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

$r_{cs^{-}}+r_{cl^{-}}=3a$

$r_{cs^{-}}+r_{cl^{-}}=\frac{3a}{2}$

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

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

Option C

In CsCl,Cl^- lie at corners of simple cube and Cs⁺ at the body centre . Hence , along the body diagonal , Cs⁺ and Cl^- touch each other so,

$r_{Cs^{+}}+r_{Cl^{-}}=2r$

Calculation of $\tau$

In $\triangle EDF$

Body centred cubic unit cell

$FD=b=\sqrt{a^{2}+a^{2}}=\sqrt{2}a$

In $\triangle AFD$ ,

$c^{2}=a^{2}+b^{2}$

$=a^{2}+(\sqrt{2}a)^{2}$

$=a^{2}+2a^{2}$

$c^{2}=3a^{2}$

$c=\sqrt{3}a$

As $\triangle AFD$ is an equilateral triangle,

$\therefore$ $\sqrt{3}a=4r$

$\Rightarrow r =\frac{\sqrt{3}a}{4}$

Hence, $r_{Cs^{+}}+r_{Cl^{-}}=2r$

= $2\times\frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{2}a$

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