JEE Chemistry Qualitative Salt Analysis MCQs Set 05

Answer: (c) blue vitriol, \( \ce{CuSO4.5H2O} \)

Explanation:
1. Blue copper sulphate pentahydrate loses its water of crystallisation on heating and becomes white anhydrous CuSO4.
2. BaSO4 and Na2SO4 are white to begin with.

Teacher's Note:
1. Hydrated copper(II) salts are blue.
2. Anhydrous CuSO4 is white and turns blue again on adding water.

Answer: (a) \( \ce{MgCO3} \)

Explanation:
1. MgCO3 decomposes to MgO and CO2 only.
2. Pb(NO3)2 gives NO2 and O2, FeSO4 gives SO2 and SO3, and ammonium dichromate gives N2 and water vapour with Cr2O3.

Teacher's Note:
1. Count the gaseous products in each decomposition.
2. Nitrates of heavy metals give NO2 and O2.

Answer: (d) \( \ce{ZnCO3} \)

Explanation:
1. ZnCO3 decomposes on heating to ZnO and CO2.
2. ZnO is yellow when hot and white when cold.
3. MgO stays white and K2CO3 and Na2CO3 do not decompose.

Teacher's Note:
1. Hot yellow, cold white is the sign of zinc oxide.
2. Lead oxide stays yellow on cooling.

Answer: (b) \( \ce{Ag2CO3} \)

Explanation:
1. Carbonates of the heavier alkali metals (Na, K, Rb) are thermally stable and do not decompose on a Bunsen flame.
2. Silver carbonate decomposes to silver oxide (and then silver) and CO2.

Teacher's Note:
1. Alkali metal carbonates (except lithium) are stable.
2. Heavy metal carbonates decompose easily.

Answer: (b) metal chlorides are volatile and give a better flame colour

Explanation:
1. Most metal chlorides are more volatile than other salts, so more metal atoms enter the flame and are excited.
2. This gives a stronger characteristic colour.

Teacher's Note:
1. Convert the salt to the chloride first.
2. Barium sulphate is not easily converted and gives a poor flame.

Answer: (d) Li

Explanation:
1. Lithium gives a crimson red flame and LiCl is deliquescent.
2. Mg gives no flame colour, Ca gives brick red and Ba gives apple green.

Teacher's Note:
1. Li crimson, Na golden yellow, K pale violet.
2. Ca brick red, Sr crimson, Ba apple green.

Answer: (a) metaborate of the metal

Explanation:
1. Borax on heating gives sodium metaborate and boron trioxide: \( \ce{Na2B4O7 -> 2NaBO2 + B2O3} \).
2. B2O3 reacts with the metal oxide to form the metal metaborate, which is coloured, for example Cu(BO2)2 (blue).

Teacher's Note:
1. Borax bead = metaborate.
2. Colour depends on the transition metal ion.

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

Explanation:
1. Zn2+ has a completely filled d subshell and gives a colourless bead.
2. Cr3+ gives a green bead, Cu2+ a blue bead and Mn2+ an amethyst (violet) bead.

Teacher's Note:
1. Coloured bead needs partly filled d orbitals.
2. d10 ions are colourless.

Answer: (d) \( \ce{Co(BO2)2} \)

Explanation:
1. Cobalt oxide reacts with boron trioxide to give cobalt metaborate, Co(BO2)2, which is blue.

Teacher's Note:
1. The bead is always a metaborate.
2. Co(BO2)2 is also the blue of Thenard's blue test.

Answer: (b) \( \ce{Mg^2+} \)

Explanation:
1. Mg2+ gives white MgO in the charcoal cavity, which on moistening with cobalt nitrate and heating gives a pink mass.
2. Zn gives a yellow-hot, white-cold residue and a green mass, and Al gives a blue mass (Thenard's blue).

Teacher's Note:
1. Zn: green, Al: blue, Mg: pink with cobalt nitrate.
2. Used for cations of groups III onwards.

Answer: (a) \( \ce{S^2-} \), \( \ce{SO3^2-} \), \( \ce{NO2^-} \)

Explanation:
1. The dilute acid group contains carbonate, sulphide, sulphite, nitrite and acetate (with dilute H2SO4).
2. NO3-, I- and Cl- do not react with dilute acids.

Teacher's Note:
1. Dilute acid gives gases from CO3 2-, S2-, SO3 2-, NO2-.
2. Nitrate needs conc. H2SO4.

