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Isolation of Elements question

2015 · Shift 1 · Q1
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Isolation of Elements question

2015 · Shift 1 · Q1

JEE AdvancedChemistryIsolation of ElementsMCQ+8 / −4
Match the anionic species given in Column-I that are present in the ore(s) given in Column-II
Column - I Column - II
(A) Carbonate (p) Siderite
(B) Sulphide (q) Malachite
(C) Hydroxide (r) Bauxite
(D) Oxide (s) Calamine
(t) Argentite
  1. A
    (A) →\to→ p,q,s; (B) →\to→ t; (C) →\to→ q,r; (D) →\to→ r
  2. B
    (A) →\to→ s; (B) →\to→ s,t; (C) →\to→ r; (D) →\to→ r,s
  3. C
    (A) →\to→ p,s; (B) →\to→ s,t; (C) →\to→ r; (D) →\to→ s
  4. D
    (A) →\to→ p,q,s; (B) →\to→ s,t; (C) →\to→ p,r; (D) →\to→ t
View written solutionFree

Correct answer: A

  1. Write the formulas of the ores in Column-II

    • Siderite =FeCO3= \mathrm{FeCO_3}=FeCO3​
    • Malachite =CuCO3⋅Cu(OH)2= \mathrm{CuCO_3\cdot Cu(OH)_2}=CuCO3​⋅Cu(OH)2​
    • Bauxite =Al2O3⋅2H2O= \mathrm{Al_2O_3\cdot 2H_2O}=Al2​O3​⋅2H2​O (hydrated alumina; contains oxide/hydroxide character)
    • Calamine =ZnCO3= \mathrm{ZnCO_3}=ZnCO3​
    • Argentite =Ag2S= \mathrm{Ag_2S}=Ag2​S

  2. Identify the anionic species present in each ore

    • Siderite FeCO3\mathrm{FeCO_3}FeCO3​ contains carbonate CO32−\mathrm{CO_3^{2-}}CO32−​

      ⇒p\Rightarrow p⇒p belongs to Carbonate

    • Malachite CuCO3⋅Cu(OH)2\mathrm{CuCO_3\cdot Cu(OH)_2}CuCO3​⋅Cu(OH)2​ contains both:

      • carbonate CO32−\mathrm{CO_3^{2-}}CO32−​
      • hydroxide OH−\mathrm{OH^-}OH−

      ⇒q\Rightarrow q⇒q belongs to Carbonate and Hydroxide

    • Bauxite Al2O3⋅2H2O\mathrm{Al_2O_3\cdot 2H_2O}Al2​O3​⋅2H2​O is a hydrated oxide, commonly treated as containing oxide and hydroxide character.

      ⇒r\Rightarrow r⇒r belongs to Hydroxide and Oxide

    • Calamine ZnCO3\mathrm{ZnCO_3}ZnCO3​ contains carbonate CO32−\mathrm{CO_3^{2-}}CO32−​

      ⇒s\Rightarrow s⇒s belongs to Carbonate

    • Argentite Ag2S\mathrm{Ag_2S}Ag2​S contains sulphide S2−\mathrm{S^{2-}}S2−

      ⇒t\Rightarrow t⇒t belongs to Sulphide

  3. Now match Column-I with Column-II

    • (A) Carbonate →p,q,s\to p, q, s→p,q,s
    • (B) Sulphide →t\to t→t
    • (C) Hydroxide →q,r\to q, r→q,r
    • (D) Oxide →r\to r→r

  4. Compare with the options

    This exactly matches Option A.

  5. Verification with stored answer

    Stored correct answer = A

    Derived answer = A

    Hence, they agree.

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