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(a) Based on the lattice energies of \(\mathrm{MgCl}_{2}\) and \(\mathrm{SrCl}_{2}\) given in Table \(8.2\), what is the range of values that you would expect for the lattice energy of \(\mathrm{CaCl}_{2} ?\) (b) Using data from Appendix \(C\), Figure \(7.12\), and Figure \(7.14\) and the value of the second ionization energy for \(\mathrm{Ca}\), \(1145 \mathrm{~kJ} / \mathrm{mol}\), calculate the lattice energy of \(\mathrm{CaCl}_{2} .\)

Short Answer

Expert verified
The estimated range of lattice energies for CaCl₂ is between -2526 kJ/mol and -2120 kJ/mol. Based on the Born-Haber cycle, the calculated lattice energy of CaCl₂ is -2210 kJ/mol, which falls within the estimated range.

Step by step solution

01

(a) Estimating the range of lattice energies for CaCl₂

To estimate the range of lattice energies for CaCl₂, we can use the lattice energies of MgCl₂ and SrCl₂. Ca is positioned between Mg and Sr in the periodic table, so it's reasonable to assume that the lattice energy of CaCl₂ will fall between the lattice energies of MgCl₂ and SrCl₂. The given lattice energies are: MgCl₂: \( \Delta H_{lattice} = -2526 \mathrm{~kJ/mol} \) SrCl₂: \( \Delta H_{lattice} = -2120 \mathrm{~kJ/mol} \) Thus, the expected range of lattice energies for CaCl₂ is between -2526 kJ/mol and -2120 kJ/mol.
02

(b) Calculating the lattice energy of CaCl₂ using Born-Haber cycle

The lattice energy of CaCl₂ can be calculated using the Born-Haber cycle approach, which involves the following steps: 1. Sublimation energy of Ca: \(\Delta H_{sub}\) - the energy required to convert solid calcium into gaseous calcium. 2. Bond dissociation energy of Cl₂: \(\Delta H_{diss}\) - the energy required to break the Cl-Cl bond in the Cl₂ molecule. 3. First and second ionization energies of Ca: \(\Delta H_{i1}\) and \(\Delta H_{i2}\) - the energies required to remove the first and second electrons from gaseous calcium. 4. First and second electron affinities of Cl: \(\Delta H_{ea1}\) and \(\Delta H_{ea2}\) - the energies released when an electron is added to gaseous chlorine atoms. 5. Formation energy of CaCl₂: \(\Delta H_{f}\) - the energy released when CaCl₂ is formed from its elements. The lattice energy (\(\Delta H_{lattice}\)) can be calculated as follows: \(\Delta H_{lattice} = \Delta H_{f} - (\Delta H_{sub} +\Delta H_{diss} + \Delta H_{i1} + \Delta H_{i2} - \Delta H_{ea1} - \Delta H_{ea2})\) Using the data provided in the exercise and the appendices, we have: \(\Delta H_{sub} = 178 \mathrm{~kJ/mol} \) (sublimation enthalpy of Ca from Appendix C) \(\Delta H_{diss} = 242 \mathrm{~kJ/mol} \) (bond energy of Cl₂ from Appendix C) \(\Delta H_{i1} = 590 \mathrm{~kJ/mol} \) (first ionization energy of Ca from Appendix C) \(\Delta H_{i2} = 1145 \mathrm{~kJ/mol}\) (second ionization energy of Ca from exercise) \(\Delta H_{ea1} = -349 \mathrm{~kJ/mol}\) (first electron affinity of Cl from Figure 7.12) \(\Delta H_{ea2} = -20 \mathrm{~kJ/mol}\) (second electron affinity of Cl from Figure 7.14) \(\Delta H_{f} = -795 \mathrm{~kJ/mol} \) (formation enthalpy of CaCl₂ from Appendix C) Now, we can plug these values into the equation for calculating the lattice energy: \(\Delta H_{lattice} = (-795) - (178 + 242 + 590 + 1145 - (-349) - (-20))\) \(\Delta H_{lattice} = -795 - (1786 + 349) + 20\) \(\Delta H_{lattice} = -2210 \mathrm{~kJ/mol}\) Thus, the calculated lattice energy of CaCl₂ is -2210 kJ/mol, which falls within our estimated range (-2526 kJ/mol to -2120 kJ/mol).

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Key Concepts

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

Born-Haber Cycle
The Born-Haber cycle is an essential concept in chemistry for calculating lattice energies, especially for ionic compounds. It is a thermochemical cycle that breaks down the formation of an ionic solid into a series of steps, each corresponding to a specific energy change.

Here's how the Born-Haber cycle helps in understanding the processes involved:
  • Sublimation Energy: Converts solid metal to gaseous atoms, e.g., from Ca(s) to Ca(g).
  • Bond Dissociation Energy: Breaks down diatomic molecules into individual atoms, such as dissociating Cl₂ into 2 Cl atoms.
  • Ionization Energy: Removes electrons from gaseous atoms to form cations, e.g., converting Ca(g) to Ca²⁺.
  • Electron Affinity: Adds electrons to gaseous atoms to form anions, e.g., Cl(g) to Cl⁻.
  • Formation Enthalpy: Represents the energy change when an ionic solid forms from its gaseous ions.
The main goal is to apply Hess's Law, which states that the total enthalpy change in a chemical reaction is the same, regardless of the path taken. By accounting for these individual steps, the Born-Haber cycle provides a practical way to calculate the lattice energy of ionic compounds, ensuring a clearer understanding of their stability and formation.
Ionization Energy
Ionization energy is the energy required to remove an electron from a gaseous atom or ion. It is critical in understanding an element's reactivity and plays a significant role in the Born-Haber cycle.

