Question

The table below lists the ionic radii for the cations and anions in three different ionic compounds.

Formula	${\mathit{r}}_{\mathbf{\text{cation}}}$	${\mathit{r}}_{\mathbf{\text{anion}}}$
${\mathbf{\text{SnO}}}_{\mathbf{2}}$	71 pm	140. pm
AIP	50.0 pm	212 pm.
BaO	135 pm	140. pm

Each compound has either the $\mathbf{\text{NaCl}}\mathbf{,}\mathbf{\text{CsCl}}$ , or ZnS type cubic structure. Predict the type of structure formed ($\mathbf{\text{NaCl}}\mathbf{,}\mathbf{\text{CsCl}}$ , or ) and the type and fraction of holes filled by the cations, and estimate the density of each compound.

Short answer

${\text{SnO}}_2=\text{NaCl}$type, density$\mathrm{density}=6.66{\text{g/cm}}^3$

One-half of octahedral holes will be filled by${\text{Sn}}^{4+}$cations.

$\text{AlP}=\text{ZnS}$type, density$\mathrm{density}=1.74{\text{g/cm}}^3$

${\text{Al}}^{3+}$ions occupy 1/2 of tetrahedral voids.

$\text{BaO}=\text{CsCl}$type, density$\mathrm{density}=7.92{\text{g/cm}}^3$

All cubic holes are occupied by ${\text{Ba}}^{2+}$.

Step-by-step solution

Use radius ratio to determine the type of structure

For${\text{SnO}}_2:\frac{r+}{r-}=\frac{71\text{pm}}{140\text{pm}}=0.51$(NaCl type structure)

For$\text{AlP :}\frac{r+}{r-}=\frac{50\text{pm}}{212\text{pm}}=0.24$(ZnS type structure)

For $\text{BaO}:\frac{r+}{r-}=\frac{135\text{pm}}{140\text{pm}}=0.964$ (CsCl type structure)

For the type of structure, find out the ions and the voids they occupy

In the NaCl type structure, cations occupy octahedral voids. So in${\text{SnO}}_2$ , there is one octahedral hole per closest packed${\text{O}}^{2-}$. One-half octahedral voids will be occupied by${\text{Sn}}^{4+}$cations.

In the ZnS type structure, cations occupy tetrahedral holes. So in AlP, there are two tetrahedral voids per closest packed${\text{P}}^{3-}$. Half tetrahedral voids are occupied by${\text{Al}}^{3+}$ions.

In the CsCl type structure, cations are present in cubic holes. So in BaO, there is one cubic hole per simple cubic packing of${\text{O}}^{2-},{\text{Ba}}^{2+}$ ions occupy the cubic holes.

Now calculate their densities

${\text{SnO}}_2$: Let the Ede length be “l.”

$$\begin{gathered} \mathrm{l}=2{\mathrm{r}}_{{\text{Sn}}^{4+}}+2{\mathrm{r}}_{{\text{O}}^{2-}} \\ =2\left(71\text{pm}\right)+2\left(140\text{pm}\right) \end{gathered}$$

Density

$\begin{array}{rcl}\mathrm{density}& =& \frac{\text{mass}}{\text{volume}}\\ & =& \frac{2{\mathrm{SnO}}_2\text{units}\times \frac{1\text{mol}{\mathrm{SnO}}_2}{6.022\times {10}^{23}\text{units}}\times \frac{150.7{\text{g SnO}}_2}{{\text{mol SnO}}_2}}{{(4.22\times {10}^{-8}\text{cm})}^3}\\ & =& 6.66{\text{g/cm}}^3\end{array}$

AIP : Body diagonal$\sqrt{3}\mathrm{l}=4{\mathrm{r}}_{{\mathrm{Al}}^{3+}}+4{\mathrm{r}}_{{\mathrm{P}}^{3-}}$

$\begin{array}{rcl}\mathrm{l}& =& \frac{4\left(50\text{pm}\right)+4\left(212\text{pm}\right)}{\sqrt{3}}\\ & =& 605\mathrm{pm}\end{array}$

Density

$\begin{array}{rcl}\mathrm{density}& =& \frac{\text{mass}}{\text{volume}}\\ & =& \frac{4\mathrm{AlP}\text{units}\times \frac{1\text{mol AlP}}{6.022\times {10}^{23}\text{units}}\times \frac{\text{57.95 g AlP}}{\text{mol AlP}}}{{(6.05\times {10}^{-8}\text{cm})}^3}\\ & =& 1.74{\text{g/cm}}^3\end{array}$

BaO : Body diagonal$\sqrt{3}\mathrm{l}=2{\mathrm{r}}_{{\mathrm{Ba}}^{2+}}+2{\mathrm{r}}_{{\mathrm{O}}^{2-}}$

$\begin{array}{rcl}\mathrm{l}& =& \frac{2\left(135\text{pm}\right)+2\left(140\text{pm}\right)}{\sqrt{3}}\\ & =& 318\mathrm{pm}\end{array}$

Density

$\begin{array}{rcl}\mathrm{density}& =& \frac{\text{mass}}{\text{volume}}\\ & =& \frac{4\mathrm{BaO}\text{units}\times \frac{1\text{mol BaO}}{6.022\times {10}^{23}\text{units}}\times \frac{\text{153.3 g BaO}}{\text{mol BaO}}}{{(3.18\times {10}^{-8}\text{cm})}^3}\\ & =& 7.92{\text{g/cm}}^3\end{array}$