Chemistry · Salt Analysis

Anion Tests — Concentrated Acid Group (Cl⁻, Br⁻, I⁻, NO₃⁻, C₂O₄²⁻)

After the dilute-acid radicals have been screened, the second stage of anion analysis treats the salt with warm concentrated sulphuric acid. Following the NCERT Laboratory Manual (Unit 7, Step-II), this group — chloride, bromide, iodide, nitrate and oxalate — reveals itself through coloured vapours and characteristic fumes, then is locked down by confirmatory tests such as the brown ring, chromyl chloride and silver-halide reactions. For NEET, this is high-yield: vapour colours, ammonia solubilities and the brown-ring mechanism are repeatedly tested.

Why a separate concentrated-acid group

Anions in qualitative analysis are not all probed in one step. The radicals that react even with dilute sulphuric acid — carbonate, sulphite, sulphide, nitrite and acetate — are screened first because they evolve gas with minimal provocation. Only when that test is negative does the analyst move to Step-II of the NCERT scheme: heating roughly 0.1 g of the salt with three to four drops of concentrated sulphuric acid, observing first in the cold and then on warming.

The five anions of this group — chloride, bromide, iodide, nitrate and oxalate — share one property: they are unreactive towards dilute acid but liberate their characteristic vapour when concentrated H₂SO₄ supplies both a non-volatile strong acid and, for the halides and nitrate, an oxidising environment. The result is a parade of colours that an experienced eye can read almost immediately, before any confirmatory test is run.

NEET Trap

Do not run this group before the dilute-acid test

A common procedural error is to reach straight for concentrated H₂SO₄. If a carbonate or sulphite is present, it will also fizz, and the brown nitrate fumes can be confused with nitrite. The NCERT sequence is deliberate: dilute acid first, concentrated acid only if that is negative.

Order matters: dilute-acid radicals (Step-I) → concentrated-acid radicals (Step-II) → sulphate/phosphate (Step-III).

Preliminary test: reading the vapours

The preliminary examination with concentrated sulphuric acid is a colour-and-smell triage. Each anion gives a distinct effect, and adding solid manganese dioxide (an oxidiser) or copper turnings intensifies the diagnostic change for the halides and nitrate respectively. The table below condenses Table 7.3 of the NCERT manual.

AnionVapour / change with conc. H₂SO₄Gas evolvedEffect of additive
Chloride, Cl⁻Colourless, pungent gas; dense white fumes when a rod dipped in NH₄OH is held at the tube mouth$\ce{HCl}$With MnO₂: greenish-yellow Cl₂
Bromide, Br⁻Reddish-brown, pungent gas; solution acquires red colour$\ce{Br2}$ vapourWith MnO₂: red fumes intensify
Iodide, I⁻Deep violet vapours; violet sublimate on the tube walls; turn starch paper blue$\ce{I2}$ vapourWith MnO₂: violet fumes denser
Nitrate, NO₃⁻Light-brown fumes; on warming with copper, dense brown fumes, solution turns blue$\ce{NO2}$With Cu turnings: excess brown fumes
Oxalate, C₂O₄²⁻Colourless, odourless gas; turns lime water milky; burns with a blue flame if ignited$\ce{CO + CO2}$Blue flame at tube mouth confirms CO

Chloride (Cl⁻)

On warming a chloride with concentrated sulphuric acid, the non-volatile acid displaces volatile hydrogen chloride. The gas is colourless with a sharp pungent smell, and the classic confirmation is to hold a glass rod moistened with ammonia solution near the mouth of the tube — dense white fumes of ammonium chloride form.

The displacement and the ammonia test are written as:

$$\ce{NaCl + H2SO4 -> NaHSO4 + HCl ^}$$

$$\ce{HCl + NH3 -> NH4Cl}\;(\text{white fumes})$$

Adding a pinch of manganese dioxide and warming oxidises the liberated chloride to greenish-yellow chlorine gas, recognised by its pungent odour and bleaching action:

$$\ce{MnO2 + 2NaCl + 2H2SO4 -> Na2SO4 + MnSO4 + 2H2O + Cl2 ^}$$

The wet test uses silver nitrate on the acidified extract: a curdy white precipitate of silver chloride forms, and crucially it dissolves readily in ammonium hydroxide as the diammine silver(I) complex.

$$\ce{NaCl + AgNO3 -> AgCl v + NaNO3}$$

$$\ce{AgCl + 2NH4OH -> [Ag(NH3)2]Cl + 2H2O}$$

Bromide and iodide (Br⁻, I⁻)

Both halides reduce concentrated sulphuric acid, and the depth of colour rises sharply from bromine to iodine. A bromide gives reddish-brown bromine vapours; the fumes intensify with MnO₂ and turn starch paper yellow.

