Question bank — Common anions — Cl⁻, Br⁻, I⁻, SO₄²⁻, NO₃⁻, CO₃²⁻ — confirmatory tests

False. BaCO₃ and BaSO₃ are also white ppts. The discriminator is dil. HCl: carbonate and sulphite react with acid (giving CO₂ / SO₂ gas) and their barium salts dissolve, whereas BaSO₄ has an extremely small Ksp ($\approx 1.1\times 10^{-10}$) and simply does not react with or dissolve in dilute acid. So only white surviving after acid confirms SO₄²⁻.

False. CO₃²⁻, SO₃²⁻ and S²⁻ each form pale or coloured silver salts (Ag₂CO₃ pale, Ag₂S black), so a "white-ish" ppt is ambiguous. You must first add dil. HNO₃, which converts carbonate to CO₂ and sulphite to SO₂ (both leave as gas) and keeps sulphide from precipitating, so that a remaining white ppt can only be AgCl.

False. Oxidising Cl⁻ to Cl₂ would require pulling an electron off chloride, but chloride holds its electron most tightly of the three halides, so Cl₂ (the oxidiser you added) has no drive to take it — no reaction, no coloured layer. That very absence of a coloured organic layer is what fingerprints Cl⁻.

False. Dissolving needs NH₃ to steal Ag⁺ into $[{\text{Ag(NH}}_3{)}_2{]}^{+}$, which only works if the salt first frees some Ag⁺ into solution. AgI's Ksp is ~$8.5\times 10^{-17}$, about a million times smaller than AgCl's ~$1.8\times 10^{-10}$, so almost no free Ag⁺ exists for NH₃ to grab — AgI stays solid while AgCl dissolves.

False. Nitrite (NO₂⁻) is also reduced to NO and forms the same brown $[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}$ complex, and it does so even with dilute acid. So a ring alone is ambiguous; nitrate demands the conc. H₂SO₄ version plus a separate step that has already destroyed any nitrite.

False. The white is insoluble CaCO₃. Extra CO₂ pushes the reaction ${\text{CaCO}}_3+{\text{CO}}_2+{\text{H}}_2\text{O}\to {\text{Ca(HCO}}_3{)}_2$, and the bicarbonate is soluble — so more gas actually clears the cloudiness. That reversal is itself a confirming clue.

False. It becomes less soluble, and the Ksp values prove it: AgCl $1.8\times 10^{-10}$, AgBr $5.4\times 10^{-13}$, AgI $8.5\times 10^{-17}$ — falling steeply. The anion grows larger and softer down the group, so it polarises the Ag⁺ more and the Ag–X bond gains covalent character, locking the salt into an ever more insoluble lattice.

False. It is the reverse, I⁻ > Br⁻ > Cl⁻. Oxidation means losing an electron; iodide's outer electron sits farthest from the nucleus and is least tightly held (largest ion), so it departs most easily. That is why Cl₂ can oxidise Br⁻ and I⁻ but nothing weaker can oxidise Cl⁻.

The dil. HCl adds Cl⁻, guaranteeing a false white AgCl. Use dil. HNO₃ instead — it destroys carbonate/sulphite without introducing any halide.

Cream (between white and yellow) points to Br⁻, not I⁻; yellow is iodide. Confirm with Cl₂ water + CCl₄: an orange organic layer (from liberated Br₂) → bromide.

If it dissolved in acid it was not BaSO₄ (which is acid-insoluble) — it was BaCO₃ or BaSO₃, an impostor. Acidify before adding BaCl₂, then insoluble white confirms sulphate.

Mixing destroys the ring. The conc. acid must be poured gently down the side so it sinks to the bottom, and the reaction happens only at the dense boundary layer, forming a localised brown ring (see the layered sketch above).

A burnt-match smell is SO₂ (from a sulphite), not the odourless CO₂ of carbonate. Carbonate's gas is colourless and odourless and turns lime water milky.

AgBr is only partially soluble in dilute NH₃ and AgI is insoluble — non-dissolution is expected for the heavier halides, not a disqualifier. Judge by ppt colour plus the oxidation/layer test.

Acid dissolves the white impostors BaCO₃ and BaSO₃ (they react with H⁺ to release gas), so only acid-insoluble BaSO₄ can remain — the acid acts as a filter that removes look-alikes.

NH₃ complexes Ag⁺ as $[{\text{Ag(NH}}_3{)}_2{]}^{+}$, but only if the salt first releases some free Ag⁺. AgCl's Ksp ($1.8\times 10^{-10}$) is large enough to supply that Ag⁺; AgI's Ksp ($8.5\times 10^{-17}$, a million-fold smaller) releases almost none, so NH₃ has nothing to grab and AgI stays solid.

Its outermost electron is furthest from the nucleus and least tightly held (largest ion), so it is removed most readily, giving I⁻ the highest ease of oxidation.

Conc. H₂SO₄ is dense and sinks, so the reduction of nitrate to NO and formation of the brown $[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}{]}^{2+}$ complex occurs only at the thin boundary where the two layers meet — the geometry of a denser layer under a lighter one forces the colour into a band.

The insoluble CaCO₃ reacts further with CO₂ and water to give soluble calcium bicarbonate, ${\text{Ca(HCO}}_3{)}_2$, redissolving the white suspension.

Increasing anion polarisability adds covalent character to the Ag–X bond, lowering the electronic energy gap so the salt absorbs more visible light — hence the shift from white (Cl) through cream (Br) to yellow (I), tracking the falling Ksp.

Liberated Br₂ and I₂ are far more soluble in the non-polar organic layer, which is also denser than water and sinks, concentrating the colour (orange for Br₂, violet for I₂) into a distinct lower band that is easy to read.

A mixed precipitate whose colour is dominated by the least soluble salt (AgI, yellow, Ksp ~$8.5\times 10^{-17}$); the halides must be separated or confirmed individually by the oxidation/layer test.

Not automatically. Several non-standard conditions can mask the effervescence: (1) too little acid or acid that is too dilute may fizz weakly and be missed; (2) very cold solution keeps CO₂ dissolved (gas is far more soluble at low temperature) so no visible bubbles escape until it is warmed; (3) a very dilute carbonate simply produces too little gas to see. Warm the mixture and use adequate dil. acid before concluding carbonate is absent.

Yes — white + fully NH₃-soluble is AgCl (Cl⁻), and Cl₂ cannot oxidise Cl⁻, so no coloured organic layer is expected. The two clues agree on chloride.

BaSO₄ is insoluble in dil. HCl but BaSO₃ dissolves in it (releasing SO₂, a burnt-match smell), so acidifying separates them. Under alkaline conditions both barium salts persist as white solids, so a raw white ppt at high pH is ambiguous — you must acidify to expose the difference. Beware too that in air sulphite slowly oxidises to sulphate, so an old sulphite sample can start giving a genuine acid-insoluble BaSO₄ as well; test fresh.

Not on its own; nitrite also rings (even with dil. acid). You must first remove/destroy nitrite (e.g. by boiling with urea or sulphamic acid in acid) so that a ring under conc. H₂SO₄ can be attributed to nitrate.

Orange indicates Br₂ (bromide). With both present, I₂ (violet) is liberated first at low Cl₂ (iodide oxidises most easily) and Br₂ (orange) only after excess Cl₂; controlled Cl₂ addition and the colour sequence distinguishes them.

It is not sulphate; the acid-solubility marks it as BaCO₃ or BaSO₃ (Ksp large enough to react away in acid). Genuine BaSO₄ (Ksp ~$1.1\times 10^{-10}$) stays white and undissolved in dilute acid.