JEE Main 2020 · 8 January · evening · Single correct · +4 / −1
Which of the following compounds is likely to show both Frenkel and Schottky defects in its crystalline form?
- (a) CsCl
- (b) AgBr
- (c) ZnS
- (d) KBr
Answer and solution
Answer (b) AgBr
Solution
Chemistry ma’am · 3 steps
Pick the solid whose ions pack tightly, yet whose small cation can still slip into the gaps.
- Schottky defect. Equal numbers of cations and anions are missing from their lattice sites. This is found in strongly ionic solids with a high coordination number, where the cation and anion are of similar size, like NaCl, KCl and CsCl.
- Frenkel defect. A small cation leaves its site and sits in an interstitial gap. This needs a big difference in ion sizes and a low coordination number, like ZnS and AgCl.
Compound Coordination What the ions are like Defect shown CsCl 8 : 8 similar sizes, strongly ionic Schottky only AgBr 6 : 6 ionic, and Ag⁺ is small and mobile both ZnS 4 : 4 Zn²⁺ much smaller than S²⁻ Frenkel only KBr 6 : 6 similar sizes, strongly ionic Schottky only - So. AgBr is ionic with 6 : 6 packing, so it shows Schottky defects. Its small Ag⁺ ion also slips into interstitial sites, so it shows Frenkel defects too. The answer is (b).
The idea to keep. Similar-sized ions with high coordination give Schottky. A small cation with low coordination gives Frenkel. AgBr is the classic exam answer for both, so remember it by name.
Answer from the NTA final answer key. Solution by PrepWiser.