Chapter 21, Problem 21.52E

Interpretation Introduction

(a)

Interpretation:

The two-particle coulombic energy of attraction is to be compared with more precise calculation of the lattice energy for the given ionic crystal.

Concept introduction:

The amount of energy released when one formula unit moles of oppositely charged gaseous ions binds together to form a crystal is known as the lattice energy. The value of lattice energy is negative. It is used as a measure for stability of a crystal.

Answer to Problem 21.52E

The two-particle coulombic energy of attraction is $-399.126\text{kJ}\cdot {\text{mol}}^{-1}$ and the calculated lattice energy is $636.2\text{kJ}\cdot {\text{mol}}^{-1}$. The magnitude of two-particle coulombic energy of attraction is less than the lattice energy.

Explanation of Solution

According to Coulomb’s law the potential energy of two oppositely charged particles that are $r$ distance from each other is given by,

$E=\frac{q_{+}{q}_{-}}{4\pi {\epsilon }_0\cdot r}$ …(1)

Where,

• $q_{+}$ and $q_{-}$ are oppositely charged ions.

• ${\epsilon }_0$ is the permittivity of free space and its value is $8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}$.

The closed distance between the opposite ions is calculated by considering the Table 21.4. The radii of ${\text{Cs}}^{+}$ and ${\text{Cl}}^{-}$ is $1.67\stackrel{\circ }{\text{A}}$ and $1.81\stackrel{\circ }{\text{A}}$ respectively. Thus, the closest distance between the opposite ions is calculated as follows:

$\begin{array}{l}r=\left(1.67+1.81\right)\stackrel{\circ }{\text{A}}\\ r=3.48\stackrel{\circ }{\text{A}}\end{array}$

Thus, the closest distance between the opposite ions is $3.48\stackrel{\circ }{\text{A}}$.

Substitute the values of charge on electron, permittivity of space and distance between two ions in equation (1).

$\begin{array}{l}E=\frac{\left(+1.602\times {10}^{-19}\text{C}\right)\left(-1.602\times {10}^{-19}\text{C}\right)}{4\pi \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\times 3.48\stackrel{\circ }{\text{A}}}\\ E=\frac{\left(+1.602\times {10}^{-19}\text{C}\right)\left(-1.602\times {10}^{-19}\text{C}\right)}{4\pi \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\times 3.48\times {10}^{-10}\text{m}}\\ E=-6.63\times {10}^{-19}\text{J}\end{array}$

For one mole of ions, the energy is multiplied by Avogadro number as shown below.

$\begin{array}{l}E=-6.63\times {10}^{-19}\text{J}\times \text{6}\text{.02}\times {\text{10}}^{23}{\text{mol}}^{-1}\\ E=-399.126\times {\text{10}}^3\text{J}\cdot {\text{mol}}^{-1}\\ E=-399.126\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Thus, the two-particle coulombic energy of attraction is $-399.126\text{kJ}\cdot {\text{mol}}^{-1}$.

The lattice energy is given by the expression as shown below.

$\text{lattice}\text{energy}=\frac{N_{\text{A}}\cdot M\cdot Z^2\cdot e^2}{4\pi {\epsilon }_0\cdot r}\left(1-\frac{\rho }{r}\right)$ …(2)

Where,

• $N_{\text{A}}$ is the Avogadro number.

• $M$ is the Madelung constant.

• $Z$ is the greatest common divisor of the magnitude of the charges on the ions.

• $e$ is the charge on electron.

• ${\epsilon }_0$ is the permittivity in free space.

• $r$ is the closest distance between the opposite ions.

• $\rho$ is the repulsive range parameter.

From the Table 21.6, the value of $M$ and $\rho$ for $\text{CsCl}$ is $1.7627$ and $0.331\stackrel{\circ }{\text{A}}$ respectively.

The $\text{CsCl}$ is $1:1$ compound. Thus, the $Z$ for $\text{CsCl}$ is $1$.

Substitute value of $Z$, $M$, $\rho$, and $r$ in equation (1).

