In the cubic crystal of CsCl $(d = 3.97\,g/c{m^3})$ ...

In the cubic crystal of CsCl (d=3.97 g/cm³) the eight corners are occupied by Cl^- with a Cs⁺ at the centre and vice-versa. Assuming anion-anion contact, calculate the distance between the neighbouring Cs⁺ and Cl^- ions. What is the radius ratio of the two ions? [At. Wt. of Cs = 132.91 and that of Cl = 35.45]

Solution

${\rm{Density}} = \frac{{{\rm{n}} \times {\rm{molecular\,weight}}}}{{{{\rm{a}}^{\rm{3}}} \times {{\rm{N}}_{\rm{a}}}}}$

$ \Rightarrow 3.97 = \frac{{1 \times 168.36}}{{{a^3} \times 6.023 \times {{10}^{23}}}}$

$ \Rightarrow a = 4.13\mathop A\limits^ \circ $

Density of cube $ = a\sqrt 3  = 7.15\mathop A\limits^ \circ  $

For b.c.c., ${\rm{2}}{{\rm{r}}^ + } + 2{r^ - } = 7.15$

$ \Rightarrow {r^ + } + {r^ - } = 3.75\mathop A\limits^ \circ  ..........*$

Also, $2{r^ - } = 4.13\mathop A\limits^ \circ  {\rm{ }}..........**$

$ \Rightarrow {r^ - } = 2.06\mathop A\limits^ \circ  $

${r^ + } = 3.57 - 2.06 = 1.51\mathop A\limits^ \circ  $

$\frac{{{r^ + }}}{{{r^ - }}} = \frac{{1.51}}{{2.06}} = 0.73$

Image Solution

* Distance between the neighbouring Cs⁺ and Cl^- ions = r⁺ + r^-
** Due to anion-anion contact