Qualitative analysis of anions is a staple of NEET's practical chemistry section. The core task: given an unknown salt, identify the anion present using systematic wet tests. Where aspirants lose marks is not in forgetting the tests themselves, but in confusing anions that produce similar observations — gases with overlapping smell or colour, precipitates with similar appearance, or confirmatory tests that share a reagent.
The systematic approach. Anion analysis follows a fixed sequence: preliminary tests (heating, flame test, borax bead), then group reagent tests with dilute H₂SO₄ and concentrated H₂SO₄, followed by confirmatory (wet) tests specific to each anion.
Group A — Anions that react with dilute H₂SO₄. CO₃²⁻ gives brisk effervescence with CO₂ (turns lime water milky). S²⁻ gives H₂S (rotten-egg smell, turns lead acetate paper black). SO₃²⁻ gives SO₂ (suffocating smell, turns acidified K₂Cr₂O₇ green). NO₂⁻ gives brown fumes of NO₂. CH₃COO⁻ gives vinegar smell.
Group B — Anions that react only with concentrated H₂SO₄. Cl⁻ gives white fumes of HCl (dense white fumes with NH₃). Br⁻ gives reddish-brown vapour of Br₂. I⁻ gives violet vapour of I₂. NO₃⁻ gives brown fumes (ring test confirms — brown ring of FeSO₄·NO at junction). C₂O₄²⁻ gives CO₂ on heating with conc. H₂SO₄ plus MnO₂.
Confirmatory tests matter. The preliminary observation narrows candidates, but the confirmatory test clinches the identity. For example, both CO₃²⁻ and HCO₃⁻ give CO₂ with acid — the distinguishing test uses MgSO₄ solution (white precipitate only with CO₃²⁻, not HCO₃⁻). SO₄²⁻ is confirmed with BaCl₂ (white precipitate of BaSO₄, insoluble in conc. HCl), while SO₃²⁻ gives BaSO₃ that dissolves in dilute HCl. PO₄³⁻ is confirmed by ammonium molybdate (canary-yellow precipitate on warming).
Watch-out for NEET: questions often list four anions and ask which gives a specific observation — they rely on you confusing SO₃²⁻ with SO₄²⁻, or CO₃²⁻ with HCO₃⁻, or Cl⁻ with Br⁻. The discriminating detail is always in the confirmatory test, not the preliminary observation.
(Ref: NCERT Chemistry Lab Manual Class 12, Unit 7, page 49 — exercise on qualitative analysis.)
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
Which anion, on treatment with dilute H₂SO₄, produces a gas that turns lime water milky?
Answer: A. CO₃²⁻ reacts with dilute H₂SO₄ to release CO₂, which turns lime water (Ca(OH)₂ solution) milky due to formation of CaCO₃. (NCERT Chemistry Lab Manual Class 12, Unit 7, page 53.)
Why B is wrong: B is wrong because SO₃²⁻ releases SO₂ with dilute H₂SO₄, which has a suffocating smell and turns acidified K₂Cr₂O₇ green — it does not turn lime water milky.
Why C is wrong: C is wrong because S²⁻ releases H₂S (rotten-egg smell) with dilute H₂SO₄, which turns lead acetate paper black, not lime water milky.
Why D is wrong: D is wrong because NO₂⁻ releases brown fumes of NO₂ with dilute H₂SO₄, not a gas that turns lime water milky.
The confirmatory test for SO₄²⁻ involves adding BaCl₂ solution. What is observed?
Answer: B. BaCl₂ + SO₄²⁻ → BaSO₄ (white precipitate). BaSO₄ is insoluble in concentrated HCl, which distinguishes it from BaSO₃ (formed with SO₃²⁻, soluble in dilute HCl). (NCERT Class 12 Chemistry, Chapter 4.)
Why A is wrong: A is wrong because a yellow precipitate is not produced; BaSO₄ is white. A yellow precipitate with a barium reagent would suggest chromate, not sulphate.
Why C is wrong: C is wrong because BaSO₄ is insoluble in dilute HCl. It is BaSO₃ (from SO₃²⁻) that dissolves in dilute HCl — confusing sulphate with sulphite is a common error.
Why D is wrong: D is wrong because SO₄²⁻ does react with BaCl₂ to give a clearly visible white precipitate; 'no visible change' would indicate the anion is not SO₄²⁻ or SO₃²⁻.
When a salt containing CH₃COO⁻ is warmed with dilute H₂SO₄, the observation is:
Answer: B. CH₃COO⁻ reacts with dilute H₂SO₄ to release acetic acid (CH₃COOH), which has the characteristic smell of vinegar. (NCERT Class 12 Chemistry, Chapter 4.)
