KCl crystallises in the same type of lattice as does NaCl. Given that `r_(Na^(+))//r_(Cl^(-))=0.55` and `r_(Na^(+))//r_(Cl^(-))=0.75`, calculate the ratio of the side of the unit cell for KCl to that of NaCl.

Options

A. 1.123
B. 0.891
C. 1.414
D. 0.414

Answer

A

Explanation

NaCl and KCl has octahedral structure
`r_(Na^(+))/(r_(Cl^(-))) 0.55 and (r_(K^(+)))/(r_(Cl^(-)))0.74`
In octahedral edge length `= r_("cation") + r_("anion")`
`(r_(Na^(+)) + r_(Cl^(-)))/(r_(Cl^(-))) = 1.55`...(1)
`(r_(K^(+)) + r_(Cl^(-)))/(r_(Cl^(-))) = 1.74`...(2)
a = edge length of KCl octahedral
`a= r^(K^(+)) + r_(Cl^(-))`
b = edge length of NaCl octahedral
`b = r_(Na^(+)) + r_(Cl^(-))`
`(a)/(b) = (r_(K^(+)) + r_(Cl^(-)))/(r_(Na^(+)) + r_(Cl^(-))) = (1.74)/(1.55) = 1.123`