Question

(i) Given that the cell parameter for calcium oxide, which crystallizes with the sodium chloride structure, is \(483 \mathrm{pm}\), calculate the lattice energy for \(\mathrm{CaO}\). (ii) Using the following data and the lattice energy calculated in part (i), construct a thermochemical cycle for the formation of calcium oxide from its elements and hence calculate the enthalpy of formation of \(\mathrm{CaO}_{(\mathrm{s})}\) : \(\Delta H_{\text {sub }}=+177 ;\) ionization potentials: first \(=+590 ;\) second \(=\) \(+1100 ; \Delta H_{\text {diss }}=+498 ;\) first \(E_{\mathrm{a}}=142 ;\) second \(E_{\mathrm{a}}=-844\left(\mathrm{~kJ} \mathrm{~mol}^{-1}\right)\) (iii) Why might you expect the answer calculated in (ii) to have an error of about \(5 \mathrm{~kJ} \mathrm{~mol}^{-1 ?}\) (iv) Why is the second ionization potential for calcium much greater than the first despite forming a stable closed-shell configuration?

Short answer

(i) The radii of Calcium (Ca\(^{2+}\)) and Oxide (O\(^{2-}\)) ions are calculated from the cell parameter using the relationship that the cell parameter is equal to twice the sum of their radii. With assumed radii ratio of 0.732 for maximum stability, \( r_{\text{Ca}^{2+}} \approx 195 \mathrm{~pm} \) and \( r_{\text{O}^{2-}} \approx 242 \mathrm{~pm} \) are obtained. Using these values, the Lattice Energy for CaO is calculated with Born-Landé equation. (ii) A Born-Haber cycle, which relates various properties for the formation of an ionic compound, is constructed using the calculated lattice energy and the provided values of ionisation potentials, dissociation enthalpy, and electron affinities. The enthalpy of formation \( ΔH_f \) of calcium oxide is then calculated using the data within the cycle. (iii) An error of about 5 kJ/mol might be expected in the calculated enthalpy of formation due to the assumptions made in the perfect ionic model and constant Born exponent used for Lattice energy calculation. Deviations in ionic character, lattice structure and ionic sizes can result from these approximations. (iv) The second ionization potential for calcium is higher than the first one as it becomes increasingly difficult to remove electrons from positively charged ions. Despite forming a stable closed-shell configuration, the increased charge density and attractive force between remaining electrons and the nucleus requires more energy to remove an electron from a positive ion's shell.

Step-by-step solution

Calculate Radius of Ca and O ions

In a sodium chloride structure, the Ca and O ions form a face-centered cubic structure. Using the given cell parameter (483 pm), we first calculate the radius of Calcium (Ca2+) and Oxide (O2-) ions from the cell parameter using the relationship that the cell parameter is equal to twice the sum of the radii of the cations and anions in that structure. This would give us \( r_{\text{Ca}^{2+}} + r_{\text{O}^{2-}} = \frac{483}{2} = 241.5 \mathrm{~pm} \)

Use the Radius Ratio Rule to Find Individual Radii

We know that a cation in coordination with 6 anions will have a Radius Ratio between 0.414 and 0.732. From this, we derive the individual radii of Calcium and Oxide ions. Assuming that the radii ratio is closest to 0.732 (for maximum stability), we find that \( r_{\text{Ca}^{2+}} = 0.732 * r_{\text{O}^{2-}} \). Solving these two equations together, we find \( r_{\text{Ca}^{2+}} \approx 195 \mathrm{~pm} \) and \( r_{\text{O}^{2-}} \approx 242 \mathrm{~pm} \).

Calculate Lattice Energy

The Born-Landé equation is used to calculate the lattice energy where \( U = \frac{N_\text{A}kz^+z^-e^2}{4πϵ_0r_{0}}*(1-\frac{1}{n}) \) where \( N_\text{A} = \text{Avogadro's number} \), \( k = \text{Born constant(= 1 for Rocksalt structure)} \), \( z^+ \) and \( z^- \) are charges of the ions, \( e \) is the charge of an electron, \( ϵ_0 \) is permittivity of vacuum, \( r_0 \) is \( r_{\text{Ca}^{2+}} + r_{\text{O}^{2-}} \) and \( n \) is the Born exponent (which is approximately 9 for smaller ions like here). Substituting the values, we can find the Lattice Energy.

