Final answer:
In the given compounds, iodine has the lowest oxidation state in NH4I (e), where the oxidation state is -1. This is because iodine is present as the iodide ion, which typically carries a -1 charge.
Step-by-step explanation:
The question asks about the oxidation state of iodine in various compounds. To determine in which compound iodine has the lowest oxidation state, we can consider each option:
- a. I₂ is molecular iodine where iodine is in its elemental form with an oxidation state of 0.
- b. LiIO₃ (Lithium iodate) - In this compound, iodine commonly has a +5 oxidation state.
- c. I₂O (Iodine monoxide) - Iodine in I₂O is likely to have an oxidation state of +1.
- d. IO₂ (Iodine dioxide) - In this oxide, iodine typically has a +4 oxidation state.
- e. NH₄I (Ammonium iodide) - Here, iodine exists as the iodide ion with an oxidation state of -1, which is the lowest oxidation state for iodine among the given options.
Therefore, the correct answer is e. NH₄I, where iodine has the lowest oxidation state of -1.