What is formula of Cuperous Iodide?

Question

Accepted Answer

The correct option is (1):

Cuprous iodide refers to copper(I) iodide. In inorganic nomenclature, "cuprous" indicates that copper is in its +1 oxidation state, meaning it has lost one electron. Iodide is an anion with a -1 charge ($I^{-}$). For a neutral compound to form, the charges of the cation (copper) and the anion (iodide) must balance. Therefore, one copper(I) ion ($Cu^{+}$) combines with one iodide ion ($I^{-}$) to form the compound CuI. This compound is a white solid, sparingly soluble in water, and is often used as a source of iodide in organic synthesis.

The incorrect option is (2):

$CuI_{2}$ would imply that copper is in the +2 oxidation state (cupric) and is bonded to two iodide ions, or that copper is in the +1 oxidation state and bonded to two iodine atoms with a net -2 charge. However, $CuI_{2}$ as a stable compound with copper in the +2 state and two iodide ions is not generally found because $Cu^{2+}$ readily oxidizes $I^{-}$ to $I_{2}$, resulting in the formation of $CuI$.

The incorrect option is (3):

$Cu_{2}I$ would imply that two copper atoms are bonded to one iodide, which would suggest a copper charge of +0.5 per copper atom if the iodide is -1. This formulation does not align with the stable and common oxidation states of copper in simple ionic compounds.

The incorrect option is (4):

$Cu_{2}I_{2}$ is essentially a dimeric form of CuI, but when writing the empirical formula for an ionic compound, we always simplify the ratio to its lowest whole numbers. Therefore, while two CuI units might associate, the formula is expressed as CuI.