Why A is wrong: A is wrong because brown fumes are characteristic of NO₂ (from NO₂⁻ with dilute H₂SO₄ or NO₃⁻ with concentrated H₂SO₄), not acetate.
Why C is wrong: C is wrong because violet vapour is characteristic of I₂ released when I⁻ reacts with concentrated H₂SO₄, not from acetate.
Why D is wrong: D is wrong because rotten-egg smell is characteristic of H₂S released from S²⁻ with dilute H₂SO₄, not from acetate.
A student adds dilute H₂SO₄ to an unknown salt and observes a gas with a suffocating smell that turns acidified potassium dichromate solution green. The anion present is:
Answer: C. SO₃²⁻ with dilute H₂SO₄ releases SO₂ (suffocating smell). SO₂ reduces Cr₂O₇²⁻ (orange) to Cr³⁺ (green) in acidified K₂Cr₂O₇, which is the confirmatory observation. (NCERT Class 12 Chemistry, Chapter 4.)
Why A is wrong: A is wrong because CO₃²⁻ releases CO₂, which is odourless and turns lime water milky — it does not have a suffocating smell and does not reduce dichromate.
Why B is wrong: B is wrong because S²⁻ releases H₂S (rotten-egg smell, not suffocating). H₂S turns lead acetate paper black but does not turn acidified K₂Cr₂O₇ green in the same characteristic way as SO₂.
Why D is wrong: D is wrong because NO₃⁻ does not react with dilute H₂SO₄ readily; it requires concentrated H₂SO₄ to release brown NO₂ fumes, which do not turn K₂Cr₂O₇ green.
Two salts, X and Y, both produce effervescence with dilute HCl. Salt X gives a white precipitate with MgSO₄ solution, but salt Y does not. The anions in X and Y are, respectively:
Answer: D. Both CO₃²⁻ and HCO₃⁻ release CO₂ with dilute acid (effervescence). MgSO₄ gives a white precipitate (MgCO₃) with CO₃²⁻ but NOT with HCO₃⁻ (Mg(HCO₃)₂ is soluble). So X = CO₃²⁻ and Y = HCO₃⁻. (NCERT Class 12 Chemistry, Chapter 4.)
Why A is wrong: A is wrong because the order is reversed: HCO₃⁻ does NOT give a precipitate with MgSO₄, so HCO₃⁻ cannot be X (the one that gives the white precipitate).
Why B is wrong: B is wrong because SO₃²⁻ releases SO₂ (suffocating smell), not CO₂, with dilute acid. The question describes effervescence (CO₂), which points to carbonate or bicarbonate.
Why C is wrong: C is wrong because SO₃²⁻ releases SO₂ with dilute acid (not effervescence in the CO₂ sense), and the MgSO₄ test distinguishes CO₃²⁻ from HCO₃⁻, not from SO₃²⁻.
A salt is treated with concentrated H₂SO₄. Dense white fumes are produced, which give thick white fumes when a glass rod dipped in NH₃ solution is brought near. The anion present is:
Answer: C. Cl⁻ with concentrated H₂SO₄ releases HCl gas (dense white fumes). HCl reacts with NH₃ to form NH₄Cl — thick white fumes at the rod. This is the chromyl chloride / HCl confirmatory sequence for chloride. (NCERT Class 12 Chemistry, Chapter 4.)
Why A is wrong: A is wrong because Br⁻ with concentrated H₂SO₄ produces reddish-brown vapour of Br₂, not white fumes. Br₂ does not give white fumes with NH₃.
Why B is wrong: B is wrong because NO₃⁻ with concentrated H₂SO₄ produces brown fumes of NO₂, not white fumes. The NH₃ test (white fumes) is specific to HCl.
Why D is wrong: D is wrong because I⁻ with concentrated H₂SO₄ produces violet vapour of I₂, not white fumes. I₂ is visually distinct and does not react with NH₃ to give white fumes.
A student performs the ring test on a salt solution. A brown ring forms at the junction of two layers. Which anion does this confirm?
Answer: A. The brown ring test is the confirmatory test for NO₃⁻. The brown ring is due to the complex [Fe(H₂O)₅(NO)]²⁺ formed at the junction of the concentrated H₂SO₄ layer and the FeSO₄ solution layer. (NCERT Class 12 Chemistry, Chapter 4.)
Why B is wrong: B is wrong because Br⁻ is confirmed by CS₂ extraction (orange layer) or chromyl chloride test (negative for Br⁻), not by the ring test.
Why C is wrong: C is wrong because NO₂⁻ also gives brown fumes with acid and can interfere, but the ring test is specifically the confirmatory test for NO₃⁻. NO₂⁻ is identified separately via its reaction with dilute H₂SO₄ (brown fumes at room temperature) and confirmed by starch-iodide paper turning blue.
