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

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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.

KCET 2025Set D-41 markMCQ

Q. In the preparation of sodium fusion extract, the purpose of fusing organic compound with a piece of sodium metal is to (A) Convert the organic compound into vapour state (B) Convert the elements of the compound from covalent form to ionic form (C) Convert the elements of the compound from ionic form to covalent form (D) Decrease the melting point of the compound

›Reveal solution

Covalently bound N, S and halogen atoms cannot respond to aqueous ionic tests, so fusion with sodium metal converts them into water-soluble ionic salts ( $\ce{NaCN}$ , $\ce{Na2S}$ , $\ce{NaX}$ ) that can then be detected.

Step 1 — The problem this test solves

Suppose we want to know whether an organic compound contains nitrogen, sulphur or a halogen. The obvious approach — dissolve it in water and add a reagent — fails completely. Why?

In an organic molecule these elements are joined to carbon by covalent bonds. They are not free ions. So:

The compound does not ionise in water.
Adding $\ce{AgNO3}$ to (say) chlorobenzene gives no precipitate, even though chlorine is certainly present — because there is no free $\ce{Cl-}$ to react with the $\ce{Ag+}$ .

All the standard qualitative tests are tests for IONS. So before we can use them, the covalently-held elements must somehow be converted into ions.

Step 2 — The solution: fuse with sodium metal

Sodium is a highly electropositive, powerfully reducing metal. Fusing (strongly heating) the organic compound with a small piece of sodium causes sodium to attack and break the covalent C–N, C–S and C–X bonds, capturing those elements as their anions in simple ionic sodium salts:

\ce{Na + C + N ->[\Delta] NaCN} \qquad \text{(sodium cyanide — for nitrogen)}

\ce{2Na + S ->[\Delta] Na2S} \qquad\ \ \text{(sodium sulphide — for sulphur)}

\ce{Na + X ->[\Delta] NaX} \qquad\ \ \ \text{(sodium halide; X = Cl, Br, I)}

\ce{Na + C + N + S ->[\Delta] NaSCN} \quad \text{(when N \textit{and} S are both present)}

The fused mass is then extracted with distilled water and filtered. The resulting solution — containing $\ce{CN-}$ , $\ce{S^{2-}}$ , $\ce{X-}$ as free ions — is the sodium fusion extract, or Lassaigne's extract.

Step 3 — Now the ionic tests work

With the elements liberated as ions, ordinary inorganic qualitative analysis applies:

Element	Ion in extract	Test	Observation
Nitrogen	$\ce{CN-}$	$\ce{FeSO4}$ , then $\ce{H2SO4}$	Prussian blue $\ce{Fe4[Fe(CN)6]3}$
Sulphur	$\ce{S^{2-}}$	Sodium nitroprusside	Violet/purple colour
Sulphur	$\ce{S^{2-}}$	Lead acetate	Black $\ce{PbS}$ precipitate
Halogen	$\ce{X-}$	$\ce{HNO3}$ + $\ce{AgNO3}$	White ( $\ce{Cl}$ ) / pale-yellow ( $\ce{Br}$ ) / yellow ( $\ce{I}$ ) ppt.

So the entire purpose of the fusion step is the covalent → ionic conversion. That is precisely option (B).

Step 4 — Eliminate the other options …

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KCET 2024Set B-21 markMCQ
$_2$
›Reveal solutionSolution
The key is the polarising power of the cation — small, highly charged cations destabilise the nitrate ion, causing it to decompose to NO₂ on heating. Among the given alkali metals, Li⁺ is the smallest, so LiNO₃ decomposes to give NO₂.

The question asks which nitrate, on heating, gives nitrogen dioxide (NO₂). Not all nitrates behave the same way. The decomposition pathway depends on the metal cation’s ability to distort the nitrate ion’s electron cloud — a concept called polarisation.

For alkali metal nitrates, the trend is clear: as you go down Group 1, the cation size increases, polarising power decreases, and the nitrate becomes more thermally stable. The smallest cation (Li⁺) has the strongest polarising effect, pulling electron density away from the nitrate ion and making it prone to break apart into NO₂ and O₂, rather than just giving nitrite and oxygen.

Let’s walk through it.

Recall the general thermal decomposition behaviour of nitrates. Most metal nitrates decompose on heating, but the products vary. For alkali metals (Group 1), the lighter ones (Li) behave differently from the heavier ones (Na, K, Rb, Cs).

