Chapter 4, Problem 4.16P

Interpretation Introduction

(a)

Interpretation:

The number of vacancies of anion per ${\text{cm}}^3$ is to be determined.

Concept introduction:

In ionic materials, most common defects found is the Schottky defect in which stoichiometric number of cations and anions are missing from the crystal lattice and the electoral neutrality of the crystal is preserved.

The number of vacancies per ${\text{cm}}^3$ is calculated by the formula;

  $n_v=\frac{\text{number of vacancies in unit cell}}{V}$ ....... (1)

Volume of a unit cell $\left(V\right)$ is calculated using the lattice parameter $\left(a_0\right)$ as:

  $V={\left(a_0\right)}^3$ ....... (2)

Answer to Problem 4.16P

The number of vacancies of anion per ${\text{cm}}^3$ is $1.61\times {10}^{21}\text{ vacancies}/{\text{cm}}^3$.

Explanation of Solution

Given information:

In every tenth unit cell of MgO there is one Schottky defect present. The crystal structure of MgO is that of the sodium chloride with lattice parameter of $0.396\text{ nm}$.

Use equation (2) to calculate the volume of a unit cell from its lattice parameter as:

  $\begin{array}{c}V={\left(a_0\right)}^3\\ ={\left(0.396\times {10}^{-7}\text{ cm}\right)}^3\\ =6.21\times {10}^{-23}{\text{ cm}}^3\end{array}$

The crystal structure of NaCl is FCC in which there are $4$

  ${\text{Na}}^{+}$ ions and $4{\text{ Cl}}^{-}$ ions. Thus, MgO will also have $4$

  ${\text{Mg}}^{2+}$ ions and $4{\text{ O}}^{2-}$ ions in a unit cell. Let the basis of the calculations be $10$ unit cells. So, there will be total ${\text{40 Mg}}^{2+}$ ions and $40{\text{ O}}^{2-}$ ions.

But it is given that for every $10$ unit cells, there is one Schottky defect present, thus, the number of cation present in the given sample will be ${\text{39 Mg}}^{2+}$ and number of anion present will be $39{\text{ O}}^{2-}$.

Using equation (1) to calculate the number of anion vacancies per ${\text{cm}}^3$ will be:

  $\begin{array}{c}{n}_v=\frac{\text{number of vacancies in unit cell}}{V}\\ =\frac{1\text{ vacancy}/\text{10 unit cells}}{6.21\times {10}^{-23}{\text{ cm}}^3/\text{unit cell}}\\ =1.61\times {10}^{21}\text{ vacancies}/{\text{cm}}^3\end{array}$

Interpretation Introduction

(b)

Interpretation:

The density of the given ceramic is to be calculated.

Concept introduction:

Volume of a unit cell $\left(V\right)$ is calculated using the lattice parameter $\left(a_0\right)$ as:

  $V={\left(a_0\right)}^3$ ....... (2)

The relationship between the theoretical density $\left(\rho \right)$ of an alloy, number of atoms of all the constituent species present in the alloy $\left(n_i\right)$ present in a unit cell and volume of a unit cell $\left(V\right)$ is:

  $\rho =\frac{{\displaystyle \sum n_i\times {\left(\text{A}\text{.M}\right)}_i}}{V\times N_{\text{A}}}$ ....... (3)

Here, ${\left(\text{A}\text{.M}\right)}_i$ is the atomic mass of the species $i$ in the alloy and $N_{\text{A}}$ is the Avogadro's number.

Answer to Problem 4.16P

The density of the given ceramic is $4.204{\text{ g/cm}}^3$.

Explanation of Solution

Given information:

In every tenth unit cell of MgO there is one Schottky defect present. The crystal structure of MgO is that of the sodium chloride with lattice parameter of $0.396\text{ nm}$.

Use equation (2) to calculate the volume of a unit cell from its lattice parameter as:

  $\begin{array}{c}V={\left(a_0\right)}^3\\ ={\left(0.396\times {10}^{-7}\text{ cm}\right)}^3\\ =6.21\times {10}^{-23}{\text{ cm}}^3\end{array}$

Let the basis of the calculations be $10$ unit cells. So, there will be total ${\text{40 Mg}}^{2+}$ ions and $40{\text{ O}}^{2-}$ ions according to the FCC crystal structure of NaCl.

It is given that for every $10$ unit cells, there is one Schottky defect present, thus, the number of cation present in the given sample will be ${\text{39 Mg}}^{2+}$ and number of anion present will be $39{\text{ O}}^{2-}$.

So, for a unit cell number of cations and anions present in a unit cell will be:

  $\begin{array}{c}{n}_{+}=\frac{39{\text{ Mg}}^{2+}\text{ ions}}{40{\text{ Mg}}^{2+}\text{ ions}}\times \frac{4{\text{ Mg}}^{2+}\text{ ions}}{1\text{ cell}}\\ =3.9{\text{ Mg}}^{2+}\text{ ions/cell}\\ n_{-}=\frac{39{\text{ O}}^{2-}\text{ ions}}{40{\text{ O}}^{2-}\text{ ions}}\times \frac{4{\text{ O}}^{2-}\text{ ions}}{1\text{ cell}}\\ =3.9{\text{ O}}^{2-}\text{ ions/cell}\end{array}$

Atomic weights of $\text{Mg and O}$ are $\text{24}\text{.312 g/mol}$ and $16\text{ g/mol}$ respectively.

Use equation (3) to calculate the density of the ceramic sample as:

  $\begin{array}{c}\rho =\frac{\left(n_{\text{+}}\times {\left(\text{A}\text{.M}\right)}_{{\text{Mg}}^{2+}}\right)+\left(n_{-}\times {\left(\text{A}\text{.M}\right)}_{{\text{O}}^{2-}}\right)}{V\times N_{\text{A}}}\\ =\frac{\left(3.9\times 24.312\right)+\left(3.9\times 16\right)}{\left(6.21\times {10}^{-23}\right)\times \left(6.022\times {10}^{23}\right)}\\ =4.204{\text{ g/cm}}^3\end{array}$