Match Column I with Column II. (More than one correlation is possible.) Column I (i) Dumas method (ii) Kjeldah

NCERT Exemplar · Q52

Q. Match Column I with Column II. (More than one correlation is possible.) Column I

(i) Dumas method

(ii) Kjeldahl's method

(iii) Carius method

(iv) Chromatography

(v) Homolysis
Column II

(a) AgNO3

(b) Silica gel

(d) Free radicals

(e) Ammonium sulphate

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Concept understanding — Lassaigne Test Chemistry

The Lassaigne Test: Why We Burn the Sample First

Imagine you have an organic compound — say, a drug, a pesticide, or a dye — and you need to know if it contains nitrogen, sulfur, or a halogen (chlorine, bromine, iodine). You can't just test the compound directly because these atoms are covalently bonded inside the molecule. They won't simply fall off and react with a reagent.

The core problem: covalent bonds are stubborn. You need to break the molecule apart and convert those atoms into simple, water-soluble ions that you can detect with standard inorganic tests. That's exactly what the Lassaigne test does.

Note

The Lassaigne test is also called the sodium fusion test. It was developed by the French chemist J.L. Lassaigne in the 19th century.

The Intuition: Fusion with Sodium

The trick is to heat the organic compound with a piece of metallic sodium. Sodium is a powerful reducing agent. When you fuse them together (heat strongly in a fusion tube), the sodium rips the molecule apart. Here's what happens to the key elements:

Nitrogen → gets converted to sodium cyanide ( $\text{NaCN}$ )
Sulfur → gets converted to sodium sulfide ( $\text{Na}_2\text{S}$ )
Halogens (Cl, Br, I) → get converted to sodium halides ( $\text{NaX}$ , where X = Cl, Br, I)

The product of this fusion is a dark, charred mass. You then extract it with distilled water, boil, and filter. The clear filtrate is called the Lassaigne extract (or sodium fusion extract). This extract now contains the ions you can test for.

Watch out

Sodium metal is extremely reactive with water and moisture. It must be handled with dry apparatus and stored under kerosene. Never let it come in contact with water directly — the fusion tube is heated, then dropped into water after cooling.

The Precise Statement

Lassaigne test: A qualitative analysis method in which an organic compound is fused with metallic sodium to convert covalently bonded nitrogen, sulfur, and halogens into their respective water-soluble inorganic sodium salts (NaCN, Na₂S, NaX). These ions are then detected in the aqueous extract using specific chemical tests.

How to Detect Each Element in the Extract

1. Detection of Nitrogen

Test: Add a few drops of freshly prepared ferrous sulfate ( $\text{FeSO}_4$ ) solution to the extract. Boil, then cool. Add dilute sulfuric acid and a drop of ferric chloride ( $\text{FeCl}_3$ ).

What happens: The cyanide ion ( $\text{CN}^-$ ) reacts with ferrous ions to form ferrous cyanide, which then reacts with ferric ions to form Prussian blue — a deep blue precipitate of $\text{Fe}_4[\text{Fe(CN)}_6]_3$ .

6\text{NaCN} + \text{FeSO}_4 \rightarrow \text{Na}_4[\text{Fe(CN)}_6] + \text{Na}_2\text{SO}_4

Result: A blue colour or precipitate confirms nitrogen.

2. Detection of Sulfur

Test: Add a few drops of sodium nitroprusside ( $\text{Na}_2[\text{Fe(CN)}_5\text{NO}]$ ) solution to the extract.

What happens: Sulfide ions ( $\text{S}^{2-}$ ) react with sodium nitroprusside to form a violet colour complex.

\text{Na}_2\text{S} + \text{Na}_2[\text{Fe(CN)}_5\text{NO}] \rightarrow \text{Na}_4[\text{Fe(CN)}_5\text{NOS}] \text{ (violet)}

Result: A violet colour confirms sulfur.

3. Detection of Halogens

Test: Acidify the extract with dilute nitric acid ( $\text{HNO}_3$ ), then add silver nitrate ( $\text{AgNO}_3$ ) solution.

What happens: Halide ions ( $\text{Cl}^-$ , $\text{Br}^-$ , $\text{I}^-$ ) form precipitates with silver ions.

