1.

CsCl has bcc arrangement, its unit cell edge length is 400pm, its inter atomic distance is

Answer»

400 pm
800 pm
`sqrt(3)XX100` pm
`(sqrt(3)/2)xx400` pm

Solution :`sqrt(3)a=r_(Cs^(+))+2r_(CL^(-))+r_(Cs^(+))`
`(sqrt(3)/2)a=(r_(Cs^(+))+r_(Cl^(-)))" " (sqrt(3)/(2))=400=` inter ionic distance


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