Answer: (c) \( \ce{CO3^2-} \)

Explanation:
1. Carbonate gives CO2, which is colourless and odourless.
2. Sulphide gives smelly H2S, chloride gives pungent HCl and nitrate gives brown NO2.

Teacher's Note:
1. CO2 has no smell.
2. Always note the colour and smell of the gas.

Answer: (c) \( \ce{SO2} \)

Explanation:
1. SO2 turns lime water milky (CaSO3) and reduces orange dichromate to green chromium(III).
2. CO2 does not reduce dichromate and H2S does not turn lime water milky.

Teacher's Note:
1. Reducing gases turn K2Cr2O7 green.
2. CO2 only turns lime water milky.

Answer: (a) \( \ce{H2S} \)

Explanation:
1. H2S reacts with lead acetate to give black PbS.
2. NO2 is brown, CO2 is odourless and SO2 smells of burning sulphur.

Teacher's Note:
1. Black lead sulphide = sulphide test.
2. Rotten egg smell is H2S.

Answer: (d) \( \ce{S^2-} \)

Explanation:
1. The rotten egg gas is H2S, which gives yellow CdS with CdCO3.
2. Sodium nitroprusside gives the violet \( \ce{[Fe(CN)5NOS]^4-} \) with sulphide ion.
3. So X is S2-.

Teacher's Note:
1. Three tests for sulphide: lead acetate (black), CdCO3 (yellow), nitroprusside (violet).
2. Sulphite does not give a rotten egg smell.

Answer: (b) \( \ce{NO2^-} \)

Explanation:
1. Nitrite in acid liberates nitrous acid, which oxidises iodide to iodine; iodine gives a blue complex with starch.
2. Sulphite is a reducing agent and would not liberate iodine.

Teacher's Note:
1. Nitrite oxidises I- to I2.
2. Blue starch test indicates oxidising agents.

Answer: (b) does not change

Explanation:
1. Nitroprusside contains Fe(II) and NO+.
2. The sulphide ion attacks the NO+ to form the purple thionitroso complex \( \ce{[Fe(CN)5NOS]^4-} \), where iron is still +2.

Teacher's Note:
1. The sulphide attacks the nitrosyl group, not the iron.
2. Fe stays +2.

Answer: (d) \( \ce{Cl^-} \) and \( \ce{NO3^-} \)

Explanation:
1. Carbonate and acetate react with dilute acid, so they belong to the dilute-acid group.
2. Chloride and nitrate need conc. H2SO4 to liberate HCl and HNO3 (brown NO2 with copper).

Teacher's Note:
1. Cl-, Br-, I-, NO3- form the conc. H2SO4 group.
2. Do dilute acid tests first.

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

Explanation:
1. Chloride gives colourless HCl gas, because HCl is not oxidised by conc. H2SO4.
2. HBr and HI are reducing agents, so they are oxidised by the acid to Br2 (red-brown) and I2 (violet).

Teacher's Note:
1. Reducing power: HI > HBr > HCl.
2. The acid is an oxidising agent.

Answer: (c) \( \ce{Na3AsO3} \)

Explanation:
1. Arsenite reacts with AgCl to give yellow silver arsenite, \( \ce{Ag3AsO3} \), displacing chloride.
2. Arsenate would give chocolate-brown silver arsenate and cyanide dissolves AgCl.

Teacher's Note:
1. Ag3AsO3 is yellow, Ag3AsO4 is chocolate brown.
2. Silver phosphate is also yellow.

Answer: (d) \( \ce{NO3^-} \)

Explanation:
1. Nitrite would give brown fumes with dilute acid.
2. Bromide fumes (Br2 with HBr) would give a yellowish precipitate with AgNO3.
3. So nitrate, whose conc. H2SO4 fumes are NO2 only.

Teacher's Note:
1. Brown vapours with conc. acid only: Br- or NO3-.
2. Silver nitrate distinguishes them.

Answer: (b) \( \ce{CrO2Cl2} \) (chromyl chloride)

Explanation:
1. This is the chromyl chloride test for chloride.
2. \( \ce{4NaCl + K2Cr2O7 + 6H2SO4 -> 2KHSO4 + 4NaHSO4 + 2CrO2Cl2 + 3H2O} \).

Teacher's Note:
1. Test is done in a dry tube.
2. Bromide and iodide do not respond.