JEE Chemistry Salt Analysis MCQs Set 01

Answer: (b) (A), (B), (C)

Explanation:
1. Carbonates of alkali metals and ammonium are soluble.
2. Calcium carbonate is insoluble.

Teacher's Note:
1. Only Na, K and NH4 carbonates dissolve in water.
2. All other metal carbonates are insoluble.

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

Explanation:
1. Nitrite is converted by acid into nitrous acid, which oxidises I\(^-\) to I\(_2\).
2. I\(_2\) gives a blue colour with starch.

Teacher's Note:
1. Nitrite oxidises iodide in acid medium.
2. Sulphite and carbonate do not do this.

Answer: (c) (A), (D)

Explanation:
1. Zn and acid reduce sulphite to H2S.
2. H2S blackens lead acetate paper (PbS) and decolourises acidified KMnO4.
3. H2S does not turn lime water milky and gives black Ag2S, not a white precipitate, with silver nitrate.

Teacher's Note:
1. Sulphite with Zn and acid gives H2S, not SO2.
2. Remember: lead acetate paper test is for H2S.

Answer: (c) \( \ce{SO3^{2-}} \)

Explanation:
1. SO2 turns baryta water turbid (BaSO3) and reduces orange dichromate to green Cr3+.
2. CO2 does not reduce dichromate.

Teacher's Note:
1. SO2 is a reducing gas.
2. Both gases turn lime or baryta water milky.

Answer: (b) \( \ce{NO} \)

Explanation:
1. NO combines with Fe\(^{2+}\) to give the brown nitrosyl complex \( [Fe(H_2O)_5NO]^{2+} \).
2. This is the basis of the brown ring test.

Teacher's Note:
1. Brown ring test uses NO.
2. N2O does not coordinate.

Answer: (b) Na3AsO3

Explanation:
1. Sodium arsenite converts AgCl into yellow silver arsenite Ag3AsO3.
2. Arsenate gives chocolate Ag3AsO4 and cyanide dissolves AgCl.

Teacher's Note:
1. Arsenite test distinguishes AgCl from AgI.
2. AgI remains unchanged.

Answer: (b) Chromyl chloride

Explanation:
1. Chloride with dichromate and conc. H\(_2\)SO\(_4\) gives volatile chromyl chloride CrO\(_2\)Cl\(_2\).
2. It is a deep red vapour.

Teacher's Note:
1. Chromyl chloride test confirms chloride.
2. Bromide and iodide do not give it.

Answer: (a) \( \ce{Cr2O7^{2-} and Cl^-} \)

Explanation:
1. Deep red fumes are of chromyl chloride, which needs both chromate or dichromate and chloride.
2. The other pairs lack the required combination.

Teacher's Note:
1. Both chloride and dichromate are needed.
2. Bromide gives brown Br2 instead.

Answer: (d) Bromide

Explanation:
1. Bromide is oxidised by conc. H2SO4 to Br2, which liberates iodine from iodide.
2. Iodine turns starch blue.

Teacher's Note:
1. Br2 is an oxidising agent.
2. HCl is not oxidised by conc. H2SO4.

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

Explanation:
1. Reddish brown gas is Br2, so the anion is bromide.
2. Na+ gives no precipitate with carbonate, which rules out CaCl2 and the barium salt.

Teacher's Note:
1. NaI would give violet vapours.
2. CaCl2 would give a carbonate precipitate.

Answer: (c) \( \ce{I^-} \)

Explanation:
1. Hypochlorite oxidises iodide to iodine, which turns starch blue-black.
2. Bromide would give brown bromine, which does not give a blue-black spot.

Teacher's Note:
1. Iodine with starch gives blue-black.
2. Hypochlorite is an oxidising agent.

Answer: (a) Iodide

Explanation:
1. Iodide gives violet I2, which is oxidised to colourless iodate by excess chlorine.
2. Bromide gives orange that only fades to pale yellow.

Teacher's Note:
1. Violet to colourless identifies iodide.
2. Chloride does not react with Cl2.

