The Prussian blue color obtained during the test of nitrogen by Lassaigne’s test is due to the formation of:
A) \[{\rm{F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3}\]
B) \[{\rm{Fe}}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]\]
C) \[{\rm{Fe}}{\left( {{\rm{CN}}} \right)_3}\]
D) \[{\rm{N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_5}} \right]{\rm{NOS}}\]
Answer
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Hint: In the correct choice, the coordination number of iron should be six.Lassaigne’s test is used for the detection of nitrogen, sulphur, halogens and phosphorus present in an organic compound.
Complete step by step answer:
The Lassaigne’s test is a qualitative test. It is used for the detection of nitrogen, sulphur, halogens and phosphorus present in an organic compound. In this test, covalent compounds are converted into ionic form. For this, these compounds are fused with sodium metal. Nitrogen and carbon present in the organic compound reacts with sodium metal to form sodium cyanide.
\[{\rm{Na + C + N }} \to {\rm{ NaCN}}\]
Sodium fusion extract is the water extract of the fused mass. Sodium cyanide formed in the above reaction is present in the sodium fusion extract. Sodium fusion extract is boiled with ferrous sulphate solution.
\[\begin{array}{l}
{\rm{FeS}}{{\rm{O}}_4}{\rm{ + 2NaOH }} \to {\rm{ Fe}}{\left( {{\rm{OH}}} \right)_2}{\rm{ + N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}\\
{\rm{6NaCN + Fe}}{\left( {{\rm{OH}}} \right)_2}{\rm{ }} \to {\rm{ N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right] + 2{\rm{NaOH}}
\end{array}\]
The solution is cooled and a little ferric chloride solution is added. Also added is excess of concentrated hydrochloric acid.
\[{\rm{3N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right] + 4{\rm{FeC}}{{\rm{l}}_3} \to {\rm{ F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3} + 12{\rm{NaCl}}\]
A prussian blue colour is obtained due to formation of \[{\rm{F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3}\].
Prussian blue colour confirms the presence of nitrogen.
Hence, the option A) \[{\rm{F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3}\] is the correct answer.
Note:
The complex for the correct answer choice will only contain the elements iron, carbon and nitrogen.
Complete step by step answer:
The Lassaigne’s test is a qualitative test. It is used for the detection of nitrogen, sulphur, halogens and phosphorus present in an organic compound. In this test, covalent compounds are converted into ionic form. For this, these compounds are fused with sodium metal. Nitrogen and carbon present in the organic compound reacts with sodium metal to form sodium cyanide.
\[{\rm{Na + C + N }} \to {\rm{ NaCN}}\]
Sodium fusion extract is the water extract of the fused mass. Sodium cyanide formed in the above reaction is present in the sodium fusion extract. Sodium fusion extract is boiled with ferrous sulphate solution.
\[\begin{array}{l}
{\rm{FeS}}{{\rm{O}}_4}{\rm{ + 2NaOH }} \to {\rm{ Fe}}{\left( {{\rm{OH}}} \right)_2}{\rm{ + N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4}\\
{\rm{6NaCN + Fe}}{\left( {{\rm{OH}}} \right)_2}{\rm{ }} \to {\rm{ N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right] + 2{\rm{NaOH}}
\end{array}\]
The solution is cooled and a little ferric chloride solution is added. Also added is excess of concentrated hydrochloric acid.
\[{\rm{3N}}{{\rm{a}}_4}\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right] + 4{\rm{FeC}}{{\rm{l}}_3} \to {\rm{ F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3} + 12{\rm{NaCl}}\]
A prussian blue colour is obtained due to formation of \[{\rm{F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3}\].
Prussian blue colour confirms the presence of nitrogen.
Hence, the option A) \[{\rm{F}}{{\rm{e}}_4}{\left[ {{\rm{Fe}}{{\left( {{\rm{CN}}} \right)}_6}} \right]_3}\] is the correct answer.
Note:
The complex for the correct answer choice will only contain the elements iron, carbon and nitrogen.
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