JEE Chemistry Qualitative Salt Analysis MCQs Set 07

Answer: (c) \( \ce{Pb^2+} \)

Explanation:
1. PbI2 is yellow and dissolves in concentrated KI forming the soluble complex \( \ce{[PbI4]^2-} \).
2. Hg2I2 is green, AgI is yellow but insoluble in KI, and Cu2+ gives white CuI with iodine.

Teacher's Note:
1. Lead iodide: golden yellow, soluble in hot water.
2. Excess iodide forms tetraiodoplumbate(II).

Answer: (a) \( \ce{Pb(NO3)2} \)

Explanation:
1. Pb(OH)2 is white and insoluble in excess NH3, PbCl2 is white and PbS is black.
2. AgNO3 gives a brown Ag2O that dissolves in excess ammonia, Hg(NO3)2 gives no precipitate with NaCl and Mn(NO3)2 does not give a precipitate with H2S in acid.

Teacher's Note:
1. Pb2+: white with NH3, Cl-, SO4 2-; black with H2S.
2. Ag+ dissolves in excess ammonia.

Answer: (d) \( \ce{Pb^2+} \)

Explanation:
1. PbCl2 is white and dissolves in hot water.
2. Pb2+ then gives yellow PbCrO4 with chromate.
3. Hg2+ and Sn2+ give no precipitate with dilute HCl and AgCl does not dissolve in hot water.

Teacher's Note:
1. Group I chlorides: Pb2+ (soluble in hot water), Ag+, Hg2 2+.
2. Chromate precipitates Pb2+ as yellow.

Answer: (b) \( \ce{Hg(NH2)Cl} \) (white)

Explanation:
1. \( \ce{Hg2Cl2 + 2NH3 -> Hg(NH2)Cl + Hg + NH4Cl} \).
2. This disproportionation gives white mercury(II) amidochloride and finely divided black mercury.

Teacher's Note:
1. Mercury(I) disproportionates in ammonia.
2. The black colour is the metal.

Answer: (b) \( \ce{HNO3} \) (aqueous)

Explanation:
1. Acid removes the free ammonia as NH4+, which shifts \( \ce{AgCl + 2NH3 <=> [Ag(NH3)2]+ + Cl-} \) to the left, so white AgCl reappears.
2. NaBr and NH4I would give pale yellow AgBr and yellow AgI instead.

Teacher's Note:
1. Neutralise the ammonia to break the complex.
2. Another halide would give a coloured precipitate.

Answer: (d) H2S is oxidised to sulphur by some acid radicals

Explanation:
1. Oxidising radicals such as nitrite, nitrate, chromate and iron(III) oxidise H2S to colloidal sulphur, which is yellow-white.

Teacher's Note:
1. The oxidation product of H2S is sulphur.
2. Group IV precipitates only in basic medium.

Answer: (a) decreases the concentration of sulphide ion

Explanation:
1. H2S is a weak acid and HCl suppresses its ionisation through the common ion effect.
2. The low [S2-] exceeds the solubility product only for the very insoluble group II sulphides (CuS, PbS, HgS) and not for group IV (ZnS, NiS, MnS, CoS).

Teacher's Note:
1. Low [S2-] gives selective precipitation.
2. In ammoniacal medium [S2-] is high.

Answer: (c) \( \ce{Hg2Cl2} \)

Explanation:
1. Tin(II) reduces Hg2+ to Hg+: \( \ce{2HgCl2 + SnCl2 -> Hg2Cl2 + SnCl4} \).
2. With excess SnCl2 the precipitate turns grey due to metallic mercury.

Teacher's Note:
1. Small amount of reductant gives calomel.
2. Excess gives mercury.

Answer: (d) \( \ce{[Cu(NH3)4]^2+} \)

Explanation:
1. A pale blue precipitate of Cu(OH)2 dissolves in excess ammonia to give the deep blue tetraamminecopper(II) ion.

Teacher's Note:
1. Cu2+ has coordination number 4 with NH3 in water.
2. Deep blue = [Cu(NH3)4]2+.

Answer: (b) \( \ce{CuCl2} \)

Explanation:
1. CuS is black; CdS is yellow and ZnS is white.
2. Ferric chloride is reduced to Fe2+ with deposition of sulphur, so no black sulphide forms in acid.

Teacher's Note:
1. Group II: CuS black, CdS yellow.
2. Group IV: ZnS white.

Answer: (a) \( \ce{CuCO3} \)

Explanation:
1. Copper carbonate dissolves with effervescence to give soluble CuCl2.
2. Silver and mercury(I) carbonates would give insoluble AgCl and Hg2Cl2, and lead carbonate gives sparingly soluble PbCl2.

Teacher's Note:
1. Check whether the chloride formed is soluble.
2. Group I chlorides are insoluble.

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

Explanation:
1. Hydrolysis: \( \ce{BiCl3 + H2O -> BiOCl + 2HCl} \).
2. The white bismuth oxychloride dissolves in HCl again.

Teacher's Note:
1. Bi3+ and Sb3+ salts hydrolyse to oxychlorides.
2. Addition of HCl reverses it.

