Each rubidium halide crystallising in the NaCl-type lattice has a unit cell length `0.30 Å` greater than that for corresponding potassium salt `(r_(k+)=1.33 Å`) of the same halogen. Hence, ionic radius of `Rb^(+)` is

Options

A. 1.03 Å
B. 1.18 Å
C. 1.48 Å
D. 1.63 Å

Answer

C

Explanation

As RbX crystallies in NaCl type lattice, unit cell length of RbX =`2(r_(Rb^+)+r_(X^-))`
Unit cell length of KX =`2(r_(K^+) + r_(X^-))`
`therefore 2(r_(Rb^+)+r_(X^-))-2(r_(K^+)+r_(X^-))`=0.30 Å
or `r_(Rb^+) - r_(K^+)`=0.15 Å
or `r_(Rb^+) = r_(K^+)`+ 0.15 Å =1.33 +1.15=1.48 Å