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- is

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 XI

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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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

Q. 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:

Q. KCl crystallizes in the same type of lattice as does NaCl. Given that

\frac{r_{N a^{+}}}{r_{C l^{-}}} = 0.5

and

\frac{r_{N a^{+}}}{r_{K^{+}}} = 0.7

Find the ratio of the sides of unit cell for KCl to that for NaCl :

K C l

crystalizes in the same type of lattice as

N a C l

.
Given that

\frac{r_{N a^{+}}}{r_{C l^{-}}} = 0.5

and

\frac{r_{K^{+}}}{r_{C l^{-}}} = 0.65

Then the ratio of the side of the unit cells for

K C l

to that of

N a C l

is:

Q. KCl crystallizes in the same type of lattice as does NaCl. Given that

\frac{r_{N a +}}{r_{C l -}} = 0.50

and

\frac{r_{N a +}}{r_{K +}} = 0.70

, calculate the ratio of the side of the unit cell for KCl to that of for NaCl.

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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+ is

The correct option is A 1.48 ∘A aRbCl−aKCl=2(rRb++rCl−)−2(r+K+rCl−)=0.30∘A⇒rRb+=rK++0.15∘A=1.33∘A+0.15 ∘A=1.48 ∘A

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

The correct option is A
$1.48 \circ A$

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