Chapter 14, Problem 14.110P

(a)

Interpretation Introduction

Interpretation:

Reason for why alkali metals have high second ionization energy has to be explained.

Concept Introduction:

First ionization energy: The ionization energy is the minimum energy required to remove the electron from an isolated atom which is in the gaseous state results to give gaseous ion with one positive charge.

Second ionization energy: The minimum energy needed to remove an electron from a unipositive gaseous ion to form a dipositive ion in the ground state is known as second ionization energy.

The outer electronic configuration of Alkali metals is ${\text{ns}}^{\text{1}}$

Explanation of Solution

The outer electronic configuration of Alkali metals is ${\text{ns}}^{\text{1}}$.  If an electron is removed from the metal, the element gets the electronic configuration of noble gas and hence removing the fourth electron is very difficult.

The energy level of the element is changed after removing an electron and the electrons are held more tightly by the nucleus in the lower level.  So a huge amount of energy is needed to remove the second electron.

(b)

Interpretation Introduction

Interpretation:

Enthalpy of reaction per mole of $CsF$ has to be determined using the given data.

Concept Introduction:

The ionization energy is the minimum energy required to remove the electron from an isolated atom which is in the gaseous state results to give gaseous ion with one positive charge.

Explanation of Solution

Given data is shown below:

  $\begin{array}{l}{\text{2CsF}}_{\text{2}}\left(s\right)\underset{}{\to }\text{ 2CsF}\left(s\right){\text{ + F}}_{\text{2}}\left(g\right)\\ \\ \Delta H_f^o\text{ of formation of CsF,}\\ \\ \text{Cs}\left(s\right)\text{ + }1/2{\text{F}}_{\text{2}}\left(g\right)\underset{}{\to }\text{ CsF}\left(s\right)\Delta H_f^o=-530\text{ kJ/mol}\\ \\ \Delta H_f^o{\text{ of formation of CsF}}_2\text{,}\\ \\ \text{Cs}\left(s\right){\text{ + F}}_{\text{2}}\left(g\right)\underset{}{\to }{\text{ CsF}}_{\text{2}}\left(s\right)\Delta H_f^o=-\text{125 kJ/mol}\end{array}$

The heat obtained during the breaking down of $Cs{F}_2\text{}\text{to CsF}$ can be obtained by combining the formation reaction of $CsF$ and formation reaction of $Cs{F}_2$ where both reactions are multiplied by $2.$

$\begin{array}{l}\bcancel{\text{2Cs}\left(s\right)}\text{ + }\bcancel{{\text{F}}_{\text{2}}\left(g\right)}\underset{}{\to }\text{ 2CsF}\left(s\right)\Delta H_f^o=2\times \left(-530\text{ kJ/mol}\right)\text{ = }-\text{1060 kJ}\\ \\ {\text{2CsF}}_{\text{2}}\left(s\right)\underset{}{\to }\bcancel{\text{2Cs}\left(s\right)}\text{ + }\bcancel{\text{2}}{\text{F}}_{\text{2}}\left(g\right)\Delta H_f^o=\text{ 2}\times \left(\text{+125 kJ/mol}\right)\text{ = 250 kJ}\\ \underset{\_}{}\\ {\text{2CsF}}_{\text{2}}\left(s\right)\underset{}{\to }\text{ 2CsF}\left(s\right){\text{ + F}}_{\text{2}}\left(g\right)\Delta H_f^o=-\text{810 kJ}\end{array}$

Heat released during the breaking down of two moles $Cs{F}_2$ to two moles of $CsF$ is $\text{810 kJ}\text{.}$

Therefore,

Enthalpy of reaction per mole of $CsF$ is given as,

  $\begin{array}{c}Enthalpyofreactionpermole\text{ = }\frac{-\text{810 kJ}}{2}\\ \\ =-\text{405 kJ/mol}\end{array}$

Enthalpy of reaction per mole of $CsF$ is $-405kJ/mol.$