JEE Chemistry Qualitative Salt Analysis MCQs Set 09

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Multiple Choice Questions

Question 1: In the group II separation of Cu2+ and Cd2+ as their tetrammine complexes with KCN, the correct stability relation that allows separation is:
(a) K3[Cu(CN)4] is more stable and K2[Cd(CN)4] is less stable
(b) K2[Cu(CN)4] is less stable and K2[Cd(CN)4] is more stable
(c) K2[Cu(CN)4] is more stable and K2[Cd(CN)4] is less stable
(d) K3[Cu(CN)4] is less stable and K2[Cd(CN)4] is more stable
Show Answer & Explanation

Answer: (a) K3[Cu(CN)4] is more stable and K2[Cd(CN)4] is less stable

Explanation:
1. KCN reduces Cu2+ to Cu+, which forms the very stable [Cu(CN)4]3−.
2. H2S cannot precipitate copper from it.
3. The cadmium complex is much less stable, so Cd2+ still gives yellow CdS.
4. Answer: (a)

Teacher's Note:
1. Identify the ion from the colour, precipitate or flame, and rule out the other options one by one.

Question 2: Sodium carbonate cannot be used in place of (NH4)2CO3 for the identification of Ca2+, Ba2+ and Sr2+ in group V because:
(a) Mg2+ ions will also be precipitated
(b) The concentration of CO32− ions is very low
(c) Sodium ions will react with the acid radicals
(d) Na+ ions will interfere with detection of Ca2+, Ba2+ and Sr2+
Show Answer & Explanation

Answer: (a) Mg2+ ions will also be precipitated

Explanation:
1. Na2CO3 ionises fully and gives a high CO32− concentration.
2. This exceeds the solubility product of MgCO3 as well, so magnesium (group VI) would precipitate with group V.
3. (NH4)2CO3 in presence of NH4Cl keeps the CO32− concentration low.
4. Answer: (a)

Teacher's Note:
1. Identify the ion from the colour, precipitate or flame, and rule out the other options one by one.

Question 3: Which of the following statements is incorrect?
(a) Fe2+ ion also gives a blood red colour with SCN− ion
(b) Fe3+ ion gives a blood red colour with SCN− ion
(c) On passing H2S into Na2ZnO2 solution, a white precipitate of ZnS is formed
(d) Cupric ion reacts with excess ammonia to give the deep blue [Cu(NH3)4]2+ ion
Show Answer & Explanation

Answer: (a) Fe2+ ion also gives a blood red colour with SCN− ion

Explanation:
1. The blood-red [Fe(SCN)]2+ complex is formed by Fe3+ only.
2. Fe2+ gives no such colour with thiocyanate.
3. The other three statements are correct.
4. Answer: (a)

Teacher's Note:
1. Identify the ion from the colour, precipitate or flame, and rule out the other options one by one.

Question 4: The salt used for performing the 'bead' test in qualitative inorganic analysis is:
(a) K2SO4·Al2(SO4)3·24H2O
(b) FeSO4·(NH4)2SO4·4H2O
(c) Na(NH4)HPO4·4H2O
(d) CaSO4·2H2O
Show Answer & Explanation

Answer: (c) Na(NH4)HPO4·4H2O

Explanation:
1. Microcosmic salt on heating forms NaPO3, which fuses to a clear glassy bead.
2. The bead combines with metal oxides to give characteristic coloured metaphosphates.
3. Answer: (c)

Teacher's Note:
1. Identify the ion from the colour, precipitate or flame, and rule out the other options one by one.

Question 5: Prussian blue is formed when:
(a) Ferrous sulphate reacts with FeCl3
(b) Ferric sulphate reacts with K4[Fe(CN)6]
(c) Ferrous ammonium sulphate reacts with FeCl3
(d) Ammonium sulphate reacts with NO2
Show Answer & Explanation

Answer: (b) Ferric sulphate reacts with K4[Fe(CN)6]

Explanation:
1. Fe3+ reacts with ferrocyanide to give Fe4[Fe(CN)6]3, ferric ferrocyanide, which is Prussian blue.
2. The other mixtures do not contain this pair.
3. Answer: (b)

Teacher's Note:
1. Identify the ion from the colour, precipitate or flame, and rule out the other options one by one.

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Multiple Choice Questions (MCQs) for JEE Chemistry Qualitative Salt Analysis

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FAQs

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