What will be the product formed when the crude metal of Zirconium is heated in an evacuated vessel with iodine?

\(ZrI_4\)

The correct answer is option 1. \(ZrI_4\).

When the crude metal of zirconium (\(Zr\)) is heated in an evacuated vessel with iodine (\(I_2\)), it undergoes a reaction to form zirconium iodides. The reaction involves the combination of zirconium with iodine. The balanced chemical equation for the reaction is:

\[ 2Zr(s) + 4I_2(g) \rightarrow 2ZrI_4(s) \]

In this reaction, zirconium reacts with iodine molecules to form zirconium tetraiodide (\(ZrI_4\)). The subscript "4" in \(ZrI_4\) indicates that each zirconium atom is bonded to four iodine atoms. The reaction is often carried out in an evacuated vessel to remove any air or other gases, ensuring a clean reaction between zirconium and iodine.

The other options (\(ZrI_2\) and \(ZrI_3\)) are not the primary products formed in this specific reaction. Zirconium iodides with different stoichiometries may be formed under various conditions or by different methods, but \(ZrI_4\) is a common product obtained when zirconium reacts with iodine.