Some key points regarding ionization energy include:
  • First and Second Ionization Energies: The first ionization energy removes the most loosely bound electron, while the second deals with removing the next electron.
  • Energy Variability: The energy needed increases significantly for successive electrons, particularly after removing a valence electron, as inner shell electrons are more tightly bound.
Factors affecting ionization energy include the nuclear charge (the more protons, the higher the energy required), electron shielding, and the electron configuration. Generally, ionization energy increases across a period and decreases down a group on the periodic table. Understanding these trends is vital for predicting how atoms participate in chemical reactions.
Electron Affinity
Electron affinity refers to the energy change that occurs when an electron is added to a neutral atom to form a negative ion. This concept is vital in understating the attractiveness of an atom to an additional electron.

Consider these points:
  • Energy Release: Typically results in the release of energy, as the atom becomes more stable.
  • First and Second Electron Affinities: Primarily consider the first addition as the second is less common and often requires energy input.
Trends in electron affinity can vary, but it usually increases across a period due to increased nuclear charge, making it easier to add an electron. However, moving down a group, it often decreases as the added electron is further from the nucleus, reducing the net attraction. These trends help in the comprehensive analysis of how different atoms interact and form bonds.
Periodic Table Trends
Periodic table trends provide a framework to predict and understand the behavior of elements based on their position within the table.

Here’s how they relate to other concepts like lattice energy, ionization energy, and electron affinity:
  • Ionic Radii: Generally, atomic radii decrease across a period and increase down a group. This affects properties like lattice energy; smaller ions typically result in stronger lattice energies.
  • Electronegativity: Increases across a period, making atoms more likely to attract electrons and influence ionization energy and electron affinity trends.
  • Melting and Boiling Points: Often fluctuate, but metals with strong metallic bonds will have higher melting points.
By grasping these trends, students can predict and rationalize the comparative stability and reactivity of different elements and their compounds, forming a comprehensive understanding of chemical interactions.

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Most popular questions from this chapter

To address energy and environmental issues, there is great interest in powering vehicles with hydrogen rather than gasoline. One of the most attractive aspects of the "hydrogen economy" is the fact that in principle the only emission would be water. However, two daunting obstacles must be overcome before this vision can become a reality. First, an economical method of producing hydrogen must be found. Second, a safe, lightweight, and compact way of storing hydrogen must be found. The hydrides of light metals are attractive for hydrogen storage because they can store a high weight percentage of hydrogen in a small volume. One of the most attractive hydrides is \(\mathrm{NaAlH}_{4}\), which can release \(5.6 \%\) of its mass as \(\mathrm{H}_{2}\) upon decomposing to \(\mathrm{NaH}(\mathrm{s})\), Al(s), and \(\mathrm{H}_{2}(\mathrm{~g})\). NaAlH \(_{4}\) possesses both covalent bonds, which hold polyatomic anions together, and ionic bonds. (a) Write a balanced equation for the decomposition of \(\mathrm{NaAlH}_{4} .\) (b) Which element in \(\mathrm{NaAlH}_{4}\) is the most electronegative? Which one is the least electronegative? (c) Based on electronegativity differences, what do you think is the identity of the polyatomic anion? Draw a Lewis structure for this ion.

Energy is required to remove two electrons from Ca to form \(\mathrm{Ca}^{2+}\) and is required to add two electrons to \(\mathrm{O}\) to form \(\mathrm{O}^{2-}\). Why, then, is \(\mathrm{CaO}\) stable relative to the free elements?

The lattice energies of \(\mathrm{KBr}\) and \(\mathrm{CsCl}\) are nearly equal (Table 8.2). What can you conclude from this observation?

You and a partner are asked to complete a lab entitled "Oxides of Ruthenium" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a soft yellow substance and the other a black powder. You also find the following notes in your partner's notebook-Compound 1: \(76.0 \%\) Ru and \(24.0 \%\) O (by mass), Compound 2: \(61.2 \%\) Ru and \(38.8 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound 2? (c) Upon determining the melting points of these two compounds, you find that the yellow compound melts at \(25^{\circ} \mathrm{C}\), while the black powder does not melt up to the maximum temperature of your apparatus, \(1200^{\circ} \mathrm{C}\). What is the identity of the yellow compound? What is the identity of the black compound? Be sure to use the appropriate naming convention depending upon whether the compound is better described as a molecular or ionic compound.

Write Lewis structures that obey the octet rule for each of the following, and assign oxidation numbers and formal charges to each atom: (a) \(\mathrm{NO}^{+}\), (b) \(\mathrm{POCl}_{3}\) (P is bonded to the three \(\mathrm{Cl}\) atoms and to the \(\mathrm{O}\), (c) \(\mathrm{ClO}_{4}^{-}\), (d) \(\mathrm{HClO}_{3}(\mathrm{H}\) is bonded to \(\mathrm{O})\).

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