$$\ce{2NaBr + 2H2SO4 -> Br2 + SO2 + Na2SO4 + 2H2O}$$

$$\ce{2NaBr + MnO2 + 2H2SO4 -> Na2SO4 + MnSO4 + 2H2O + Br2}$$

An iodide is an even stronger reducing agent. Heating with concentrated acid gives deep violet iodine vapours and a violet sublimate on the tube walls; the vapour turns starch paper blue. Because iodide reduces the acid in stages, sulphur dioxide, free sulphur and even hydrogen sulphide appear alongside the iodine.

$$\ce{2NaI + 2H2SO4 -> Na2SO4 + SO2 + 2H2O + I2}$$

$$\ce{6NaI + 4H2SO4 -> 3I2 + S + 3Na2SO4 + 4H2O}$$

$$\ce{8NaI + 5H2SO4 -> 4I2 + H2S + 4Na2SO4 + 4H2O}$$

For both halides the cleanest confirmation avoids concentrated acid: add chlorine water to the neutralised extract with a layer of chloroform or carbon tetrachloride. Chlorine, being a stronger oxidant, displaces the free halogen, which dissolves in the organic layer — orange-brown for bromine, violet for iodine.

$$\ce{2NaBr + Cl2 -> 2NaCl + Br2}\qquad \ce{2NaI + Cl2 -> 2NaCl + I2}$$

Build the full anion picture

The other half of anion analysis lives in the dilute-acid anion group — carbonate, sulphite, sulphide, nitrite and acetate.

Silver halides and ammonia solubility

The single most reliable way to tell the three halides apart is the colour of the silver precipitate together with its solubility in ammonia. Silver chloride is white and dissolves easily; silver bromide is pale yellow and dissolves only with difficulty; silver iodide is yellow and does not dissolve at all. This solubility trend is a recurring NEET discriminator.

Figure 1 — Halide confirmation flow Acidified extract + AgNO₃ Curdy white AgCl Pale yellow AgBr Yellow AgI Cl⁻ soluble in NH₃ (readily) Br⁻ soluble in NH₃ (with difficulty) I⁻ insoluble in NH₃

Silver halide colour deepens and ammonia solubility falls down the group: AgCl (white, soluble) → AgBr (pale yellow, sparingly soluble) → AgI (yellow, insoluble).

NEET Trap

"Yellow precipitate" alone does not mean iodide

Both AgBr and AgI look yellowish, and silver phosphate is also yellow. The ammonia test is the deciding factor: AgBr dissolves slowly in ammonia while AgI does not dissolve at all. Never call iodide on colour alone.

Confirm I⁻ by: yellow AgI insoluble in NH₄OH + violet layer with chlorine water + blue starch.

Nitrate (NO₃⁻) and the brown-ring test

A nitrate heated with concentrated sulphuric acid alone gives only light-brown fumes; the diagnostic intensification comes from adding copper turnings, which reduces the liberated nitric acid to nitric oxide that is air-oxidised to brown nitrogen dioxide, while copper sulphate turns the solution blue.

$$\ce{NaNO3 + H2SO4 -> NaHSO4 + HNO3}$$

$$\ce{2NaNO3 + 4H2SO4 + 3Cu -> 3CuSO4 + Na2SO4 + 4H2O + 2NO}$$

$$\ce{2NO + O2 -> 2NO2}\;(\text{brown fumes})$$

The confirmatory brown-ring test is gentler. To an aqueous solution of the salt, freshly prepared ferrous sulphate is added; concentrated H₂SO₄ is then poured carefully down the side of the tilted tube so it sinks below the lighter aqueous layer. At the junction, nitric acid produced by the acid oxidises some Fe²⁺ while being reduced to NO, and the NO combines with surplus ferrous sulphate to form the brown nitroso-ferrous complex.

$$\ce{6FeSO4 + 3H2SO4 + 2HNO3 -> 3Fe2(SO4)3 + 4H2O + 2NO}$$

$$\ce{FeSO4 + NO -> [Fe(NO)]SO4}\;(\text{brown})$$

Figure 2 — Brown-ring test layering aqueous salt + FeSO₄ brown ring [Fe(NO)]SO₄ conc. H₂SO₄ (dense, lower)

Concentrated acid is run down the side so it forms a dense lower layer; the brown nitroso-ferrous complex appears as a ring exactly at the junction of the two liquids.

