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

Short Answer

Expert verified
The trends in lattice energy for the given ionic compounds can be explained as follows: (a) NaCl > RbBr > CsBr: Lattice energy decreases as the size of the cations and anions increases, which is the case in this series. (b) BaO > KF: Lattice energy is higher for BaO due to the greater positive charge of Ba虏鈦 and the greater negative charge of O虏鈦 compared to K鈦 and F鈦. (c) SrO > SrCl鈧: SrO has a higher lattice energy than SrCl鈧 because the greater negative charge of O虏鈦 ion leads to stronger electrostatic attraction with Sr虏鈦.

Step by step solution

01

Explaining the trend in the lattice energy of NaCl, RbBr, and CsBr

In this series, NaCl has the highest lattice energy, and CsBr has the lowest. This can be attributed to the difference in the size of the cations and anions involved. Both Na鈦 and Cl鈦 ions are smaller than Rb鈦 and Br鈦 ions, which in turn are smaller than Cs鈦 and Br鈦 ions. Since lattice energy is inversely proportional to ion size, the lattice energy of NaCl will be higher than that of RbBr, which in turn will have a higher lattice energy than CsBr.
02

Explaining the trend in the lattice energy of BaO and KF

In this case, BaO has a higher lattice energy than KF. This can be explained by looking at the charges of the ions. The Ba虏鈦 ion has a greater positive charge than K鈦, and the O虏鈦 ion has a greater negative charge than F鈦. As a result, the electrostatic attraction between Ba虏鈦 and O虏鈦 is stronger than that between K鈦 and F鈦, leading to a higher lattice energy in BaO compared to KF.
03

Explaining the trend in the lattice energy of SrO and SrCl鈧

In this series, SrO has a higher lattice energy than SrCl鈧. Since Sr虏鈦 is a common cation in both compounds, the difference in lattice energy must be attributed to the anions involved. O虏鈦 ion has a higher negative charge than Cl鈦 ions, and due to the greater electrostatic attraction between Sr虏鈦 and O虏鈦, SrO has a higher lattice energy compared to SrCl鈧. In conclusion, the trends in lattice energy covered in this exercise can be explained by considering the charge and size of the ions involved in the ionic compounds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionic Compounds
Ionic compounds are substances composed of positive and negative ions held together by the strong electrostatic forces between oppositely charged particles. These compounds typically form when a metal reacts with a non-metal, transferring electrons from the metal to the non-metal.

Consider sodium chloride (NaCl) as an example. Sodium (Na) loses one electron to become Na鈦, while chlorine (Cl) gains an electron to become Cl鈦. They're then bonded together to form NaCl, an ionic compound. The structure of these compounds is often a repeating 3D lattice where each ion is surrounded by ions of opposite charge, optimizing the ionic bonds and overall stability.

When comparing ionic compounds like NaCl, RbBr, and CsBr, their differences in lattice energy 鈥 the energy released when the ions come together to form a solid 鈥 can be explained by specific properties of the ions involved. These include ionic size and the magnitude of charges on the ions, both of which are crucial in determining the strength of the attraction within the compound.
Ion Size and Charge
Ion size and charge play a pivotal role in determining the lattice energy of ionic compounds. Generally, smaller ions can get closer together, resulting in a stronger attraction and higher lattice energy. Conversely, larger ions result in weaker attractions and lower lattice energies.

Here's how we apply this concept to explain variances in lattice energies:
  • NaCl vs. RbBr vs. CsBr: Sodium has a smaller ionic radius compared to rubidium and cesium, leading to a stronger electrostatic attraction between Na鈦 and Cl鈦 ions, resulting in a higher lattice energy for NaCl. Additionally, the smaller size of Cl鈦 compared to Br鈦 contributes further to NaCl's higher lattice energy.
  • BaO vs. KF: In BaO, the ions involved have higher charges compared to those in KF. Since Ba虏鈦 is smaller than K鈦 and has a double positive charge, it can attract O虏鈦, which has a double negative charge, more strongly than K鈦 can attract F鈦. This results in a stronger electrostatic force and a higher lattice energy for BaO compared to KF.
Electrostatic Attraction
Electrostatic attraction is the force that draws together positively charged cations and negatively charged anions in an ionic compound. It explains how ions with opposite charges can form stable, solid structures despite their individual charge-based tendencies to repel same-charged species.

The strength of this electrostatic attraction is governed by Coulomb's law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between their centers. This means that ions with higher charges will attract each other more strongly, while increased distance caused by larger ionic sizes would weaken this attraction.

Considering SrO vs. SrCl鈧 as an example, the higher lattice energy of SrO is due to the electrostatic attraction between Sr虏鈦 and O虏鈦 being stronger than that between Sr虏鈦 and two Cl鈦 ions. This is because while Sr虏鈦 remains constant in both compounds, the doubly charged O虏鈦 creates a stronger attraction than the singly charged Cl鈦, following Coulomb's law.

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Most popular questions from this chapter

Based on Lewis structures, predict the ordering of \(\mathrm{N}-\mathrm{O}\) bond lengths in \(\mathrm{NO}^{+}, \mathrm{NO}_{2}^{-},\) and \(\mathrm{NO}_{3}^{-}\).

Consider benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) in the gas phase. (a) Write the reaction for breaking all the bonds in \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\), and use data in Appendix \(\mathrm{C}\) to determine the enthalpy change for this reaction. (b) Write a reaction that corresponds to breaking all the carbon-carbon bonds in \(\mathrm{C}_{6} \mathrm{H}_{6}(g) .\) (c) By combining your answers to parts (a) and (b) and using the average bond enthalpy for \(\mathrm{C}-\mathrm{H}\) from Table \(8.4,\) calculate the average bond enthalpy for the carbon-carbon bonds in \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\). (d) Comment on your answer from part (c) as compared to the values for \(\mathrm{C}-\mathrm{C}\) single bonds and \(\mathrm{C}=\mathrm{C}\) double bonds in Table 8.4

List the individual steps used in constructing a Born-Haber cycle for the formation of \(\mathrm{Bal}_{2}\) from the elements. Which of the steps would you expect to be exothermic?

(a) Write one or more appropriate Lewis structures for the nitrite ion, \(\mathrm{NO}_{2}^{-}\). (b) With what allotrope of oxygen is it isoelectronic? (c) What would you predict for the lengths of the bonds in \(\mathrm{NO}_{2}^{-}\) relative to \(\mathrm{N}-\mathrm{O}\) single bonds and double bonds?

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