Each rubidium halide crystallising in the N aCl type lattice has a unit cell length 0-30 A greater than the corresponding potassium salt r K - 1-33 A of the same halogen- Hence- ionic radius of Rb -is-A- 1-03 AB- 1-18 AC- 1-48 A -D- 1-63 A

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Ionic Compounds and Radius Ratios

Each rubidium...

Question

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:

$1.03 \circ A$

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1.18 \circ A

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1.48 \circ A

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1.63 \circ A

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Solution

The correct option is C $1.48 \circ A$
As $R b X$ crystallises in an $N a C l$ type lattice, the unit cell length of $R b X = 2 (r_{R b^{+}} + r_{X^{-}})$
Unit cell length of $K X = 2 (r_{K^{+}} + r_{X^{-}})$
$∴ 2 (r_{R b^{+}} + r_{X^{-}}) - 2 (r_{K^{+}} + r_{X^{-}}) = 0.30 \circ A$
Or, $r_{R b^{+}} - r_{K^{+}} = 0.15 \circ A$
Or, $r_{R b^{+}} = r_{K^{+}} + 0.15 \circ A$
Or, $r_{R b^{+}} = 1.33 \circ A + 0.15 \circ A$
Or, $r_{R b^{+}} = 1.48 \circ A$

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Similar questions

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

K C l

crystallises in the same type of lattice as

N a C l

. Given that

r_{N a^{+}} / r_{C l^{-}} = 0.50

and

r_{N a^{+}} / r_{K^{+}} = 0.70

, calculate the ratio of the side of the unit cell of

K C l

to that of

N a C l

K C l

crystallises in the same type of lattice as does

N a C l

. Given that

r_{N a^{+}} / r_{C l^{-}} = 0.55

and

r_{K^{+}} / r_{C l^{-}} = 0.74

. Calculate the ratio of the side of the unit cell for

K C l

to that of

N a C l

K C l

crystallises in the same type of lattice as

N a C l

. Given that

r_{N a^{+}} / r_{C l^{-}} = 0.50

and

r_{N a^{+}} / r_{K^{+}} = 0.70

, calculate the ratio of the side of the unit cell of

K C l

to that of

N a C l

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Each rubidium halide crystallising in the NaCl type lattice has a unit cell length 0.30 ∘A greater than the corresponding potassium salt (rK+=1.33 ∘A) of the same halogen. Hence, ionic radius of Rb+ is:

The correct option is C 1.48 ∘AAs RbX crystallises in an NaCl type lattice, the unit cell length of RbX=2(rRb++rX−) Unit cell length of KX=2(rK++rX−) ∴2(rRb++rX−)−2(rK++rX−)=0.30 ∘A Or, rRb+−rK+=0.15 ∘A Or, rRb+=rK++0.15 ∘A Or, rRb+=1.33 ∘A+0.15 ∘A Or, rRb+=1.48 ∘A

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: