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).

Short answer

The electron configurations for the most stable ions of the elements are: - Al³⁺: \(1s^2 2s^2 2p^6\) - Ba²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) - Se²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) - I⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\)

Step-by-step solution

(Identify the atomic numbers and electron configurations)

(First, we need to find the atomic numbers of Al, Ba, Se, and I from the periodic table. The atomic numbers are: Al - 13, Ba - 56, Se - 34, I - 53. The electron configurations for the neutral atoms are as follows: - Al (Z=13): \(1s^2 2s^2 2p^6 3s^2 3p^1\) - Ba (Z=56): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2\) - Se (Z=34): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^4\) - I (Z=53): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^5\))

(Determine the most common ion formed)

(By looking at the periodic table and group numbers, the most stable ions for each element are: - Al (Group 13): Al³⁺ (loses 3 electrons) - Ba (Group 2): Ba²⁺ (loses 2 electrons) - Se (Group 16): Se²⁻ (gains 2 electrons) - I (Group 17): I⁻ (gains 1 electron))

(Write the electron configurations for the most stable ion)

(Write the electron configurations for the most stable ions of the elements: - Al³⁺: \(1s^2 2s^2 2p^6\) (loses 3s²3p¹ electrons) - Ba²⁺: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (loses 6s² electrons) - Se²⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) (gains 2 electrons to complete 4p orbitals) - I⁻: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (gains 1 electron to complete 5p orbitals))