Question

The main reason alkali metal dihalides \(\left(\mathrm{MX}_{2}\right)\) do \(\mathrm{not}\) form is the high \(\mathrm{IE}_{2}\) of the metal. (a) Why is IE \(_{2}\) so high for alkali metals? (b) The IE \(_{2}\) for Cs is \(2255 \mathrm{~kJ} / \mathrm{mol}\), low enough for \(\mathrm{CsF}_{2}\) to form exothermically \(\left(\Delta H_{i}^{\circ}=-125 \mathrm{~kJ} / \mathrm{mol}\right) .\) This compound cannot be synthesized, however, because CsF forms with a much greater release of heat \(\left(\Delta H_{\mathrm{f}}^{\circ}=-530 \mathrm{~kJ} / \mathrm{mol}\right) .\) Thus, the breakdown of \(\mathrm{CsF}_{2}\) to CsF happens readily. Write the equation for this breakdown, and calculate the enthalpy of reaction per mole of CsF.

Short answer

The IE2 for alkali metals is high due to the stability of the noble gas configuration after the first electron is removed. The reaction for the breakdown is CsF2 (s) -> CsF (s) + F2 (g). The enthalpy of reaction is -935 kJ/mol.

Step-by-step solution

Understand the Concept of Ionization Energy (IE)

Ionization energy is the energy required to remove an electron from a gaseous atom or ion. The first ionization energy (IE1) is for removing the first electron, and the second ionization energy (IE2) is for removing the second electron.

Explain Why IE2 is High for Alkali Metals

For alkali metals, the IE2 is high because, after the first electron is removed, the resulting ion has a noble gas electron configuration. Removing one more electron from this stable configuration requires significantly more energy, making IE2 very high.

Examine the Given Data and Formulate the Breakdown Reaction

Given that the enthalpy of formation of CsF2 (ΔH_i^∘ = -125 kJ/mol) and CsF (ΔH_f^∘ = -530 kJ/mol), we need to write the equation for the breakdown of CsF2 into CsF. The reaction is: CsF2 (s) -> CsF (s) + F2 (g)

Calculate the Enthalpy of Reaction

To calculate the enthalpy of reaction per mole of CsF, subtract the enthalpy of formation of CsF2 from the twice of the enthalpy of formation of CsF: ΔH_reaction = 2(ΔH_f^∘(CsF)) - ΔH_i^∘(CsF2) ΔH_reaction = 2(-530 kJ/mol) - (-125 kJ/mol) ΔH_reaction = -1060 kJ/mol + 125 kJ/mol ΔH_reaction = -935 kJ/mol

Key concepts

Ionization Energy

Ionization energy refers to the energy needed to remove an electron from a gaseous atom or ion. The first ionization energy (IE1) is for taking off the first electron, whereas the second ionization energy (IE2) is for taking off the second electron. The reason the IE2 is so high for alkali metals is because, after losing the first electron, they achieve a noble gas configuration. This stable state makes it much harder to remove another electron; hence, a lot of energy is required.

Alkali Metals

Alkali metals are a group of elements in the first column of the periodic table. They include lithium (Li), sodium (Na), potassium (K), rubidium (Rb), cesium (Cs), and francium (Fr). These metals are characterized by having a single electron in their outermost shell, which makes them highly reactive and eager to lose that electron to reach a stable state. This single electron also makes the first ionization energy relatively low, but the second ionization energy is extremely high because the removal of the first electron results in a stable noble gas configuration.

Enthalpy of Formation

The enthalpy of formation (ΔH_f^∘) is the heat change that occurs when one mole of a compound is formed from its elements in their standard states. For example, when cesium fluoride (CsF) forms, it releases a significant amount of heat, making the process exothermic. In the context of the given exercise, the greater release of heat for the formation of CsF (ΔH_f^∘ = -530 kJ/mol) compared to CsF2 (ΔH_i^∘ = -125 kJ/mol) demonstrates why CsF is more stable. This difference in enthalpy values explains why CsF2 breaks down readily to form CsF.

Chemical Reactions

Chemical reactions involve the rearrangement of atoms to form new substances, and they often involve changes in energy. For example, in the breakdown of CsF2 into CsF and F2, the reaction can be written as: CsF2 (s) -> CsF (s) + F2 (g). To calculate the enthalpy of this reaction per mole of CsF, we use the enthalpy values provided. The formula is: ΔH_reaction = 2(ΔH_f^∘(CsF)) - ΔH_i^∘(CsF2), which results in: ΔH_reaction = 2(-530 kJ/mol) - (-125 kJ/mol) = -1060 kJ/mol + 125 kJ/mol = -935 kJ/mol. This negative value indicates that the breakdown is highly exothermic, releasing a considerable amount of energy.