$\begin{array}{rll}\text{lattice}\text{energy}& =& \frac{\left(6.02\times {10}^{23}{\text{mol}}^{-1}\right)\left(1.7627\right)\cdot {\left(1\right)}^2{\left(1.602\times {10}^{-19}\text{C}\right)}^2}{\left[4\pi \times \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\right]\cdot \left(3.48\times {10}^{-10}\text{m}\right)}\left(1-\frac{0.331\stackrel{\circ }{\text{A}}}{3.48\stackrel{\circ }{\text{A}}}\right)\\ & =& 7.03\times {10}^5\left(0.905\right)\text{J}\cdot {\text{mol}}^{-1}\\ & =& 6.362\times {10}^5\text{J}\cdot {\text{mol}}^{-1}\\ & =& 636.2\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Thus, the calculated lattice energy is $636.2\text{kJ}\cdot {\text{mol}}^{-1}$.

Therefore, the magnitude of two-particle coulombic energy of attraction is less than the lattice energy.

Conclusion

The two-particle coulombic energy of attraction is $-399.126\text{kJ}\cdot {\text{mol}}^{-1}$ and the calculated lattice energy is $636.2\text{kJ}\cdot {\text{mol}}^{-1}$. The magnitude of two-particle coulombic energy of attraction is less than the lattice energy.

Interpretation Introduction

(b)

Interpretation:

The two-particle coulombic energy of attraction is to be compared with more precise calculation of the lattice energy for the given ionic crystal.

Concept introduction:

The amount of energy released when one formula unit moles of oppositely charged gaseous ions binds together to form a crystal is known as the Lattice energy. The value of lattice energy is negative. It is used as the measure for stability of a crystal.

Answer to Problem 21.52E

The two-particle coulombic energy of attraction is $-1020.992\text{kJ}\cdot {\text{mol}}^{-1}$ and the calculated lattice energy is $5268.4\text{kJ}\cdot {\text{mol}}^{-1}$. The magnitude of two-particle coulombic energy of attraction is less than the lattice energy.

Explanation of Solution

According to Coulomb’s law the potential energy of two oppositely charged particles that are $r$ distance from each other is given by,

$E=\frac{q_{+}{q}_{-}}{4\pi {\epsilon }_0\cdot r}$ … (1)

Where,

• $q_{+}$ and $q_{-}$ are oppositely charged ions.

• ${\epsilon }_0$ is the permittivity of free space and its value is $8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}$.

The closed distance between the opposite ions is calculated by considering the Table 21.4. The radii of ${\text{Zn}}^{2+}$ and ${\text{S}}^{2-}$ is $0.74\stackrel{\circ }{\text{A}}$ and $1.84\stackrel{\circ }{\text{A}}$ respectively. Thus, the closest distance between the opposite ions is calculated as follows:

$\begin{array}{l}r=\left(0.74+1.84\right)\stackrel{\circ }{\text{A}}\\ r=1.36\stackrel{\circ }{\text{A}}\end{array}$

Thus, the closest distance between the opposite ions is $1.36\stackrel{\circ }{\text{A}}$.

Substitute the values of charge on electron, permittivity of space and distance between two ions in equation (1).

$\begin{array}{l}E=\frac{\left(+1.602\times {10}^{-19}\text{C}\right)\left(-1.602\times {10}^{-19}\text{C}\right)}{4\pi \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\times 1.36\stackrel{\circ }{\text{A}}}\\ E=\frac{\left(+1.602\times {10}^{-19}\text{C}\right)\left(-1.602\times {10}^{-19}\text{C}\right)}{4\pi \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\times 1.36\times {10}^{-10}\text{m}}\\ E=-1.696\times {10}^{-18}\text{J}\end{array}$

For one mole of ions, the energy is multiplied by Avogadro number as shown below.

$\begin{array}{l}E=-1.696\times {10}^{-18}\text{J}\times \text{6}\text{.02}\times {\text{10}}^{23}{\text{mol}}^{-1}\\ E=-1020.992\times {\text{10}}^3\text{J}\cdot {\text{mol}}^{-1}\\ E=-1020.992\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Thus, the two-particle coulombic energy of attraction is $-1020.992\text{kJ}\cdot {\text{mol}}^{-1}$.