Why D is wrong: D is wrong because I⁻ is confirmed by starch solution turning blue (I₂ released), or by CS₂ extraction (violet layer), not by the ring test.
An unknown salt gives (i) no reaction with dilute H₂SO₄, (ii) a canary-yellow precipitate when warmed with ammonium molybdate and concentrated HNO₃, and (iii) a white precipitate with magnesia mixture (MgCl₂ + NH₄Cl + NH₄OH). The anion is:
Answer: D. Step 1: No reaction with dilute H₂SO₄ eliminates Group A anions (CO₃²⁻, S²⁻, SO₃²⁻, NO₂⁻, CH₃COO⁻). Step 2: Canary-yellow precipitate with ammonium molybdate is the confirmatory test for PO₄³⁻ (ammonium phosphomolybdate). Step 3: White precipitate with magnesia mixture (MgNH₄PO₄) further confirms PO₄³⁻. (NCERT Class 12 Chemistry, Chapter 4.)
Why A is wrong: A is wrong because SO₄²⁻ is confirmed by BaCl₂ (white precipitate of BaSO₄ insoluble in conc. HCl), not by ammonium molybdate. SO₄²⁻ does not give a canary-yellow precipitate.
Why B is wrong: B is wrong because Cl⁻ requires concentrated H₂SO₄ to release HCl and is confirmed by the chromyl chloride test or AgNO₃ (white curdy precipitate of AgCl). It does not give a canary-yellow precipitate with ammonium molybdate.
Why C is wrong: C is wrong because CO₃²⁻ reacts with dilute H₂SO₄ (effervescence), but observation (i) says no reaction with dilute H₂SO₄, eliminating CO₃²⁻ at the first step.
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Given
• No reaction with dilute H₂SO₄.• Reddish-brown vapour with concentrated H₂SO₄.• Pale-yellow precipitate with AgNO₃, sparingly soluble in NH₃.
Required
Identify the anion present.
Concept
Anion analysis follows a two-stage approach: (a) preliminary test with dilute then concentrated H₂SO₄ to narrow the group, (b) confirmatory wet test to identify the specific anion. The colour and solubility of the silver halide precipitate distinguishes Cl⁻, Br⁻, and I⁻.
Formula / Rule
• Group A anions (CO₃²⁻, S²⁻, SO₃²⁻, NO₂⁻, CH₃COO⁻) react with dilute H₂SO₄ → eliminated.• Group B: Cl⁻ → white fumes of HCl; Br⁻ → reddish-brown vapour of Br₂; I⁻ → violet vapour of I₂; NO₃⁻ → brown fumes of NO₂.• AgNO₃ confirmatory: AgCl = white, soluble in dilute NH₃; AgBr = pale yellow, sparingly soluble in NH₃; AgI = yellow, insoluble in NH₃.
Substitution / Application
• No gas with dilute H₂SO₄ → anion is NOT in Group A.• Reddish-brown vapour with concentrated H₂SO₄ → candidate is Br⁻ (not Cl⁻ which gives white fumes, not I⁻ which gives violet vapour).• Pale-yellow precipitate with AgNO₃, sparingly soluble in NH₃ → matches AgBr.
Reasoning
Both observations converge on Br⁻. The reddish-brown vapour test narrows to Br⁻, and the AgNO₃ confirmatory test (pale yellow, sparingly soluble in NH₃) clinches it. If the precipitate were white and freely soluble in dilute NH₃, the anion would be Cl⁻; if deep yellow and insoluble, it would be I⁻.
Final answer
The anion is Br⁻ (bromide).
Common trap
Confusing the colour of AgBr (pale yellow) with AgI (yellow). Also confusing the NH₃ solubility: AgCl dissolves freely, AgBr dissolves sparingly, AgI is insoluble. Misremembering this solubility order causes mis-identification.
Similar NEET-style question
"An unknown salt gives violet vapour with concentrated H₂SO₄ and a yellow precipitate with AgNO₃ that is insoluble in NH₃. Identify the anion." (Answer: I⁻ — violet vapour = I₂, yellow AgI insoluble in NH₃.)
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Carbonate (CO3²⁻): dil. HCl → effervescence + lime water turns milky. Sulphate (SO4²⁻): BaCl2 → white ppt insoluble in dil. HCl. Sulphite (SO3²⁻): dil. HCl → SO2 (smell) + decolourised KMnO4. Halides (Cl⁻/Br⁻/I⁻): AgNO3 → white/cream/yellow ppt; Cl⁻ soluble in NH4OH. Nitrate (NO3⁻): brown-ring test (FeSO4 + conc. H2SO4 layered).
-- NCERT Chemistry Lab Manual Class 12, Unit 7, p. 58More in Practical and Analytical Chemistry: 6 exam traps and mistakes · 2 formulas · 1 question pattern from its other lessons.
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