KCl crystallises in the same type of lattices as does NaCl. Given that ` r_(Na^(+)) //r_(Cl^(-)) = 0.55 and r_(K^(+)) //r_(Cl^(-)) = 0.74 ` . Calculate the ratio of the side of the unit cell of KCl to that of NaCl.

Options

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

Answer

A

Explanation

We aim at :`(r_(K^+)+r_(Cl^-))//(r_(Na^+)+r_(Cl^-))`
Given `r_(Na^+)/r_(Cl^-)=0.55` and `r_(K^+)/r_(Cl^-)`=0.74
`r_(Na^+)/r_(Cl^-)+1=1.55` and `r_(K^+)/r_(Cl^-)+1=1.74`
i.e., `(r_(Na^+)+r_(Cl^-))/(r_(Cl^(-)))=1.55`
`(r_(K^+)+r_(Cl^-))/(r_(Cl^(-)))=1.74`
Dividing (ii) by (i) , `(r_(K^+)+r_(Cl^-))/(r_(NA^+)+r_(Cl^-))=1.74/1.55`=1.123