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Chemistry · Group 13 to Group 18 Elements

+\boldsymbol{Y} \boldsymbol{C l}_{2} \longrightarrow \boldsymbol{X} \boldsymbol{

\( \boldsymbol{X} \boldsymbol{C l}_{2}(\boldsymbol{e} \boldsymbol{x} \boldsymbol{c e s s})+\boldsymbol{Y} \boldsymbol{C l}_{2} \longrightarrow \boldsymbol{X} \boldsymbol{C l}_{4}+ \) \( \boldsymbol{Y} \downarrow \) \( \boldsymbol{Y} \boldsymbol{O} \frac{\boldsymbol{\Delta}}{>400} \frac{1}{2} \boldsymbol{O}_{2}+\boldsymbol{Y} \) Ore of \( Y \) would be:

  • A. Siderite
  • B. Malachite
  • C. Hornsilver
  • D. cinnabar

Step-by-step solution

The reaction XCl2 (excess) + YCl2 → XCl4 + Y↓ is characteristic of SnCl2 reducing HgCl2 to Hg, with SnCl2 oxidizing to SnCl4. Thus X = Sn, Y = Hg. The decomposition YO → (1/2)O2 + Y at >400°C matches HgO → Hg + (1/2)O2. The ore of mercury is cinnabar (HgS).
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