The elements present in organic compounds are carbon and hydrogen. In addition to these, they may also contain oxygen, nitrogen, sulphur, halogens and phosphorus.
-- NCERT Class 11 Chemistry, Ch. 8, p. 284Qualitative N, S, P and Halogens
8 MCQs9-step worked example
Source: NCERT Purification and Characterisation of Organic CompoundsPYQ coverage: NEET 2023, 2025Official key: NTA-verifiedLast updated: 27 Sep 2026
Qualitative N, S, P and Halogens, explained for NEET
The fusion test's most reliable failure is not a colour you misread — it is a colour that never forms. When a compound contains both nitrogen and sulphur, sodium fusion produces sodium thiocyanate rather than sodium cyanide, and the ferric test then gives a blood-red solution, not Prussian blue. An aspirant who has memorised "nitrogen → Prussian blue" reads the blood-red as "no nitrogen" and loses the mark twice over, since sulphur is present too.
Lassaigne's test itself is simple in principle: organic N, S, P and halogens are covalently bound and invisible to ionic reagents, so fusion with sodium metal converts them into the ionic sodium salts NaCN, Na₂S, NaX and Na₃PO₄. The hot melt is extinguished in distilled water and filtered; the filtrate is the sodium fusion extract, and every subsequent test is run on it. NCERT Class 11 Chemistry Chapter 8, pages 284–285, sets out each detection.
The second high-cost step is the acid boil. Before adding silver nitrate for a halogen, the extract must be acidified with dilute nitric acid and boiled whenever N or S may be present — otherwise cyanide and sulphide ions survive into the test tube and precipitate as AgCN and Ag₂S, both white-to-black solids that masquerade as a halide result. Boiling drives off HCN and H₂S and leaves the halide alone to answer.
The halide answer is then read off the precipitate: AgCl white and freely soluble in ammonium hydroxide, AgBr pale yellow and only sparingly soluble, AgI yellow and insoluble in ammonium hydroxide. Phosphorus is separate — the extract is boiled with concentrated nitric acid and treated with ammonium molybdate to give a yellow ammonium phosphomolybdate precipitate.
Watch-out: NEET framers set traps at the observation-to-conclusion hop, not at the chemistry. If a stem states blood-red, ask what pair of elements that implies; if it states a silver halide dissolving in ammonia, ask which halide that rules out.
Can you answer these Qualitative N, S, P and Halogens MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In Lassaigne's test, the organic compound is fused with sodium metal in order to:
Show answer and why every option is right or wrong
Answer: B. B is correct. Nitrogen, sulphur and halogens in an organic molecule are covalently bound and give no reaction with ionic reagents; fusion with sodium converts them into NaCN, Na₂S and NaX, which are ionic and detectable — NCERT Class 11 Chemistry Chapter 8, page 284. Phosphorus is tested separately: heating with sodium peroxide oxidises it to phosphate (page 285).
Why A is wrong: A is wrong because complete oxidation to CO₂ and H₂O is combustion analysis, the quantitative estimation route, not the sodium fusion that precedes qualitative detection.
Why C is wrong: C is wrong because sodium fusion does not strip out the carbon skeleton; it converts carbon and nitrogen together into cyanide ion, and the carbon is what makes NaCN possible.
Why D is wrong: D is wrong because the sodium salts formed are water-soluble — they are dissolved into the fusion extract, and precipitation happens only later with the specific test reagent.
MCQ 2Direct ApplicationPractice
An organic compound known to contain both nitrogen and sulphur is subjected to sodium fusion, and the extract is treated with ferric chloride solution. The observation is:
Show answer and why every option is right or wrong
Answer: C. C is correct. When N and S are both present in the same compound, sodium fusion yields sodium thiocyanate, and Fe³⁺ then gives the blood-red ferric thiocyanate complex instead of Prussian blue — NCERT Class 11 Chemistry Chapter 8, page 285.
Why A is wrong: A is wrong because Prussian blue requires free cyanide ion in the extract; with sulphur also present the nitrogen leaves the fusion as thiocyanate, so no ferrocyanide forms.
Why B is wrong: B is wrong because the black precipitate belongs to the lead acetate test for sulphide, a separate test on a separate portion; ferric chloride does not give it.
Why D is wrong: D is wrong because no cancellation occurs — a definite and characteristic blood-red colour appears, and reading it as a negative result is the trap this question tests.
