Question

Cesium bromide crystallises in a cubic system. Its unit cell has $C{s}^{+}$ ion at the body centre and $B{r}^{-}$ ion at each corner. Its density is $4.44gmc{m}^{-3}.$ Determine
(a) each length of the unit cell.
(b) size of $B{r}^{-}$ ion if the radius of $C{s}^{+}$ is 174 pm, and
(c) fraction of volume occupied by molecules in the unit cell.

[$Cs=132.90g.mo{l}^{-1}$ and $Br=79.9g.mo{l}^{-1}$ ]

Answer 1

In $CsBr$, $B{r}^{-}$ ions are present at each corner and $C{s}^{+}$ ion is at body center. So, there is $1$ formula unit of $CsBr$ per unit cell.

Theoretical density of crystal, $\rho =\frac{nM}{N_o{a}^3}g/c{m}^3$

So, $n=1;M=132.9+79.9=212.8g/mol;\rho =4.44g/c{m}^3$

$\Rightarrow a^3=\frac{nM}{\rho N_o}$

$\Rightarrow a^3=\frac{1\times 212.8}{4.44\times 6.022\times {10}^{23}}$

$\Rightarrow a=(79.6{)}^{1/3}\times ({10}^{-24}{)}^{1/3}cm$

$\Rightarrow a=4.3\times {10}^{-8}cm=430pm$

So, edge length of unit cell $=430pm$

So, on the body diagonal, $C{s}^{+}$ ion at body center and $B{r}^{-}$ ion at the corner are touching each other. So, if $r_{+}$ is radius of $C{s}^{+}$ and $r_{-}$ is radius of $B{r}^{-}$. Then,

$2(r_{+}+r_{-})=\sqrt{3}a$

$\Rightarrow r_{+}+r_{-}=\frac{\sqrt{3}}{2}\times 430pm$

$\Rightarrow r_{-}=372.4-174=198.4pm$

Total volume of unit cell $=a^3=(4.3\times {10}^{-8}{)}^3=79.6\times {10}^{-24}c{m}^3$

Volume occupied, $V=$ Volume occupied by $B{r}^{-}$ ions $+$ Volume occupied by $C{s}^{+}$ ions.

$V=\frac{4}{3}\pi (r_{-}^3+r_{+}^3)$

$V=\frac{4}{3}\pi \left(\right(1.98{)}^3+\left(1.74{)}^3\right)\times {10}^{-24}c{m}^3$

$V=54.6\times {10}^{-24}c{m}^3$

Fraction of volume occupied $=\frac{54.6\times {10}^{-24}}{79.6\times {10}^{-24}}=0.68$