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  • Tell me? If distance between Cs+ and Cl- in CsCl structure is 200 pm , then the edge length of unit cell is -

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

  • Option 1)

    220 pm

  • Option 2)

    231 pm

  • Option 3)

    242 pm

  • Option 4)

    251 pm

 

Answers (1)

best_answer

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

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

                                                =231pm

 

 

CsCl type structure -

Cl^{-} located at all corners

Cs^{+} located at body centre

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

Coordination number = 8:8

- wherein

Number of Cl^{-} = 1

Number of Cs^{+} = 1

Number of CsCl molecule per unit cell = 1

 

 


Option 1)

220 pm

This is incorrect

Option 2)

231 pm

This is correct

Option 3)

242 pm

This is incorrect

Option 4)

251 pm

This is incorrect

Posted by

prateek

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