1.

KCl crystallises in the same type of lattice as does NaCl. Given (r_(Na^(+)))/(r_(Cl^(-)))=0.5 and (r_(Na^(+)))/(r_(K^(+)))=0.7The ratio of the side of the unit cell for NaCl to that for KCl is

Answer»

`1:1.172`
`1:1.1143`
`1:1.1413`
`1:1.732`

SOLUTION :NACL crystallises in the face centred cubic unit cell such that
`r_(NA^(+))+r_(Cl^(-))=(a)/(2)`
Given,`(r_(Na^(+)))/(r_(Cl^(-)))=0.5,(r_(Na^(+)))/(r_(K^(+)))=0.7`
Thus we have ,`(r_(Na^(+))+r_(Cl^(-)))/(r_(Cl^(-)))=1.5`
`(r_(K^(+)))/(r_(Cl^(-)))=(r_(K^(+)))/(r_(Na^(+))//0.5)=(0.5)/(r_(Na^(+))//r_(K^(+)))=(0.5)/(0.7)`
Therefore,`(r_(K^(+))+r_(Cl^(-)))/(r_(Na^(+))+r_(Cl^(-)))=(1.2)/(0.7)xx(1)/(1.5)`
`:.(a_(KCl//2))/(a_(NaCl//2))=(1.2)/(0.7xx1.5)=(a_(KCl))/(a_(NaCl))=1.143`
or`(a_(NaCl))/(a_(KCl))=1:1.143`


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