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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


Option: 2


Option: 3


Option: 4


Answers (1)


As we have learnt,


  • Here Clions 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 Csand Cl- is 8 and here the number of formula units per unit cell is 1. re + r= √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



Posted by

manish painkra

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