The problem asks which of the given reagents, when heated with copper(I) sulfide ($Cu_2S$), will produce copper metal ($Cu$). This process involves a redox reaction where $Cu_2S$ acts as a reducing agent. The sulfide ion ($S^{2-}$) in $Cu_2S$ gets oxidized, typically to $SO_2$, while copper ions in the reagent (and in $Cu_2S$ itself) get reduced to elemental copper ($Cu^0$). This type of reaction is known as auto-reduction or self-reduction in metallurgy.

Let's analyze each option:

1. Option A: $CuFeS_2$ (Chalcopyrite) Heating $CuFeS_2$ with $Cu_2S$ involves a mixture of metal sulfides. There is no oxidizing agent (like oxygen) present to convert the sulfide to $SO_2$. For copper metal to be formed, copper ions must be reduced. This would require the oxidation of another species. In a mixture of sulfides, such a redox reaction to produce free metal is not favorable under simple heating. The extraction of copper from chalcopyrite requires roasting in air, not reaction with $Cu_2S$. Therefore, this option is incorrect.

2. Option B: CuO (Copper(II) oxide) When copper(II) oxide is heated with copper(I) sulfide, a redox reaction occurs. The sulfide ion is oxidized, and copper ions ($Cu^{2+}$ from $CuO$ and $Cu^+$ from $Cu_2S$) are reduced to copper metal. The balanced chemical equation is: $Cu_2S(s) + 2CuO(s) \xrightarrow{\Delta} 4Cu(l) + SO_2(g)$ In this reaction:

• Oxidation: $S^{2-} \rightarrow S^{4+} + 6e^-$
• Reduction: $2Cu^+ + 2e^- \rightarrow 2Cu$ and $2Cu^{2+} + 4e^- \rightarrow 2Cu$ Overall, 6 electrons are transferred. This reaction is thermodynamically feasible at high temperatures. Thus, copper metal is produced. This option is correct.

3. Option C: $Cu_2O$ (Copper(I) oxide) This is the classic auto-reduction reaction that occurs during the extraction of copper in a Bessemer converter. Copper(I) sulfide reacts with copper(I) oxide, which is formed by the partial oxidation of $Cu_2S$. The balanced chemical equation is: $Cu_2S(s) + 2Cu_2O(s) \xrightarrow{\Delta} 6Cu(l) + SO_2(g)$ In this reaction:

• Oxidation: $S^{2-} \rightarrow S^{4+} + 6e^-$
• Reduction: $6Cu^+ + 6e^- \rightarrow 6Cu$ This is a key step in producing blister copper. Thus, copper metal is produced. This option is correct.

4. Option D: $CuSO_4$ (Copper(II) sulfate) Heating copper(II) sulfate with copper(I) sulfide also results in a redox reaction producing copper metal. This is analogous to the self-reduction process seen in the metallurgy of lead ($PbS + PbSO_4 \rightarrow 2Pb + 2SO_2$). The balanced chemical equation is: $Cu_2S(s) + CuSO_4(s) \xrightarrow{\Delta} 3Cu(s) + 2SO_2(g)$ In this reaction, there is a comproportionation of sulfur ($S^{-2}$ and $S^{+6}$ both form $S^{+4}$), and reduction of copper ($Cu^+$ and $Cu^{2+}$ both form $Cu^0$). Let's check the redox balance:

• Reactants: Total charge from Cu is $(2 \times +1) + (+2) = +4$. Total charge from S is $(-2) + (+6) = +4$.
• Products: Total charge from Cu is $3 \times 0 = 0$. Total charge from S is $2 \times (+4) = +8$.
• The change in oxidation state for Cu is $0 - 4 = -4$ (reduction). The change in oxidation state for S is $+8 - 4 = +4$ (oxidation). The reaction is balanced in terms of redox changes. Thus, copper metal is produced. This option is correct.

Based on the analysis, reagents CuO, $Cu_2O$, and $CuSO_4$ give copper metal upon heating with $Cu_2S$.