Answer
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Hint: The brown ring test is a test for detection for nitrate ions in a solution. The nitrate ions form a complex with iron (II) sulphate in presence of sulphuric acid, which is the brown ring complex. Iron (II) sulphate is a reducing agent, it reduces the nitrate ions and oxidises itself which forms the brown ring.
Complete step by step answer:
We carry out the brown ring test, also known as the nitrate ring test for the determination of presence of nitrate ions in a given chemical solution.
For this test, we need a solution of nitrate, iron (II) sulphate and sulphuric acid.
We add the iron (II) sulphate to the nitrate solution and then we add the concentrated sulphuric acid solution slowly so that the acid forms a layer below the solution. At the junction of the two liquids, if a brown ring appears, it indicates the presence of nitrate ions in the solution.
The brown coloured ring is formed due to the formation of a nitrosyl complex from the iron (II) sulphate and nitric oxide in the solution. The formula of the brown ring complex is-${{\left[ Fe{{\left( {{H}_{2}}O \right)}_{5}}NO \right]}^{2+}}$.
As we know, Iron (II) sulphate is a reducing agent, it reduces nitrate ion to nitric oxide and itself is oxidised to iron (III). We can write the reaction as-
\[\begin{align}
&2HN{{O}_{3}}+2{{H}_{2}}S{{O}_{4}}+6FeS{{O}_{4}}\to 3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2NO+4{{H}_{2}}O \\
&\left[Fe{{({{H}_{2}}O)}_{6}}\right]S{{O}_{4}}+NO\to\left[Fe{{({{H}_{2}}O)}_{5}}NO\right]S{{O}_{4}}+{{H}_{2}}O \\
\end{align}\]
This complex is formed due to the reduction of nitrate ions to nitric oxide.
Hence, the correct answer is option A.
Note:
Besides the brown ring test, there are other tests too which are used for the detection of nitrate ions in a solution like copper turnings test, Devarda’s test and diphenylamine test but the brown ring test is very common and is used in school laboratories.
However, the presence of nitrite ions along with nitrate ions will affect the test by turning the whole solution brown instead of just a brown ring.
Complete step by step answer:
We carry out the brown ring test, also known as the nitrate ring test for the determination of presence of nitrate ions in a given chemical solution.
For this test, we need a solution of nitrate, iron (II) sulphate and sulphuric acid.
We add the iron (II) sulphate to the nitrate solution and then we add the concentrated sulphuric acid solution slowly so that the acid forms a layer below the solution. At the junction of the two liquids, if a brown ring appears, it indicates the presence of nitrate ions in the solution.
The brown coloured ring is formed due to the formation of a nitrosyl complex from the iron (II) sulphate and nitric oxide in the solution. The formula of the brown ring complex is-${{\left[ Fe{{\left( {{H}_{2}}O \right)}_{5}}NO \right]}^{2+}}$.
As we know, Iron (II) sulphate is a reducing agent, it reduces nitrate ion to nitric oxide and itself is oxidised to iron (III). We can write the reaction as-
\[\begin{align}
&2HN{{O}_{3}}+2{{H}_{2}}S{{O}_{4}}+6FeS{{O}_{4}}\to 3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2NO+4{{H}_{2}}O \\
&\left[Fe{{({{H}_{2}}O)}_{6}}\right]S{{O}_{4}}+NO\to\left[Fe{{({{H}_{2}}O)}_{5}}NO\right]S{{O}_{4}}+{{H}_{2}}O \\
\end{align}\]
This complex is formed due to the reduction of nitrate ions to nitric oxide.
Hence, the correct answer is option A.
Note:
Besides the brown ring test, there are other tests too which are used for the detection of nitrate ions in a solution like copper turnings test, Devarda’s test and diphenylamine test but the brown ring test is very common and is used in school laboratories.
However, the presence of nitrite ions along with nitrate ions will affect the test by turning the whole solution brown instead of just a brown ring.
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