CsCl crystallises in body centred cubic lattice. If 'a' is its edge length then which of the following expression is correct?

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CsCl crystallises in body centred cubic lattice. If 'a' is its edge length then which of the following expression is correct?

Engineering

Chemistry

HCP and CCP Parameters

Empirical Formula

Structure of Some Important Compounds

Question

CsCl crystallises in body centred cubic lattice. If 'a' is its edge length then which of the following expression is correct?

A

$r_{{Cs}^{+}} + r_{{Cl}^{-}} = \frac{3 a}{2}$

B

$r_{{Cs}^{+}} + r_{{Cl}^{-}} = \sqrt{3} a$

C

$r_{{Cs}^{+}} + r_{{Cl}^{-}} = 3a$

D

$r_{{Cs}^{+}} + r_{{Cl}^{-}} = \frac{\sqrt{3} a}{2}$

$\sqrt{3} a = 2 (r_{{Cs}^{+}} + r_{{Cl}^{-}})$

CsCl crystallizes in a body-centered cubic (BCC) lattice. In this structure, the cesium ion (Cs⁺) is at the body center and the chloride ions (Cl⁻) are at the corners of the cube, or vice versa. The edge length of the cube is denoted by 'a'.

To find the correct expression for the sum of the ionic radii (r_Cs⁺ + r_Cl⁻), we need to consider the distance between the body center and a corner of the cube. This distance is half of the body diagonal of the cube.

Step 1: Calculate the body diagonal of the cube. The body diagonal runs from one corner to the opposite corner through the center of the cube. Its length is given by:

$Body diagonal = \sqrt{a^{2} + a^{2} + a^{2}} = \sqrt{3 a^{2}} = \sqrt{3} a$

Step 2: The distance from the body center to a corner is half of the body diagonal. Therefore:

$Distance = \frac{\sqrt{3} a}{2}$

Step 3: In the CsCl structure, this distance is equal to the sum of the ionic radii of Cs⁺ and Cl⁻, because the ions touch each other along the body diagonal. So:

$r_{{Cs}^{+}} + r_{{Cl}^{-}} = \frac{\sqrt{3} a}{2}$

Thus, the correct expression is the fourth option.

Related Topics and Formulae

Body-Centered Cubic (BCC) Lattice: In a BCC unit cell, atoms are present at the corners and at the body center. The coordination number is 8. The relationship between the atomic radius (r) and the edge length (a) for a BCC lattice is $r = \frac{\sqrt{3} a}{4}$ . However, for ionic compounds like CsCl, which has a BCC-like arrangement but with two different ions, the sum of the radii is derived as above.

CsCl Structure: It is often considered a simple cubic lattice for Cl⁻ ions with Cs⁺ at the body center (or vice versa). The ionic compounds crystallizing in this structure have a 8:8 coordination.