A brown ring is formed in the ring test for \[\text{NO}_{3}^{-}\] ion. It is due to the formation of:
A ) \[\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\text{ }\]
B ) \[\text{FeS}{{\text{O}}_{4}}\text{N}{{\text{O}}_{2}}\]
C ) \[{{\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{4}}{{\left( \text{NO} \right)}_{2}} \right]}^{2+}}\]
D ) \[\text{FeS}{{\text{O}}_{4}}\text{HN}{{\text{O}}_{3}}\]
Answer
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Hint: The brown ring test for nitrate ion confirms the presence of nitrate ion. A freshly prepared ferrous sulphate solution is added to the solution containing nitrate ions. A ring containing a brown coloured complex is formed.
Complete answer:
The brown ring test for nitrate ion is a qualitative test that confirms the presence of nitrate ion. In the ring test for \[\text{NO}_{3}^{-}\] ion, a freshly prepared ferrous sulphate solution is added to the solution containing nitrate ion. A ring containing a brown coloured complex is formed. The formation of this brown coloured complex confirms the ring test for the presence of nitrate ion. The brown coloured complex is \[\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\].
Write the balanced chemical equation for the brown ring test for nitrate ions.
${\text{NO}_{3}^{-}\text{ + 3F}{{\text{e}}^{2+}}\text{ + 4 }{{\text{H}}^{+}}\text{ }\to \text{ NO + 3F}{{\text{e}}^{3+}}\text{ +2 }{{\text{H}}_{2}}\text{O}}$
${{{\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{6}} \right]}^{2+}}\text{ + NO }\to \text{ }\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\text{ + }{{\text{H}}_{2}}\text{O}}$
In the first step, ferrous ions react with nitrate ions in presence of protons to form ferric ions and nitric oxide. In the second step, nitric oxide reacts with hydrated ferrous ions to form a brown ring complex. In the first step, the oxidation state of nitrogen decreases by 3 . In the second step for the reaction, there is no change in the oxidation state of ferrous ions.
Hence, the option A ) \[\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\] is the correct option.
Note: A ring is formed at the junction of two layers which is brown in colour. This indicates the presence of nitrate ions.
Complete answer:
The brown ring test for nitrate ion is a qualitative test that confirms the presence of nitrate ion. In the ring test for \[\text{NO}_{3}^{-}\] ion, a freshly prepared ferrous sulphate solution is added to the solution containing nitrate ion. A ring containing a brown coloured complex is formed. The formation of this brown coloured complex confirms the ring test for the presence of nitrate ion. The brown coloured complex is \[\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\].
Write the balanced chemical equation for the brown ring test for nitrate ions.
${\text{NO}_{3}^{-}\text{ + 3F}{{\text{e}}^{2+}}\text{ + 4 }{{\text{H}}^{+}}\text{ }\to \text{ NO + 3F}{{\text{e}}^{3+}}\text{ +2 }{{\text{H}}_{2}}\text{O}}$
${{{\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{6}} \right]}^{2+}}\text{ + NO }\to \text{ }\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\text{ + }{{\text{H}}_{2}}\text{O}}$
In the first step, ferrous ions react with nitrate ions in presence of protons to form ferric ions and nitric oxide. In the second step, nitric oxide reacts with hydrated ferrous ions to form a brown ring complex. In the first step, the oxidation state of nitrogen decreases by 3 . In the second step for the reaction, there is no change in the oxidation state of ferrous ions.
Hence, the option A ) \[\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\] is the correct option.
Note: A ring is formed at the junction of two layers which is brown in colour. This indicates the presence of nitrate ions.
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