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# In a crystal oxide ions are arranged in fcc and A$^{+2}$ ions are at 1/8th of the tetrahedral voids, and ions $B^{+3}$ occupied $\large\frac{1}{2}$ of the octahedral voids. Calculate the packing fraction of the crystal if O-2 of the removed from alternate corner and $A^{+2}$ is being place at 2 of the corners.

Since oxide ions are fcc so $4O^{-2}$ / unit cell
$A^{+2}$ are at 1/8th of the tetrahedral so 1$A^{+2}$/unitcell
$B^{+3}$ occupies $\large\frac{1}{2}$ of the octahedral voids $\therefore 2B^{+2}$/unit cell
After removing $O^{-2}$ ions
Oxide ion / unit cell =$\large\frac{4}{8}$$+3=3.5 A^{+2} ions/ unit cell =\large\frac{2}{8}$$+1=\large\frac{10}{8}$$=1.25 B^{+3} ions/unit cell = 2 We know that, a =\large\frac{4r_-}{\sqrt 2} \large\frac{r_A^{+2}}{r_-}$$=0.225$
$\therefore r_A^+=0.225r_-$
$\large\frac{r_B^{+2}}{r_-}$$=0.414$
$\therefore r_B^{+3}=0.414r_-$
Putting all the values
P.F = 0.676
Hence (B) is the correct answer.