Partial Roasting of chalcopyrite produces:

\(Cu_2O\) and \(FeO\)

The correct answer is option 2. \(Cu_2O\) and \(FeO\).

Partial roasting is a process used in the metallurgical extraction of metals from their sulfide ores. It involves heating the ore in the presence of oxygen at temperatures below its melting point. During partial roasting, only a portion of the sulfur in the sulfide ore is oxidized, while the metal component remains partly in the sulfide form.

In the case of chalcopyrite (\(CuFeS_2\)), the reaction during partial roasting can be represented as follows:

\[4CuFeS_2 + 9O_2 \rightarrow 2Cu_2S + 2FeO + 4SO_2\]

Here's a breakdown of the reaction:

The chalcopyrite ore (\(CuFeS_2\)) is heated in the presence of oxygen (\(O_2\)).

A portion of the sulfur in chalcopyrite is oxidized to form sulfur dioxide (\(SO_2\)).

The copper in chalcopyrite is partly oxidized to form copper sulfide (\(Cu_2S\)) and copper oxide (\(Cu_2O\)).

The iron in chalcopyrite is partly oxidized to form iron oxide (\(FeO\)).

The reaction results in the formation of copper oxide (\(Cu_2O\)) and iron oxide (\(FeO\)) as the main products of partial roasting of chalcopyrite.

Copper oxide (\(Cu_2O\)) can be further processed to obtain metallic copper, while iron oxide (\(FeO\)) is usually considered as waste or undergoes further processing depending on the specific extraction process.

Therefore, the correct answer is option 2: \(Cu_2O\) and \(FeO\).