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Chapter 8: Problem 26

Which of the following trends in lattice energy is due to differences in ionic radii? (a) \(\mathrm{LiF}>\mathrm{NaF}>\mathrm{CsF},(\mathbf{b}) \mathrm{CaO}>\mathrm{KCl}\), (c) \(\mathrm{PbS}>\mathrm{Li}_{2} \mathrm{O}\).

Short Answer

Expert verified
(a) \\(LiF > NaF > CsF\\) is due to differences in ionic radii.

Step by step solution

01

Understanding Lattice Energy

Lattice energy is the energy required to separate one mole of an ionic solid into its gaseous ions. It is influenced by the charges on the ions and the distance between the ions, which is related to their radii.
02

Identifying Ionic Radii Influence

The smaller the ionic radii, the shorter the distance between ions and the greater the lattice energy. Therefore, trends based on differences in ionic radii will show higher lattice energies for compounds with smaller ionic radii.
03

Analyzing Trend (a): Lithium, Sodium, Cesium Fluorides

In \(LiF> NaF> CsF\), the trend shows decreasing lattice energies. As you move from Li to Na to Cs in the alkali metals, the ionic radii increase. Hence, this trend can be explained by differences in ionic radii.
04

Analyzing Trend (b): Calcium Oxide and Potassium Chloride

In \(CaO > KCl\), Calcium and Oxygen ions both have higher charges compared to Potassium and Chlorine ions. The trend is likely due to the higher charges rather than just ionic radii, making this not primarily based on ionic radii differences.
05

Analyzing Trend (c): Lead Sulfide and Lithium Oxide

In \(PbS > Li_2O\), the comparison involves different cations with different charges and sizes. The trend here does not clearly result from ionic radii differences alone.
06

Conclusion

After analyzing the trends, the trend \(LiF > NaF > CsF\) is consistent with lattice energy differences due mainly to differences in ionic radii, as smaller ions result in higher lattice energies.

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Key Concepts

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

Ionic Radii
Ionic radii refer to the effective size of an ion in a crystal lattice. The size of ions can greatly influence a compound's physical and chemical properties, such as lattice energy.

When ions form, metals tend to lose electrons to become cations, while non-metals gain electrons to become anions. Losing electrons results in a smaller radius due to reduced electron-electron repulsion, whereas gaining electrons increases repulsion and makes anions larger in radius.

The size of an ionic radius is crucial in determining how closely ions can pack together in a solid. Smaller ions can pack more closely, reducing the distance between them, and often result in higher lattice energies.
  • *Trends in a Period:* As you move across a period on the periodic table, ionic radii tend to decrease for cations and slightly increase for anions. This is due to increasing nuclear charge, which attracts electrons more strongly.
  • *Trends in a Group:* Moving down a group, ionic radii generally increase. This trend occurs because each successive element has an additional electron shell, increasing the overall size of the ion.
Ionic Compounds
Ionic compounds are made up of ions held together by ionic bonds, which are formed by the complete transfer of electrons from one atom to another. These compounds are typically composed of a metal and a non-metal. The metal atom loses one or more electrons, becoming a positively charged ion (cation), while the non-metal gains those electrons, becoming a negatively charged ion (anion).

Characteristics of ionic compounds include
  • *High Melting and Boiling Points:* The strong ionic bonds in the lattice require significant energy to break, leading to high melting and boiling points.
  • *Electrical Conductivity:* In the solid state, ionic compounds do not conduct electricity as ions are fixed in place. However, when dissolved in water or melted, the ions are free to move, allowing them to conduct electricity.
  • *Solubility:* Many ionic compounds are soluble in water, as the polar nature of water molecules can overcome the electrostatic forces holding the ions together.

The stability and properties of ionic compounds largely depend on the balance between the attractive forces of the ions and the repulsive forces between them.
Ionic Solids
Ionic solids are a class of compounds characterized by the formation of a lattice structure consisting of alternating positive and negative ions. These solids are held together by strong electrostatic forces, which significantly influence their physical properties.

This organized lattice structure results in some distinct features:
  • *Rigidity and Brittleness:* The fixed position of ions in the lattice makes ionic solids hard, but they are brittle because shifting layers leads to repulsion between like-charged ions, causing the material to shatter.
  • *Poor Electrical Conductivity in Solid State:* In their solid form, ionic solids do not conduct electricity due to the immobility of ions. However, once melted or dissolved in water, they can conduct electricity as the ions are free to move.
  • *High Melting and Boiling Points:* Due to the strong attraction between the ions, significant energy is required to break these bonds, leading to high melting and boiling points.

Ionic solids form crystalline structures that can sometimes be identified by their geometric shapes. The strength and characteristic properties of these solids are mainly due to the strong ionic bonds and their arrangement in the lattice.

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Most popular questions from this chapter

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 \(\mathrm{O}^{2-}\). Yet \(\mathrm{CaO}\) is stable relative to the free elements. Which statement is the best explanation? (a) The lattice energy of \(\mathrm{CaO}\) is large enough to overcome these processes. (b) \(\mathrm{CaO}\) is a covalent compound, and these processes are irrelevant. (c) CaO has a higher molar mass than either Ca or O. (d) The enthalpy of formation of \(\mathrm{CaO}\) is small. \((\mathbf{e}) \mathrm{CaO}\) is stable to atmospheric conditions.

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 \(30.4 \% \mathrm{~N}\). Measurements of its molecular mass yield a value of \(184.3 \mathrm{~g} / \mathrm{mol}\). 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 \(S-S\) distance in the \(S_{8}\) ring is 205 pm.) \((\mathbf{d})\) The enthalpy of formation of the compound is estimated to be \(480 \mathrm{~kJ} / \mathrm{mol}^{-1} . \Delta H_{f}^{\circ}\) of \(\mathrm{S}(g)\) is \(222.8 \mathrm{~kJ} / \mathrm{mol}\). Estimate the average bond enthalpy in the compound.

(a) Use Lewis symbols to represent the reaction that occurs between Li and O atoms. (b) What is the chemical formula of the most likely product? (c) How many electrons are transferred? (d) Which atom loses electrons in the reaction?

(a) Which of these compounds is an exception to the octet rule: carbon dioxide, water, ammonia, phosphorus trifluoride, or arsenic pentafluoride? (b) Which of these compounds or ions is an exception to the octet rule: borohydride \(\left(\mathrm{BH}_{4}^{-}\right)\), borazine \(\left(\mathrm{B}_{3} \mathrm{~N}_{3} \mathrm{H}_{6},\right.\) which is analogous to benzene with alternating \(\mathrm{B}\) and \(\mathrm{N}\) in the ring), or boron trichloride?

What is the Lewis symbol for each of the following atoms or ions? \((\mathbf{a}) \mathrm{Be},(\mathbf{b}) \mathrm{Rb},(\mathbf{c}) \mathrm{I}^{-},(\mathbf{d}) \mathrm{Se}^{2-} .\)
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