JEE Chemistry Qualitative Salt Analysis MCQs Set 06

Answer: (c) a sulphite

Explanation:
1. Sulphite is a reducing agent and reduces purple MnO4- to colourless Mn2+: \( \ce{5SO3^2- + 2MnO4- + 6H+ -> 2Mn^2+ + 5SO4^2- + 3H2O} \).
2. Carbonate, nitrate and bicarbonate cannot reduce permanganate.

Teacher's Note:
1. Reducing anions decolourise acidified KMnO4.
2. Sulphite, sulphide, nitrite and iodide do this.

Answer: (a) the bromide

Explanation:
1. Nitrate gives NO2, which dissolves in water to give colourless nitric and nitrous acids.
2. Bromide with MnO2 and conc. H2SO4 gives bromine, which dissolves in water to give a reddish brown solution.

Teacher's Note:
1. NO2 reacts with water; Br2 only dissolves.
2. Bromine water is orange-brown.

Answer: (d) \( \ce{NH4NO2} \)

Explanation:
1. Nitrite in acid forms nitrous acid, which oxidises iodide to iodine; iodine gives a blue colour with starch.
2. Sulphite is a reducing agent, and carbonate and ammonium chloride do not oxidise iodide.

Teacher's Note:
1. Only the oxidising anion nitrite liberates iodine.
2. NH4NO2 is an ammonium salt of nitrite.

Answer: (b) calcium salt

Explanation:
1. Calcium gives a brick red flame, strontium a crimson one, sodium golden yellow, and cobalt salts give no characteristic flame.

Teacher's Note:
1. Ca brick red, Sr crimson.
2. Do not confuse the two reds.

Answer: (b) sulphite

Explanation:
1. BaSO3 is white and dissolves in HCl with release of SO2, a colourless pungent gas.
2. SO2 and sulphites are both used as bleaching agents in the textile industry.
3. Carbonate would give odourless CO2 and acetate gives no precipitate.

Teacher's Note:
1. Ba2+ + SO3 2- gives BaSO3.
2. SO2 bleaches by reduction.

Answer: (d) any of \( \ce{SO3^2-} \), \( \ce{NO2^-} \) or \( \ce{S^2-} \)

Explanation:
1. Sulphite is oxidised to sulphate, nitrite is oxidised to nitrate and sulphide is oxidised to sulphur (a yellow solid), and in every case MnO4- is reduced to Mn2+.
2. None of these reactions evolves a gas, so all three radicals fit.

Teacher's Note:
1. All reducing anions decolourise acidified KMnO4.
2. Check that no gas is evolved in each case.

Answer: (a) \( \ce{NO} \) gas is liberated and iodine is set free

Explanation:
1. \( \ce{2NO2- + 2I- + 4H+ -> 2NO + I2 + 2H2O} \).
2. Nitrite is reduced to nitric oxide while iodide is oxidised to iodine.

Teacher's Note:
1. Nitrite is the oxidising agent here.
2. NO turns brown in air.

Answer: (c) \( \ce{NH3} \)

Explanation:
1. Ammonia dissolves in water to form the weak base NH4OH, which turns red litmus blue.
2. HCl, H2S and SO2 form acidic solutions.

Teacher's Note:
1. Only basic gas: NH3.
2. It also gives white fumes with HCl.

Answer: (d) none; silver does not react appreciably

Explanation:
1. The standard reduction potentials are Ag+/Ag = +0.80 V, Cl2/Cl- = +1.36 V and H+/H2 = 0.00 V.
2. Silver cannot be oxidised by H+ (no hydrogen), and Cl- cannot be oxidised by anything weak, so there is no reaction beyond a trace of surface AgCl.

Teacher's Note:
1. Silver is below hydrogen in the electrochemical series.
2. It dissolves in oxidising acids like HNO3.

Answer: (b) sodium thiosulphate

Explanation:
1. Thiosulphate decomposes in acid: \( \ce{S2O3^2- + 2H+ -> SO2 + S + H2O} \).
2. Sulphide gives H2S, sulphite gives only SO2 and sulphate does not react.

Teacher's Note:
1. Sulphur turbidity with SO2 is the thiosulphate test.
2. Sulphite gives no turbidity.

Answer: (a) \( \ce{HCO3^-} \)

Explanation:
1. Carbonate gives MgCO3 immediately in the cold.
2. Bicarbonate forms soluble Mg(HCO3)2 in the cold, but on heating it decomposes: \( \ce{Mg(HCO3)2 -> MgCO3 + CO2 + H2O} \) and MgCO3 is precipitated.

Teacher's Note:
1. Heating converts bicarbonate to carbonate.
2. This distinguishes HCO3- from CO3 2-.

Answer: (c) Mn

Explanation:
1. Manganese gives an amethyst (violet) bead in the oxidising flame and a colourless bead in the reducing flame.
2. Iron gives yellow or brown, cobalt gives blue and lead gives no characteristic bead.

Teacher's Note:
1. Mn: amethyst; Cr: green; Co: blue.
2. Fe: yellow in oxidising flame.

