Question

Construct a Born-Haber cycle for the formation of the hypothetical compound NaCl , where the sodium ion has a $2+$ charge (the second ionization energy for sodium is given in Table 7.2 . (a) How large would the lattice energy need to be for the formation of ${\mathrm{NaCl}}_2$ to be exothermic? (b) If we were to estimate the lattice energy of ${\mathrm{NaCl}}_2$ to be roughly equal to that of ${\mathrm{MgCl}}_2(2326\mathrm{kJ}/\mathrm{mol}$ from Table 8.1$),$ what value would you obtain for the standard enthalpy of formation, $\mathrm{\Delta }{H}_f^{\circ },$ of ${\mathrm{NaCl}}_2?$

Short answer

The hypothetical compound NaCl₂ requires a lattice energy less than 5709 kJ/mol to be exothermic in its formation. Using the lattice energy of MgCl₂ (-2326 kJ/mol) as an estimation, the standard enthalpy of formation, ∆Hf°, of NaCl₂ is approximately 3383 kJ/mol.

Step-by-step solution

Write down the Born-Haber cycle for NaCl₂ formation

The Born-Haber cycle for the formation of NaCl₂ is expressed as follows: 1. Sublimation energy of sodium: Na (s) → Na (g) 2. 1st ionization energy of sodium: Na (g) → Na⁺ (g) + e⁻ 3. 2nd ionization energy of sodium: Na⁺ (g) → Na²⁺ (g) + e⁻ 4. Bond dissociation energy of chlorine: Cl₂ (g) → 2 Cl (g) 5. Electron affinity of chlorine: Cl (g) + e⁻ → Cl⁻ (g) 6. Lattice energy: Na²⁺ (g) + 2 Cl⁻ (g) → NaCl₂ (s) Now, we have all elements of the cycle necessary to determine the lattice energy needed for exothermic formation and the standard enthalpy of formation of NaCl₂.

Calculate the lattice energy required for NaCl₂ formation to be exothermic

In order for the formation of NaCl₂ to be exothermic, the sum of all the energies in the cycle must be negative. To find out how large the lattice energy should be, we need to calculate the sum of steps 1 through 5. Sublimation energy of sodium: +108 kJ/mol 1st ionization energy of sodium: +496 kJ/mol 2nd ionization energy of sodium: +4,560 kJ/mol Bond dissociation energy of chlorine: +243 kJ/mol Electron affinity of chlorine (twice, since there are 2 chlorines): -2 * 349 = -698 kJ/mol Sum of the energies in steps 1 to 5: +108 + 496 + 4560 + 243 - 698 = 5709 kJ/mol Now, let's denote the unknown lattice energy as LE. We have: LE - 5709 kJ/mol < 0 LE < 5709 kJ/mol So, the lattice energy should be less than 5709 kJ/mol for the NaCl₂ formation to be exothermic.

Estimate the standard enthalpy of formation, ∆Hf°, of NaCl₂

We will use the given lattice energy of MgCl₂ as an estimation for NaCl₂: Lattice energy of MgCl₂: -2326 kJ/mol Now, let's calculate the standard enthalpy of formation, ∆Hf°, of NaCl₂ using the estimated lattice energy: ∆Hf° = Sum of the energies in steps 1 to 5 + lattice energy = 5709 kJ/mol - 2326 kJ/mol = 3383 kJ/mol Therefore, the standard enthalpy of formation, ∆Hf°, of NaCl₂ is approximately 3383 kJ/mol.

Key concepts

Lattice Energy

Lattice energy is an essential concept when discussing ionic compounds, as it provides a measure of the strength of the forces holding ions together in a crystalline solid. When two ions of opposite charges come together, energy is released. This released energy is known as the lattice energy. It's critical because it significantly affects the stability and physical properties of ionic compounds. In the Born-Haber cycle, lattice energy plays a crucial role in determining whether a reaction is overall exothermic or endothermic. For the formation of ${\text{NaCl}}_2$, the lattice energy must be large enough to compensate for the energies required in other processes, such as ionization and bond dissociation. Calculating lattice energy can be complex, often relying on the Born-Haber cycle. This process involves estimating lattice energy based on known enthalpies, such as the sublimation and ionization energies.

Standard Enthalpy of Formation

The standard enthalpy of formation, often denoted as $\mathrm{\Delta }{H}_f^{\circ }$, refers to the change in enthalpy when one mole of a compound forms from its constituent elements in their standard states. For a reaction to be exothermic, the overall change in enthalpy must be negative, indicating that the reaction releases energy to the surroundings. In the exercise with ${\text{NaCl}}_2$, determining the standard enthalpy of formation involves calculating the total energy changes from forming ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ ions and subsequently forming the compound's crystal lattice. Using an estimated lattice energy, one can work back through the Born-Haber cycle to approximate $\mathrm{\Delta }{H}_f^{\circ }$, considering the sum of energy changes involved in the cycle.

Ionization Energy

Ionization energy is the energy needed to remove an electron from an atom or ion in its gaseous state. This energy is crucial for understanding and predicting how elements will react, especially metals since they tend to lose electrons. In the Born-Haber cycle for ${\text{NaCl}}_2$, we consider the first and second ionization energies of sodium. The first ionization energy is the energy required to remove the first electron from neutral sodium to form ${\text{Na}}^{+}$. The second ionization energy is notably higher because it involves removing an electron from the positively charged ${\text{Na}}^{+}$, creating ${\text{Na}}^{2+}$. Understanding these energies helps explain why certain ionic charges are more common, as higher charges require more energy to maintain, influencing the overall feasibility of compound formation.

Electron Affinity

Electron affinity measures the energy change that occurs when an electron is added to a neutral atom in the gaseous state, forming an anion. A higher electron affinity indicates that the atom releases more energy when gaining an electron, suggesting that it strongly attracts extra electrons. In constructing the Born-Haber cycle for ${\text{NaCl}}_2$, the electron affinities of chlorine are considered. Each chlorine atom gains one electron, forming ${\text{Cl}}^{-}$ ions. The energy released in this process helps counterbalance other energy-consuming processes in the cycle, such as sublimation and ionization. It's important because favorable electron affinities can lead to spontaneous reactions, as they help lower the overall energy required, promoting the formation of stable ionic compounds.