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Q.
The face diagonal length of $f. c. c$. cubic cell is $660 \sqrt{2}$ pm. If the radius of the cation is $110\, pm$, the radius of the anion in $( pm )$ is
The Solid State
Solution:
$\sqrt{2} a=660 \sqrt{2}\, pm$
so $a=660 \,pm$
Now if tetrahedral void is occupied by cations than
$\frac{\sqrt{3}}{4} a=\left(r_{+}+r_{-}\right)$
$r_{-}=\left(\frac{\sqrt{3} \times 660}{4}-110\right)=110\left[\frac{3}{2} \sqrt{3}-1\right]=1.598$
so $\frac{r_{+}}{r_{-}}=\frac{1}{1.598} \simeq \frac{1}{1.6}=\frac{10}{16}=0.625$
but $\frac{r_{+}}{r_{-}}>0.414$ so it must not be occupying tetrahedral
void then
$a=2\left(r_{+}+r_{-}\right) $
$\Rightarrow 330=r_{+}+r_{-} $
$r_{-}=220 \,pm$
$\left\{\frac{r_{+}}{r_{-}}=0.5\right.$ it can occupy octahedral void $\}$