Book Cover Chemistry 8th Edition / Chang
Student Study Guide

Chapter 9: Chemical Bonding I: Basic Concepts


Index | 9.1 – 9.2 | 9.3 | 9.4 & 9.6 | 9.5 | 9.7 | 9.8 | 9.9 | 9.10 |

IONIC BONDING AND LATTICE ENERGY (9.3)

STUDY OBJECTIVES

  1. Predict the relative strength of ionic bonds using Coulomb's law.
  2. List three energy terms that influence the tendency for two elements to form an ionic compound.
  3. Use the Born-Haber cycle to calculate the magnitude of the lattice energies of ionic solids.

Coulomb's Law. The force that gives rise to the ionic bond is the electrical attraction existing between a positive ion and a negative ion. Chemical bonding results when the energy of two interacting atoms (or ions) is lowered. Coulomb's law states that the potential energy (E) of interaction of two ions is directly proportional to the product of their charges and inversely proportional to the distance between them:

where Q+ and Q– are the charges of the two ions, r is their distance of separation, and k is a proportionality constant (its value will not be needed). When one ion is positive and the other negative, E will be negative. Bringing two oppositely charged particles closer together lowers their energy. The lower the value of the potential energy, the more stable is the pair of ions. Energy would need to be added to separate the two ions.

The factors that govern the stability of ion pairs are the magnitude of their charges, and the distance between the ion centers. The distance between ionic centers (r) is the sum of the ionic radii of the individual ions.

r = r+ + r–

The energy of attraction between two oppositely charged ions depends:

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Lattice Energy. In an ionic crystal the cations and anions are arranged in an orderly three-dimensional array, as illustrated in Figure 2.11 of the textbook. In a sodium chloride or rock salt type of crystal, the anion has six nearest neighbor cations. The ions are packed in such a way as to maximize attraction and minimize repulsion.

The lattice energy provides a measure of the attraction between ions and the strength of the ionic bond. The lattice energy is the energy required to separate the ions in 1 mole of a solid ionic compound into gaseous ions. Gaseous ions are far enough apart from one another that they do not interact. For example, the lattice energy of NaF(s) is equal to 908 kJ/mol. This means 908 kJ are required to vaporize one mole of NaF(s) and form one mole of Na+ ions and one mole of F– ions in the gas phase.

NaF(s) Na+(g) + F–(g)     lattice energy = 908 kJ

The lattice energy is related to the stability of the ionic solid. The larger the lattice energy, the more stable the solid.

The value of the lattice energy depends on the charges of the ions and the ionic radii, in accordance with Coulomb's law. This can be seen in Table 9.1 by comparing the sodium halides. As the sum of the two ionic radii increases in going from NaF to NaI, the separation of ionic centers (r = r+ + r–) increases and the lattice energy decreases. The effect of ionic charge on lattice energy can be seen by comparing the lattice energies of NaF and CaO in Table 9.1. The larger value for CaO is due to the stronger attraction of doubly charged cations (Ca2+) for doubly charged anions (O).

The lattice energies are reflected in the melting points of ionic crystals. During melting the ions gain enough kinetic energy to overcome the potential energy of attraction, and they move away from each other. The higher the melting point, the more energy the ions need to separate from one another. Therefore as the lattice energy increases so does the melting point.

Table 9.1 Lattice Energies of Several Ionic Compounds

r+ + r–*
(pm)
Lattice Energy
(kJ/mol)
Melting Point
(°C)
NaF 231 908 1012
NaCl 276 788 801
NaBr 290 736 747
NaI 314 686 660
CaO 239 3540 2580

* See visual aid Figure 8.9 (textbook) for ionic radii values.

Factors Favoring Formation of Ionic Bonds. It is important to identify any properties of atoms that affect their ability to form ionic compounds. In the formation of ions from atoms, the metal atom loses an electron and the nonmetal atom gains an electron.

