Question

$AgBr$ shows both Schottky and Frenkel defect because

• A. size of $Ag>>>>>Br$
• B. size of $Ag<<<<<Br$
• C. $AgBr$ is highly ionic
• D. Both B and C

Answer 1

The correct option is A size of $Ag>>>>>Br$

$AgBr$ shows both Frenkel and Schottky defects because the radius ratio for $AgBr$ is intermediate. Ions have Schottky defects when their anions and cations are both absent from the crystal lattice.In $AgBr$, the $A{g}^{+}$ ions and corresponding $B{r}^{-}$ ions are absent from the crystal lattice causing Schottky defects. However, $A{g}^{+}$ ions are exceptionally mobile and they have a tendency to move about inside the lattice. So they complement the Schottky defects with Frenkel defects with the $A{g}^{+}$ ions trapped in interstitial spaces. Because of the similar size of $A{g}^{+}$ ions and $B{r}^{‒}$ ions and the small size of $A{g}^{+}$ ions and the fact that $A{g}^{+}$ ions have a property to be mobile, both Schottky and Frenkel defects occur.