Halogen	Precipitate	Colour	Solubility in $\text{NH}_3$
Chlorine	$\text{AgCl}$	White	Soluble
Bromine	$\text{AgBr}$	Pale yellow	Partially soluble
Iodine	$\text{AgI}$	Yellow	Insoluble

Watch out

If nitrogen or sulfur is present, you must remove them before testing for halogens. Why? Because $\text{NaCN}$ and $\text{Na}_2\text{S}$ also react with $\text{AgNO}_3$ to form precipitates ( $\text{AgCN}$ and $\text{Ag}_2\text{S}$ ), giving false positives. To remove them, boil the extract with dilute $\text{HNO}_3$ — this converts $\text{CN}^-$ to $\text{HCN}$ gas and $\text{S}^{2-}$ to $\text{H}_2\text{S}$ gas, both of which escape.

Common Mistakes Students Make …

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The matching pairs are based on the core principle or reagent of each analytical or reaction method: Dumas measures nitrogen as $\ce{N2}$ , Kjeldahl traps it as $(\ce{NH4})_2\ce{SO4}$ , Carius uses $\ce{AgNO3}$ to detect halogens, chromatography uses silica gel as a stationary phase, and homolysis produces free radicals.

Dumas method – This is a quantitative method for estimating nitrogen in an organic compound. The sample is heated with $\ce{CuO}$ , converting all nitrogen into nitrogen gas ( $\ce{N2}$ ), which is then measured by volume. The key product is nitrogen gas.

→ (i) matches (c).
Kjeldahl's method – Also for nitrogen estimation, but here the sample is digested with concentrated $\ce{H2SO4}$ in the presence of a catalyst. The nitrogen is converted into ammonium sulphate, $(\ce{NH4})_2\ce{SO4}$ . This is then distilled with alkali to liberate ammonia, which is titrated.

→ (ii) matches (e).
Carius method – Used for the estimation of halogens (or sulphur/phosphorus). The organic compound is heated with fuming $\ce{HNO3}$ in a sealed tube in the presence of $\ce{AgNO3}$ . The halogen precipitates as the corresponding silver halide (e.g., $\ce{AgCl}$ , $\ce{AgBr}$ ). The reagent $\ce{AgNO3}$ is central.

→ (iii) matches (a). …

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

AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
$\underline{E}$
›Reveal solutionSolution
This tests detection of phosphorus in organic compounds; a yellow precipitate with ammonium molybdate is the signature confirmatory test for phosphate, so $E = \mathrm{P}$ .

Concept and Intuition

When an organic compound containing phosphorus is fused/oxidized (here via $\mathrm{Na_2O_2}$ , a strong oxidizing fusion mixture, followed by boiling with $\mathrm{HNO_3}$ ), the phosphorus is converted to phosphate ion ( $\mathrm{PO_4^{3-}}$ ). Treating this phosphate solution with ammonium molybdate reagent gives a characteristic canary-yellow precipitate of ammonium phosphomolybdate, $\mathrm{(NH_4)_3PO_4 \cdot 12MoO_3}$ (or written as $(NH_4)_3[PMo_{12}O_{40}]$ ). This is the standard qualitative test for phosphorus taught alongside Lassaigne's test for N, S, halogens.

Step-by-Step Solution

The organic compound containing extra element E is oxidatively fused with $\mathrm{Na_2O_2}$ , converting E to its highest stable oxidation-state sodium salt.

Boiling with $\mathrm{HNO_3}$ further oxidizes/acidifies this to give the corresponding oxyanion in solution.

Adding ammonium molybdate solution to this solution produces a yellow precipitate — this reaction is specific and diagnostic for phosphate ion. …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$CH_3CH_2N(CH_3)_2$
›Reveal solutionSolution
Only compounds containing BOTH carbon and nitrogen can be converted to $NaCN$ during Lassaigne fusion; 4 of the 7 listed compounds qualify.

Concept and Intuition

Lassaigne's test detects nitrogen in organic compounds by fusing the sample with sodium metal, which converts covalently-bound C and N into sodium cyanide, $NaCN$ . This is then treated with $FeSO_4$ and acidified with dilute $HCl$ / $H_2SO_4$ ; if $CN^-$ was formed, it reacts (via $Fe(CN)_6^{4-}$ / $Fe^{3+}$ ) to give the deep blue precipitate Prussian blue. Crucially, the compound must supply both an organic carbon skeleton and nitrogen bonded within/near it — nitrogen alone (with no carbon in the molecule) cannot form $NaCN$ .

