When Pb2+ reacts with 2I- ions, what is formed as a solid?

When lead ions (Pb²⁺) react with iodide ions (I⁻), the reaction leads to the formation of lead(II) iodide, which is represented by the formula PbI₂. In this reaction, the lead ions combine with two iodide ions to form a neutral compound because the positively charged lead ion balances the negative charges from the two iodide ions.

Lead(II) iodide is notable for its low solubility in water, meaning that when these ions come together in solution, they will precipitate out to form a solid. This precipitation is consistent with the characteristics of many ionic compounds, where the formation of a solid indicates a successful chemical reaction between ions in solution.

The other options do not reflect the proper stoichiometry or the nature of the compounds formed in this reaction. For instance, PbI₄ and Pb₂I do not correspond to the usual products formed by the combination of lead(II) ions with iodide ions under standard conditions, and iodine (I₂) is a diatomic element that does not result from this reaction. Thus, the formation of solid PbI₂ is the correct interpretation of the chemical interaction occurring in this scenario.