MCQ 3Easy RecallPractice
Before testing the sodium fusion extract for a halogen with silver nitrate, the extract is acidified with dilute nitric acid and boiled. Boiling is necessary because:
Show answer and why every option is right or wrong
Answer: C. C is correct. If the compound contains N or S, acidification alone leaves cyanide and sulphide in solution, and both give silver precipitates that confuse the halogen result; boiling drives off HCN and H₂S first — NCERT Class 11 Chemistry Chapter 8, page 285.
Why A is wrong: A is wrong because silver nitrate is perfectly stable in cold acidified solution and precipitates silver halide readily without heating; the boil targets the interfering ions, not the reagent.
Why B is wrong: B is wrong because the silver halides are deliberately insoluble — that insolubility is the whole basis of the test — and raising their solubility would destroy the observation.
Why D is wrong: D is wrong because sodium halides are fully ionised in aqueous solution at room temperature; the halide ion is already free before any heating.
MCQ 4Direct ApplicationPractice
A sodium fusion extract is acidified and treated with silver nitrate. A pale yellow precipitate forms which is only sparingly soluble in ammonium hydroxide. The halogen present is:
Show answer and why every option is right or wrong
Answer: A. A is correct. Silver bromide is the pale yellow precipitate and is described as sparingly soluble in ammonium hydroxide, which distinguishes it from white AgCl (freely soluble) and yellow AgI (insoluble) — NCERT Class 11 Chemistry Chapter 8, page 285.
Why B is wrong: B is wrong because chlorine gives a white precipitate of AgCl that dissolves freely in ammonium hydroxide, not a pale yellow one of limited solubility.
Why C is wrong: C is wrong because silver fluoride is soluble in water and gives no precipitate at all in this test, so fluorine cannot be detected by the silver nitrate method.
Why D is wrong: D is wrong because iodine gives a yellow AgI precipitate that is insoluble in ammonium hydroxide; the stated partial solubility rules it out.
MCQ 5Easy RecallPractice
Phosphorus in an organic compound is detected by heating the fused mass with concentrated nitric acid and then adding ammonium molybdate. The positive observation is:
Show answer and why every option is right or wrong
Answer: B. B is correct. The compound is heated with sodium peroxide to convert phosphorus to phosphate, and ammonium molybdate in nitric acid then gives the yellow ammonium phosphomolybdate precipitate — NCERT Class 11 Chemistry Chapter 8, page 285.
Why A is wrong: A is wrong because blood-red is the ferric thiocyanate colour seen when nitrogen and sulphur are both present, and it has nothing to do with the molybdate reagent.
Why C is wrong: C is wrong because the black precipitate in this chapter is lead sulphide from the lead acetate test for sulphur; lead phosphate is not the phosphorus test product.
Why D is wrong: D is wrong because the violet colouration with sodium nitroprusside is the alternative test for sulphide ion, not for phosphate.
MCQ 6Direct ApplicationPractice
A portion of a sodium fusion extract is treated with sodium nitroprusside and a violet colouration appears. A second portion, acidified with acetic acid and treated with lead acetate, gives a black precipitate. These two observations together establish the presence of:
Show answer and why every option is right or wrong
Answer: D. D is correct. Both the violet nitroprusside colouration and the black lead sulphide precipitate are tests for sulphide ion in the extract; they confirm sulphur and say nothing about the other elements — NCERT Class 11 Chemistry Chapter 8, pages 284–285.
Why A is wrong: A is wrong because neither reagent responds to cyanide ion; nitrogen would require the ferrous sulphate and ferric chloride sequence giving Prussian blue.
Why B is wrong: B is wrong because halogens are detected with acidified silver nitrate after boiling; lead acetate and nitroprusside give no halide reaction.
Why C is wrong: C is wrong because a joint N-and-S result would announce itself as the blood-red thiocyanate colour with ferric chloride, which was not observed here.
MCQ 7CalculationPractice
An organic compound gives a blood-red colouration on treating the sodium fusion extract with ferric chloride. A fresh portion of the same extract is then acidified with dilute nitric acid, boiled, and treated with silver nitrate, giving a yellow precipitate insoluble in ammonium hydroxide. The elements present in the compound are:
Show answer and why every option is right or wrong
Answer: A. A is correct. Blood-red with ferric chloride means thiocyanate, which forms only when nitrogen and sulphur are both present; the yellow silver precipitate insoluble in ammonium hydroxide is AgI, so iodine is the third element — NCERT Class 11 Chemistry Chapter 8, pages 284–285.
Why B is wrong: B is wrong because it discards nitrogen, and bromine's AgBr is pale yellow and sparingly soluble in ammonium hydroxide rather than yellow and insoluble.
