# NEET Chemistry The Solid State Questions Solved

In a face centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral and octahedral voids ?

(A) 0.543

(B) 0.732

(C) 0.414

(D) 0.637

(A). Consider one tetrahedral void. If the side length of the tetrahedron is l and if d is the distance from the centre to the corner, then

${l}^{2}=2{d}^{2}-2{d}^{2}\mathrm{cos}\theta =2{d}^{2}\left(1+\frac{1}{3}\right)=\frac{4}{3}×2{d}^{2}$

but if a is the side length of the unit cell

$l=\frac{a}{\sqrt{2}}$ and $d={r}_{1}+{r}_{2}$, where ${r}_{1}$ is the radius of the corner atom and ${r}_{2}$ the radius of the void.

$\therefore \sqrt{\frac{8}{3}}\left({r}_{1}+{r}_{2}\right)=\frac{a}{\sqrt{2}},$

$\therefore {r}_{1}+{r}_{2}=\frac{a}{\sqrt{2}}×\frac{\sqrt{3}}{\sqrt{8}}=\frac{\sqrt{3}a}{4}$

but $l=2{r}_{1}.$

$\therefore {r}_{2}=\frac{\sqrt{3}a}{4}-\frac{\sqrt{2}a}{4}=\frac{a}{4}\left(\sqrt{3}-\sqrt{2}\right)$

For the octahedral void, if the radius of the void is ${r}_{3}$, then $2\left({r}_{1}+{r}_{3}\right)=a$

$=\frac{a}{2}-\frac{a}{2\sqrt{2}}=\frac{2a}{4}-\frac{\sqrt{2}a}{4}=\frac{\left(2-\sqrt{2}\right)a}{4}$

$\therefore$ratio of the sizes of tetrahedral to octahedral voids

$\frac{\frac{a}{4}\left(\sqrt{3}-\sqrt{2}\right)}{\frac{a}{4}\left(2-\sqrt{2}\right)}=\frac{1.732-1.414}{2-1.414}=\frac{0.318}{0.586}=0.543$

Difficulty Level:

• 37%
• 23%
• 34%
• 6%