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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: - Al鲁鈦�: \(1s^2 2s^2 2p^6\) - Ba虏鈦�: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) - Se虏鈦�: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) - I鈦�: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\)

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): Al鲁鈦� (loses 3 electrons) - Ba (Group 2): Ba虏鈦� (loses 2 electrons) - Se (Group 16): Se虏鈦� (gains 2 electrons) - I (Group 17): I鈦� (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: - Al鲁鈦�: \(1s^2 2s^2 2p^6\) (loses 3s虏3p鹿 electrons) - Ba虏鈦�: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (loses 6s虏 electrons) - Se虏鈦�: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6\) (gains 2 electrons to complete 4p orbitals) - I鈦�: \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6\) (gains 1 electron to complete 5p orbitals))

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Most popular questions from this chapter

Identify the five compounds of \(\mathrm{H}, \mathrm{N},\) and \(\mathrm{O}\) described as follows. For each compound, write a Lewis structure that is consistent with the information given. a. All the compounds are electrolytes, although not all of them are strong electrolytes. Compounds \(\mathrm{C}\) and \(\mathrm{D}\) are ionic and compound \(\mathrm{B}\) is covalent. b. Nitrogen occurs in its highest possible oxidation state in compounds \(\mathrm{A}\) and \(\mathrm{C}\); nitrogen occurs in its lowest possible oxidation state in compounds \(\mathrm{C}, \mathrm{D}\), and \(\mathrm{E}\). The formal charge on both nitrogens in compound \(\mathrm{C}\) is \(+1\); the formal charge on the only nitrogen in compound \(\mathrm{B}\) is 0. c. Compounds \(\mathrm{A}\) and \(\mathrm{E}\) exist in solution. Both solutions give off gases. Commercially available concentrated solutions of compound A are normally 16\(M .\) The commercial, concentrated solution of compound \(E\) is 15\(M .\) d. Commercial solutions of compound E are labeled with a misnomer that implies that a binary, gaseous compound of nitrogen and hydrogen has reacted with water to produce ammonium ions and hydroxide ions. Actually, this reaction occurs to only a slight extent. e. Compound \(D\) is 43.7\(\% \mathrm{N}\) and 50.0\(\%\) O by mass. If compound D were a gas at STP, it would have a density of 2.86 \(\mathrm{g} / \mathrm{L} .\) f. A formula unit of compound \(\mathrm{C}\) has one more oxygen than a formula unit of compound \(\mathrm{D}\). Compounds \(\mathrm{C}\) and \(\mathrm{A}\) have one ion in common when compound \(\mathrm{A}\) is acting as a strong electrolyte. g. Solutions of compound \(\mathrm{C}\) are weakly acidic; solutions of compound \(\mathrm{A}\) are strongly acidic; solutions of compounds \(\mathrm{B}\) and \(\mathrm{E}\) are basic. The titration of 0.726 g compound \(\mathrm{B}\) requires 21.98 \(\mathrm{mL}\) of 1.000 \(\mathrm{M} \mathrm{HCl}\) for complete neutralization.

Nitrous oxide \(\left(\mathrm{N}_{2} \mathrm{O}\right)\) has three possible Lewis structures: $$: \mathrm{N}=\mathrm{N}=\dot{\mathrm{O}}\longleftrightarrow: \mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{Q}} : \longleftrightarrow : \dot{\mathrm{N}}-\mathrm{N} \equiv \mathrm{O}$$ Given the following bond lengths, \(\mathrm{N}-\mathrm{N} \qquad 167 \mathrm{pm} \quad \mathrm{N}=\mathrm{O} \quad 115 \mathrm{pm}\) \(\mathrm{N}=\mathrm{N} \qquad 120 \mathrm{pm} \quad \mathrm{N}-\mathrm{O} \quad 147 \mathrm{pm}\) \(\mathrm{N} \equiv \mathrm{N} \quad 110 \mathrm{pm}\) rationalize the observations that the N-N bond length in \(\mathrm{N}_{2} \mathrm{O}\) is 112 \(\mathrm{pm}\) and that the \(\mathrm{N}-\mathrm{O}\) bond length is 119 \(\mathrm{pm}\) . Assign formal charges to the resonance structures for \(\mathrm{N}_{2} \mathrm{O}\) . Can you eliminate any of the resonance structures on the basis of formal charges? Is this consistent with observation?

Without using Fig. 8.3, predict the order of increasing electronegativity in each of the following groups of elements. a. \(\mathrm{Na}, \mathrm{K}, \mathrm{Rb} \quad\) c. \(\mathrm{F}, \mathrm{Cl}, \mathrm{Br}\) b. \(\mathrm{B}, \mathrm{O}, \mathrm{Ga} \qquad\) d. \(\mathrm{S}, \mathrm{O}, \mathrm{F}\)

The most common type of exception to the octet rule are compounds or ions with central atoms having more than eight electrons around them. PF_. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3}\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

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. \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}\)
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