Question 1Level 2 · ExamSingle correct
A colourless aqueous solution X contains one or more of Ag+, Pb2+ and Hg2+ (as nitrates) and no other metal ion. Use the standard scheme: Group I, dil. HCl; Group II, H2S in dil. HCl; Group III, NH4Cl + NH4OH; Group IV, H2S in ammoniacal medium; Group V, (NH4)2CO3 in the presence of NH4Cl + NH4OH. Dilute HCl gives a white precipitate, which dissolves completely in hot water; the hot solution gives a yellow precipitate with K2CrO4. The filtrate from the HCl step, still acidic, gives a black precipitate when saturated with H2S. Which conclusion is justified? Exactly one option is correct.
AHg2+ must be present, because the black Group II precipitate can only be HgS.
BPb2+ is present and Ag+ is absent; the black precipitate may be PbS from Pb2+ left over from Group I, so Hg2+ can be reported only after a confirmatory test for mercury on the Group II precipitate.
CBoth Ag+ and Pb2+ are present.
DPb2+ is absent, because a lead precipitate would not dissolve in hot water.
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Answer: B
Complete solubility of the Group I precipitate in hot water rules out AgCl (insoluble in hot water), so Ag+ is absent. The hot extract gives yellow PbCrO4, so Pb2+ is present.
PbCl2 is only sparingly soluble, so Pb2+ is never removed completely in Group I; the remainder is precipitated as black PbS in Group II. A black Group II precipitate is therefore not a unique observation: Hg2+ may be reported only if the Group II precipitate gives a confirmatory test for mercury (black residue insoluble in hot 50% HNO3, which after dissolution in aqua regia gives white turning grey with SnCl2).
(A) over-concludes; (C) contradicts the complete dissolution in hot water; (D) ignores the chromate test.
Difficulty: HardQuestion 2Level 3 · AdvancedMatching list
Each entry of List-I describes what is seen when aqueous KI is added (dropwise and then in excess where stated) to a dilute aqueous solution of the nitrate of a single cation, together with one further observation. Match each with the cation in List-II. Exactly one option is correct.
List-I- (P) Scarlet precipitate that dissolves in excess KI to give a colourless solution; the original solution gives a yellow precipitate with NaOH.
- (Q) White precipitate in a brown solution; adding Na2S2O3 removes the brown colour and leaves the white precipitate.
- (R) No precipitate; the solution turns brown and then blue on adding starch; the original solution turns blood-red with KSCN.
- (S) Yellow precipitate that dissolves on boiling with water and separates as shining golden flakes on cooling.
List-II- (1) Ag+
- (2) Cu2+
- (3) Pb2+
- (4) Hg2+
- (5) Fe3+
AP→4, Q→2, R→1, S→3
BP→2, Q→4, R→5, S→3
CP→4, Q→5, R→2, S→1
DP→4, Q→2, R→5, S→3
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Answer: D
(P) Scarlet HgI2 dissolves in excess iodide as colourless [HgI4]2−; the yellow precipitate with NaOH is HgO. → Hg2+ (4).
(Q) 2Cu2++4I−→2CuI↓+I2: white CuI in a brown (iodine/triiodide) solution; thiosulphate reduces the iodine and leaves white CuI. → Cu2+ (2).
(R) 2Fe3++2I−→2Fe2++I2: no precipitate, only liberated iodine (blue with starch); blood-red with KSCN confirms iron(III). → Fe3+ (5).
(S) Yellow PbI2 dissolves in hot water and recrystallises as golden spangles on cooling. → Pb2+ (3).
Ag+ (1) gives yellow AgI, which does not dissolve in hot water or in excess dilute KI to a colourless solution, and gives no brown solution; it is the distractor.
Code: P→4, Q→2, R→5, S→3.
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
A black Group II precipitate (from H2S in dil. HCl) is boiled with 50% HNO3; a black residue remains. The residue is dissolved in aqua regia, the excess acid is boiled off, the residue is taken up in water and SnCl2 solution is added dropwise. Which description of the residue and the SnCl2 test is correct? Exactly one option is correct.
AThe residue is CuS; SnCl2 gives a deep blue solution.
BThe residue is PbS; SnCl2 gives a white precipitate of PbCl2 that stays white.
CThe residue is HgS; SnCl2 first gives white Hg2Cl2 and then grey mercury, being itself oxidised to tin(IV).
DThe residue is HgS; SnCl2 is reduced by Hg2+ to metallic tin, which appears grey.
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Answer: C
Of the Group II sulphides in the syllabus list, CuS and PbS dissolve in hot 50% HNO3; HgS does not, so the black residue is HgS. Aqua regia converts it to HgCl2.
SnCl2 reduces it in two steps: 2HgCl2+SnCl2→Hg2Cl2↓(white)+SnCl4, then Hg2Cl2+SnCl2→2Hg↓(grey/black)+SnCl4. Tin(II) is oxidised to tin(IV).
(A) CuS would have dissolved in the nitric acid; (B) PbS also dissolves in it; (D) reverses the redox roles.
Difficulty: Very hard