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Standard XII

Chemistry

CFSE in Octahedral and Tetrahedral Complexes

Calculate the...

Question

Calculate the ionic radius of a $C s^{+}$ ion assuming that the cell edge length for $C s C l$ is 0.4123 nm and that the ionic radius of a $C l^{-}$ ion is 0.181 nm.

0.176 nm

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0.231 nm

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0.357 nm

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0.116 nm

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Solution

The correct option is C 0.176 nm
$r_{C s} + r_{C l^{-}} = \frac{d_{b o d y}}{2} = \frac{0.7141}{2} = 0.3571 n m$

$r_{C s^{+}} = 0.3571 - 0.181 = 0.176 n m$

Hence, the correct option is A.

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Similar questions

Calculate the ionic radius of a Cs⁺ ion assuming the cell edge length for CsCl is 0.4123 nm and that the ionic radius of a Cl^- ion is 0.181 nm.

Q. The ionic radius of

M n^{2 +}

ion is

0.082 n m

and that of

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0.186 n m

. The coordination number of the cation is:

Q. The ionic radii of Cs and Cl are

0.165

nm and

0.181

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133

amu and

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C s C l

has cubic structure of ions in which

C s^{+}

ions is present in the body-centre of the cube. If density is

3.99 g c m^{- 3}

.
(a) Calculate the length of the edge of a unit cell.
(b) What is the distance between

C s^{+}

and

C l^{-}

ions?
(c) What is the radius of

C s^{+}

ion if the radius of

C l^{-}

ion is

180 p m

Q. If the length of body diagonal for

CsCl

, which crystallises into a cubic structure with

{Cl}^{-}

ions at the corners and

{Cs}^{+}

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Calculate the ionic radius of a Cs+ ion assuming that the cell edge length for CsCl is 0.4123 nm and that the ionic radius of a Cl− ion is 0.181 nm.

The correct option is C 0.176 nmrCs+rCl−=dbody2=0.71412=0.3571nmrCs+=0.3571−0.181=0.176nmHence, the correct option is A.

Calculate the ionic radius of a $C s^{+}$ ion assuming that the cell edge length for $C s C l$ is 0.4123 nm and that the ionic radius of a $C l^{-}$ ion is 0.181 nm.

The correct option is C 0.176 nm
$r_{C s} + r_{C l^{-}} = \frac{d_{b o d y}}{2} = \frac{0.7141}{2} = 0.3571 n m$

$r_{C s^{+}} = 0.3571 - 0.181 = 0.176 n m$

Hence, the correct option is A.