Problem 45 - Chemistry An Atoms First Approach

Question

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

Step-by-Step Solution

Answer

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⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\)

Step 1: Find the atomic numbers of the elements and their most common oxidation states

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

Step 2: Write the ground state electron configuration for each element

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: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^5\)

Step 3: Write the electron configurations for the most stable ions

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⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^10 4s^2 4p^6 4d^10 5s^2 5p^6\)