KCI cyrstallizes in the saem type of lattice as does NaCl. Given that `(r_(Na^(o+))/(r_(Cl_(-)))=0.5` and `(r_(Na^(o+)))/(r_(K^(o+)))=0.7` Calculate the ratio of side of the unit cell for KCl to that for NaCl.

Options

A. 0.6
B. 1.1
C. 2.2
D. 3.3

Explanation

NaCl crystallizes in the face-centered cubic unit cell, such that `r_(Na^(o+))+r_(Cl^(-))=a//2`
where a is the edge length of unit cell. Now since
`r_(Na^(o+))//r_(Cl^(-))=0.5` and `r_(Na^(o+))//r_(K^(o+))=0.7` we will have
`(r_(Na^(o+))+r_(Cl^(-)))/(r_(Cl^(-)))=(r_(K^(o+)))/(r_(Na^(o+))//0.5)=(0.5)/(r_(Na^(o+))//r_(K^(o+)))(0.5)/(0.7)`
[Add 1 to both sides] or `(r_(K^(o+)))/(r_(Na^(o+))+r_(Cl^(-)))=(1.2)/(0.7)xx(1)/(1.5)`
or `(a_(KCl)//2)/(a_(NaCl)//2)=(1.2)/(0.7xx1.5)` or `(a_(KCl))/(a_(NaCl))=(1.2)/(1.05)=1.143`.