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Question

Which of the following compounds is likely to show both Frenkel and Schottky defects in its crystalline form?


A

ZnS

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B

CsCl

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C

KBr

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D

AgBr

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Solution

The correct option is D

AgBr


  • The explanation for the correct option:

D) AgBr

  • Frenkel defect is a point defect that arises when the smaller ion leaves its position and enters an interstitial site forming a vacancy.
  • For example - AgBr,ZnS,etc.,
  • In this case, the cation which is Ag+ is smaller in size as compared to the anionBr-.
  • There is a large difference between the size of the cation and the anion.
  • Therefore, the cation leaves its position and enters an interstitial site showing a Frenkel defect.
  • A Schottky defect is a point defect in which an equal number of cations and anions are missing.
  • This defect happens when there is a small difference in the size of cations and anions.
  • An equal number of Ag+ions and Br- ions leave their normal lattice sites.
  • So, AgBr show both Frenkel and Schottky defects.

The explanation for the incorrect options:
A) ZnS

  • In this case, the cation which isZn2+ is smaller in size as compared to the anion Br-.
  • Therefore, the cation leaves its position and enters an interstitial site showing a Frenkel defect.
  • It does not show Schottky defect as there is a huge difference in the size of the cation and anion.

B)CsCl

  • Here, the size of the cation and anion i.e., Cs+andCl- are nearly similar.
  • Therefore, it only shows Schottky's defect and not Frenkel's defect.

C) KBr

  • Here, the size of the cation and anion i.e., K+andBr- are nearly similar.
  • Therefore, it only shows Schottky's defect and not Frenkel's defect.

Therefore, D) AgBris correct. Only AgBrshows both Frenkel and Schottky defects.


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