Li(g) Li+(g) + e–
e– + F(g) F–(g)

The overall change is:

Li(g) + F(g) Li+(g) + F–(g)

One factor favoring the formation of an ionic compound is the ease with which the metal atom loses an electron. A second factor is the tendency of the nonmetal atom to gain an electron. Thus ionic compounds tend to form between elements of low ionization energy and those of high electron affinity. The alkali metals and alkaline earth metals have low ionization energies. They tend to form ionic compounds with the halogens and Group 6A elements both of which have high electron affinities.

A third factor is the lattice energy. Electrostatic attraction of the ions results in large amounts of energy being released when two kinds of gaseous ions are brought together to form a crystal lattice.

Li+(g) + F–(g) LiF(s)

In general, the smaller the ionic radius and the greater the ionic charge, the greater the lattice energy.

Calculation of the Lattice Energy Using the Born-Haber Cycle. Lattice energies cannot be measured directly and must be calculated using Hess's law. In Chapter 6 you learned that if a reaction can be broken down into a series of steps, the overall enthalpy of reaction is equal to the sum of the enthalpy changes for the individual steps. The series of steps used to calculate the lattice energies of ionic solids is called the Born-Haber cycle. Each step is one you've seen previously in relation to the properties of atoms. Here we will illustrate the calculation of the lattice energy for potassium chloride.

KCl(s) K+(g) +Cl–(g)     lattice energy = ?

The Born-Haber cycle starts by taking the overall equation to be the reaction in which the ionic compound is formed from the elements in their standard states. In this example the enthalpy change is the same as the enthalpy of formation of potassium chloride from potassium and chlorine.

This reaction is then envisioned to occur by a number of steps.

  1. First potassium sublimes:

  2. Next, diatomic chlorine is dissociated into Cl atoms. The required energy is the bond energy for 1/2 mole of Cl—Cl bonds.

Up until now the focus has been on making the atoms of the elements potassium and chlorine. Next these will be made into the appropriate ions.

  1. Now ionize 1 mole of potassium atoms:

  2. To form a Cl– ion, an electron is added to the chlorine atom. This will release an energy equal to the electron affinity of chlorine, EA. Because energy is given off, H has the opposite sign of the electron affinity.

  3. Finally, 1 mole of gaseous K+ ions, and 1 mole of gaseous Cl– ions are combined to make 1 mole of KCl(s). An amount of energy equal to the lattice energy will be released. The lattice energy must have the same magnitude as , but an opposite sign.

The summation of the steps gives the overall reaction above.

Therefore, in general:

Hof(KCl) =

or

Hooverall = Ho1 + Ho2 + Ho3 + Ho4 + Ho5

where Ho5 = –lattice energy

–lattice energy = Ho5 = Hooverall – [Ho1 + Ho2 + Ho3 + Ho4]

EXAMPLE Predicting Melting Points

Which member of the pair will have the higher melting point?

  1. NaCl or
    CaO
  2.          

  3. NaCl or
    NaI
  4.          


EXAMPLE The Born-Haber Cycle

Given the following data calculate the lattice energy of potassium chloride:

Enthalpy of sublimation of potassium = 90.0 kJ
Bond energy (BE) of Cl—Cl = 242.7 kJ
Ionization energy (I) of K = 419 kJ
Electron affinity (EA) of Cl = +349 kJ
DHof(KCl) = –435.9 kJ

Lattice energy =
         


OBJECTIVE CHECK

Complete the following questions to check your understanding of the material. Select the check button to see if you answered correctly.

  1. Write a chemical equation for the process that corresponds to the lattice energy of MgO.
  2. Which member of the pair will have the higher melting point?
    1. KCl or
      CaCl2
    2. RbI or
      NaI
  3. Which member of the pair will have the higher lattice energy?
    1. NaCl or
      CaO
    2. KI or
      KCl
  4. Calculate the lattice energy of sodium bromide from the following information.

    Hsubl (Na) = 109 kJ
    Ionization Energy (Na) = 496 kJ
    Bond Energy (Br—Br) = 192 kJ
    Electron Affinity (Br) = 324 kJ
    Hof(NaBr) = –359 kJ kJ



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