Problem 26
Question
Which of the following trends in lattice energy is due to differences in ionic radii? (a) \(\mathrm{NaCl}>\mathrm{RbBr}>\mathrm{CsBr}\), (b) \(\mathrm{BaO}>\mathrm{KF}\), (c) \(\mathrm{SrO}>\mathrm{SrCl}_{2}\).
Step-by-Step Solution
Verified Answer
The lattice energy trend in option (a) \(\mathrm{NaCl}>\mathrm{RbBr}>\mathrm{CsBr}\) is due to differences in ionic radii, as these compounds have similar charges on their ions and the size of the cations increases as we move down the group. Options (b) and (c) have lattice energy trends that are not solely dictated by the ionic radii.
1Step 1: Option (a) Analysis
In option (a), we have \(\mathrm{NaCl}\), \(\mathrm{RbBr}\), and \(\mathrm{CsBr}\). They are group 1 alkali metal halide compounds with similar charges on their ions (+1 for the metal and -1 for the halide). Thus, the difference in lattice energy depends on the ionic radii of these ions. As we move down the group, the size of the cations increases, decreasing the electrostatic interactions, and hence the lattice energy. This means \(\mathrm{NaCl}>\mathrm{RbBr}>\mathrm{CsBr}\) in terms of lattice energy can be explained by differences in ionic radii.
2Step 2: Option (b) Analysis
For option (b), we have \(\mathrm{BaO}\) and \(\mathrm{KF}\). \(\mathrm{BaO}\) has a divalent metal cation (\(\mathrm{Ba}^{2+}\)), while \(\mathrm{KF}\) has a monovalent cation (\(\mathrm{K}^+\)). Even if the ionic radii for \(\mathrm{Ba}^{2+}\) and \(\mathrm{K}^+\) are different, the fact that \(\mathrm{BaO}\) has a higher charge on the cation plays a crucial role in the lattice energy. So, we cannot attribute the lattice energy trend purely to ionic radii in this case.
3Step 3: Option (c) Analysis
For option (c), we have \(\mathrm{SrO}\) and \(\mathrm{SrCl}_{2}\). Both compounds contain the divalent metal cation \(\mathrm{Sr}^{2+}\), but the anions are different (\(\mathrm{O}^{2-}\) in \(\mathrm{SrO}\) and \(\mathrm{Cl}^-_{2}\) in \(\mathrm{SrCl}_{2}\)). The difference in the charge of the anions plays an important role in the lattice energy, and therefore, we cannot attribute their lattice energy trend solely to differences in the ionic radii.
4Step 4: Conclusion
Based on our analysis, the lattice energy trend in option (a) \(\mathrm{NaCl}>\mathrm{RbBr}>\mathrm{CsBr}\) is due to differences in ionic radii. Options (b) and (c) have lattice energy trends that are not solely dictated by the ionic radii.
Key Concepts
Ionic RadiiAlkali Metal HalidesElectrostatic Interactions
Ionic Radii
Ionic radii play a crucial role in determining the properties of ionic compounds. They refer to the effective size of an ion in a crystal lattice. When atoms gain or lose electrons to form ions, their size changes - cations become smaller, while anions become larger. This change in size affects how ions pack together in a solid structure.
In a group of the periodic table, as you move down, the ionic radii increase because additional electron shells are added. This increase in size affects the lattice energy of compounds. The larger the ion, the less densely they can pack, and the weaker the electrostatic forces between them.
In a group of the periodic table, as you move down, the ionic radii increase because additional electron shells are added. This increase in size affects the lattice energy of compounds. The larger the ion, the less densely they can pack, and the weaker the electrostatic forces between them.
- Smaller ions, such as in NaCl, allow for stronger interactions, hence higher lattice energy.
- Larger ions, as seen descending a group, generally exhibit lower lattice energies.
Alkali Metal Halides
Alkali metal halides are a group of compounds consisting of alkali metals partnered with halogens. Some common examples include NaCl, KBr, and LiF. These compounds illustrate the nature of ionic bonding really well.
Each alkali metal has exactly one electron in its outer shell, which it readily donates to achieve a stable electron configuration similar to the noble gases. Similarly, halogens need just one electron to complete their valence shell, so they form ions that are negatively charged.
Each alkali metal has exactly one electron in its outer shell, which it readily donates to achieve a stable electron configuration similar to the noble gases. Similarly, halogens need just one electron to complete their valence shell, so they form ions that are negatively charged.
- The combination results in ions with opposite charges, held together by strong electrostatic forces.
- Despite similar charge states, the varying sizes of the alkali metals (like Na+, Rb+) lead to differences in lattice energies.
Electrostatic Interactions
Electrostatic interactions are the forces that occur between charged particles. In the context of ionic solids, these interactions occur between positively charged cations and negatively charged anions.
The strength of these interactions directly influences the lattice energy of ionic compounds. Stronger electrostatic interactions result in higher lattice energies. The magnitude of these forces depends on:
The strength of these interactions directly influences the lattice energy of ionic compounds. Stronger electrostatic interactions result in higher lattice energies. The magnitude of these forces depends on:
- The charge on the ions: Higher charges lead to stronger interactions.
- The distance between ions: Smaller distances (or smaller ionic radii) lead to stronger interactions.
Other exercises in this chapter
Problem 21
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(a) Does the lattice energy of an ionic solid increase or decrease (i) as the charges of the ions increase, (ii) as the sizes of the ions increase? (b) Arrange
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Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+}\), and energy is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm
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List the individual steps used in constructing a Born-Haber cycle for the formation of \(\mathrm{BaI}_{2}\) from the elements. Which of the steps would you expe
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