Problem 88 - Chemistry

Question

The following ions all have the same number of electrons: \(\mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-} .\) Order them according to their expected sizes, and explain your answer. 6.88 Calculate overall energy changes in kilojoules per mole for the formation of \(\mathrm{MgF}(s)\) and \(\mathrm{MgF}_{2}(s)\) from their elements. In light of your answers, which compound is more likely to form in the reaction of magnesium with fluorine, \(\mathrm{MgF}\) or \(\mathrm{MgF}_{2}\) ? The following data are needed: \(E_{\mathrm{ea}}\) for \(\mathrm{F}(\mathrm{g})=-328 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 1}\) for \(\mathrm{Mg}(g)=+737.7 \mathrm{~kJ} / \mathrm{mol}\) \(E_{i 2}\) for \(\mathrm{Mg}(g)=+1450.7 \mathrm{~kJ} / \mathrm{mol}\) Heat of sublimation for \(\mathrm{Mg}(s)=+147.7 \mathrm{~kJ} / \mathrm{mol}\)Bond dissociation energy for \(\mathrm{F}_{2}(\mathrm{~g})=+158 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{MgF}_{2}(s)=+2952 \mathrm{~kJ} / \mathrm{mol}\) Lattice energy for \(\mathrm{MgF}(s)=930 \mathrm{~kJ} / \mathrm{mol}\) (estimated)

Step-by-Step Solution

Answer

\(\mathrm{MgF}_2\) is more stable due to stronger lattice energy, making it more likely to form.

Step 1: Understanding Electron Configuration

All given ions \(\mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-}\) have 18 electrons matching Argon's electron configuration. Since they all have the same number of electrons, they are considered isoelectronic.

Step 2: Analyze Nuclear Charge and Size

The ionic size is affected by the effective nuclear charge. \(\mathrm{S}^{2-}\) has the smallest nuclear charge (16 protons), while \(\mathrm{Ti}^{4+}\) has the highest (22 protons). Higher nuclear charge pulls electrons closer, decreasing size. Thus, from largest to smallest: \(\mathrm{S}^{2-} > \mathrm{Ca}^{2+} > \mathrm{Sc}^{3+} > \mathrm{Ti}^{4+}\).

Step 3: Calculate Formation Energy of MgF

To find the energy change for forming \(\mathrm{MgF}\), follow these steps: 1. Convert \(\mathrm{Mg}(s)\) to \(\mathrm{Mg}(g)\) \(147.7 \mathrm{~kJ/mol}\).2. Ionize \(\mathrm{Mg}(g)\) to \(\mathrm{Mg}^+(g)\) \(737.7 \mathrm{~kJ/mol}\).3. Dissociate \(\mathrm{F}_2(g)\) bond: \(158\mathrm{~kJ/mol}\) divided by 2, as one mole of \(\mathrm{F_2}\) releases two \(\mathrm{F}\).4. Electron affinity of \(\mathrm{F}(g)\) \(-328 \mathrm{~kJ/mol}\).5. Combine \(\mathrm{Mg}^+(g)\) and \(\mathrm{F}^-(g)\) for \(\mathrm{MgF}(s)\): Ionization and lattice formation \(-930 \mathrm{~kJ/mol}\)Add steps (1), (2), half of (3), (4), and subtract lattice energy (5).

Step 4: Calculate Formation Energy of MgF2

Forming \(\mathrm{MgF}_2\):1. Mg(s) to Mg(g): \(147.7 \mathrm{~kJ/mol}\)2. First ionization: \(737.7 \mathrm{~kJ/mol}\)3. Second ionization: \(1450.7 \mathrm{~kJ/mol}\)4. F2 bond dissociation: \(79 \mathrm{~kJ/mol}\) use half of \(158\)5. Two electron affinities: 2 \(\times\)-328 \(= -656 \mathrm{~kJ/mol}\)6. Formation of \(\mathrm{MgF}_2\) lattice: \(-2952 \mathrm{~kJ/mol}\)Sum energies: Add steps (1), (2), (3), (4), (minus twice (5)), plus lattice energy (6).

Step 5: Compare Formation Energies

Compare energies for \(\mathrm{MgF}\) (estimated using previous steps) and \(\mathrm{MgF}_2\). \(\mathrm{MgF}_2\) has significantly lower energy, favoring formation due to greater lattice energy compensating for higher ionization.

Key Concepts

Ionic Size

The concept of ionic size is essential in understanding how ions compare to each other, especially when they share the same number of electrons, a condition known as being isoelectronic. The ions \(\mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-}\)\ all have 18 electrons, the same as the noble gas Argon. This alignment of electron count doesn't mean they are the same size.
The size of an ion greatly depends on its nuclear charge, which is the number of protons in the nucleus. A higher nuclear charge pulls electrons closer to the nucleus, making the ion smaller.

• \(\mathrm{Ti}^{4+}\) has the highest number of protons among the given ions, so it's the smallest.
• \(\mathrm{S}^{2-}\) has the lowest, resulting in the largest ionic size.

Therefore, the order from largest to smallest ionic size is \(\mathrm{S}^{2-}, \mathrm{Ca}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}\). Understanding ionic size helps in predicting behaviors in chemical reactions and bonding.

Energy Changes

Energy changes are crucial when forming compounds as they determine stability and the likelihood of a reaction occurring. The formation energy, also called the energy change, involves all energies absorbed and released during the transformation of elements to compounds.
For instance, in forming \(\mathrm{MgF}\)

• The energy is absorbed to sublimate magnesium and ionize it into \(\mathrm{Mg}^+\).
• Additionally, energy is required to dissociate the fluorine molecule.
• Conversely, energy is released during the electron affinity process of fluorine and during lattice formation.

After summing all these energy changes, more energy is released than absorbed for stable compounds, indicating a spontaneous and favorable process. For the proposed task, creating \(\mathrm{MgF}_2\) is more energy favorable than \(\mathrm{MgF}\), mostly attributed to the significant lattice energy.

Magnesium Fluoride

Magnesium fluoride as a compound appears in two stoichiometric forms in this context: \(\mathrm{MgF}\) and \(\mathrm{MgF}_2\). Each structure has its own formation energy and stability profile.
\(\mathrm{MgF}_2\) is deemed more stable due to larger lattice energy compared to \(\mathrm{MgF}\). Lattice energy is the energy released when ions combine to form a solid, and it correlates directly with the strength of ionic bonds.
The large lattice energy of \(\mathrm{MgF}_2\) far outweighs the additional ionization energy required to create \(\mathrm{Mg}^{2+}\) from magnesium. As a result, the \(\mathrm{MgF}_2\) compound forms preferentially, demonstrating the principle that stability and energy release guide the formation of ionic compounds.

Electron Configuration

The foundation of understanding why ions behave the way they do in reactions lies in electron configuration. This concept describes how electrons are distributed in an atom's orbitals.
For ions like \(\mathrm{Ti}^{4+}, \mathrm{Sc}^{3+}, \mathrm{Ca}^{2+}, \mathrm{S}^{2-}\), achieving the electron configuration of a noble gas provides stability. All these ions possess the electron configuration of Argon, which explains their greater stability and reactivity patterns.
Electron configuration affects how ions interact and bond:

• It governs the ionic size as previously discussed.
• It influences ionization energy, which is the energy required to remove electrons.

Understanding these configurations helps predict chemical behaviors and the structure of compounds formed from such ions.