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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⁺ has a charge of +1, Br⁻ has a charge of -1, and the ionic radii of these ions are around 152 pm (K⁺) and 182 pm (Br⁻). - CsCl: Cs⁺ has a charge of +1, Cl⁻ has a charge of -1, and the ionic radii of these ions are around 167 pm (Cs⁺) and 181 pm (Cl⁻).
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⁺ ion compared to the K⁺ ion is offset by the slightly smaller size of the Cl⁻ ion compared to the Br⁻ 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

Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of \(69.6 \% \mathrm{~S}\) and \(30.4 \%\) N. Measurements of its molecular mass yield a value of \(184.3 \mathrm{~g} \mathrm{~mol}^{-1} .\) The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond distances between the atoms in the ring. (Note: The \(\mathrm{S}-\mathrm{S}\) distance in the \(S_{8}\) ring is \(2.05 \AA\).) (d) The enthalpy of formation of the compound is estimated to be \(480 \mathrm{~kJ} \mathrm{~mol}^{-1}\). \(\Delta H_{f}^{\circ}\) of \(S(g)\) is \(222.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Estimate the average bond enthalpy in the compound.

(a) What is the trend in electronegativity going from left to right in a row of the periodic table? (b) How do electronegativity values generally vary going down a column in the periodic table? (c) How do periodic trends in electronegativity relate to those for ionization energy and electron affinity?

Predict the ordering of the \(\mathrm{C}-\mathrm{O}\) bond lengths in \(\mathrm{CO}\), \(\mathrm{CO}_{2}\), and \(\mathrm{CO}_{3}^{2-}\)

You and a partner are asked to complete a lab entitled "Oxides of Ruthenium" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a soft yellow substance and the other a black powder. You also find the following notes in your partner's notebook-Compound 1: \(76.0 \%\) Ru and \(24.0 \%\) O (by mass), Compound 2: \(61.2 \%\) Ru and \(38.8 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound 2? (c) Upon determining the melting points of these two compounds, you find that the yellow compound melts at \(25^{\circ} \mathrm{C}\), while the black powder does not melt up to the maximum temperature of your apparatus, \(1200^{\circ} \mathrm{C}\). What is the identity of the yellow compound? What is the identity of the black compound? Be sure to use the appropriate naming convention depending upon whether the compound is better described as a molecular or ionic compound.

Using Lewis symbols, diagram the reaction between magnesium and oxygen atoms to give the ionic substance \(\mathrm{MgO}\).

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