Oxalate (C₂O₄²⁻)

An oxalate heated with concentrated sulphuric acid is dehydrated and decarboxylated, releasing a mixture of carbon dioxide and carbon monoxide. The CO₂ turns lime water milky; the CO is recognised because it burns with a blue flame at the mouth of the tube. The combination of the two gases is itself a strong indication of oxalate.

$$\ce{(COONa)2 + conc.\,H2SO4 -> Na2SO4 + H2O + CO2 ^ + CO ^}$$

Confirmation rests on the reducing power of oxalate. The sodium carbonate extract is acidified with acetic acid and treated with calcium chloride to precipitate white calcium oxalate, which is insoluble in oxalic acid. That precipitate is then dissolved in dilute H₂SO₄ and warmed with dilute potassium permanganate — the pink colour is discharged as oxalate is oxidised to carbon dioxide.

$$\ce{CaCl2 + Na2C2O4 -> CaC2O4 v + 2NaCl}$$

$$\ce{2KMnO4 + 3H2SO4 + 5H2C2O4 -> 2MnSO4 + K2SO4 + 8H2O + 10CO2}$$

NEET Trap

Oxalate vs carbonate — both give CO₂

Carbonate effervesces with dilute acid and gives only CO₂. Oxalate is silent with dilute acid and gives CO and CO₂ with concentrated acid; the tell-tale extra is the blue flame of carbon monoxide. The KMnO₄ decolourisation finally separates oxalate, since carbonate does not reduce permanganate.

Blue flame of CO at the tube mouth + warm acidified KMnO₄ decolourised ⇒ oxalate.

Chromyl chloride test for chloride

The chromyl chloride test is the most specific confirmation for chloride and is a NEET favourite. A chloride is mixed with solid potassium dichromate and concentrated sulphuric acid and heated; deep red vapours of volatile chromyl chloride, $\ce{CrO2Cl2}$, are evolved. Bromides and iodides do not form an analogous stable volatile compound, so the red vapour is diagnostic of chloride alone.

$$\ce{4NaCl + K2Cr2O7 + 6H2SO4 -> 2CrO2Cl2 ^ + 2KHSO4 + 4NaHSO4 + 3H2O}$$

The red vapour is passed into sodium hydroxide to give a yellow sodium chromate solution. Acidifying part of this with acetic acid and adding lead acetate precipitates yellow lead chromate, completing the confirmation.

$$\ce{CrO2Cl2 + 4NaOH -> Na2CrO4 + 2NaCl + 2H2O}$$

$$\ce{(CH3COO)2Pb + Na2CrO4 -> PbCrO4 v + 2CH3COONa}$$

NEET Trap

Chromyl chloride is for chloride only

The test must use a minimum amount of substance to limit Cr(III) pollution, and it succeeds only with chloride. Examiners pair it with a bromide or iodide and ask which gives the red vapour — the answer is always chloride.

Red CrO₂Cl₂ vapour → yellow Na₂CrO₄ in NaOH → yellow PbCrO₄ with lead acetate = chloride confirmed.

Observation → inference master table

The whole concentrated-acid group can be navigated from a single decision table. Read the vapour colour first, then run the confirmatory wet test that matches.

ObservationInferenceConfirmatory test
Colourless pungent gas; white fumes with NH₃ rodCl⁻Curdy white AgCl soluble in NH₄OH; chromyl chloride red vapour
Reddish-brown vapour; solution turns red; starch paper yellowBr⁻Pale-yellow AgBr sparingly soluble in NH₄OH; orange-brown organic layer with Cl₂ water
Deep violet vapour; violet sublimate; starch paper blueI⁻Yellow AgI insoluble in NH₄OH; violet organic layer with Cl₂ water
Light-brown fumes; dense brown fumes + blue solution with CuNO₃⁻Brown-ring test with FeSO₄ and conc. H₂SO₄
Colourless odourless gas; lime water milky; CO burns blueC₂O₄²⁻White CaC₂O₄ insoluble in oxalic acid; warm KMnO₄ decolourised
Quick Recap

Concentrated-acid anion group at a glance

  • Tested only after the dilute-acid group is negative; reagent is warm concentrated H₂SO₄.
  • Vapour colours: Cl⁻ colourless HCl (white fumes with NH₃), Br⁻ reddish-brown, I⁻ violet, NO₃⁻ brown NO₂, C₂O₄²⁻ colourless CO + CO₂ (blue flame).
  • Silver-halide ammonia solubility: AgCl readily soluble > AgBr sparingly soluble > AgI insoluble.
  • Brown ring = nitroso-ferrous complex [Fe(NO)]SO₄ at the junction of FeSO₄ and dense conc. H₂SO₄.
  • Chromyl chloride (red CrO₂Cl₂) confirms chloride only — bromide and iodide do not respond.
  • Oxalate decolourises warm acidified KMnO₄; carbonate does not, separating the two CO₂-givers.

NEET PYQ Snapshot — Anion Tests (Concentrated Acid Group)

No NEET PYQ has yet targeted this specific anion subtopic; the cards below are concept-level questions in the exam's style.