The lattice energy is given by the expression as shown below.

$\text{lattice}\text{energy}=\frac{N_{\text{A}}\cdot M\cdot Z^2\cdot e^2}{4\pi {\epsilon }_0\cdot r}\left(1-\frac{\rho }{r}\right)$… (2)

Where,

• $N_{\text{A}}$ is the Avogadro number.

• $M$ is the Madelung constant.

• $Z$ is the greatest common divisor of the magnitude of the charges on the ions.

• $e$ is the charge on electron.

• ${\epsilon }_0$ is the permittivity in free space.

• $r$ is the closest distance between the opposite ions.

• $\rho$ is the repulsive range parameter.

From the Table 21.6, the value of $M$ and $\rho$ for $\text{ZnS}$ is $1.6381$ and $0.289\stackrel{\circ }{\text{A}}$ respectively.

$\text{ZnS}$ is $2:2$ compound. Thus, the $Z$ for $\text{ZnS}$ is $2$.

Substitute value of $Z$, $M$, $\rho$, and $r$ in equation (1).

$\begin{array}{rll}\text{lattice}\text{energy}& =& \frac{\left(6.02\times {10}^{23}{\text{mol}}^{-1}\right)\left(1.6381\right)\cdot {\left(2\right)}^2{\left(1.602\times {10}^{-19}\text{C}\right)}^2}{\left[4\pi \times \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\right]\cdot \left(1.36\times {10}^{-10}\text{m}\right)}\left(1-\frac{0.289\stackrel{\circ }{\text{A}}}{1.36\stackrel{\circ }{\text{A}}}\right)\\ & =& 6.69\times {10}^6\left(0.7875\right)\text{J}\cdot {\text{mol}}^{-1}\\ & =& 5.2684\times {10}^6\text{J}\cdot {\text{mol}}^{-1}\\ & =& 5268.4\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Thus, the calculated lattice energy is $5268.4\text{kJ}\cdot {\text{mol}}^{-1}$.

Therefore, the magnitude of two-particle coulombic energy of attraction is less than the lattice energy.

Conclusion

The two-particle coulombic energy of attraction is $-1020.992\text{kJ}\cdot {\text{mol}}^{-1}$ and the calculated lattice energy is $5268.4\text{kJ}\cdot {\text{mol}}^{-1}$. The magnitude of two-particle coulombic energy of attraction is less than the lattice energy.

Interpretation Introduction

(c)

Interpretation:

The two-particle coulombic energy of attraction is to be compared with more precise calculation of the lattice energy for the given ionic crystal.

Concept introduction:

The amount of energy released when one formula unit moles of oppositely charged gaseous ions binds together to form a crystal is known as the Lattice energy. The value of lattice energy is negative. It is used as the measure for stability of a crystal.

Answer to Problem 21.52E

The two-particle coulombic energy of attraction is $-697.72\text{kJ}\cdot {\text{mol}}^{-1}$ and the calculated lattice energy is $5874\text{kJ}\cdot {\text{mol}}^{-1}$. The magnitude of two-particle coulombic energy of attraction is less than the lattice energy.

Explanation of Solution

According to Coulomb’s law the potential energy of two oppositely charged particles that are $r$ distance from each other is given by,

$E=\frac{q_{+}{q}_{-}}{4\pi {\epsilon }_0\cdot r}$…(1)

Where,

• $q_{+}$ and $q_{-}$ are oppositely charged ions.

• ${\epsilon }_0$ is the permittivity of free space and its value is $8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}$.