Construct the Born-Haber cycle

The Born-Haber cycle is a thermochemical cycle that relates various physical and chemical properties for the formation of an ionic compound. Construct the Born-Haber cycle using the calculated lattice energy and the provided values of ionisation potentials, dissociation enthalpy, and electron affinities.

Calculate Enthalpy of Formation

Using the data in the Born-Haber cycle, you can then calculate the enthalpy of formation \( ΔH_f \) of calcium oxide using the equation: \( ΔH_f = ΔH_{sub} + I.P._1 + I.P._2 + D_{H_2} + (EA_1 + EA_2) + U \) . Substitute the given and calculated values into the equation to find \( ΔH_f \) of calcium oxide.

Reasoning on the Error in Calculation

Bear in mind that the perfect ionic model and constant Born exponent used for Lattice energy calculation are assumptions. In reality, ionic character, lattice structure and ionic sizes can result in deviations. Hence, this approximation could result in an error of about 5 kJ/mol.

Discuss the Ionization Potentials

The second ionization potential is higher than the first one due to the fact that it is increasingly difficult to remove electrons from positively charged ions. Even though forming a stable closed-shell configuration, removing an electron from the shell of a positive ion requires more energy due to the increased charge density and attractive force between the remaining electrons and the nucleus.

Key concepts

Born-Haber cycle

The Born-Haber cycle is a hypothetical series of steps that represent the formation of an ionic compound from its constituent elements. To understand this concept with a real-world example, let's consider the formation of calcium oxide (\text{CaO}).

To determine the lattice energy, which is the energy released when gaseous ions combine to form an ionic solid, we conduct the cycle in reverse. We start with \text{CaO} and imagine breaking it down into its gaseous ions, \text{Ca}^{2+} and \text{O}^{2-}. Next, we account for the energy to form these ions from the elements in their standard states - this involves the enthalpy of sublimation for calcium, the dissociation energy for oxygen, ionization energies for calcium, and electron affinities for oxygen.

The visual representation of the Born-Haber cycle ultimately leads us to the enthalpy of formation. Each step accounts for specific energy changes according to Hess's law, allowing you to calculate the overall \(\text{enthalpy of formation}\) from these individual values. It's similar to a balance sheet that informs you of how the energy is consumed or released through each phase of the transition from elements to compound.

Enthalpy of formation

The enthalpy of formation, \(\text{ΔH}_f\), measures the change in enthalpy when one mole of a compound is formed from its elements in their standard states. For our example, calcium oxide, we’re interested in the energy change when calcium and oxygen combine to form solid \text{CaO} under standard conditions.

If we think of thermochemistry as a financial transaction, the \(\text{enthalpy of formation}\) is like the final cost or gain from producing a substance. In the Born-Haber cycle context, the \(\text{ΔH}_f\) for \text{CaO} is the sum of the energies involved in sublimating calcium, dissociating diatomic oxygen, ionizing calcium atoms (done twice, for the first and second ionization energies), adding electrons to the oxygen atom (electron affinities), and, finally, the 'cash-back' or energy released represented by the negative lattice energy. Simplified, it's expressed as \(\text{ΔH}_f = \text{ΔH}_{\text{sub}} + \text{I.P.}_1 + \text{I.P.}_2 + \text{D}_{\text{H}_2} + (\text{EA}_1 + \text{EA}_2) + \text{U}\), where 'U' is the lattice energy.

Ionization potential

Ionization potential (or ionization energy) is the amount of energy required to remove an electron from the outer shell of an atom in its gaseous state. For elements that form positive ions, such as calcium in our example, the process of ionization can occur in successive stages, each requiring a different amount of energy.

Calcium requires one amount of energy to remove the first electron (\(\text{I.P.}_1\)), and a larger amount for the second (\(\text{I.P.}_2\)). This is because, after losing the first electron, the resulting ion has a higher charge density, making the remaining electrons more tightly bound to the nucleus, and thus requiring more energy to remove the next one. Despite achieving a stable electron configuration after the first ionization, overcoming the electrostatic attraction in the second ionization demands considerably higher energy, which explains the significant difference between \(\text{I.P.}_1\) and \(\text{I.P.}_2\).