Concept

A salt warmed with concentrated H₂SO₄ gives a colourless pungent gas that forms dense white fumes with a rod dipped in ammonia. The anion is most likely:

  • (1) Br⁻
  • (2) NO₃⁻
  • (3) Cl⁻
  • (4) C₂O₄²⁻
Answer: (3) Cl⁻

Concentrated H₂SO₄ displaces volatile HCl, which gives white NH₄Cl fumes with ammonia: $\ce{HCl + NH3 -> NH4Cl}$.

Concept

Three salts give silver precipitates: white soluble in ammonia, pale yellow sparingly soluble, and yellow insoluble. The anions in order are:

  • (1) Cl⁻, Br⁻, I⁻
  • (2) I⁻, Br⁻, Cl⁻
  • (3) Br⁻, Cl⁻, I⁻
  • (4) Cl⁻, I⁻, Br⁻
Answer: (1) Cl⁻, Br⁻, I⁻

Ammonia solubility falls down the group: AgCl readily soluble, AgBr sparingly soluble, AgI insoluble.

Concept

The brown ring in the nitrate confirmatory test is due to the formation of:

  • (1) FeSO₄·7H₂O
  • (2) [Fe(NO)]SO₄
  • (3) Fe₂(SO₄)₃
  • (4) Fe(OH)₃
Answer: (2) [Fe(NO)]SO₄

NO from reduced nitric acid combines with surplus FeSO₄ to give the brown nitroso-ferrous complex: $\ce{FeSO4 + NO -> [Fe(NO)]SO4}$.

Concept

Heating a salt with K₂Cr₂O₇ and concentrated H₂SO₄ gives deep red vapours. This confirms the presence of:

  • (1) iodide
  • (2) bromide
  • (3) chloride
  • (4) nitrate
Answer: (3) chloride

The red vapour is chromyl chloride, CrO₂Cl₂; only chloride forms it, making the test specific.

FAQs — Anion Tests (Concentrated Acid Group)

Quick answers to the recurring doubts on the concentrated-H₂SO₄ anion group.

Why is the concentrated-acid group tested only after the dilute-acid group?
The dilute sulphuric acid group (carbonate, sulphite, sulphide, nitrite, acetate) reacts even with dilute acid, so it is screened first. The concentrated-acid group — chloride, bromide, iodide, nitrate and oxalate — gives no positive result with dilute acid; these radicals release their characteristic vapours only when heated with concentrated H₂SO₄. Testing in this order avoids wasting the more reactive concentrated acid on radicals already identified.
How do silver halide solubilities in ammonia distinguish chloride, bromide and iodide?
Add silver nitrate to the acidified extract. Chloride gives a curdy white AgCl that is readily soluble in ammonium hydroxide (forming the diammine silver(I) complex). Bromide gives a pale-yellow AgBr that dissolves only with difficulty in ammonia. Iodide gives a yellow AgI that is insoluble in ammonia. Solubility therefore falls sharply down the group: AgCl > AgBr > AgI.
What causes the brown ring in the nitrate confirmatory test?
Concentrated H₂SO₄ liberates nitric acid from the nitrate, which oxidises some Fe²⁺ to Fe³⁺ and is itself reduced to nitric oxide, NO. The NO then combines with unreacted ferrous sulphate to form the nitroso-ferrous complex [Fe(NO)]SO₄, which is brown. Because the acid layer sits below the aqueous layer, the complex forms only at the junction, appearing as a dark brown ring.
Why does the chromyl chloride test confirm chloride but not bromide or iodide?
Heating a chloride with potassium dichromate and concentrated H₂SO₄ produces volatile red chromyl chloride, CrO₂Cl₂. Bromides and iodides do not form a corresponding stable volatile chromyl compound under these conditions, so they give no red vapour. The chromyl chloride is therefore specific to chloride; the vapour passed into NaOH gives a yellow chromate that, with lead acetate, yields a yellow lead chromate precipitate.
How is oxalate distinguished from carbonate when both can release CO₂?
Oxalate is part of the concentrated-acid group and gives a mixture of carbon monoxide and carbon dioxide with concentrated H₂SO₄; the CO burns with a blue flame at the tube mouth, whereas carbonate (a dilute-acid radical) gives only CO₂ with dilute acid. Confirmation: oxalate decolourises warm acidified KMnO₄, being oxidised to CO₂, while the calcium oxalate precipitate is insoluble in oxalic acid.
Why does heating an iodide with concentrated H₂SO₄ also release sulphur and H₂S?
Iodide is a strong reducing agent. Beyond liberating violet iodine vapour, it reduces concentrated H₂SO₄ progressively — to SO₂, then to sulphur, and finally to hydrogen sulphide — so the reaction mixture darkens and smells of rotten eggs. This is why the silver nitrate and chlorine-water confirmatory tests, which avoid concentrated acid, give cleaner evidence for iodide.