Why C is wrong: C is wrong on both counts: it drops the sulphur that the blood-red colour requires, and chlorine would give a white AgCl precipitate that dissolves freely in ammonium hydroxide.
Why D is wrong: D is wrong because it reads the blood-red as evidence against nitrogen; thiocyanate contains nitrogen by definition, so the colour confirms nitrogen rather than excluding it.
MCQ 8CalculationPractice
A student tests a compound containing nitrogen and a halogen. The sodium fusion extract is acidified with dilute nitric acid but not boiled, and silver nitrate is added at once. A white precipitate appears, and the student reports chlorine. The report is unsafe because:
Show answer and why every option is right or wrong
Answer: D. D is correct. When nitrogen is present, the acidified extract retains cyanide ion unless it is boiled; silver cyanide is a white precipitate that cannot be told apart from AgCl by appearance, so the halogen conclusion is unsupported — NCERT Class 11 Chemistry Chapter 8, page 285, has the extract boiled with concentrated nitric acid first for exactly this reason.
Why A is wrong: A is wrong because fluorine simply gives no precipitate, silver fluoride being water-soluble; the difficulty here is an extra precipitate, not a failure to discriminate.
Why B is wrong: B is wrong because silver sulphide is black, not white, and the stem specifies nitrogen and a halogen rather than sulphur.
Why C is wrong: C is wrong because nitric acid does not alter bromide's identity, and AgBr is pale yellow whether or not the extract was boiled.
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How do you solve a Qualitative N, S, P and Halogens question? A worked example
- 1
Given.
An organic compound gives these results on Lassaigne's test:• Sodium fusion extract + FeSO₄, then H₂SO₄ and FeCl₃ → blood-red colouration.• A second portion + sodium nitroprusside → violet colouration.• A third portion, acidified with dilute HNO₃ and boiled, + AgNO₃ → pale yellow precipitate, sparingly soluble in NH₄OH.
- 2
Required.
Name every element detected, and state the ionic species responsible for each observation.
- 3
Concept.
Sodium fusion converts covalent heteroatoms into ionic sodium salts in a single melt. Because all the heteroatoms enter the same melt, their products can combine: N and S together give thiocyanate rather than separate cyanide and sulphide. Each subsequent test is therefore read as a statement about the melt's composition, not about one element in isolation.
- 4
Formula.
No algebraic formula applies. The governing relations are the three reaction identities:• Fe³⁺ + SCN⁻ → blood-red [Fe(SCN)]²⁺ (N and S present)• S²⁻ + sodium nitroprusside → violet complex (S present)• Ag⁺ + Br⁻ → pale yellow AgBr, sparingly soluble in NH₄OH (Br present)
- 5
Substitution.
Match each observation to its identity:• Blood-red with Fe³⁺ → thiocyanate ion present → nitrogen and sulphur both present.• Violet with nitroprusside → sulphide ion present → sulphur confirmed independently.• Pale yellow AgX, sparingly soluble in NH₄OH → bromide ion → bromine present.
- 6
Calculation.
Count the distinct elements implied. Thiocyanate SCN⁻ carries one N and one S, so it accounts for two elements at once; the nitroprusside result adds no new element but corroborates sulphur; the silver test adds bromine. Total: three elements. Note that the subscripts in SCN⁻, Fe³⁺ and AgBr are exact counting integers and the charges are exact — none of them is a measured quantity, so no significant-figure reasoning applies anywhere in this problem.
- 7
Final answer.
Nitrogen, sulphur and bromine are present. Nitrogen and sulphur are established jointly by the blood-red ferric thiocyanate; sulphur is separately confirmed by the violet nitroprusside colour; bromine is established by the pale yellow AgBr that is only sparingly soluble in ammonium hydroxide.
- 8
Common trap.
Two traps sit in this single problem. The first is reading blood-red as "nitrogen absent" — it is in fact the only positive nitrogen result available when sulphur shares the molecule. The second is the halogen colour ladder: a candidate who has memorised "yellow = iodine" may see "pale yellow" and answer iodine, ignoring the stated solubility. Ammonium-hydroxide behaviour is the discriminator NEET leans on: freely soluble is chloride, sparingly soluble is bromide, insoluble is iodide.
- 9
Similar NEET-style question.
