Step-by-step Solution:

1. Understand the Reactant and Process: The question asks for the products of the partial roasting of chalcopyrite. The chemical formula for chalcopyrite is CuFeS₂. Roasting is a metallurgical process that involves heating a sulfide ore in the presence of air (oxygen). "Partial" roasting implies that the reaction is carried out with a limited supply of air or under controlled conditions to achieve a specific set of products, rather than complete oxidation.

2. Analyze the Components of Chalcopyrite: Chalcopyrite (CuFeS₂) can be considered as a mixed sulfide containing copper, iron, and sulfur. When heated in the presence of oxygen, both the metal sulfides can be oxidized.

3. Apply Principles of Metallurgy: In the extraction of copper, the goal of the initial roasting step is to remove iron as an impurity. Iron has a higher affinity for oxygen than copper. Therefore, the iron sulfide component of the ore is preferentially oxidized over the copper sulfide component.

4. Write the Chemical Reactions: The partial roasting of chalcopyrite proceeds as follows:

   • Chalcopyrite (CuFeS₂) is heated in a limited supply of air. The iron sulfide is oxidized to ferrous oxide (FeO), while the copper sulfide is converted to cuprous sulfide (Cu₂S). Sulfur is oxidized to sulfur dioxide gas (SO₂).
   • The overall balanced chemical equation for this process is: $2\text{CuFeS}_2(s) + 4\text{O}_2(g) \rightarrow \text{Cu}_2\text{S}(s) + 2\text{FeO}(s) + 3\text{SO}_2(g)$
   • The solid product is a mixture of cuprous sulfide (Cu₂S) and ferrous oxide (FeO), which is called "matte". The sulfur dioxide (SO₂) escapes as a gas.

5. Evaluate the Options: Based on the reaction products derived above, let's examine the given options:

   • A: Cu₂S and FeO: This option correctly identifies the main solid products of the partial roasting of chalcopyrite. Cuprous sulfide and ferrous oxide form the matte.
   • B: Cu₂O and FeO: This is incorrect. During partial roasting, copper is not oxidized to Cu₂O. It remains in the sulfide form (Cu₂S). The conversion of Cu₂S to Cu₂O occurs later during the Bessemerization stage for self-reduction.
   • C: CuS and Fe₂O₃: This is incorrect. CuFeS₂ gets converted to the more stable cuprous sulfide (Cu₂S), not CuS. Also, under the controlled conditions of partial roasting, iron is primarily oxidized to ferrous oxide (FeO), not ferric oxide (Fe₂O₃), as FeO is required for slag formation (FeSiO₃) in the next step.
   • D: Cu₂O and Fe₂O₃: This is incorrect for the reasons mentioned for options B and C. It represents a more complete oxidation, which is not the goal of this specific step.

6. Conclusion: The partial roasting of chalcopyrite (CuFeS₂) yields cuprous sulfide (Cu₂S) and ferrous oxide (FeO) as the main solid products. Thus, option A is the correct answer.