CsCI crystallizes in body centred cubic lattice. If 'a' its edge length then which of the following expressions is correct

A
$r_{{Cs}^{+}} + r_{{CI}^{-}} = 3 a$
B
$r_{{Cs}^{+}} + r_{{Cl}^{-}} = \frac{3 a}{2}$
C
$r_{{Cs}^{+}} + r_{{Cl}^{-}} = \frac{\sqrt{3}}{2} a$
D
$r_{{Cs}^{+}} + r_{{Cl}^{-}} = \sqrt{3} a$

Solution:

In $CsCl, {Cl}^{-}$ lie at corners of simple cube and ${Cs}^{+}$ at the body centre. Hence. Along the body diagonal, ${Cs}^{+}$ and $C I^{-}$ touch each other so.

$\frac{\sqrt{3} a}{2} = r_{{Cs}^{4}} + r_{{CI}^{-}}$

Solution:

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