Answer: (c) \( \ce{NO2} \)

Explanation:
1. Some nitrates (heavy metal nitrates) give brown NO2 on heating.
2. NO2 is a strong oxidising agent, so it bleaches moist litmus; SO2 and N2O are colourless and iodine vapour does not bleach.

Teacher's Note:
1. Coloured and bleaching: NO2 or Cl2.
2. SO2 is colourless.

Answer: (a) 4

Explanation:
1. Sulphide gives H2S, thiosulphate gives SO2 and sulphur, nitrite gives brown NO2 and sulphite gives SO2 with dilute HCl.
2. Nitrate and sulphate do not react with dilute HCl.

Teacher's Note:
1. Dilute acid group: CO3, S, SO3, S2O3, NO2, CH3COO.
2. Nitrate needs conc. H2SO4.

Answer: (d) 2

Explanation:
1. Zinc and acid generate nascent hydrogen, which reduces sulphite to H2S: \( \ce{SO3^2- + 3Zn + 8H+ -> H2S + 3Zn^2+ + 3H2O} \).
2. H2S has two lone pairs on sulphur.

Teacher's Note:
1. The gas has a rotten egg smell.
2. Sulphur has 6 valence electrons, 2 are used in bonds.

Answer: (b) 3

Explanation:
1. Cr3+ (green), Cu2+ (blue) and Mn2+ (amethyst) give coloured beads.
2. Zn2+ is d10 and gives a colourless bead.

Teacher's Note:
1. d10 ions give no bead colour.
2. Count the coloured ones.

Answer: (b) 3

Explanation:
1. Sodium carbonate, sodium nitrate and ammonium sulphite are soluble.
2. Lead carbonate and copper(I) chloride are insoluble.

Teacher's Note:
1. Alkali metal and ammonium salts are soluble.
2. CuCl and PbCO3 are insoluble.

Answer: (d) 8

Explanation:
1. Gas A is HCl, which is diatomic, so y = 2.
2. Gas B is Cl2, which has 3 lone pairs on each chlorine atom, so x = 6.
3. x + y = 8.

Teacher's Note:
1. Permanganate oxidises Cl- to Cl2.
2. Count lone pairs on both atoms.

Answer: (a) \( \ce{PbCl2} \)

Explanation:
1. Lead chloride is sparingly soluble in cold water but dissolves well in hot water; this is used to separate it from AgCl and Hg2Cl2 in group I.
2. BiCl3 and SnCl4 hydrolyse and dissolve in water, and AgCl is insoluble even in hot water.

Teacher's Note:
1. Group I chlorides: PbCl2 (soluble in hot water), AgCl, Hg2Cl2.
2. Use hot water to dissolve PbCl2.

Answer: (c) \( \ce{K2CO3} \)

Explanation:
1. Lilac flame means potassium.
2. The solution is alkaline due to hydrolysis of carbonate, which turns red litmus blue; KNO3 is neutral.

Teacher's Note:
1. K flame is lilac, Na is yellow.
2. Carbonates hydrolyse to give alkaline solutions.

Answer: (d) sulphur (sulphide ion)

Explanation:
1. Alkaline nitroprusside gives a violet colour with sulphide ion (and with sulphur in Lassaigne's extract).

Teacher's Note:
1. Nitroprusside: violet with S2-.
2. Nitrogen is detected as Prussian blue.

Answer: (b) \( \ce{Cl2} \) > \( \ce{H2SO4} \) > \( \ce{Br2} \) > \( \ce{I2} \)

Explanation:
1. H2SO4 oxidises HBr and HI but not HCl, so it is a stronger oxidising agent than Br2 and I2 but weaker than Cl2.
2. Order: Cl2 > H2SO4 > Br2 > I2.

Teacher's Note:
1. If the acid oxidises the halide, the acid is the stronger oxidant.
2. Cl- is not oxidised by conc. H2SO4.

Answer: (a) \( \ce{Na2S2O3} \)

Explanation:
1. Silver thiosulphate is white and dissolves in excess thiosulphate as a complex.
2. On heating with water it forms black Ag2S and sulphate (hence BaSO4 with barium nitrate): \( \ce{Ag2S2O3 + H2O -> Ag2S + H2SO4} \).

Teacher's Note:
1. White to yellow to brown to black is the thiosulphate sequence.
2. Sulphide gives a black precipitate immediately.

Answer: (c) \( \ce{NH4HS} \)

Explanation:
1. With excess H2S the ammonium hydrogen sulphide is formed: \( \ce{NH3 + H2S -> NH4HS} \).
2. (NH4)2S would form only with excess ammonia.

Teacher's Note:
1. Saturation with H2S means excess acid gas.
2. NH4HS: ammonium hydrosulphide.

Answer: (c) \( \ce{Zn(NO3)2} \)

Explanation:
1. Zinc nitrate gives NO2 (brown) and O2 with a residue of ZnO, which is yellow when hot and white when cold.
2. Zn2+ gives white ZnS with ammonium sulphide in ammoniacal solution.

Teacher's Note:
1. Hot yellow cold white = ZnO.
2. Al3+ gives white Al(OH)3 but its oxide is white throughout.