Question

A solid made up of ions of A and B has an edge length of 0.564 nm has four formula units. Among the two ions, the smaller ones occupy the interstitial voids and the larger ions occupy the space lattice with ccp type of arrangement. The solid has a higher tendency to show Frenkel defects than Schottky defects. (Assume anions are in contact with each other)
The ionic radius of the positive ion is :

• A. $0.755\stackrel{\circ }{A}$
• B. $0.625\stackrel{\circ }{A}$
  
• C. $0.826\stackrel{\circ }{A}$
  
• D. $0.449\stackrel{\circ }{A}$

Answer 1

The correct option is D $0.449\stackrel{\circ }{A}$

Since the unit cell has four formula units, it could be NaCl or ZnS type. As the solid has a higher tendency to show Frenkel defects than Schottky defects, it should be the ZnS type.
Unit cell edge length = 0.564 nm,
$a=2\sqrt{2}{r}^{-}=0.564nm\Rightarrow r^{-}=0.1994nm\Rightarrow r^{+}=0.225r^{-}$ (for tetrahedral void) $\Rightarrow r^{+}=0.0449nm$