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Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory
Chapter 6, Problem 22

Which of the following spheres is likely to represent a metal atom and which a nonmetal atom? Which sphere in the products represents a cation and which an anion? Four colored spheres representing metal and nonmetal atoms, indicating cations and anions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Metal and Nonmetal Properties

Metals are typically characterized by their ability to conduct electricity and heat, malleability, ductility, and a shiny appearance. They tend to lose electrons during chemical reactions, forming cations. Nonmetals, on the other hand, are generally poor conductors, brittle in solid form, and can gain or share electrons, forming anions. Understanding these properties helps in identifying which spheres represent metals and nonmetals.
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Ionic Compounds and Charge

Ionic compounds are formed when metals react with nonmetals, resulting in the transfer of electrons. Metals lose electrons to become positively charged cations, while nonmetals gain electrons to become negatively charged anions. This charge distinction is crucial for determining the identity of the spheres in the products, as cations and anions will have different colors or sizes based on their electron configurations.
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Visual Representation of Atoms

In the provided image, different colored spheres represent various atoms, with size and color often indicating their type and charge. Typically, larger spheres may represent metals (cations) due to their lower electronegativity and tendency to lose electrons, while smaller spheres may represent nonmetals (anions) that gain electrons. Analyzing these visual cues is essential for correctly identifying the nature of each atom in the context of the question.
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Related Practice
Textbook Question
For a multielectron atom, a 3s orbital lies lower in energy than a 3p orbital because (LO 5.16) (a) a 3p orbital has more nodal surfaces than a 3s orbital. (b) an electron in a 3p orbital has a higher probability of being closer to the nucleus than an electron in a 3s orbital. (c) inner electrons shield electrons in a 3p orbital more effec-tively than electrons in a 3s orbital. (d) the energy of the electron can be spread between three 3p orbitals instead of only one 3s orbital.
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Textbook Question

Given the following information, construct a Born–Haber cycle to calculate the lattice energy of CaCl2(s). (LO 6.13)

Net energy change for the formation of CaCl2(s) form Ca(s) and Cl2(g) = -795.4 kJ/mol

Heat of sublimation for Ca(s) = +178 kJ/mol

Ei1 for Ca(s) = +590 kJ/mol

Ei2 for Ca(g) = +1145 kJ/mol

Bond dissociation energy for Cl2(g) = +243 kJ/mol

Eea1 for Cl(g) = -348.6 kJ/mol

(a) 2603 kJ/mol (b) 2254 kJ/mol (c) 2481 kJ/mo (d) 1663 kJ/mol

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Textbook Question
Which element has the largest atomic radius? (LO 5.20) (a) Rb (b) Co (c) Mg d) As
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Textbook Question

Three binary compounds are represented on the following drawing: red with red, blue with blue, and green with green. Give a likely formula for each compound.

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Textbook Question

Given the following values for steps in the formation of CaO(s) from its elements, draw a Born–Haber cycle similar to that shown in Figure 6.7. Eea1 for O1g2 = -141 kJ/mol Eea2 for O1g2 = 745.1 kJ/mol Heat of sublimation for Ca1s2 = 178 kJ/mol Ei1 for Ca1g2 = 590 kJ/mol Ei1 for Ca1g2 = 1145 kJ/mol Bond dissociation energy for O21g2 = 498 kJ/mol Lattice energy for CaO1s2 = 3401 kJ/mol

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Textbook Question

What is the difference between a molecule and an ion?

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