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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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hey everyone in this example, we need to determine which element between gallium and titanium will have the larger atomic radius. We should recall that atomic radius is going to increase on our periodic tables as we go down on our periodic table. So we would find larger radius is further down on our periodic table. So what we can do is go ahead and identify where gallium is on our periodic tables first. And we would identify that gallium is located across period four In Group three A. Now we want to compare this to our location for titanium on our periodic tables, where we would see titanium falls across period six on our periodic tables. And it's also located in group three A. Which is the road down below gallium. So we can say that these two atoms both have the same group number but they're in different periods. And again our period is the group that falls across our periodic tables. So they're the rose across our periodic tables. And as we said above, we recalled that atomic radius will increase the further down the periodic table that you go. So because Our titanium is in period six, we would go ahead and say that therefore it's lower on the periodic table because it's at a higher period number and therefore will have a higher atomic radius or we can say larger atomic radius. And so to complete this example, we would confirm that titanium is the atom between gallium and itself that has the larger atomic radius. So I hope that everything I reviewed is clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.
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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