Answer:
the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
Explanation:
From the diagram shown below :
The anion [tex]Cl^-[/tex] is located at the corners
The cation [tex]Cs^+[/tex] is located at the body center
The Body diagonal length = [tex]\sqrt{3 \ a }[/tex]
∴ [tex]2 \ r^+ \ + 2r^- \ = \sqrt{3 \ a} \\ \\ r^+ +r^- = \frac{\sqrt{3}}{2} a[/tex]
Given that :
[tex]\frac{r^+}{r^-} =0.84[/tex] (i.e the ratio of the ionic radius of the cation to the ionic radius of
the anion )
[tex]0.84r^- \ + r^- \ = \frac{\sqrt{3}}{2}a \\ \\ 1.84 r^- = \frac{3}{2}a \\ \\ r^- = \frac{\sqrt{3}}{2*1.84}a[/tex]
Also ; a = 664 pm
Then :
[tex]r^- = \frac{\sqrt{3} }{2*1.84}*664 \ pm\\ \\ r^- = 312.52 \ pm[/tex]
Therefore, the ionic radius of the anion [tex]r^- = 312.52 \ pm[/tex]
The ionic radius of the anion [tex]r^-=312.52pm[/tex]
The anion is placed on the corners and the cation is placed on the frame center.
The Body diagonal length = [tex]\sqrt{3a}[/tex]
[tex]2r^++2r^-=\sqrt{3a} \\\\r^++r^-=\sqrt{3}/2a }[/tex]
Given:
Ratio= 0.840
[tex]\frac{r^+}{r^-}=0.840[/tex]
[tex]0.84r^-+r^-=\sqrt{3}/2a \\\\1.84r^-=3/2a\\\\r^-=\sqrt{3}/2*1.84a[/tex]
Also ; a = 664 pm
Then : [tex]r^-[/tex] =312.52 pm
Therefore, the ionic radius of the anion = 312.52 pm
Find more information about Cubic structure here:
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