Discussion Forum

1)

$XCl_{2}(excess)+ YCl_{2}\rightarrow XCl_{4}+Y\downarrow$

$YO$    $\underrightarrow{\triangle}$    $\frac{1}{2}O_{2}+Y$

          >400°C

 Ore of Y would be

Answer:

Option D

Explanation:

$SnCl_{2}+HgCl_{2}\rightarrow SnCl_{4}+Hg$

   $(xCl_{2})$  $(yCl_{2})$    $(xCl_{2})$   (y)

 $HgO$    $\underrightarrow{\triangle}$   $Hg+ \frac{1}{2}O_{2}$

                >400° c   (y) 

          

 So ore of  $\gamma$ is HgS , i.e, cinnabar