Chapter 11.13, Problem 71AAP

Calculate the ionic packing factor for (a) MnO and (b) SrO. Ionic radii are Mn²⁺ = 0.091 nm, Sr²⁺ = 0.127 nm, and O²⁻ = 0.132 nm.

To determine

The ionic packing factor for manganese oxide $\left(\text{MnO}\right)$.

Answer to Problem 71AAP

The ionic packing factor for manganese oxide $\left(\text{MnO}\right)$ is $0.577$.

Explanation of Solution

Write the expression to calculate lattice constant of manganese oxide structure $\left(a\right)$.

 $a=2\left(r_{{\text{Mn}}^{2+}}+R_{{\text{O}}^{2-}}\right)$                                                                                                    (I)

Here, ionic radius of ${\text{Mn}}^{2+}$ ion is $r_{{\text{Mn}}^{2+}}$ and ionic radius of ${\text{O}}^{2-}$ ion is $R_{{\text{O}}^{2-}}$.

Write the expression to ionic packing factor for manganese oxide $\left(\text{IPF}\right)$.

 $\begin{array}{c}\text{IPF}=\frac{\text{Volume of ions per unit cell}}{\text{Volume of unit cell}}\\ =\frac{\left(\frac{4\pi }{3}\right)\left[n_{{\text{Mn}}^{2+}}{r}_{{\text{Mn}}^{2+}}^3+n_{{\text{O}}^{2-}}{R}_{{\text{O}}^{2-}}^3\right]}{a^3}\end{array}$                                                                         (II)

Here, number of atoms of ${\text{Mn}}^{2+}$ ion is $n_{{\text{Mn}}^{2+}}$ and number of atoms of ${\text{O}}^{2-}$ ion is $n_{{\text{O}}^{2-}}$.

Conclusion:

Substitute $0.091\text{nm}$ for $r_{{\text{Mn}}^{2+}}$ and $0.132\text{nm}$ for $R_{{\text{O}}^{2-}}$ in Equation (I).

 $\begin{array}{c}a=2\left(0.091\text{nm}+0.132\text{nm}\right)\\ =0.446\text{nm}\end{array}$

There will be four manganese ions and four oxygen ions present in a single unit cell of manganese oxide.

Substitute $4$ for $n_{{\text{Mn}}^{2+}}$, $0.091\text{nm}$ for $r_{{\text{Mn}}^{2+}}$, $4$ for $n_{{\text{O}}^{2-}}$, $0.132\text{nm}$ for $R_{{\text{O}}^{2-}}$ and $0.446\text{nm}$ for $a$ in Equation (II).

 $\begin{array}{c}\text{IPF}=\frac{\left(\frac{4\pi }{3}\right)\left[\left(4\right){\left(0.091\text{nm}\right)}^3+\left(4\right){\left(0.132\text{nm}\right)}^3\right]}{{\left(0.446\text{nm}\right)}^3}\\ =0.577\end{array}$

Thus, the ionic packing factor for manganese oxide $\left(\text{MnO}\right)$ is $0.577$.

To determine

The ionic packing factor for strontium oxide $\left(\text{SrO}\right)$.

Answer to Problem 71AAP

The ionic packing factor for strontium oxide $\left(\text{SrO}\right)$ is $0.524$.

Explanation of Solution

Write the expression to calculate lattice constant of strontium oxide structure $\left(a\right)$.

 $a=2\left(r_{{\text{Sr}}^{2+}}+R_{{\text{O}}^{2-}}\right)$                                                                                                 (III)

Here, ionic radius of ${\text{Sr}}^{2+}$ ion is $r_{{\text{Sr}}^{2+}}$.

Write the expression to ionic packing factor for strontium oxide $\left(\text{IPF}\right)$.

 $\begin{array}{c}\text{IPF}=\frac{\text{Volume of ions per unit cell}}{\text{Volume of unit cell}}\\ =\frac{\left(\frac{4\pi }{3}\right)\left[n_{{\text{Sr}}^{2+}}{r}_{{\text{Sr}}^{2+}}^3+n_{{\text{O}}^{2-}}{R}_{{\text{O}}^{2-}}^3\right]}{a^3}\end{array}$                                                                  (IV)

Here, number of atoms of ${\text{Sr}}^{2+}$ ion is $n_{{\text{Sr}}^{2+}}$.

Conclusion:

Substitute $0.127\text{nm}$ for $r_{{\text{Sr}}^{2+}}$ and $0.132\text{nm}$ for $R_{{\text{O}}^{2-}}$ in Equation (III).

 $\begin{array}{c}a=2\left(0.127\text{nm}+0.132\text{nm}\right)\\ =0.518\text{nm}\end{array}$

There will be 4 strontium ions and 4 oxygen ions present in a single unit cell of strontium oxide.

Substitute $4$ for $n_{{\text{Sr}}^{2+}}$, $0.127\text{nm}$ for $r_{{\text{Sr}}^{2+}}$, $4$ for $n_{{\text{O}}^{2-}}$, $0.132\text{nm}$ for $R_{{\text{O}}^{2-}}$ and $0.518\text{nm}$ for $a$ in Equation (IV).

 $\begin{array}{c}\text{IPF}=\frac{\left(\frac{4\pi }{3}\right)\left[\left(4\right){\left(0.127\text{nm}\right)}^3+\left(4\right){\left(0.132\text{nm}\right)}^3\right]}{{\left(0.518\text{nm}\right)}^3}\\ =0.524\end{array}$

Thus, the ionic packing factor for strontium oxide $\left(\text{SrO}\right)$ is $0.524$.