Each rubidium halide crystallising in the NaCl type lattice has a unit cell length 0-30 -A greater than that for corresponding potassium salt- rK-1-33 -A of the same halogen- Hence- ionic radius of Rb- is-

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Standard XII

Chemistry

Crystal Lattice and Unit Cell

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 that for corresponding potassium salt. $(r_{(K^{+})} = 1.33 \circ A)$ of the same halogen. Hence, ionic radius of $R b^{+}$ is:

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Solution

The unit cell length increases by $0.30 \circ A$ in case of rubidium halide crystal. This is due to two atoms present at the corners.

Thus $r_{R b^{+}} = 1.33 + \frac{0.30}{2} \circ A = 1.48 \circ A$

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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 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

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 that for corresponding potassium salt. (r(K+)=1.33 ∘A) of the same halogen. Hence, ionic radius of Rb+ is:

The unit cell length increases by 0.30 ∘A in case of rubidium halide crystal. This is due to two atoms present at the corners. Thus rRb+=1.33+0.302 ∘A=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 that for corresponding potassium salt. $(r_{(K^{+})} = 1.33 \circ A)$ of the same halogen. Hence, ionic radius of $R b^{+}$ is:

The unit cell length increases by $0.30 \circ A$ in case of rubidium halide crystal. This is due to two atoms present at the corners.

Thus $r_{R b^{+}} = 1.33 + \frac{0.30}{2} \circ A = 1.48 \circ A$