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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?

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Hey everyone we're asked referred to the illustration below, identify the sphere that depicts a metal atom and the sphere that depicts a non metal atom. Among the products, identify the spheres that depict a caddy on and the sphere that depicts an an ion. We've learned that a medal that tends to lose electrons forms a cat A on while a nonmetal tends to gain electrons forms and an ion. So for a cat ian, this will have less electrons than our parent adam. In comparison to an an ion, this will have more electrons than the parent adam. So why is this important? This is important because this will tell us which of these fears is going to be the an ion and the cat eon. So since Acadian will have less electrons than the parent atom, this will have the smaller radius. Looking at our spheres. This means that the green sphere is going to be our caddy on And this is also going to be our medal in comparison to an an ion since it has more electrons than the parent atom, This will have a larger radius. And comparing our spheres, we can see that the red sphere has a larger radius. So this is our an ion and the red sphere is also going to be our non metal and this is going to be our final answers. Now, I hope that made sense. And let us know if you have any questions
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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