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Match the anionic species in Column I with the ore(s) in Column II that contain them. Column I: (A) Carbonate (B) Sulphide (C) Hydroxide (D) Oxide Column II: (P) Siderite (Q) Malachite (R) Bauxite (S) Calamine (T) Argentite

1. (A)-(P), (B)-(T), (C)-(R), (D)-(S)
2. (A)-(Q), (B)-(T), (C)-(R), (D)-(S)
3. (A)-(Q), (B)-(S), (C)-(R), (D)-(T)
4. (A)-(P), (B)-(S), (C)-(R), (D)-(T)

Correct answer: (A)-(Q), (B)-(T), (C)-(R), (D)-(S)

Solution

Ore compositions: Siderite = FeCO3 (carbonate) Malachite = Cu2(OH)2CO3 = CuCO3.Cu(OH)2 (carbonate + hydroxide) Bauxite = Al2O3.2H2O (oxide/hydroxide — listed as hydroxide here) Calamine = ZnCO3 (carbonate) OR ZnSiO3 in some sources; traditionally ZnCO3 Argentite = Ag2S (sulphide) Matching: (A) Carbonate: Malachite (Q) contains carbonate; Siderite (P) also contains carbonate; Calamine (S) contains carbonate. But the options only give one ore per species. Looking at the option (B): (A)-(Q): Malachite contains carbonate. (B)-(T): Argentite is Ag2S, sulphide. (C)-(R): Bauxite contains hydroxide (Al(OH)3 or Al2O3.2H2O). (D)-(S): Calamine = ZnCO3... that would make D carbonate, not oxide. Hmm. Calamine is sometimes ZnCO3 (carbonate), sometimes used for a different ore. Let me reconsider: if Calamine is taken as carbonate (ZnCO3), then D-oxide has no clear assignment except... Siderite (FeCO3) is also carbonate. Standard classification used in Indian textbooks: Siderite = FeCO3 (carbonate), Malachite = Cu(OH)2.CuCO3 (carbonate + hydroxide), Bauxite = Al2O3.xH2O (oxide), Calamine = ZnCO3 (carbonate), Argentite = Ag2S (sulphide). Option (B): (A)-(Q) malachite-carbonate, (B)-(T) argentite-sulphide, (C)-(R) bauxite-hydroxide... but bauxite is oxide! And (D)-(S) calamine-oxide? Calamine is carbonate. This is contradictory. In JEE context, Bauxite (Al2O3.2H2O) is classified as an oxide ore (Al2O3). Calamine could be ZnSiO3 (silicate) or ZnCO3. Most likely intended answer in JEE: (A) Carbonate -> Malachite (Q) [CuCO3.Cu(OH)2]; (B) Sulphide -> Argentite (T) [Ag2S]; (C) Hydroxide -> Bauxite (R) is debatable but if Al2O3.2H2O treated as hydroxide; (D) Oxide -> Siderite? No. Actually if Calamine = ZnCO3 and Siderite = FeCO3: neither is oxide. Possibly the question uses: Bauxite = oxide (Al2O3), and the hydroxide is Malachite. But the option (B) places Malachite under carbonate and Bauxite under hydroxide. Given the standard answer for JEE questions of this type is option B.

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