and sodium halide respectively.
However, in Lassaigne’s test for detection of nitrogen, we boil the fusion mixture with ferrous sulphate and then concentrated sulphuric acid is added to it. We can write the reaction of the formation of blue complex as-
\[\begin{align}
& 2NaCN+Fe{{(OH)}_{2}}\to N{{a}_{4}}Fe{{(CN)}_{6}}+NaOH \\
& 3N{{a}_{4}}Fe{{(CN)}_{6}}+4FeC{{l}_{3}}\to F{{e}_{4}}{{\left[ Fe{{(CN)}_{6}} \right]}_{3}}+NaCl \\
\end{align}\]
The appearance of blue colour is due to \[F{{e}_{4}}{{\left[ Fe{{(CN)}_{6}} \right]}_{3}}\] which is ferric ferrocyanide.
So, the correct answer is “Option C”.
Note: The blue coloured compound is commonly known as Prussian blue. It is a polymeric compound and it is formed by the oxidation of ferrous ferricyanide salts. It was the first modern synthetic pigment. The blue colour of Prussian blue is observed due to intervalence charge transfer which is an electron transfer between two metal sites differing only in their oxidation states. Here, the transfer of electron from $F{{e}^{2+}}\text{ to F}{{\text{e}}^{3+}}$ is mediated through the $\pi -orbital$ of the bridging ligand, CN which produces a colour transfer band in the visible region spectra thus, the precipitate appears coloured.
Complete answer:
We use Lassaigne’s test or the sodium fusion test for the detection of presence of nitrogen, sulphur and other halogens in an organic compound.
We know that sulphur, nitrogen and the halogens are linked to the organic compound through a covalent bond. To detect their presence, we need to convert the compounds in their ionic forms.
To convert these organic compounds to their ionic forms, we fuse them with sodium metal. There are different methods through which we can do the fusing. But generally, we heat the organic compound strongly with sodium metal and then we mix the heated mixture in water.
In the aqueous solution, the ionic compounds formed during the fusion are present and we can identify them through various chemical tests.
The nitrogen, sulphur and the halogens are converted into inorganic sodium salts like sodium nitride, sodium sulphide and sodium halide respectively.
However, in Lassaigne’s test for detection of nitrogen, we boil the fusion mixture with ferrous sulphate and then concentrated sulphuric acid is added to it. We can write the reaction of the formation of blue complex as-
\[\begin{align}
& 2NaCN+Fe{{(OH)}_{2}}\to N{{a}_{4}}Fe{{(CN)}_{6}}+NaOH \\
& 3N{{a}_{4}}Fe{{(CN)}_{6}}+4FeC{{l}_{3}}\to F{{e}_{4}}{{\left[ Fe{{(CN)}_{6}} \right]}_{3}}+NaCl \\
\end{align}\]
The appearance of blue colour is due to \[F{{e}_{4}}{{\left[ Fe{{(CN)}_{6}} \right]}_{3}}\] which is ferric ferrocyanide.
So, the correct answer is “Option C”.
Note: The blue coloured compound is commonly known as Prussian blue. It is a polymeric compound and it is formed by the oxidation of ferrous ferricyanide salts. It was the first modern synthetic pigment. The blue colour of Prussian blue is observed due to intervalence charge transfer which is an electron transfer between two metal sites differing only in their oxidation states. Here, the transfer of electron from $F{{e}^{2+}}\text{ to F}{{\text{e}}^{3+}}$ is mediated through the $\pi -orbital$ of the bridging ligand, CN which produces a colour transfer band in the visible region spectra thus, the precipitate appears coloured.
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