Take 1 mL of an aqueous solution of the nitrate salt and mix the freshly prepared ferrous sulphate solution thoroughly. Now, add 2 mL conc. H 2 SO 4 slowly along the sides of the test tube dropwis

Question

Take 1 mL of an aqueous solution of the nitrate salt and mix the freshly prepared ferrous sulphate solution thoroughly. Now, add 2 mL conc. $H_{2} SO_{4}$ slowly along the sides of the test tube dropwise so that it forms a layer on the top of the liquid already present in the test tube. A dark brown compound is formed at the junction of the two solutions

Select INCORRECT statement :-

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A. The brown compound is nitrosoferrous sulfate, i.e., $F e (NO) S O_{4}$ .

This statement is CORRECT. The brown ring complex is indeed nitrosoferrous sulfate, which is more accurately represented as $[F e (H_{2} O)_{5} (NO)] S O_{4}$ . The NO ligand is typically considered to be $N O^{+}$ in this complex, making the iron in a +1 oxidation state ( $F e^{+}$ ). The formula $F e (NO) S O_{4}$ is a simplified representation.

B. The observation is the same if, alternatively, concentrated $H_{2} S O_{4}$ is first added to the cold solution, and then $F e S O_{4}$ (aq.) is added dropwise along the side of the test tube.

This statement is CORRECT. The order of addition of reagents can be varied. The key is to have the nitrate solution, ferrous sulfate, and concentrated sulfuric acid present in a way that allows the reaction to occur at the interface. Adding concentrated $H_{2} S O_{4}$ first to the cold nitrate solution and then adding $F e S O_{4}$ along the sides will still create the necessary conditions for the brown ring to form at the junction. The cold solution helps prevent the decomposition of the complex and ensures the reaction occurs at the interface rather than throughout the solution.

C. The brown compound formed has a spin magnetic moment $= 15$ B.M.

This statement is CORRECT. In the brown ring complex, $[F e (H_{2} O)_{5} (NO)]^{2 +}$ , the iron is formally in the $F e^{+}$ oxidation state, and NO acts as $N O^{+}$ .

$F e^{+}$ has an electronic configuration of $[A r] 3 d^{7}$ .

In the presence of $N O^{+}$ (a strong field ligand) and $H_{2} O$ (a weak field ligand), the $3 d^{7}$ electrons are arranged. The $N O^{+}$ ligand is a strong $π$ -acceptor, which significantly affects the d-electron configuration.

The complex is considered to have 3 unpaired electrons.

For 3 unpaired electrons, the spin magnetic moment ( $μ_{s}$ ) is calculated using the formula:

$μ_{s} = n (n + 2)$

where $n$ is the number of unpaired electrons.

Here, $n = 3$ .

$μ_{s} = 3 (3 + 2) = 3 \times 5 = 15 B . M .$

This value is consistent with experimental observations for the brown ring complex.

D. The brown compound is not formed if the aqueous salt is treated with sulfamic acid before the test.

This statement is INCORRECT. Sulfamic acid ( $H S O_{3} N H_{2}$ ) reacts with nitrite ions ( $N O_{2}^{-}$ ) to remove them, as nitrite ions can also give a brown ring with $F e S O_{4}$ .

$N O_{2}^{-} + H S O_{3} N H_{2} ⟶ N_{2} (g) + S O_{4}^{2 -} + H^{+} + H_{2} O$

However, sulfamic acid does not react with nitrate ions ( $N O_{3}^{-}$ ). Therefore, if the aqueous salt contains nitrate ions, the brown ring test will still be positive even after treatment with sulfamic acid. This treatment is used to confirm that the brown ring is due to nitrate and not nitrite.

The final answer is D

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