Answer: (c) Soluble in water

Explanation:
1. Nitrates are generally soluble in water.
2. Hg and Bi nitrates hydrolyse to give insoluble basic salts.

Teacher's Note:
1. All nitrates are soluble, with Hg and Bi as basic salts.
2. Nitrates decompose on strong heating.

Answer: (c) (A), (B)

Explanation:
1. Sodium arsenite turns AgCl yellow but not AgI.
2. Dilute ammonia dissolves AgCl but not AgI.
3. Both dissolve in KCN, and neither dissolves in dilute HNO3, so these cannot distinguish them.

Teacher's Note:
1. AgI is far less soluble than AgCl.
2. KCN dissolves all silver halides.

Answer: (b) K\(_2\)CrO\(_4\)

Explanation:
1. K2CrO4 gives a yellow precipitate of BaCrO4 only with Ba2+.
2. AgNO3 gives a white precipitate with all chlorides, including HCl and CdCl2.

Teacher's Note:
1. A specific test for Ba2+ is yellow chromate.
2. Cl- tests cannot single out BaCl2.

Answer: (b) \( \ce{K2HgI4 + KOH} \)

Explanation:
1. Nessler's reagent is alkaline potassium tetraiodomercurate(II), \( K_2[HgI_4]+KOH \).
2. It gives a brown precipitate or colour with NH3 or NH4+.

Teacher's Note:
1. The alkaline medium is essential.
2. The brown product is iodide of Millon's base.

Answer: (d) (A) and (C) both

Explanation:
1. NH\(_4^+\) gives yellow \( (NH_4)_2[PtCl_6] \) with chloroplatinic acid and yellow \( (NH_4)_3[Co(NO_2)_6] \) with sodium cobaltinitrite.
2. Both are correct, so the answer is both.

Teacher's Note:
1. Same yellow ppts with K+.
2. HgCl2 gives white ppt with NH3.

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

Explanation:
1. PbI\(_2\) is yellow and dissolves in concentrated KI as \( [PbI_4]^{2-} \).
2. Hg\(_2^{2+}\) gives green Hg\(_2\)I\(_2\), Ag+ gives AgI that stays and Cu2+ gives CuI.

Teacher's Note:
1. PbI2 also dissolves in hot water, forming golden spangles on cooling.
2. Check colour and solubility.

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

Explanation:
1. Pb(OH)2 is white and insoluble in excess ammonia, PbCl2 is white and PbS is black.
2. AgNO3 dissolves in excess ammonia, Hg(NO3)2 gives HgS, which is black, but no PbCl2-type chloride precipitate, and Mn2+ gives no sulphide with H2S in acid.

Teacher's Note:
1. Lead fits all three observations.
2. Silver is excluded by the ammonia result.

Answer: (b) HCl (aq)

Explanation:
1. HCl precipitates Ag+ as white AgCl while Cu2+ remains as soluble CuCl2.
2. H2S would precipitate both as sulphides.

Teacher's Note:
1. Group I reagent separates Ag+.
2. Cu2+ is group II.

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

Explanation:
1. PbCl2 is white and soluble in hot water.
2. Pb2+ then gives yellow PbCrO4.

Teacher's Note:
1. AgCl is insoluble in hot water.
2. Hg2+ gives no chloride precipitate.

Answer: (c) Of the oxidation of H2S gas by some acid radicals

Explanation:
1. Oxidising anions such as nitrite, chromate or iodate oxidise H2S to colloidal sulphur.
2. The yellow turbidity is sulphur.

Teacher's Note:
1. Oxidising radicals convert H2S to S.
2. This is not a sulphide precipitate.

Answer: (c) HCl decreases the concentration of S2-

Explanation:
1. HCl suppresses ionisation of H2S by the common ion effect, lowering [S2-].
2. Only the very insoluble group II sulphides have Ksp low enough to precipitate.

Teacher's Note:
1. Group II: low Ksp. Group IV: higher Ksp.
2. Basic medium raises [S2-].