In a cubic crystal of CsCl (density = 3.97 g/cm³), the eight corners are occupied by Cl^- ions with Cs⁺ at the center and vice versa. Calculate the distance between the neighboring Cs⁺ and Cl^- ions. The radius ratio of two ions is?

As we known 

$\mathrm{Density} \left(\mathrm{\rho }\right)=\frac{\mathrm{n}\times {\mathrm{M}}_{\mathrm{m}}}{{\mathrm{N}}_{\mathrm{A}}\times {\mathrm{a}}^3} \mathrm{here}, \mathrm{n} = 1$

$M_{\mathrm{m}}=132.9+35.5=168.4\mathrm{g}$

or, $a^3=\frac{1\times 168.4}{6.023\times {10}^{23}\times 3.97}$

or, $a=4.13\times {10}^{-8}=4.13\text{Å}$

As it is bcc with Cs⁺ at center (radius r⁺) and Cl- at corner (radius r^-)

So, body diagonal $=\sqrt{3}a\text{ or, }2r^{+}+2r^{-}=\sqrt{3}a$

$\text{ or, }{r}^{+}+r^{-}=\frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{2}\times 4.13\text{Å}$

$r^{+}+r^{-}=3.57\text{Å}$

Now assume that two Cl^- ions touch each other 

That is, $r^{-}+r^{-}=a$

$$\begin{gathered} \mathrm{or}, \begin{array}{r}\frac{{\mathrm{r}}^{+}+{\mathrm{r}}^{-}}{{\mathrm{r}}^{-}}=\frac{3.57}{2.065}\\ \end{array} \\ \text{ or, }\frac{{\mathrm{r}}^{+}}{{\mathrm{r}}^{-}}+1=\frac{3.57}{2.065}=1.728 \\ \text{ or, }\frac{{\mathrm{r}}^{+}}{{\mathrm{r}}^{-}}=0.728 \end{gathered}$$