Question

Spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has 1/8 of the tetrahedral holes occupied by one type of metal ion and 1/2 of the octahedral holes occupied by another type of metal ion. Such a spine is formed by $Z{n}^{2+},A{l}^{3+}$ and $O^{2-}$ with $Z{n}^{2+}$ in the tetrahedral holes. If CCP arrangement of oxide ions remains undistorted in the presence of all the cations, formula of spinel and percentage of the packing fraction of crystal are respectively:

• A. $ZnA{l}_2{O}_4$, 77%
• B. $ZnA{l}_2{O}_4$, 74%
• C. $Z{n}^{2}Al{O}_4$, 74%
• D. $Z{n}_{3}A{l}_2{O}_6$, 74%

Answer 1

The correct option is A $ZnA{l}_2{O}_4$, 77%

Given that spinel is arranged in FCC or CCP and given $Z{n}^{+2}$ occupies in the $\frac{1}{8}$ of the tetrahedral holes and $A{l}^{3+}$ occupies in the $\frac{1}{2}$ of the octahedral holes and $O^{2-}$ is at the lattice points.

In CCP $4$ octahedral holes and $8$ tetrahedral holes are present.

So, $Z{n}^{2+}$ will be in $1$ tetrahedral hole and $A{l}^{3+}$ will be in $2$ tetrahedral hole. So, there are $1$ $Z{n}^{2+}$ atom and two $A{l}^{3+}$ atoms and $A{O}^{2-}$ in CCP unit cell.

The formula of spinel is $ZnA{l}_2{O}_4$

We know that,

$\frac{r_{Z{n}^{2+}}}{r}=0.225$

$\frac{r_{A{l}^{3+}}}{r}=0.414$

where $r$ is radius of anion and $4r=\sqrt{2}a$

where $a$ is edge length

So, $⟹r=\frac{a}{2\sqrt{2}}$

$PackagingFraction=\frac{Totalvolumeofparticlespresentinoneunit}{Volumeofoneunitcell}$

$=\frac{\left(\frac{4}{3}\pi r_{Z{n}^{2+}}^3\right)+\left(\frac{2\times 4}{3}\pi r_{A{l}^{2+}}^3\right)+\left(\frac{4\times 4}{3}\pi r_{O^{2-}}^3\right)}{a^3}$
$=\frac{\left(\frac{4}{3}\pi \frac{a^3(0.225{)}^3}{(2\sqrt{2}{)}^3}\right)+(2\times \frac{4}{3}\pi \frac{a^2(0.414{)}^3}{(2\sqrt{2}{)}^3})+(4\times \frac{4}{3}\pi \frac{a^3}{(2\sqrt{2}{)}^3})}{a^3}=0.77$

$=0.77$

Packaging Fraction $=0.77$