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Q.

Each rubidium halide crystallising in the RbCl type lattice has a unit cell length 0.30A^o greater than that for corresponding potassium salt $(r_{(K^{+})} = 1 .33 Å)$ of the same halogen. Hence, ionic radius of Rb+ is

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a

1.63A⁰

b

$1.18 A^{0}$

c

1.48A⁰

d

1.03A⁰

answer is B.

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Detailed Solution

RbCl type : $2 {Rb}^{+} + 2 {Cl}^{-} =$ Edge length
KCl type : $2 K^{+} + 2 {Cl}^{-} =$ Edge length
Edge length of RbCl > KCl by 0.3A⁰
$\begin{array}{l} 2 {Rb}^{+} + 2 {Cl}^{-} = x + 0 .3 \dots . . (2) \\ 2 K^{+} + 2 {Cl}^{-} = x \\ (2) - (1) \Rightarrow 2 {Rb}^{+} - 2 K^{+} = 0 .3; \\ 2 {Rb}^{+} - 2 \times 1 .33 = 0 .3, 2 {Rb}^{+} = 2 .96 A^{0} \\ {Rb}^{+} = 1 .48 A^{0} \end{array}$

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