The closed distance between the opposite ions is calculated by considering the Table 21.4. The radii of ${\text{Ti}}^{4+}$ and ${\text{O}}^{2-}$ is $0.68\stackrel{\circ }{\text{A}}$ and $1.31\stackrel{\circ }{\text{A}}$ respectively. Thus, the closest distance between the opposite ions is calculated as follows:

$\begin{array}{l}r=\left(0.68+1.31\right)\stackrel{\circ }{\text{A}}\\ r=1.99\stackrel{\circ }{\text{A}}\end{array}$

Thus, the closest distance between the opposite ions is $1.99\stackrel{\circ }{\text{A}}$.

Substitute the values of charge on electron, permittivity of space and distance between two ions in equation (1).

$\begin{array}{l}E=\frac{\left(+1.602\times {10}^{-19}\text{C}\right)\left(-1.602\times {10}^{-19}\text{C}\right)}{4\pi \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\times 1.99\stackrel{\circ }{\text{A}}}\\ E=\frac{\left(+1.602\times {10}^{-19}\text{C}\right)\left(-1.602\times {10}^{-19}\text{C}\right)}{4\pi \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\times 1.99\times {10}^{-10}\text{m}}\\ E=-1.159\times {10}^{-18}\text{J}\end{array}$

For one mole of ions, the energy is multiplied by Avogadro number as shown below.

$\begin{array}{l}E=-1.159\times {10}^{-18}\text{J}\times \text{6}\text{.02}\times {\text{10}}^{23}{\text{mol}}^{-1}\\ E=-697.72\times {\text{10}}^3\text{J}\cdot {\text{mol}}^{-1}\\ E=-697.72\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Thus, the two-particle coulombic energy of attraction is $-697.72\text{kJ}\cdot {\text{mol}}^{-1}$.

The lattice energy is given by the expression as shown below.

$\text{lattice}\text{energy}=\frac{N_{\text{A}}\cdot M\cdot Z^2\cdot e^2}{4\pi {\epsilon }_0\cdot r}\left(1-\frac{\rho }{r}\right)$ … (2)

Where,

• $N_{\text{A}}$ is the Avogadro number.

• $M$ is the Madelung constant.

• $Z$ is the greatest common divisor of the magnitude of the charges on the ions.

• $e$ is the charge on electron.

• ${\epsilon }_0$ is the permittivity in free space.

• $r$ is the closest distance between the opposite ions.

• $\rho$ is the repulsive range parameter.

From the Table 21.6, the value of $M$ and $\rho$ for $\text{ZnS}$ is $1.6381$ and $0.289\stackrel{\circ }{\text{A}}$ respectively.

${\text{TiO}}_2$ is $2:1$ compound. Thus, the $Z$ for ${\text{TiO}}_2$ is $2$.

Substitute value of $Z$, $M$, $\rho$, and $r$ in equation (1).

$\begin{array}{rll}\text{lattice}\text{energy}& =& \frac{\left(6.02\times {10}^{23}{\text{mol}}^{-1}\right)\left(2.408\right)\cdot {\left(2\right)}^2{\left(1.602\times {10}^{-19}\text{C}\right)}^2}{\left[4\pi \times \left(8.854\times {10}^{-12}{\text{C}}^2/\text{J}\cdot \text{m}\right)\right]\cdot \left(1.99\times {10}^{-10}\text{m}\right)}\left(1-\frac{0.250\stackrel{\circ }{\text{A}}}{1.99\stackrel{\circ }{\text{A}}}\right)\\ & =& 6.721\times {10}^6\left(0.874\right)\text{J}\cdot {\text{mol}}^{-1}\\ & =& 5.874\times {10}^6\text{J}\cdot {\text{mol}}^{-1}\\ & =& 5874\text{kJ}\cdot {\text{mol}}^{-1}\end{array}$

Thus, the calculated lattice energy is $5874\text{kJ}\cdot {\text{mol}}^{-1}$.

Therefore, the magnitude of two-particle coulombic energy of attraction is less than the lattice energy.

Conclusion

The two-particle coulombic energy of attraction is $-697.72\text{kJ}\cdot {\text{mol}}^{-1}$ and the calculated lattice energy is $5874\text{kJ}\cdot {\text{mol}}^{-1}$. The magnitude of two-particle coulombic energy of attraction is less than the lattice energy.