Study the following steps:
${\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}\xrightarrow{{{\text{roast in air}}}}{\text{(X)}}\xrightarrow{{{\text{roast without air}}}}{\text{(Y)}}$
Based on the above steps identify X and Y.

Option	X	Y
A	Mixture of ${\text{Cu}}$ and ${\text{CuO}}$	Mixture of ${\text{Cu}}$ and ${\text{S}}{{\text{O}}_2}$
B	Mixture of ${\text{C}}{{\text{u}}_{\text{2}}}{\text{O}}$ and ${\text{S}}{{\text{O}}_{\text{2}}}$	Mixture of ${\text{Cu}}$ and ${\text{S}}{{\text{O}}_2}$
C	Mixture of ${\text{Cu}}$ and ${\text{S}}{{\text{O}}_{\text{2}}}$	Mixture of ${\text{CuO}}$ and ${\text{Cu}}$
D	Mixture of ${\text{Cu}}$ and ${\text{CuO}}$	Mixture of ${\text{CuO}}$ and ${\text{Cu}}$

Hint:In the above question, \[{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}\] is roasted in air and hence we can say that oxidation roasting takes place. Then the product of the reaction will again react with remaining \[{\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}\] to carry out reduction reaction.

Complete step by step answer:
In the above question, the equation given is:
${\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}\xrightarrow{{{\text{roast in air}}}}{\text{(X)}}\xrightarrow{{{\text{roast without air}}}}{\text{(Y)}}$
The word roast indicates the roasting process is being carried out. Roasting is a process in which the ore of the material is either heated alone or in the presence of some substance so as to convert the impurities in its volatile form.
To find out X:
Since ${\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}$ is roasted with air, it indicates the ore is heated in the presence of \[{{\text{O}}_{\text{2}}}\] to get converted into its oxides and impurities are converted into volatile form ,so that it can escape.
$C{u_2}S{\text{ + }}{{\text{O}}_2} \to {\text{ C}}{{\text{u}}_2}{\text{O + S}}{{\text{O}}_2}$
where ${\text{C}}{{\text{u}}_{\text{2}}}{\text{O}}$ is oxide form of ${\text{Cu}}$ and ${\text{S}}{{\text{O}}_{\text{2}}}$ is the volatile form of ${\text{S}}$.
By balancing the above equation, we get:
$2C{u_2}S{\text{ + 3}}{{\text{O}}_2} \to {\text{ 2C}}{{\text{u}}_2}{\text{O + 2S}}{{\text{O}}_2}$
To find out Y:
Roasting without air indicates that it is a reduction reaction. Here, the product ${\text{C}}{{\text{u}}_{\text{2}}}{\text{O}}$ reacts with remaining ${\text{C}}{{\text{u}}_{\text{2}}}{\text{S}}$.
${\text{C}}{{\text{u}}_{\text{2}}}{\text{S + C}}{{\text{u}}_{\text{2}}}{\text{O }} \to {\text{ Cu + S}}{{\text{O}}_{\text{2}}}$
By balancing the above equation, we get:
${\text{2C}}{{\text{u}}_{\text{2}}}{\text{S + C}}{{\text{u}}_{\text{2}}}{\text{O }} \to {\text{ 6Cu + S}}{{\text{O}}_{\text{2}}}$
As here the product reacts with its reactant and gets reduced therefore, this is a self-reduction reaction.
As X is a mixture of ${\text{C}}{{\text{u}}_{\text{2}}}{\text{O}}$ and ${\text{S}}{{\text{O}}_2}$, Y is a mixture of ${\text{Cu}}$ and ${\text{S}}{{\text{O}}_{\text{2}}}$.

Therefore, option B is correct.

Note:
We need to remember that not all metals are capable of undergoing self-reduction processes. Only the less electropositive metals like Hg, Pb, Cu etc. can take part.
The sulphide ores of the electropositive metals like Hg, Pb, Cu etc. are heated in the presence of air to convert the part of the ore into the oxide or sulphate which then reacts with sulphide ore to obtain the metal and sulphur dioxide. In this process, no reducing agent is used.