Lattice Energy & Ionic Bonds:

Problem 6.58: Order the following compounds according to their expected lattice energies: LiCl, KCl, KBr, MgCl₂.

The potential energy between two ions is given (here as a proportion) roughly by

If the signs on the charges are oppositive, we have a negative energy corresponding to attraction. If the charges have the same sign, we have a positive energy of repulsion. In a crystal lattice containing ions, all these attractions and repulsions add up to the lattice energy. For a given crystal structure these can be determined by the above formula with an inclusion of what is called a Madelung constant. We won't discuss that but do tell you that it is a straightforward, albeit complicated and tedious, calculation!
The two effects we look at are the charge sizes and the ionic radii. The closer two ions are, the stronger the attraction and repulsion. The larger the charges, the larger the interaction. Using those trends as "barometers" for the size of the energies, we can easily order the above species as

Problem 6.59: Order the following compounds according to their expected lattice energies: AlBr₃, MgBr₂, LiBr, CaO.

We follow the same rules here as in the previous problem. Also, note that, with aluminum bromide, we have four ions interacting, three with magnesium bromide, and just two with the other two substances. However, with CaO both ions of charges of magnitude 2. Putting those ideas together gives us, then,

Be sure to study this one carefully until you completely understand it!

Problem 6.61: Cesium has the smallest ionization energy of all elements (376 kJ/mol) and chlorine has the most negative electron affinity (-349 kJ/mol). Will a cesium atom donate an electonn to a chlorine atom to form isolated Cs⁺ and Cl^- ions?

To understand this, let us add the energies to get the total energy.

The total energy is positive and, hence, the process is not favored!

Problem 6.63: Look up the lattice energies in Table 6.3 and calculate the energy change (in kJ/mol)
for the formation of the following substances from their elements:

First, we present the table:

Now, we are ready to do the calculations. Note that the "reactions" sum in the same way as do the energies. This is an important point to note in that we shall be doing many things like this in many different chapters! These are done without diagrams; you may wish to draw your own as a valuable exercise if you are uncomfortable visualizing things with just numbers.

(a)	LiF (The sublimation energy for Li is +159.4 kJ/mol, the E_i for Li is 520 kJ/mol, the E_ea for F is -328 kJ/mol, and the bond dissociation energy of F₂ is +158 kJ/mol.)

(b)	CaF₂ (The sublimation energy for Ca is +178.2 kJ/mol, E_a1 = +589.8 kJ/mol, and E_a2 = +1145 kJ/mol.)

Here we do things in much the same way EXCEPT that we have two ionization energies for the calcium. Otherwise, things are pretty much straightforward and easy to pHollow. Also, I am trying a different format for this sort of thing, so bear with me...

Problem 6.65: Calculate a lattice energy for CaH₂ (in kJ/mol) using the following information.

*Property*	*Value (kJ/mol)*
E_ea for H	-72.8
E_i₁for Ca	+589.8
E_i₂for Ca	+1145
Heat of sublimation for Ca	+178.2
Bond dissociation energy for H₂	+435.9
Net energy change for the formation of CaH₂ from its elements	-186.2

Here goes another rather involved calculation. However, it involves nothing more difficult than addition and subtraction with one multiplication to give you some practice... . Below, we let "U" denote the lattice energy; note that this is reported as a positive number (whereas the formation of a lattice is actually negative); hence, the handling of the minus sign on "U." The lattice energy is released when the lattice is formed. As a positive number, that means it is the energy needed to break up the lattice!

Why do things dissolve? Is it maybe because of the relative sizes of the solvation and lattice energies?

Problem 6.66: Calculate the overall energy change (in kJ/mol) for the formation of CsF from its elements using the following data:

*Property*	*Value (kJ/mol)*
E_ea for F	-328
E_i₁for Cs	+375.7
E_i₂for Cs	+2422
Heat of sublimation for Cs	+76.1
Bond dissociation energy for F₂	+158
Lattice energy for CsF	+740

From the number, you can tell that this is one Devil of a problem. Note that this is interesting in the context of its companion, which follows...

Problem 6.67: The estimated lattice energy for CsF₂ is +2347 kJ/mol. Use the data given in Problem 6.66 to calculate an overall energy change (in kJ/mol) for the formation of CsF₂ from its elements. Does the overall reaction absorb energy or release it? In light of your answers to Problem 6.66, which compound is more likely to form in the reaction of cesium with fluorine, CsF or CsF₂?

The lattice energy for this molecule is large. But is it large enough for the molecule to form? We now work out the steps and discuss our conclusions.

For this reaction, we see that the overall energy is positive. For the reaction in Problem 6.66, the number was rather large and negative. Thus, formation of CsF is very much favored over the formation of CsF₂. Congratulations! You have just solved what amounts to a good theoretical problem.

Problem 6.71: Use the data and the result in Problem 6.63a to draw a Born-Haber cycle for the formation of LiF from its elements.

Usually, I do all the problems myself and eschew use of the instructor manual. In this case, however, I feel it wise to spare you my horrid art work (there are examples of that elsewhere if you really want to see it). This is the answer as printed in the instructor's manual.
The diagram makes the following few points clear:

Vaporization of the lithium, splitting of the fluorine molecule, and stripping of an electron from the lithium atom all require that we provide energy to the system. You might call this part of the total energy our "expenses."

The electron affinity and the lattice energy are given off by the system. You might call this part of the total energy our "dividends."

The net result is a stable compound in crystalline form. For this, the total energy is negative. However, we receive this energy (we are, after all, the surroundings!). Thus, from our viewpoint, the net result for the formation of a stable compound is, for us, profits!

Several of these points will be made more apparent in chapter 8. For now, we just look at how the flow of energy into and out of a system pretty much defines the system in terms of stability. To emphasize this point, you should probably draw diagrams for problems 6.66 and 6.67.