Question

If R is the radius of the octahedral void and r is the radius of the atom in close packing, then $(\frac{R}{r}+1{)}^2$ is equal to :

Answer 1

Octahedral void : All the anions touch each other and co-ordination number is $6$.

In $△ABC,A{C}^2=A{B}^2+B{C}^2$

$=a^2+a^2$

$AC=\sqrt{2}a$

$2r^{+}+2r^{-}=AC=\sqrt{2}a$

$BC=a=2r^{-}$

$\frac{2r^{+}+2r^{-}}{2r^{-}}=\frac{\sqrt{2}\times 2r^{-}}{2r^{-}}$

$\frac{r^{+}}{r^{-}}+1=\sqrt{2}$

$(\frac{r^{+}}{r^{-}}+1{)}^2=2$