For testing halogens in an organic compound with AgNO3 solution, sodium extract (Lassaigne's test) is acidifie

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Q. For testing halogens in an organic compound with AgNO3 solution, sodium extract (Lassaigne's test) is acidified with dilute HNO3. What will happen if a student acidifies the extract with dilute H2SO4 in place of dilute HNO3?

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Concept understanding — Lassaigne Test Principle

The Lassaigne Test: Why We Fuse Sodium with the Sample

Imagine you have an organic compound — say, a drug, a pesticide, or a dye — and you need to find out if it contains nitrogen, sulfur, or a halogen (chlorine, bromine, iodine). You cannot simply test the original compound directly because these elements are covalently bonded inside the molecule. They are "hidden" — not present as free ions.

The Lassaigne test solves this by breaking the covalent bonds and converting these elements into water-soluble ionic salts. That is the entire point: turn invisible atoms into detectable ions.

The Intuition

Think of the organic molecule as a locked safe. Inside the safe are the atoms you want to detect (N, S, X). The safe is the covalent framework. To get the atoms out, you need to smash the safe open. Sodium metal is the sledgehammer.

When you heat the organic compound with a piece of sodium metal, the sodium reacts violently. It rips apart the carbon‑carbon and carbon‑heteroatom bonds. In the process, the sodium itself gets oxidised (loses an electron) and the atoms from the compound get reduced (gain electrons). The result is a set of simple, stable, ionic compounds:

Nitrogen (from the compound) combines with sodium and carbon to form sodium cyanide ( $\text{NaCN}$ ).
Sulfur forms sodium sulfide ( $\text{Na}_2\text{S}$ ).
Halogens form sodium halides ( $\text{NaX}$ , where X = Cl, Br, I).

These are all water‑soluble. Once you dissolve the fused mass in water, you have a solution containing $\text{CN}^-$ , $\text{S}^{2-}$ , and $\text{X}^-$ ions — ready for standard qualitative tests.

Note

If both nitrogen and sulfur are present in the same compound, they can also form sodium thiocyanate ( $\text{NaSCN}$ ) during fusion. This is important because it can interfere with the test for sulfur — more on that later.

The Precise Statement

Lassaigne’s test (or sodium fusion test): A small piece of dry sodium metal is heated with a sample of the organic compound in a fusion tube until the tube is red hot. The fused mass is then extracted with distilled water, boiled, cooled, and filtered. The filtrate (called the Lassaigne’s extract) is used to test for the presence of nitrogen, sulfur, and halogens.

The chemical principle is:

\text{Organic compound (containing N, S, X)} + \text{Na (excess)} \xrightarrow{\Delta} \text{NaCN} + \text{Na}_2\text{S} + \text{NaX} + \text{residue}

Why This Works (and What Can Go Wrong)

For nitrogen: The cyanide ion ( $\text{CN}^-$ ) is detected by adding ferrous sulfate and ferric chloride — a deep blue precipitate of Prussian blue ( $\text{Fe}_4[\text{Fe(CN)}_6]_3$ ) confirms nitrogen.

For sulfur: The sulfide ion ( $\text{S}^{2-}$ ) gives a purple/violet colour with sodium nitroprusside ( $\text{Na}_2[\text{Fe(CN)}_5\text{NO}]$ ). Alternatively, it forms a black precipitate of lead sulfide ( $\text{PbS}$ ) with lead acetate.

For halogens: The halide ion ( $\text{X}^-$ ) is detected by acidifying the extract with nitric acid and adding silver nitrate — a curdy white precipitate (AgCl), pale yellow (AgBr), or yellow (AgI) indicates the halogen. …

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Acidifying the sodium extract with dilute $\mathrm{H_2SO_4}$ instead of dilute $\mathrm{HNO_3}$ will produce a white precipitate of $\mathrm{Ag_2SO_4}$ that interferes with the halide test, making it impossible to confirm the presence of halogens reliably.

Why We Acidify in Lassaigne's Test

When an organic compound containing halogens is fused with sodium metal, the halogen converts to sodium halide ( $\mathrm{NaX}$ , where $\mathrm{X} = \mathrm{Cl, Br, I}$ ) in the aqueous extract. To detect these halides, we add $\mathrm{AgNO_3}$ , which precipitates them as silver halides:

\mathrm{NaX + AgNO_3 \to AgX \downarrow + NaNO_3}

The precipitate colour tells us which halogen is present: $\mathrm{AgCl}$ is white, $\mathrm{AgBr}$ is pale yellow, and $\mathrm{AgI}$ is yellow.

