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

Options

A. 0.124
B. 1.123
C. 0.891
D. 1.414

Answer

B

Explanation

Given that,
`r_(Na)+r_(Cl)=0.55`
`r_(K^(+))//C_(Cl^(-))=0.75`
`(r_(KCl))/(r_(NaCl))=?`
`because (r_(Na^(+)))/(r_(Cl^(-)))=0.55`
`therefore (r_(Na^(+)))/(r_(Cl^(-)))+1=0.55+1`
`(r_(Na^(+))+r_(Cl^(-)))/(r_(Cl^(-)))=1.55" ….(i)"`
`therefore (r_(K^(+)))/(r_(Cl^(-)))=0.74`
`therefore (r_(K^(+)))/(r_(Cl^(-)))+1=0.74+1`
`(r_(K^(+))+r_(Cl^(-)))/(r_(Cl^(-)))=1.74" .....(ii)"`
Dividing equations (ii) by (i)
`(r_(K^(+))+r_(Cl^(-)))/(r_(Na^(+))+r_(Cl^(-)))=(1.74)/(1.55) =1.1226`