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NTA Abhyas JEE Main2020ChemistryGeneral Principles and Processes of Isolation of MetalsPractice

X C l 2 e x c e s s + Y C l 2 → X C l 4 + Y ↓ YO → > 400 o C Δ 1 2 O 2 +Y , Ore of Y would be

Options

  1. ASiderite
  2. BCinnabar
  3. CMalachite
  4. DHornsilver

Correct answer

B. Cinnabar

Step-by-step solution

The reaction sequence is as follows S n C l 2 ( X C l 2 ) + H g C l 2 ( Y C l 2 ) → S n C l 4 ( X C l 4 ) + H g ( Y ) HgO → >   400 o C Δ Hg   +   1 2 O 2 HgS → Cinnabar

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