Question

CsCl has cubic structure. Suppose its density is $4.37g/c{m}^3$. What is the distance between $C{s}^{+}$ and $C{l}^{-}$ ions?
(Atomic mass of $Cs$ is 133). $pm$

Answer 1

For CsCl, $Z_{eff}=1$
because total number of $C{l}^{-}$ ions per unit cell are $\frac{1}{8}\times 8=1$ and total number of $C{s}^{+}$ ions per unit cell are $=1$, so total number of $CsCl$ in a unit cell is 1 or $Z_{eff}=1$
So, density is given as,
$\rho =\frac{Z_{eff}\times M}{a^3\times {10}^{-30}\times N_A}$
or $a^3=\frac{Z_{eff}\times M}{\rho \times {10}^{-30}\times N_A}$
$a^3=\frac{1\times (133+35.5)}{4.37\times {10}^{-30}\times 6.02\times {10}^{23}}$

$a=(64.00{)}^{\frac{1}{3}}\times {10}^2$
$a=400pm$

Interionic distance $=\frac{\sqrt{3}a}{2}$
$=\frac{1.732}{2}\times 400$
$=346.4pm$