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Chapter 8: Problem 68

The lattice energies of \(\mathrm{FeCl}_{3}, \mathrm{FeCl}_{2},\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) are (in no particular order) \(-2631,-5359,\) and \(-14,774 \mathrm{kJ} / \mathrm{mol}\) . Match the appropriate formula to each lattice energy. Explain.

Short Answer

Expert verified
Fe鈧侽鈧� has a lattice energy of -14,774 kJ/mol, because it has the highest product of charges (+6). FeCl鈧� has a lattice energy of -2,631 kJ/mol, as it has the lowest product of charges (+2). FeCl鈧� has a lattice energy of -5,359 kJ/mol, with an intermediate product of charges (+3).

Step by step solution

01

Identify the charges of the ions in each compound

We need to determine the charges of the ions in each compound: FeCl鈧�: Fe has a charge of +3 and Cl has a charge of -1. FeCl鈧�: Fe has a charge of +2 and Cl has a charge of -1. Fe鈧侽鈧�: Fe has a charge of +3 and O has a charge of -2.
02

Calculate the product of the charges and distances

We don't need to calculate the actual distances between ions since we're only comparing the lattice energies. We just need to focus on the differences in the magnitude of ionic charges. FeCl鈧�: The ionic charges are +3 and -1, so their product is +(3脳1) = +3. FeCl鈧�: The ionic charges are +2 and -1, so their product is +(2脳1) = +2. Fe鈧侽鈧�: The ionic charges are +3 and -2, so their product is +(3脳2) = +6. Now we can analyze the magnitudes of these products to determine which lattice energy value corresponds to which compound.
03

Match the products of charges with the lattice energy values

Since lattice energy is directly proportional to the product of charges, the compound with the highest product of charges should have the highest lattice energy value, and the compound with the lowest product of charges should have the lowest lattice energy value. - Fe鈧侽鈧�: The product of charges is +6. This compound should have the highest lattice energy value, which is -14,774 kJ/mol. - FeCl鈧�: The product of charges is +2. This compound should have the lowest lattice energy value, which is -2,631 kJ/mol. - FeCl鈧�: The product of charges is +3. This compound should have the intermediate lattice energy value, which is -5,359 kJ/mol. So, we can conclude that: Fe鈧侽鈧� has a lattice energy of -14,774 kJ/mol. FeCl鈧� has a lattice energy of -2,631 kJ/mol. FeCl鈧� has a lattice energy of -5,359 kJ/mol.

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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 an essential part of chemistry, particularly in the study of materials. These compounds are formed when atoms transfer electrons to achieve a more stable, energetically favorable electronic configuration. Essentially, they involve ions鈥攁toms possessing an electric charge due to the loss or gain of electrons.

In ionic compounds:
  • Atoms become ions by either losing or gaining electrons: metals typically lose electrons, becoming positively charged cations, whereas non-metals generally gain electrons, becoming negatively charged anions.
  • Cations and anions interact to form a strong electrostatic attraction, known as an ionic bond, which holds the compound together in a structured lattice.
  • Examples include NaCl (table salt), where sodium (Na) loses an electron to become Na鈦� and chlorine (Cl) gains an electron to become Cl鈦�.
Ionic compounds are characterized by their high melting and boiling points, as well as their ability to conduct electricity when molten or dissolved in water.
charge of ions
In the world of ionic compounds, the charge of ions plays a key role in determining many of their properties, including lattice energy. Ions are atoms or molecules that have gained or lost electrons, resulting in a net charge.

Understanding ion charges:
  • A metal atom loses electrons and forms a positively charged ion, known as a cation. For example, iron (Fe) can lose electrons to form Fe虏鈦� or Fe鲁鈦�.
  • A non-metal atom gains electrons and forms a negatively charged ion, known as an anion. Chlorine (Cl), for instance, gains an electron to form Cl鈦�.
  • The number of electrons lost or gained determines the magnitude of the ion's charge.
The charge of ions directly affects the formation and strength of ionic bonds, as well as the resultant compound's lattice energy, due to the electrostatic forces between oppositely charged ions.
relationship between charges and lattice energy
Lattice energy is a measure of the strength of the forces holding ions together in an ionic solid. This is a crucial concept in understanding the stability of ionic compounds.