Answer: (c) \( \ce{HgI2} \) and \( \ce{BiI3} \)

Explanation:
1. HgI2 is red and dissolves in excess KI to give colourless \( \ce{[HgI4]^2-} \).
2. BiI3 is black and dissolves in excess KI to give orange \( \ce{[BiI4]-} \).
3. Hg2I2 is green, CuI is white, AgI is yellow, CdI2 is white and PbI2 is yellow.

Teacher's Note:
1. Excess iodide gives tetraiodo complexes.
2. Know the colours of metal iodides.

Answer: (a) lead

Explanation:
1. PbCrO4 is yellow and dissolves in NaOH as plumbite and chromate.
2. Silver chromate is brick red and mercury and iron give no yellow chromate precipitates under these conditions.

Teacher's Note:
1. Lead chromate: yellow; silver chromate: red.
2. PbCrO4 dissolves in hot NaOH.

Answer: (d) \( \ce{Sb2S3} \)

Explanation:
1. Arsenic, antimony and tin sulphides (group IIB) form soluble thio-anions with yellow ammonium sulphide.
2. HgS, PbS and Bi2S3 belong to group IIA and do not dissolve.

Teacher's Note:
1. IIB sulphides dissolve in yellow ammonium sulphide.
2. IIA sulphides do not.

Answer: (b) \( \ce{OH^-} \) concentration decreases

Explanation:
1. NH4Cl supplies NH4+, a common ion, which suppresses the ionisation of the weak base NH4OH.
2. So [OH-] falls; this is why NH4Cl is added in group III so that only the hydroxides of Fe3+, Al3+ and Cr3+ (very small Ksp) precipitate.

Teacher's Note:
1. Common ion effect.
2. Low [OH-] avoids precipitating group IV and V hydroxides.

Answer: (b) \( \ce{AlCl3} \)

Explanation:
1. AlCl3 fumes in moist air (hydrolysis to HCl).
2. Al(OH)3 precipitates with NH4OH and dissolves in NaOH, and Al3+ does not form a sulphide with H2S.
3. FeCl3 gives a precipitate that does not dissolve in NaOH, and ZnCl2 gives no precipitate with NH4OH in presence of NH4Cl.

Teacher's Note:
1. Amphoteric hydroxide: Al(OH)3.
2. Al2S3 is hydrolysed in water.

Answer: (d) \( \ce{Fe^3+} \)

Explanation:
1. Fe3+ gives a brown colouration with ferricyanide and red-brown iron(III) acetate (basic acetate) on adding sodium acetate.

Teacher's Note:
1. Fe3+: red with SCN-, blue with ferrocyanide.
2. Fe2+: blue with ferricyanide.

Answer: (a) \( \ce{NaOH / H2O2} \)

Explanation:
1. Hydrogen peroxide in NaOH oxidises Cr(OH)3 to soluble yellow chromate, while Fe(OH)3 stays as a brown precipitate.
2. Ammonia and acids do not separate them cleanly.

Teacher's Note:
1. Chromium(III) is oxidised to chromate (VI).
2. Fe(OH)3 is not amphoteric.

Answer: (c) \( \ce{Mn^2+} \)

Explanation:
1. Mn(OH)2 is white and rapidly oxidises in air to brown MnO(OH)2.
2. It is incompletely precipitated by ammonia and dissolves in ammonium salts.
3. Zn2+ and Al3+ are colourless and Ni2+ gives a green precipitate.

Teacher's Note:
1. Mn2+ is pale pink.
2. White turning brown in air is the manganese(II) test.

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

Explanation:
1. MnS is buff or pink and dissolves even in acetic acid.
2. ZnS is insoluble in acetic acid but dissolves in dilute HCl, and CoS and NiS are insoluble in dilute acids once formed.

Teacher's Note:
1. MnS is the most acid-soluble sulphide of group IV.
2. NiS and CoS need aqua regia.

Answer: (b) \( \ce{HCl} \) (aq)

Explanation:
1. Acid removes ammonia as NH4+, which shifts \( \ce{Zn^2+ + 4NH3 <=> [Zn(NH3)4]^2+} \) to the left and frees Zn2+.
2. Adding NH3 does the opposite.

Teacher's Note:
1. Le Chatelier's principle.
2. Dilution has a smaller effect.

Answer: (a) Zn

Explanation:
1. Zn(OH)2 dissolves in excess ammonia as \( \ce{[Zn(NH3)4]^2+} \), and ZnS is white.
2. Al(OH)3 does not dissolve in excess ammonia, and Ca2+ and Ba2+ do not precipitate with ammonia.

Teacher's Note:
1. White ZnS is the only white sulphide in group IV.
2. Ni gives red complex with dmg.

Answer: (c) Ni

Explanation:
1. Nickel(II) forms the bright red bis(dimethylglyoximato)nickel(II) complex, stabilised by intramolecular hydrogen bonds.
2. The other metals do not give this complex.

Teacher's Note:
1. Ni2+: dmg red; Co2+: nitroso-naphthol or NH4SCN blue.
2. The complex is square planar.

Answer: (c) \( \ce{Ni^2+} \)

Explanation:
1. PbS, Bi2S3 and CuS have very small solubility products and form in acid solution.
2. NiS is a group IV sulphide that needs a basic medium.

Teacher's Note:
1. Group II precipitates in dilute HCl.
2. Group IV needs ammoniacal medium.