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Ch.6 - Ionic Compounds: Periodic Trends and Bonding Theory
Chapter 6, Problem 15

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

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Key Concepts

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

Atomic Radius

Atomic radius is a measure of the size of an atom, typically defined as the distance from the nucleus to the outermost shell of electrons. It can vary based on the element's position in the periodic table, with larger atoms generally found towards the bottom left. Understanding atomic radius is crucial for predicting how atoms interact and bond with one another.
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Periodic Trends

Periodic trends refer to the predictable patterns observed in the properties of elements as you move across or down the periodic table. Atomic radius generally increases down a group due to the addition of electron shells, while it decreases across a period from left to right due to increased nuclear charge. Recognizing these trends helps in determining the atomic size of different elements.
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Group and Period Position

The position of an element in the periodic table, defined by its group (column) and period (row), significantly influences its atomic radius. Elements in the same group share similar properties, including atomic size, while those in the same period exhibit a decrease in atomic radius from left to right. This knowledge is essential for comparing the atomic sizes of the elements listed in the question.
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Related Practice
Textbook Question

Which molecular scale image best represents the ionic com-pound that forms between cesium and chlorine? (Cesium is represented by red circles, and chlorine is represented by blue circles.) (LO 6.12) (a)

(b)

(c)

(d)

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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 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?

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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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