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Chapter 8: Problem 46

What noble gas has the same election configuration as each of the ions in the following compounds? a. cesium sulfide b. strontium fluoride c. calcium nitride d. aluminum bromide

Short Answer

Expert verified
The noble gases with the same electron configuration as the ions in the given compounds are as follows: a. Cesium sulfide: Krypton (Kr) b. Strontium fluoride: Strontium ion (Sr虏鈦�) - Krypton (Kr), Fluorine ion (F鈦�) - Helium (He) c. Calcium nitride: Calcium ion (Ca虏鈦�) - Argon (Ar), Nitrogen ion (N鲁鈦�) - Helium (He) d. Aluminum bromide: Aluminum ion (Al鲁鈦�) - Neon (Ne), Bromine ion (Br鈦�) - Krypton (Kr)

Step by step solution

01

a. Cesium sulfide

Cesium (Cs) is an alkali metal in group 1 and forms a 1+ ion (Cs+). Sulfur (S) is a non-metal in group 16 and forms a 2- ion (S虏鈦�). When combined, these two ions form the compound cesium sulfide (Cs鈧係). Cesium loses one electron when forming a Cs+ ion, and sulfur gains two electrons when forming an S虏鈦� ion. Thus, both ions will have the same electron configuration: Cs+: [Kr] S虏鈦�: [Kr] The noble gas with the same electron configuration as both these ions is krypton (Kr).
02

b. Strontium fluoride

Strontium (Sr) is an alkaline earth metal in group 2 and forms a 2+ ion (Sr虏鈦�). Fluorine (F) is a halogen in group 17 and forms a 1- ion (F 鈦�). The compound strontium fluoride (SrF鈧�) is formed by these two ions. Strontium loses two electrons when forming an Sr虏鈦� ion, and fluorine gains one electron when forming an F鈦� ion. The electron configuration of both ions is as follows: Sr虏鈦�: [Kr] F鈦�: [He] The noble gas with the same electron configuration as Sr虏鈦� is krypton (Kr) and as F鈦� is helium (He).
03

c. Calcium nitride

Calcium (Ca) is an alkaline earth metal in group 2 and forms a 2+ ion (Ca虏鈦�). Nitrogen (N) is a non-metal in group 15 and forms a 3- ion (N鲁鈦�). These ions make up the compound calcium nitride (Ca鈧僋鈧�). Calcium loses two electrons when forming an Ca虏鈦� ion, and nitrogen gains three electrons when forming an N鲁鈦� ion. The electron configurations are as follows: Ca虏鈦�: [Ar] N鲁鈦�: [He] The noble gas with the same electron configuration as Ca虏鈦� is argon (Ar), and as N鲁鈦� is helium (He).
04

d. Aluminum bromide

Aluminum (Al) is a metal in group 13 and forms a 3+ ion (Al鲁鈦�). Bromine (Br) is a halogen in group 17 and forms a 1- ion (Br鈦�). The compound aluminum bromide (AlBr鈧�) consists of these ions. Aluminum loses three electrons when forming an Al鲁鈦� ion, and Bromine gains one electron when forming a Br鈦� ion. We can write the electron configurations as: Al鲁鈦�: [Ne] Br鈦�: [Kr] The noble gas with the same electron configuration as the Al鲁鈦� ion is neon (Ne), and as the Br鈦� ion is krypton (Kr).

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Most popular questions from this chapter

Oxidation of the cyanide ion produces the stable cyanate ion, \(\mathrm{OCN}^{-}\). The fulminate ion, \(\mathrm{CNO}^{-}\), on the other hand, is very unstable. Fulminate salts explode when struck; \(\mathrm{Hg}(\mathrm{CNO})_{2}\) is used in blasting caps. Write the Lewis structures and assign formal charges for the cyanate and fulminate ions. Why is the fulminate ion so unstable? (C is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{N}\) is the central atom in \(\mathrm{CNO}^{-} .\).)

Compare and contrast the bonding found in the \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{HF}(\mathrm{g})\) molecules with that found in \(\operatorname{NaF}(s)\).

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part \(\mathrm{c}\), why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about \(\mathrm{MgO}\) compared to \(\mathrm{MgO}_{2} ?\) \(\mathrm{Mg}_{2} \mathrm{O} ?\)

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

Identify the five compounds of \(\mathrm{H}, \mathrm{N}\), and \(\mathrm{O}\) described below. 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 A and E exist in solution. Both solutions give off gases. Commercially available concentrated solutions of compound \(\mathrm{A}\) are normally \(16 M .\) The commercial, concentrated solution of compound \(\mathrm{E}\) is \(15 M\). d. Commercial solutions of compound \(\mathrm{E}\) 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 \(\mathrm{D}\) is \(43.7 \% \mathrm{~N}\) and \(50.0 \% \mathrm{O}\) 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 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 \(0.726 \mathrm{~g}\) compound \(\mathrm{B}\) requires \(21.98 \mathrm{~mL}\) of \(1.000 M \mathrm{HCl}\) for complete neutralization.
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