Here's how charge affects lattice energy:
  • Lattice energy is influenced by both the magnitude of the charges and the distance between the ions.
  • The greater the charges on the ions, the stronger the attractive forces, and hence, the higher the lattice energy.
  • For instance, compounds with higher charged ions like Fe鈧侽鈧� (Fe鲁鈦� and O虏鈦�) have higher lattice energies compared to FeCl鈧� (Fe虏鈦� and Cl鈦�).
Essentially, lattice energy increases with increasing ion charge, making it a direct proportional relationship. This understanding helps in predicting and comparing the stability and properties of different ionic compounds.
comparison of lattice energies
Comparing lattice energies helps us understand differences in stability and strength of ionic bonds in various compounds. The comparison is primarily based on the product of ion charges within each compound.

Understanding lattice energy comparisons:
  • The higher the product of the charges, the more negative the lattice energy, indicating a stronger bond and greater lattice stability.
  • For example, among FeCl鈧� (+2 脳 -1), FeCl鈧� (+3 脳 -1), and Fe鈧侽鈧� (+3 脳 -2), Fe鈧侽鈧� has the highest product (+6), hence the most negative lattice energy (-14,774 kJ/mol).
  • This comparison directly traces back to the electrostatic principles governing ionic interactions.
Therefore, when comparing lattice energies of ionic compounds, the greater the charged products, the stronger and more stable the ionic structure.

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Most popular questions from this chapter

Use the following data to estimate \(\Delta H_{f}^{\circ}\) for potassium chloride. $$\mathrm{K}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{KCl}(s)$$ \(\begin{array}{l}{\text { Lattice energy }} & {-690 . \mathrm{kJ} / \mathrm{mol}} \\ {\text { Ionization energy for } \mathrm{K}} & \quad{419 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Electron affinity of } \mathrm{Cl}} & {-349 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Bond energy of } \mathrm{Cl}_{2}} & \quad {239 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Enthalpy of sublimation for } \mathrm{K}} & \quad {90 . \mathrm{kJ} / \mathrm{mol}}\end{array}\)

Oxidation of the cyanide ion produces the stable cyanate ion, \(\mathrm{OCN}^{-}\) . The fulminate ion, \(\mathrm{CNO}^{-}\), on the other hand, is very unstable. Fulminate salts explode when struck; \(\mathrm{Hg}(\mathrm{CNO})_{2}\) is used in blasting caps. Write the Lewis structures and assign formal charges for the cyanate and fulminate ions. Why is the fulminate ion so unstable? (C is the central atom in \(\mathrm{OCN}^{-}\) and \(\mathrm{N}\) is the central atom in \(\mathrm{CNO}^{-}\) )

Predict the molecular structure for each of the following. (See Exercises 115 and 116.) a. \(\mathrm{BrFI}_{2} \quad\) b. \(\mathrm{XeO}_{2} \mathrm{F}_{2} \quad\) c. \(\mathrm{TeF}_{2} \mathrm{Cl}_{3}^{-}\) For each formula there are at least two different structures that can be drawn using the same central atom. Draw all possible structures for each formula.

Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for barium bromide. $$\mathrm{Ba}(s)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{BaBr}_{2}(s)$$ \(\begin{array}{ll}{\text { Lattice energy }} & {-1985 \mathrm{kJ} / \mathrm{mol}} \\ {\text { First ionization energy of Ba }} & \quad {503 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Second ionization energy of } \mathrm{Ba}} & \quad {965 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Electron affinity of } \mathrm{Br}} & {-325 \mathrm{kJ} / \mathrm{mol}}\\\\{\text { Bond energy of } \mathrm{Br}_{2}} & \quad {193 \mathrm{kJ} / \mathrm{mol}} \\\ {\text { Enthalpy of sublimation of } \mathrm{Ba}} & \quad {178 \mathrm{kJ} / \mathrm{mol}}\end{array}\)

Place the species below in order of the shortest to the longest nitrogen鈥搊xygen bond. $$\mathrm{H}_{2} \mathrm{NOH}, \quad \mathrm{N}_{2} \mathrm{O}, \quad \mathrm{NO}^{+}, \quad \mathrm{NO}_{2}^{-}, \quad \mathrm{NO}_{3}^{-}$$ \(\left(\mathrm{H}_{2} \mathrm{NOH} \text { exists as } \mathrm{H}_{2} \mathrm{N}-\mathrm{OH} .\right)\)
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