Final answer:
The electron configuration of the iodide ion (I⁻) is 1s² 2s² 2p²³ 3s² 3p²³ 4s² 3d⁶ 4p²³ 5s² 4d⁶ 5p⁴, corresponding to the configuration of xenon, the nearest noble gas.
Step-by-step explanation:
The electron configuration of the iodide ion, which is an anion of iodine, corresponds to the configuration of a neutral iodine atom gaining one additional electron. A neutral iodine atom has 53 electrons, and the iodide ion (I⁻) has 54 electrons as it gains an extra electron to achieve a stable, full outer electron shell.
Step 1 in understanding this process is to write out the electron configuration for a neutral iodine atom. Iodine's atomic number is 53, so its electron configuration up to the 4d subshell is 1s² 2s² 2p²³ 3s² 3p²³ 4s² 3d⁶ 4p²³. The electrons continue to fill into the 5s and 4d subshells, leading us to 5s² 4d⁶. Finally, the remaining electrons go into the 5p subshell, leading to a neutral iodine configuration of 5p³.
For the iodide ion, an additional electron is added to the 5p subshell because ions strive to achieve the most stable, noble gas electron configuration. Therefore, the electron configuration for the iodide ion is 1s² 2s² 2p²³ 3s² 3p²³ 4s² 3d⁶ 4p²³ 5s² 4d⁶ 5p⁴, which corresponds to the same electron configuration as xenon (Xe), the nearest noble gas.