/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 The lattice energies of \(\mathr... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 23

The lattice energies of \(\mathrm{KBr}\) and \(\mathrm{CsCl}\) are nearly equal (Table 8.2). What can you conclude from this observation?

Short Answer

Expert verified
From the observation that the lattice energies of KBr and CsCl are nearly equal, we can conclude that the charges and distances between the ions (ionic radii) in the two compounds are similar, resulting in a similar strength of ionic bonds in both compounds. The slight differences in the ionic radii of the cations and anions in KBr and CsCl offset each other, leading to nearly equal lattice energies.

Step by step solution

01

Understand the Lattice Energy and Contributing Factors

Lattice energy is the amount of energy required to break one mole of an ionic solid into individual ions in the gas phase. It is a measure of the strength of the ionic bond between the cations and anions in the compound. There are two primary factors that determine the lattice energy of an ionic compound: the charge of the ions and the distances between the ions (ionic radii). The lattice energy can be calculated using the Born-Haber equation, or it can be qualitatively predicted using Coulomb's law: \[E_l \propto \frac{Q_1Q_2}{r}\] where \(E_l\) is the lattice energy, \(Q_1\) and \(Q_2\) are the charges of the ions, and \(r\) is the distance between the ions (ionic radii).
02

Compare the Charges and Ionic Radii of KBr and CsCl

Now let's look at the charges and ionic radii of the ions in KBr and CsCl: - KBr: K鈦� has a charge of +1, Br鈦� has a charge of -1, and the ionic radii of these ions are around 152 pm (K鈦�) and 182 pm (Br鈦�). - CsCl: Cs鈦� has a charge of +1, Cl鈦� has a charge of -1, and the ionic radii of these ions are around 167 pm (Cs鈦�) and 181 pm (Cl鈦�).
03

Analyze the Factors Contributing to the Equal Lattice Energy

From the charges and ionic radii of the ions in KBr and CsCl, we can see that both compounds have similar charges and similar ionic radii. The slightly larger size of the Cs鈦� ion compared to the K鈦� ion is offset by the slightly smaller size of the Cl鈦� ion compared to the Br鈦� ion. Since the charges and ionic radii of the ions in KBr and CsCl are similar, this results in nearly equal lattice energies for the two compounds.
04

Conclusion

From the observation that the lattice energies of KBr and CsCl are nearly equal, we can conclude that the charges and distances between the ions (ionic radii) in the two compounds are similar, resulting in a similar strength of ionic bonds in both compounds. The slight differences in the ionic radii of the cations and anions in KBr and CsCl offset each other, leading to nearly equal lattice energies.

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Most popular questions from this chapter

What is the Lewis symbol for each of the following atoms or ions: (a) \(\mathrm{Ca}\), (b) \(\mathrm{P}\), (c) \(\mathrm{Mg}^{2+}\), (d) \(S^{2-}\) ?

(a) Determine the formal charge on the chlorine atom in the hypochlorite ion, \(\mathrm{ClO}^{-}\), and the perchlorate ion, \(\mathrm{ClO}_{4}^{-}\), using resonance structures where the \(\mathrm{Cl}\) atom has an octet. (b) What are the oxidation numbers of chlorine in \(\mathrm{ClO}^{-}\) and in \(\mathrm{ClO}_{4}^{-} ?(\mathrm{c})\) Is it uncommon for the formal charge and the oxidation state to be different? Explain. (d) Perchlorate is a much stronger oxidizing agent than hypochlorite. Would you expect there to be any relationship between the oxidizing power of the oxyanion and either the oxidation state or the formal charge of chlorine?

Using Lewis symbols and Lewis structures, diagram the formation of \(\mathrm{SiCl}_{4}\) from \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms.

Explain the following trends in lattice energy: (a) \(\mathrm{CaF}_{2}>\mathrm{BaF}_{2} ;\) (b) \(\mathrm{NaCl}>\mathrm{RbBr}>\mathrm{CsBr} ;\) (c) \(\mathrm{BaO}>\mathrm{KF}\).

Use Lewissymbols and Lewis structures to diagram the formation of \(\mathrm{PF}_{3}\) from \(\mathrm{P}\) and \(\mathrm{F}\) atoms.
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