1.

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

Answer»

400 pm
800 pm
`sqrt3xx"100 pm"`
`((SQRT3)/(2))xx"400 pm"`

Solution :Hint : `sqrt3a=r_(Cs^(+))+2r_(CL^(-))+r_(Cs^(+))`
`((sqrt3)/(2))a=(r_(Cs^(+))+r_(Cl^(-)))((sqrt3)/(2))400=" inter ionic distance"`


Discussion

No Comment Found

Related InterviewSolutions