Step-by-Step Solution

$CH_3CH_2N(CH_3)_2$ — a tertiary amine, has C and N bonded → gives Prussian blue.

$CH_3Cl$ — has carbon but no nitrogen at all → the N-test doesn't apply/gives no blue color.

$N_2H_4$ (hydrazine) — has nitrogen but no carbon in the molecule → cannot form $NaCN$ → negative.

$(CH_3)_2C{=}N{-}OH$ (an oxime) — C=N bond present → gives Prussian blue.

$C_6H_5CN$ (benzonitrile) — already has a C≡N (nitrile) group → strongly gives Prussian blue. …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$\mathrm{FeCl_3}$
›Reveal solutionSolution
When both N and S are present, sodium fusion produces sodium thiocyanate; reacting this with $\mathrm{FeCl_3}$ gives the deep blood-red ferric thiocyanate complex ion, $[\mathrm{Fe(SCN)}]^{2+}$ .

Concept and Intuition

Lassaigne's test converts covalently bound heteroatoms (N, S, halogens) in an organic compound into detectable inorganic ions by fusing the sample with sodium metal. When both nitrogen and sulphur are present in the same compound, they combine during fusion to give sodium thiocyanate ( $\mathrm{NaSCN}$ ) rather than separate cyanide and sulphide ions. Adding ferric chloride to the extract then forms the characteristic blood-red ferric thiocyanate complex.

Step-by-Step Solution

Sodium fusion of an N,S-containing compound: $\mathrm{Na + C + N + S \rightarrow NaSCN}$ (thiocyanate forms preferentially over separate $\mathrm{NaCN}$ and $\mathrm{Na_2S}$ ). …
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AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
$[Fe(CN)_5NOS]^{4-}$
›Reveal solutionSolution
This tests the chemistry behind Lassaigne's confirmatory blood-red test for sulfur (with nitrogen) in organic qualitative analysis; the colour is due to ferric thiocyanate, $[Fe(SCN)]^{2+}$ .

Concept and Intuition

Sodium fusion converts covalently bound N and S in an organic compound into ionic $CN^-$ and $S^{2-}$ in the extract. When BOTH nitrogen and sulfur are present in the original compound, some of the fusion also directly produces thiocyanate ion, $SCN^-$ (sodium sulphocyanide/thiocyanate, NaSCN), because carbon, nitrogen and sulfur are all fused together in the same molecule during the sodium fusion. Testing this extract with ferric chloride solution then gives a deep blood-red colouration due to the formation of the ferric-thiocyanate complex.

Step-by-Step Solution

Organic compound containing C, N, S is fused with sodium metal; sodium fusion breaks these into simple inorganic ions, including thiocyanate ( $SCN^-$ ) when N and S are both present in proximity in the molecule.

The Lassaigne extract is then filtered and tested by adding a few drops of freshly prepared ferric chloride ( $FeCl_3$ ) solution.

Thiocyanate ions react with $Fe^{3+}$ to form the deep blood-red complex ion: $Fe^{3+} + SCN^- \rightarrow [Fe(SCN)]^{2+} \text{ (blood red)}$ …
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AP EAPCET 2021Set ap-2021-09-06-AN1 markMCQ
$NaNH_2$
›Reveal solutionSolution
Fusing an organic compound with sodium converts its nitrogen (together with

carbon from the compound) into sodium cyanide, $NaCN$ — the species tested

for via the Prussian-blue colour reaction.

Concept and Intuition

Lassaigne's/sodium fusion test converts covalently bound heteroatoms (N, S,

halogens) in an organic compound into simple, water-soluble, easily-testable

sodium salts by fusing the compound with molten sodium metal at high

temperature. Nitrogen specifically combines with carbon and sodium (both

originally present in the organic molecule and the metal) to give sodium

cyanide, $Na^+ CN^-$ .

Step-by-Step Solution

Organic compound + Na (fusion, high temperature) $\rightarrow$ Na fuses with C and N present in the molecule.

The nitrogen is captured as the cyanide ion, giving $NaCN$ in the Lassaigne extract.

