Which of the following is the correct electron configuration for the iodide ion (I−)?

• A. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^4
• B. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6
• C. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^5
• D. 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10

Answer: (B)

Verified step by step guidance

1. Identify the atomic number of iodine (I), which is 53, meaning a neutral iodine atom has 53 electrons.
2. Write the electron configuration for neutral iodine by filling orbitals in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, and 5p, until all 53 electrons are accounted for.
3. Recognize that the iodide ion (I⁻) has gained one extra electron compared to neutral iodine, so it has 54 electrons in total.
4. Add one electron to the next available orbital in the neutral iodine configuration, which is the 5p orbital, increasing the number of electrons in 5p from 5 to 6.
5. Write the full electron configuration for I⁻, ensuring the 5p orbital is fully filled with 6 electrons, reflecting the extra electron gained.