Q22 / 180 · COMEDK 2022 · Chemistry
Which of the following show both Frenkel and Schottky defect?
- A$\mathrm{ZnS}$
- Correct answerB$\mathrm{AgBr}$
- C$\mathrm{NaCl}$
- D$\mathrm{AgCl}$
Solution
Frenkel defect arises due to the displacement of an ion from its lattice site to an interstitial site, typically occurring in ionic crystals where there is a large difference in size between the cation and the anion. Schottky defect arises due to the missing of equal numbers of cations and anions from the lattice sites to maintain electrical neutrality, typically occurring in ionic crystals with high coordination numbers and small differences in ionic sizes.
Silver bromide ($\mathrm{AgBr}$) is a unique case in solid state chemistry. Due to the intermediate size of the $\mathrm{Ag}^{+}$ ion, it is capable of showing both Frenkel defect (due to the small size of $\mathrm{Ag}^{+}$) and Schottky defect (due to the ionic nature and crystal structure).
$\mathrm{ZnS}$ primarily exhibits Frenkel defect. $\mathrm{NaCl}$ and $\mathrm{AgCl}$ primarily exhibit Schottky defect.
Answer: $\mathrm{AgBr}$
Silver bromide ($\mathrm{AgBr}$) is a unique case in solid state chemistry. Due to the intermediate size of the $\mathrm{Ag}^{+}$ ion, it is capable of showing both Frenkel defect (due to the small size of $\mathrm{Ag}^{+}$) and Schottky defect (due to the ionic nature and crystal structure).
$\mathrm{ZnS}$ primarily exhibits Frenkel defect. $\mathrm{NaCl}$ and $\mathrm{AgCl}$ primarily exhibit Schottky defect.
Answer: $\mathrm{AgBr}$
Hence, the correct answer is option: 'B'.