RbCl has NaCl type lattice and its unit cell length is 0-30 A greater than that for KCl- If rk-1-33 A- the ionic radius of Rb -isA- None of theseB- 1-63 AC- 1-03 AD- 1-48 A

The correct option is D 1.48 ∘A a_RbCl a_KCl = 2 (r_Rb^+ + r_Cl^ ) 2(r_K^+ + r_Cl^ ) = 0.30 ∘A ⇒ r_Rb^+ = r_K^+ + 0.15 ∘A = 1.33 ∘A + 0.15 ∘A = 1.48 ∘A ...

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Byju's Answer

Standard XII

Chemistry

Relation Between r and a

RbCl has NaCl...

Question

RbCl has NaCl type lattice and its unit cell length is $0.30 \circ A$ greater than that for KCl. If $r_{k +} = 1.33 \circ A$ , the ionic radius of $R b^{+}$ is

$1.48 \circ A$

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$1.63 \circ A$

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$1.03 \circ A$

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$N o n e o f t h e s e$

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Solution

The correct option is A
$1.48 \circ A$

$a_{R b C l} - a_{K C l} = 2 (r_{R b^{+}} + r_{C l^{-}}) - 2 (r_{K}^{+} + r_{C l^{-}}) = 0.30 \circ A \Rightarrow r_{R b^{+}} = r_{K^{+}} + 0.15 \circ A = 1.33 \circ A + 0.15 \circ A = 1.48 \circ A$

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RbCl has NaCl type lattice and its unit cell length is $0.30 \circ A$ greater than that for KCl. If $r_{k +} = 1.33 \circ A$ , the ionic radius of $R b^{+}$ is

Each rubidium halide crystallising in the $N a C l$ type lattice has a unit cell length $0.30 \circ A$ greater than the corresponding potassium salt $(r_{K^{+}} = 1.33 \circ A)$ of the same halogen. Hence, ionic radius of $R b^{+}$ is: