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Choose the Best Answer · Q10

Q.CsCl has bcc arrangement, its unit cell edge length is 400pm, its inter atomic distance is

(a) 400pm
(b) 800pm
(c) 3×100\sqrt{3} \times 100pm
(d) (32)×400\left(\dfrac{\sqrt{3}}{2}\right) \times 400pm
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Step 1. CsCl adopts a bcc-type arrangement: Cl−Cl^- ions at the corners and Cs+Cs^+ at the body centre (or vice versa), and the two ions touch exactly along the body diagonal of the cube.

Step 2. The full body diagonal of a cube of edge a is 3 a\sqrt{3}\,a, and this diagonal spans 4 ionic radii (4r=3a4r = \sqrt{3}a). …

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