Question

An interpenetrating primitive cubic structure like that of CsCl with anions in the corners has an edge length of 664 pm. If the ratio of the ionic radius of the cation to the ionic radius of the anion is 0.840, what is the ionic radius of the anion

Answer 1

Answer:

the ionic radius of the anion $r^- = 312.52 \ pm$

Explanation:

From the diagram shown below :

The anion $Cl^-$ is located at the corners

The cation $Cs^+$ is located at the body center

The Body diagonal length =  $\sqrt{3 \ a }$

∴ $2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a$

Given that :

$\frac{r^+}{r^-} =0.84$   (i.e the  ratio of the ionic radius of the cation to the ionic radius of

                 the anion )

$0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a$

Also ; a =  664 pm

Then :

$r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm$

Therefore,  the ionic radius of the anion $r^- = 312.52 \ pm$