Write electron configurations for the most stable ion formed by each of the elements Al, Ba, Se, and I (when in stable ionic compounds).

Look up the atomic numbers and the most common oxidation states for the elements Al, Ba, Se, and I. For reference purposes, we can use a periodic table. Al (Aluminium) - Atomic number: 13, Most common oxidation state: +3 Ba (Barium) - Atomic number: 56, Most common oxidation state: +2 Se (Selenium) - Atomic number: 34, Most common oxidation state: -2 I (Iodine) - Atomic number: 53, Most common oxidation state: -1

Using the atomic numbers, write the electron configurations for the neutral atoms of Al, Ba, Se, and I. Al: \(1s^2 2s^2 2p^6 3s^2 3p^1\) Ba: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6 6s^2\) Se: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^4\) I: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^5\)

Based on the most common oxidation states, we can determine the electron configurations for the most stable ions formed by each element. Al: As it tends to lose 3 electrons, the most stable ion is Al³⁺ (Al+3). Its electron configuration is: \(1s^2 2s^2 2p^6\) Ba: As it tends to lose 2 electrons, the most stable ion is Ba²⁺ (Ba+2). Its electron configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\) Se: As it tends to gain 2 electrons, the most stable ion is Se²⁻ (Se-2). Its electron configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6\) I: As it tends to gain 1 electron, the most stable ion is I⁻ (I-1). Its electron configuration is: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\) In conclusion, the electron configurations for the most stable ions formed by the elements Al, Ba, Se, and I are: Al³⁺: \(1s^2 2s^2 2p^6\) Ba²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\) Se²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6\) I⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\)