Match the anionic species given in Column I that are present in the ore(s) given in Column II.
← JEE Advanced 2015 · Paper 1
ChemistryMatrix match+8 / -4Question 39 of 60
JEE Advanced 2015 · Paper 1, Chemistry question 39
Column I
- ACarbonate
- BSulphide
- CHydroxide
- DOxide
Column II
- PSiderite
- QMalachite
- RBauxite
- SCalamine
- TArgentite
| P | Q | R | S | T | |
|---|---|---|---|---|---|
| A | |||||
| B | |||||
| C | |||||
| D |
Answer and worked solution
Official answer: A → P, Q, S; B → T; C → Q, R; D → R
- Siderite: FeCO₃.
- Malachite: CuCO₃·Cu(OH)₂.
- Bauxite: AlOₓ(OH)₃₋₂ₓ, hydrated alumina.
- Calamine: ZnCO₃.
- Argentite: Ag₂S.
So:
- carbonate → siderite, malachite, calamine;
- sulphide → argentite;
- hydroxide → malachite, bauxite;
- oxide → bauxite.
Answer: A → P, Q, S; B → T; C → Q, R; D → R