To a solution of an unknown compound, a little dilute tetraoxosulphate (VI) acid was added with some freshly…

Analysis of the Question

This question tests your knowledge of qualitative analysis in chemistry, specifically the confirmatory test for a particular anion. You are asked to identify the anion that produces a "brown ring" when reacted with iron(II) tetraoxosulphate(VI) ($FeSO_4$) and concentrated tetraoxosulphate(VI) acid ($H_2SO_4$).

Concept Explanation: The Brown Ring Test

The Brown Ring Test is a classic laboratory procedure used to detect the presence of the nitrate ion ($NO_3^-$).

1. The Reagents: The test requires the unknown solution, freshly prepared iron(II) sulphate ($FeSO_4$), and concentrated tetraoxosulphate(VI) acid ($H_2SO_4$).
2. The Process: When $FeSO_4$ is added to the solution containing nitrate ions, and concentrated $H_2SO_4$ is carefully poured down the side of the test tube, a brown ring forms at the junction of the two layers.
3. The Chemistry: The concentrated $H_2SO_4$ reduces the nitrate ion to nitrogen(II) oxide ($NO$). The $NO$ then reacts with the excess $Fe^{2+}$ ions to form a brown-colored complex compound known as nitrosonium iron(II) tetraoxosulphate(VI), which has the formula $[Fe(H_2O)_5(NO)]SO_4$.

Step-by-Step Solution

• Step 1: Identify the reagents provided in the question. The use of $FeSO_4$ and concentrated $H_2SO_4$ is the standard procedure for the Brown Ring Test.
• Step 2: Evaluate the options provided: • A (Carbonate $CO_3^{2-}$): Tested by adding dilute acids, which causes effervescence (bubbles of $CO_2$). • B (Chloride $Cl^-$): Tested by adding silver trioxonitrate(V) ($AgNO_3$) solution, which forms a white precipitate. • C (Sulphite $SO_3^{2-}$): Tested by adding dilute acids, which releases $SO_2$ gas that turns acidified potassium tetraoxomanganate(VII) from purple to colorless. • D (Nitrate $NO_3^-$): This is the specific ion that undergoes the Brown Ring Test. (Note: The option in your question is written as $NO_3^-$, though it appears as $NO_3^2-$ in your text, it refers to the nitrate ion).

Conclusion

The formation of the brown ring is the definitive confirmatory test for the nitrate ion ($NO_3^-$). Therefore, the presence of the brown ring confirms the presence of $NO_3^-$.

Exam Tip: Always remember that the Brown Ring Test must be performed with freshly prepared iron(II) sulphate. If the $FeSO_4$ is old, it will have oxidized to iron(III) ($Fe^{3+}$) and will not be able to form the brown complex, leading to a false negative result.