Question

The roasting of ore of metal usually results in the conversion of the metal to the oxide, but in the roasting of cinnabar $\left( {HgS} \right)$ , produce metallic mercury rather than oxide of mercury. If true enter 1 else 0.

Answer 1

We know for most metals in the process of roasting the metal ore is converted to the metal oxide, i.e. $MX + {O_2}\xrightarrow{{}}MO +$ other byproducts. But in case of $Hg$ the electrochemical potential of $Hg$ is so low that after being converted to $HgO$ , it is further reduced to $Hg$ , i.e. $2HgS + Heat\xrightarrow{{}}HgO + S{O_2}\xrightarrow{{}}2Hg + {O_2}$

Complete step by step answer:
Let’s start by discussing roasting.
Roasting is a basic step in the process of metallurgy (the process of extraction of pure metals from metal ore). In this process, the ore is converted into its oxide by heating it to its melting point (or above its melting point). The ore is heated in the presence of excess air (excess air means a large number of ${O_2}$ molecules as air contains ${O_2}$ molecules).
The reaction of metal ore with excess air is shown below
$MX + {O_2}\xrightarrow{{}}MO +$ Other byproducts
Here $X$ any unwanted elements that may be bonded to the metal atom. The other byproducts of this reaction are removed to obtain pure metal oxide.
Now, in the problem given to us, we are asked to discuss why in the process of roasting of cinnabar we obtain pure $Hg$ instead of $HgO$ .
So, in the process of roasting of cinnabar ore all the steps till the production of $HgO$ are very similar to the steps involved in the roasting of any metal ore, but after this, we see a step that is unique to cinnabar ores.
So, the electrochemical potential of $Hg$ is so low that after being converted to $HgO$ , it is further reduced to $Hg$ as shown in the reaction below
$2HgS + Heat\xrightarrow{{}}HgO + S{O_2}\xrightarrow{{}}2Hg + {O_2}$
So in this reaction, we obtain $Hg$ instead of $HgO$ without having to perform any extra processes.
So, the statement given in the problem is true, hence enter 1.

Note: In the solution above when we were discussing the roasting of metals we discussed that during roasting $MO$ is formed. To obtain pure metal after this an additional process or step is added which involves heating the metal oxide with a reducing agent which reduces the metal oxide to pure metal.