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  • If distance (in pm) between Cs+ and Cl- in CsCl structure is 200 pm, then the edge length of unit cell is-Option: 1 220</di

If distance (in pm) between Cs+ and Cl- in CsCl structure is 200 pm, then the edge length of unit cell is-

Option: 1

220


Option: 2

231


Option: 3

242


Option: 4

251


Answers (1)

best_answer

As we have learnt,

 

CsCl
  • Here Cl- ions are at the comers of a cube and Cs+ ions are in the cubic void (centred position) i.e. BCC like structure.
  • Here co-ordination number of both Cs+ and Cl- is 8 and here the number of formula units per unit cell is 1. re + ra = √3a/2.
  • Here rCs+/rCl+ should be 0.732 but it is 0.93, e.g. CsX, TiBr, NH4Cl, NH4Br

-

Edge length in CaCl structure =\frac{2}{\sqrt{3}}(r_{Cs^+}+r_{Cl^-})

                                                =\frac{2}{\sqrt{3}}\times 200

                                                =231pm

 

Posted by

manish painkra

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