Final answer:
In rubidium iodide, iodine atoms are larger than rubidium atoms, taking into account their ionic forms, with iodide ions (I−) being larger than the rubidium ions (Rb+).
Step-by-step explanation:
When drawing a particle diagram of rubidium iodide, it is essential to consider the relative sizes of rubidium (Rb) and iodine (I) atoms. Rubidium iodide is an ionic compound consisting of Rb+ and I− ions. Based on their positions in the periodic table, we know that as you move down a group, the atomic radii increase due to additional electron shells. Therefore, rubidium (Rb), which is below iodine (I) in the periodic table, will have a larger atomic radius. In the ionic state, Rb+ ions are smaller than Rb atoms due to the loss of an electron resulting in a decreased electron-electron repulsion in the outer shell. On the other hand, I− ions are larger than I atoms as the addition of an electron increases repulsion among electrons in the valence shell.
Iodine generally has a larger atomic radius than bromine because it has more electron shells. When it forms an ion by gaining an extra electron to become I−, the iodide ion's radius is even larger due to increased electron-electron repulsion. When comparing ionic radii, the rubidium ion (Rb+) will have a smaller radius compared to the iodide ion (I−).
Based on this information, the correct particle diagram option is: b) I atoms larger than Rb atoms, in the context of their ionic forms in rubidium iodide.