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

`(r_(Na^(+)))/(r_(Cl^(-)))=0.55," "(r_(K^(+)))/(r_(Cl^(-)))=0.74`
`or (r_Na^(+))/(r_(Cl^(-)))+1=0.55+1," "(r_(K^(+)))/(r_(Cl^(-)))=0.74+1`
`(r_(K^(+))+r_(Cl^(-)))/(r_(Cl^(-)))=1.55," "(r_(K^(+))+r_(Cl^(-)))/(r_(Cl^(-)))=1.74`
Edge length ratio is
`(r_(K^(+))+r_(Cl^(-)))/(r_(Cl^(-)))xx(r_(Cl^(-)))/(r_(Na^(+))+r_(Cl^(-)))=(1.74)/(1.55)rArr (r_(K^(+))+r_(Cl^(-)))/(r_(Na^(+))+r_(Cl^(-)))=1.122`