Heavier alkali metal nitrates (NaNO₃, KNO₃, RbNO₃) decompose to give the nitrite and oxygen:

$2\text{MNO}_3 \xrightarrow{\Delta} 2\text{MNO}_2 + \text{O}_2$

Lithium nitrate, due to the small size of Li⁺, decomposes further to give the oxide, nitrogen dioxide, and oxygen:

$4\text{LiNO}_3 \xrightarrow{\Delta} 2\text{Li}_2\text{O} + 4\text{NO}_2 + \text{O}_2$

Why does LiNO₃ behave differently? The nitrate ion (NO₃⁻) is large and has a diffuse negative charge. A small, highly charged cation like Li⁺ strongly polarises the nitrate ion — it pulls electron density toward itself. This weakens the N–O bonds, making the ion unstable at high temperature. Instead of simply losing one oxygen (to form nitrite), the nitrate ion breaks apart more completely, releasing NO₂. …
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COMEDK 2024Set 2024-E1 markMCQ
$\mathrm{X}$
›Reveal solutionSolution
The problem involves estimating the percentage of an element (sulfur) from the mass of BaSO₄ precipitated, and identifying the violet complex formed in Lassaigne’s test with sodium nitroprusside. The calculated percentage is 20.6%, and the violet compound is Na₄[Fe(CN)₅NOS], so the correct option is (B).

Concept & Intuition

The key idea: when an organic compound containing sulfur is fused with sodium (Lassaigne’s test), sulfur forms sodium sulfide. On treatment with sodium nitroprusside, a violet complex is formed — this is a classic test for sulfur. Separately, in the estimation via fuming HNO₃ and BaCl₂, sulfur is oxidized to sulfate, which precipitates as BaSO₄. The mass of BaSO₄ gives the mass of sulfur, from which the percentage in the original compound is found. The violet complex’s formula is well-known: Na₄[Fe(CN)₅NOS].

Step-by-step solution

Identify element X from the chemical tests

The compound is heated with fuming HNO₃ (oxidation), then treated with BaCl₂ to give a BaSO₄ precipitate. This is the standard method for estimating sulfur in an organic compound. Also, the Lassaigne’s extract with sodium nitroprusside gives a violet colour — a confirmatory test for sulfur. Hence, element X is sulfur (S).

Calculate the mass of sulfur from BaSO₄

Molar mass of BaSO₄ = 137.3 (Ba) + 32.1 (S) + 64 (O₄) = 233.4 g/mol.

Mass of BaSO₄ obtained = 1.2 g.

Mass of sulfur in this BaSO₄ = (atomic mass of S / molar mass of BaSO₄) × mass of BaSO₄

= (32.1 / 233.4) × 1.2 ≈ 0.165 g.

Calculate the percentage of sulfur in the compound

Mass of organic compound taken = 0.8 g.

Percentage of S = (mass of S / mass of compound) × 100

= (0.165 / 0.8) × 100 ≈ 20.625%.

Rounded to one decimal place: 20.6%.

Identify the violet compound from Lassaigne’s test

When sulfur is present, the Lassaigne’s extract contains Na₂S. On adding sodium nitroprusside (Na₂[Fe(CN)₅NO]), a violet complex forms:
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COMEDK 2023Set 2023-M1 markMCQ
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$
›Reveal solutionSolution

The blue colour of the nitrogen test is Prussian blue, ferric ferrocyanide $\mathrm{Fe_4[Fe(CN)_6]_3}$ .

During sodium fusion, nitrogen (and carbon) in the organic compound form sodium cyanide: $\mathrm{Na + C + N \rightarrow NaCN}$ .

On treating the fusion extract with a freshly prepared $\mathrm{FeSO_4}$ solution, $\mathrm{Fe^{2+}}$ reacts with cyanide to give sodium hexacyanoferrate(II):

$\mathrm{6\,CN^- + Fe^{2+} \rightarrow [Fe(CN)_6]^{4-}}$ …
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KCET 2018Set A-11 markMCQ
$Q. VERSION: 23-A 58. The reaction of Benzenediazonium chloride with aniline yields yellow dye. The name of the yellow dye is (A) p-Hydroxyazobenzene (B) p-Aminoazobenzene (C) p-Nitroazobenzene (D) o-Nitroazobenzene$
›Reveal solutionSolution

This is a diazo coupling reaction: the weakly electrophilic diazonium ion attacks the strongly activated para position of aniline, giving $p$ -aminoazobenzene (a yellow azo dye).

Step 1 — Recognise the reaction type.

A diazonium ion $\ce{C6H5-N2+}$ is a weak electrophile. It can only attack rings that are strongly activated — i.e. phenols and aromatic amines. This electrophilic substitution is called coupling, and the product contains the chromophoric $\ce{-N=N-}$ (azo) linkage, which is why the products are coloured dyes.

Step 2 — Where does it attack?

The $\ce{-NH2}$ group of aniline is a powerful $o,p$ -director. Because the diazonium ion is a bulky electrophile, attack occurs almost exclusively at the sterically free para position.
$\ce{C6H5N2+Cl- + C6H5NH2 -> C6H5-N=N-C6H4-NH2 (para) + HCl}$
Step 3 — Name the product. …
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Q.Match Column I with Column II. (More than one correlation is possible.)
Column I

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