An organic compound gives Prussian blue with ferric chloride after the standard ferrous sulphate treatment, no colouration with sodium nitroprusside, and a white precipitate with acidified boiled extract and silver nitrate which dissolves readily in ammonium hydroxide. Identify the elements present and state why the extract had to be boiled before the silver nitrate test. *(Answer: nitrogen and chlorine; boiling expelled HCN so that no white AgCN could be mistaken for AgCl.)*
What to remember before solving Qualitative N, S, P and Halogens questions
Key Fact
Detection of carbon and hydrogen
Carbon and hydrogen are detected by heating the compound with copper(II) oxide. Carbon present in the compound is oxidised to carbon dioxide (tested with lime-water, which develops turbidity) and hydrogen to water (tested with anhydrous copper sulphate, which turns blue). C + 2CuO --(heat)--> 2Cu + CO2 H2 + CuO --(heat)--> Cu + H2O CO2 + Ca(OH)2 -----> CaCO3(precipitate) + H2O 5H2O + CuSO4 -----> CuSO4.5H2O (White) (Blue)
-- NCERT Class 11 Chemistry, Ch. 8, p. 284Nitrogen, sulphur, halogens and phosphorus present in an organic compound are detected by "Lassaigne's test". The elements present in the compound are converted from covalent form into the ionic form by fusing the compound with sodium metal. Following reactions take place: Na + C + N --(heat)--> NaCN 2Na + S --(heat)--> Na2S Na + X --(heat)--> Na X (X = Cl, Br or I) C, N, S and X come from organic compound. Cyanide, sulphide and halide of sodium so formed on sodium fusion are extracted from the fused mass by boiling it with distilled water. This extract is known as sodium fusion extract.
-- NCERT Class 11 Chemistry, Ch. 8, p. 284Key Fact
Test for nitrogen
The sodium fusion extract is boiled with iron(II) sulphate and then acidified with concentrated sulphuric acid. The formation of Prussian blue colour confirms the presence of nitrogen. Sodium cyanide first reacts with iron(II) sulphate and forms sodium hexacyanidoferrate(II). On heating with concentrated sulphuric acid some iron(II) ions are oxidised to iron(III) ions which react with sodium hexacyanidoferrate(II) to produce iron(III) hexacyanidoferrate(II) (ferriferrocyanide) which is Prussian blue in colour. 6CN- + Fe2+ -> [Fe(CN)6]4- 3[Fe(CN)6]4- + 4Fe3+ --(xH2O)--> Fe4[Fe(CN)6]3.xH2O Prussian blue
-- NCERT Class 11 Chemistry, Ch. 8, p. 285Key Fact
Test for sulphur
(a) The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur. S2- + Pb2+ -----> PbS Black (b) On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour further indicates the presence of sulphur. S2- + [Fe(CN)5NO]2- -----> [Fe(CN)5NOS]4- Violet In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free cyanide ions. Na + C + N + S -----> NaSCN Fe3+ + SCN- -----> [Fe(SCN)]2+ Blood red If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide. These ions give their usual tests. NaSCN + 2Na -----> NaCN + Na2S
-- NCERT Class 11 Chemistry, Ch. 8, p. 285Key Fact
Test for halogens
The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine. X- + Ag+ -----> AgX X represents a halogen - Cl, Br or I. If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne's test. These ions would otherwise interfere with silver nitrate test for halogens.
-- NCERT Class 11 Chemistry, Ch. 8, p. 285Key Fact
Test for phosphorus
The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus. Na3PO4 + 3HNO3 -----> H3PO4 + 3NaNO3 H3PO4 + 12(NH4)2MoO4 + 21HNO3 -----> (NH4)3PO4.12MoO3 + 21NH4NO3 + 12H2O Ammonium molybdate Ammonium phosphomolybdate
-- NCERT Class 11 Chemistry, Ch. 8, p. 285Qualitative N, S, P and Halogens questions from past NEET papers
2 questions from NEET 2023, 2025. Answers verified against NTA official keys.
NEET 2025
Which one of the following reactions does NOT belong to “Lassaigne’s test”?
12CuO + C —[Δ]→ 2Cu + CO₂
2Na + C + N —[Δ]→ NaCN
32Na + S —[Δ]→ Na₂S
4Na + X —[Δ]→ NaX
NTA Answer: Option 1(final)
NEET 2023
1NaSCN
2[Fe(CN)₅NOS]⁴⁻
3[Fe(SCN)]²⁺
4Fe₄[Fe(CN)₆]₃·xH₂O
NTA Answer: Option 3(final)
All 7 past-paper questions from Purification and Characterisation of Organic Compounds →
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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