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Compare the electron affinity of fluorine to the ionization energy of sodium. Is the process of an electron being “pulled” from the sodium atom to the fluorine atom exothermic or endothermic? Why is NaF a stable compound? Is the overall formation of NaF endothermic or exothermic? How can this be?

Short Answer

Expert verified
The electron affinity of fluorine is -328 kJ/mol, and the ionization energy of sodium is +496 kJ/mol. The process of electron transfer from sodium to fluorine is endothermic, as more energy is required to remove the electron from sodium than is released by fluorine's electron affinity. However, NaF is a stable compound due to the release of lattice energy during the formation of ionic bonds between Na+ and F- ions. This energy release compensates for the endothermic electron transfer, making the overall formation of NaF exothermic.

Step by step solution

01

Understand Electron Affinity and Ionization Energy

Electron affinity is the energy released when an electron is added to a neutral atom, resulting in the formation of a negatively charged ion. A higher electron affinity indicates that an atom more readily accepts an electron. Ionization energy, on the other hand, is the energy required to remove an electron from a neutral atom, forming a positively charged ion. A higher ionization energy indicates that an atom has a greater resistance to losing an electron.
02

Determine the electron affinity of fluorine and ionization energy of sodium

The electron affinity of fluorine (F) is -328 kJ/mol, which means that fluorine releases 328 kJ of energy per mole to gain an electron. The ionization energy of sodium (Na) is +496 kJ/mol, meaning that 496 kJ of energy per mole is required to remove an electron from a sodium atom.
03

Analyze the electron transfer process

Since fluorine has a high electron affinity value (-328 kJ/mol), it suggests that fluorine has a strong ability to gain an electron. Sodium, on the other hand, has a lower ionization energy (+496 kJ/mol) than the electron affinity of fluorine, indicating that it is relatively easier for sodium to lose an electron. Thus, when an electron is transferred from sodium to fluorine, the energy released due to the gain of an electron by fluorine (-328 kJ/mol) is less than the energy required to remove an electron from sodium (+496 kJ/mol). This makes the overall process endothermic, as energy is absorbed overall in this electron transfer.
04

Stability of NaF and overall formation

Despite the endothermic electron transfer, NaF is a stable compound because the formation of ionic bonds between the positively charged Na+ and negatively charged F- ions releases energy (called lattice energy), which compensates for the energy absorbed during the electron transfer. The overall formation of NaF is exothermic, as the energy released due to the formation of the ionic bond outweighs the energy absorbed in the electron transfer process. This leads to the stabilization of NaF as a compound. So to summarize: 1. The electron affinity of fluorine is -328 kJ/mol, and the ionization energy of sodium is +496 kJ/mol. 2. The process of an electron being transferred from sodium to fluorine is endothermic since the energy required for ionization is greater than the energy released by electron affinity. 3. NaF is a stable compound because the formation of ionic bonds between Na+ and F- ions releases energy, compensating for the energy absorbed during the electron transfer. 4. The overall formation of NaF is exothermic because the energy released during the formation of ionic bonds outweighs the energy absorbed in the electron transfer process.

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Key Concepts

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

Ionization Energy
Ionization energy refers to the amount of energy needed to remove an electron from a neutral atom, resulting in a positively charged ion. This is a critical concept in chemistry as it helps determine how easily an atom can lose an electron to become positively charged. When comparing elements, atoms with higher ionization energy make for poor electron donors, since they hold onto their electrons more tightly. For example, sodium’s ionization energy is 496 kJ/mol, indicating that significant energy is required to strip an electron from it.

Understanding ionization energy is crucial when investigating reactions involving electron transfer. Elements with low ionization energy, like sodium, often participate in reactions as donors, losing electrons to become cations. This is an essential aspect of the formation of ionic bonds. Remember, atoms with lower ionization energies tend to form positive ions more readily.
Ionic Bonds
Ionic bonds are a type of chemical bond where electrons are transferred from one atom to another, resulting in the formation of ions. These ions are then attracted to each other by electrostatic forces, creating a stable compound. This type of bond typically occurs between metal and non-metal elements.

