Question

Which one of the following compounds show both Schottky and Frenkel defects?

• A. $NaBr$
• B. $AgBr$
• C. $KBr$
• D. $KCl$

Answer 1

The correct option is B $AgBr$

In Schottky defect equal number of cations and anions are missing from their normal lattice sites. This defect usually observed in strongly ionic compounds, which have cations and anions of almost same size.
So, $NaBr$, $KBr$, $KCl$, $AgBr$ usually have schottky defect.
In Frenkel defect is caused if some of the ions of the lattice occupy intersticial sites leaving a corresponding number of normal lattice sites vacant. For this defect the size of anion should be bigger than cation. Thus this defect is mainly found in transition metal halides like $AgCl$ $AgBr$ etc.
Thus $AgBr$ shows both Schottky and Frenkel defect.