Confirmatory test: the extract is boiled with $FeSO_4$ , treated with dilute $H_2SO_4$ / $HCl$ , giving a Prussian blue precipitate ( $Fe_4[Fe(CN)_6]_3$ ), confirming the presence of nitrogen (as cyanide) in the original compound. …
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AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
$NaOH$
›Reveal solutionSolution
Cyanide is converted to thiocyanate ( $CN^-\to SCN^-$ ) by boiling with sulphur source under alkaline conditions; after acidification, $Fe^{3+}$ reacts with $SCN^-$ to give the well-known blood-red $[Fe(SCN)]^{2+}$ complex.

Concept and Intuition

This is the standard qualitative test that distinguishes cyanide ( $CN^-$ ) from thiocyanate ( $SCN^-$ ) and confirms cyanide indirectly: cyanide, boiled with excess sulphur (here supplied via thiosulphate) in alkaline medium, is converted to thiocyanate:
$CN^- + S_2O_3^{2-} \xrightarrow{\Delta,\ NaOH} SCN^- + SO_3^{2-}$
Thiocyanate ion, once the solution is acidified, gives the classic deep blood-red colour with ferric ions due to formation of the complex ion $[Fe(SCN)]^{2+}$ (or $[Fe(SCN)_6]^{3-}$ at higher $SCN^-$ concentration) — this is the standard confirmatory test for $Fe^{3+}$ / $SCN^-$ in qualitative analysis.

Step-by-Step Solution

$KCN$ + $NaOH$ (alkaline) + thiosulphate, boiled: cyanide is converted to thiocyanate ion, $SCN^-$ .

Cool and acidify with $HCl$ : neutralises excess alkali, doesn't destroy $SCN^-$ . …
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AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
$Mg_3(PO_4)_2$
›Reveal solutionSolution
This tests the classical gravimetric estimation of phosphorus in organic compounds — the element is finally weighed as magnesium pyrophosphate, $Mg_2P_2O_7$ .

Concept and Intuition

Quantitative estimation of phosphorus (and several other elements) in organic compounds relies on converting the element into a stable, well-defined, easily weighable inorganic compound. Phosphorus cannot be weighed directly as a precipitate that is chemically indefinite, so the analytical method carries it through a series of conversions until it lands on a stoichiometrically fixed, non-hygroscopic solid.

Step-by-Step Solution

The organic compound containing phosphorus is oxidised completely (e.g., fusion with sodium peroxide) so that phosphorus is converted entirely to phosphate ( $PO_4^{3-}$ ).

The phosphate is first precipitated as ammonium phosphomolybdate, $(NH_4)_3PO_4\cdot 12MoO_3$ , to separate/detect phosphorus and to concentrate it.

For accurate gravimetric estimation, this precipitate is redissolved and reprecipitated as magnesium ammonium phosphate, $MgNH_4PO_4$ .

On ignition (strong heating), magnesium ammonium phosphate loses ammonia and water and converts to the stable, stoichiometric solid magnesium pyrophosphate: …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$NaCN$
›Reveal solutionSolution
When an organic compound contains both nitrogen and sulphur, Lassaigne's sodium fusion test converts them jointly into sodium thiocyanate, $NaCNS$ , not into separate $NaCN$ and $Na_2S$ .

Concept and Intuition

The Lassaigne test fuses an organic compound with sodium metal to convert covalently bound heteroatoms (N, S, halogens) into simple, water-soluble ionic sodium salts that can then be detected by specific tests (Prussian blue test for N, lead acetate/nitroprusside test for S, silver nitrate test for halogens).

When nitrogen alone is present, sodium fusion converts it into sodium cyanide, $NaCN$ . When sulphur alone is present, it forms sodium sulphide, $Na_2S$ . However, when BOTH nitrogen and sulphur are present in the same molecule, they do not form separately — instead the carbon, nitrogen, and sulphur atoms combine during fusion to give sodium thiocyanate, $Na^+ SCN^-$ (written $NaCNS$ or $NaSCN$ ). This is a classic exception students must remember.

Step-by-Step Solution

Identify that the question specifies BOTH N and S present in the compound (not separately).

Recall the individual conversions: N alone → $NaCN$ ; S alone → $Na_2S$ .

Recall the joint-presence exception: when N and S coexist, sodium fusion instead yields $NaCNS$ (sodium thiocyanate), because C, N, and S combine into one thiocyanate ion rather than forming two separate salts. …
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Q.Match Column I with Column II. (More than one correlation is possible.)
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