For example, sodium (a metal) and fluorine (a nonmetal) form an ionic bond in which sodium loses an electron and forms a positive ion ( ext{Na}^+), while fluorine gains an electron forming a negative ion ( ext{F}^-). Despite the energy cost associated with transferring electrons, the attraction between ext{Na}^+ and ext{F}^- releases energy, known as lattice energy. This energy released makes the compound stable, counteracting the energy required to ionize the atoms initially.

Ionic bonds are characterized by their strength, which results from the attraction between oppositely charged ions. This often leads to substances with high melting and boiling points, as the energy to break these bonds is substantial.
Lattice Energy
Lattice energy is the energy released when one mole of an ionic crystalline compound is formed from gaseous ions. This is a key concept because it represents the strength of the forces holding the ions together in an ionic bond. The magnitude of lattice energy indicates how much energy is released upon the formation of a solid lattice structure from the gaseous ions.

For an example like NaF, though the transfer of electrons appears energy costly, the lattice energy compensates for this. Lattice energy contributes to the overall stability and exothermic nature of forming compounds like NaF, where the energy released upon the formation of the crystal lattice surpasses the energy absorbed during the electron transfer.
  • Strength of interaction increases with greater charge on ions.
  • Smaller ionic radii lead to higher lattice energy due to closer proximity of charges.
Lattice energy therefore plays a vital role in offsetting the ionization energy, providing overall energy gain in reactions.
Exothermic Processes
Exothermic processes refer to reactions or changes where energy is released to the surroundings, usually in the form of heat. This release of energy results in a temperature increase of the environment surrounding the reacting system.

When examining the formation of compounds like NaF, we find it to be exothermic. Initially, the act of removing an electron from sodium and attaching it to fluorine is endothermic. However, the net process of forming ionic bonds releases substantial energy, making the overall process exothermic. This happens because the lattice energy released during the formation of the ionic compound is greater than the energy required for ionization and electron affinity processes.
  • Exothermic reactions often involve bond formation.
  • They are generally favorable, driving many natural processes.
The exothermic nature of reactions plays a crucial role in the energetic stability and feasibility of chemical compound formations.

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Most popular questions from this chapter

Consider the following: Li(s) \(+\frac{1}{2} \mathrm{I}_{2}(g) \rightarrow\) Liil(s) \(\Delta H=\) \(-292 \mathrm{kJ} . \mathrm{LiI}(s)\) has a lattice energy of \(-753 \mathrm{kJ} / \mathrm{mol} .\) The ionization energy of Li(g) is \(520 . \mathrm{kJ} / \mathrm{mol},\) the bond energy of \(\mathrm{I}_{2}(g)\) is 151 \(\mathrm{kJ} / \mathrm{mol}\) , and the electron affinity of \(\mathrm{I}(g)\) is \(-295 \mathrm{kJ} / \mathrm{mol}\) . Use these data to determine the heat of sublimation of Li(s).

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part c, why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about MgO compared to \(\mathrm{MgO}_{2} ? \mathrm{Mg}_{2} \mathrm{O} ?\)

Consider the following bond lengths: $$\mathrm{C}-\mathrm{O} \quad 143 \mathrm{pm} \quad \mathrm{C}=\mathrm{O} \quad 123 \mathrm{pm} \quad \mathrm{C} \equiv \mathrm{O} \quad 109 \mathrm{pm}$$ In the \(\mathrm{CO}_{3}^{2-}\) ion, all three \(\mathrm{C}-\mathrm{O}\) bonds have identical bond lengths of 136 \(\mathrm{pm} .\) Why?

Write electron configurations for a. the cations \(\mathrm{Mg}^{2+}, \mathrm{K}^{+},\) and \(\mathrm{Al}^{3+}\) . b. the anions \(\mathrm{N}^{3-}, \mathrm{O}^{2-}, \mathrm{F}^{-},\) and \(\mathrm{Te}^{2-}\)

For each of the following, write an equation that corresponds to the energy given. a. lattice energy of \(\mathrm{NaCl}\) b. lattice energy of \(\mathrm{NH}_{4} \mathrm{Br}\) c. lattice energy of \(\mathrm{MgS}\) d. \(\mathrm{O}=\mathrm{O}\) double bond energy beginning with \(\mathrm{O}_{2}(g)\) as a reactant

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