But the sodium extract is alkaline (from excess sodium and the formation of $\mathrm{NaOH}$ , $\mathrm{Na_2CO_3}$ , $\mathrm{NaCN}$ , etc.). If we add $\mathrm{AgNO_3}$ directly to this alkaline solution, silver hydroxide and silver oxide precipitate:

\mathrm{AgNO_3 + NaOH \to AgOH \downarrow + NaNO_3}

\mathrm{2AgOH \to Ag_2O \downarrow + H_2O}

These brown/black precipitates mask the halide test. So we acidify first to neutralise the alkalinity.

Why Nitric Acid, Not Sulphuric Acid?

The choice of acid matters because we need one that does not itself interfere with the silver ion.

With dilute $\mathrm{HNO_3}$ : Nitric acid neutralises the alkaline impurities ( $\mathrm{NaOH}$ , $\mathrm{Na_2CO_3}$ , $\mathrm{NaCN}$ , $\mathrm{Na_2S}$ ) without forming any precipitate with $\mathrm{Ag^+}$ . The nitrate ion ( $\mathrm{NO_3^-}$ ) does not precipitate silver, so the only precipitate that forms is from the halide ions—exactly what we want to observe.
With dilute $\mathrm{H_2SO_4}$ : Sulphuric acid also neutralises the alkaline species, but it introduces sulphate ions ( $\mathrm{SO_4^{2-}}$ ) into the solution. When $\mathrm{AgNO_3}$ is added, these sulphate ions react:

$\mathrm{2AgNO_3 + H_2SO_4 \to Ag_2SO_4 \downarrow + 2HNO_3}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2026Set 2026-M1 markMCQ
$\mathrm{NH}_2-\mathrm{NH}_2 \mathrm{HCl}$
›Reveal solutionSolution
Lassaigne’s test for nitrogen fails when the nitrogen atoms are already bonded to each other (as in hydrazine derivatives), because the sodium fusion cannot break the N–N bond to form NaCN. The correct option is (A).

Concept and Intuition

Lassaigne’s test (sodium fusion test) works by fusing an organic compound with sodium metal. If nitrogen is present, it combines with carbon (from the compound) to form sodium cyanide (NaCN). This NaCN is then detected by converting it to Prussian blue (Fe₄[Fe(CN)₆]₃) or by other standard tests.

The key requirement: the nitrogen must be able to form a cyanide ion (CN⁻). That means the nitrogen must be attached to a carbon atom in the original compound, so that during fusion, the carbon and nitrogen can combine. If the nitrogen atoms are bonded only to each other (as in hydrazine, NH₂–NH₂) or to hydrogen, there is no carbon atom nearby to form CN⁻. In such cases, the nitrogen escapes as ammonia or nitrogen gas, and the test fails.

Step-by-Step Reasoning

Identify the structural feature that matters

For Lassaigne’s test to succeed, the compound must contain at least one carbon–nitrogen bond. During fusion, the C–N bond breaks and reforms as the cyanide ion. If the nitrogen is only bonded to hydrogen or to another nitrogen, no cyanide can form.

Examine each option

(A) NH₂–NH₂·HCl (hydrazine hydrochloride)

Here, the two nitrogens are bonded to each other, and there is no carbon atom in the molecule at all. No C–N bond exists. Fusion will produce N₂ or NH₃, but no NaCN. Test fails.

(B) NH₂CONHNH₂·HCl (semicarbazide hydrochloride)

This molecule contains a carbonyl group (C=O) with a nitrogen attached to carbon (the amide part). There is a C–N bond. During fusion, that carbon can combine with nitrogen to form NaCN. Test succeeds.

(C) C₆H₅NHNH₂·HCl (phenylhydrazine hydrochloride)

The phenyl ring provides carbon atoms directly bonded to the first nitrogen (C₆H₅–NH–). That C–N bond is present. Fusion will produce NaCN. Test succeeds.

(D) NH₂CONH₂ (urea) …
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KCET 2025Set D-41 markMCQ
$_3^-$
›Reveal solutionSolution

Concentrated HNO $_3$ is boiled with the fusion extract to expel the sulphide and cyanide ions, which would otherwise give their own silver precipitates and give a false halogen test.

