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Chapter 8: Problem 45

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

Expert verified
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

01

(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): \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^5\))
02

(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))
03

(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))

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Most popular questions from this chapter

Write electron configurations for a. the cations \(\mathrm{Mg}^{2+}, \mathrm{K}^{+},\) and \(\mathrm{Al}^{3+}\) . b. the anions \(\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-},\) and \(\mathrm{Te}^{2-}\)

In general, the higher the charge on the ions in an ionic compound, the more favorable the lattice energy. Why do some stable ionic compounds have \(+1\) charged ions even though \(+4,+5,\) and \(+6\) charged ions would have a more favorable lattice energy?

Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Al}\) and \(\mathrm{Cl} \quad\) c. \(\mathrm{Sr}\) and \(\mathrm{F}\) b. \(\mathrm{Na}\) and \(\mathrm{O} \quad\) d. \(\mathrm{Ca}\) and \(\mathrm{Se}\)

Arrange the atoms and/or ions in the following groups in order of decreasing size. a. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) b. \(\mathrm{Fe}^{2+}, \mathrm{Ni}^{2+}, \mathrm{Zn}^{2+}\) c. \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}\)

One type of exception to the octet rule are compounds with central atoms having fewer than eight electrons around them. BeH_ and \(\mathrm{BH}_{3}\) are examples of this type of exception. Draw the Lewis structures for \(\mathrm{BeH}_{2}\) and \(\mathrm{BH}_{3} .\)
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