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

Options

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

Answer

B

Explanation

RbCl type: `2Rb^(+) + 2Cl^(-)` = Edge length
KCl type : `2K^(+) + 2Cl^(-)` = Edge length
Edge length of `RbCl gt KCl` by `0.3 A`
`2Rb^(+) + 2Cl^(-)= x+0.3 , 2Rb^(+) -2 xx 1.33 = 0.3`
`2Rb^(+) = 2.96 A : Rb^(+) = 1.48 A`