Step 1 — What the sodium fusion extract contains.

Fusing the organic compound with sodium converts each hetero-element into a simple sodium salt:
$\text{N} \rightarrow \text{NaCN}, \qquad \text{S} \rightarrow \text{Na}_2\text{S}, \qquad \text{X} \rightarrow \text{NaX}\ (\text{X = Cl, Br, I})$
Boiling the fused mass with distilled water gives the sodium fusion extract, which is alkaline and may contain $\mathrm{CN^-}$ , $\mathrm{S^{2-}}$ and $\mathrm{X^-}$ all together.

Step 2 — Why the halogen test needs cleaning up.

The halogen test is done by acidifying and adding AgNO $_3$ :
$\mathrm{Ag^+ + X^- \rightarrow AgX \downarrow}$
But $\mathrm{Ag^+}$ also precipitates the other two ions:
$\mathrm{2Ag^+ + S^{2-} \rightarrow Ag_2S \downarrow \ (black)}$ $\mathrm{Ag^+ + CN^- \rightarrow AgCN \downarrow \ (white)}$
AgCN is a white precipitate — indistinguishable at a glance from AgCl. So a compound containing only N or S could be wrongly reported as containing a halogen.

Step 3 — What boiling with conc. HNO $_3$ does.

$\mathrm{CN^-}$ and $\mathrm{S^{2-}}$ are the conjugate bases of volatile weak acids. On boiling with a strong acid they are protonated and driven off as gases:
$\mathrm{NaCN + HNO_3 \rightarrow NaNO_3 + HCN \uparrow}$
$\mathrm{Na_2S + 2HNO_3 \rightarrow 2NaNO_3 + H_2S \uparrow}$ …
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KCET 2019Set A-11 markMCQ
$+4$
›Reveal solutionSolution
In the reaction of gold with aqua regia, nitrogen in nitric acid (oxidation state $+5$ ) is reduced to nitrogen monoxide (oxidation state $+2$ ). The correct option is (B).

The question asks about the change in oxidation state of nitrogen when gold reacts with aqua regia. Aqua regia is a 3:1 mixture of concentrated hydrochloric acid (HCl) and concentrated nitric acid (HNO₃). The key here is to identify which nitrogen-containing species is involved and what it becomes.

Nitric acid is the oxidizing agent in aqua regia. In the reaction, gold (Au) is oxidized to gold(III) chloride (AuCl₃) or the complex chloroauric acid (HAuCl₄), while nitric acid is reduced. The nitrogen in HNO₃ has an oxidation state of $+5$ (since H is $+1$ , O is $-2$ , and the molecule is neutral: $+1 + x + 3(-2) = 0 \Rightarrow x = +5$ ). The typical reduction product of nitric acid with a metal like gold is nitrogen monoxide (NO), where nitrogen has an oxidation state of $+2$ (O is $-2$ , so $x - 2 = 0 \Rightarrow x = +2$ ).

Let’s confirm the overall reaction to be thorough.

Write the balanced reaction. Gold does not dissolve in HCl or HNO₃ alone, but in aqua regia the reaction is:

$\text{Au} + \text{HNO}_3 + 3\text{HCl} \rightarrow \text{AuCl}_3 + \text{NO} + 2\text{H}_2\text{O}$
(Often written with the complex: $\text{Au} + 4\text{HCl} + \text{HNO}_3 \rightarrow \text{HAuCl}_4 + \text{NO} + 2\text{H}_2\text{O}$ , but the nitrogen chemistry is the same.)

Identify the nitrogen species.

On the left: nitrogen is in $\text{HNO}_3$ (oxidation state $+5$ ).

On the right: nitrogen is in $\text{NO}$ (oxidation state $+2$ ).

Determine the change.

The oxidation state goes from $+5$ to $+2$ . This is a reduction (gain of electrons), as nitric acid acts as the oxidizing agent. …
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Q.For testing halogens in an organic compound with AgNO3 solution, sodium extract (Lassaigne's test) is acidified with dilute HNO3. What will happen if a student acidifies the extract with dilute H2SO4 in place of dilute HNO3?

Why We Acidify in Lassaigne's Test

Why Nitric Acid, Not Sulphuric Acid?

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