Question

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

• A. $r_{C{s}^{+}}+r_{C{l}^{-}}=3a$
• B. $r_{C{s}^{+}}+r_{C{l}^{-}}=\frac{3a}{2}$
• C. $r_{C{s}^{+}}+r_{C{l}^{-}}=\frac{\sqrt{3}}{2}a$
• D. $r_{C{s}^{+}}+r_{C{l}^{-}}=\sqrt{3}a$

Answer 1

Answer: (C)

In $CsCl,C{l}^{-}$lies at corners of simple cube and $C{s}^{+}$at the body centre. Hence, along the body diagonal, $C{s}^{+}$and $C{l}^{-}$touch each other so
$r_{C{s}^{+}}+r_{C{l}^{-}}=2r$
Calculation of $r$ In $\Delta EDF$,

Body centred cubic unit cell
$FD=b=\sqrt{a^2+a^2}=\sqrt{2}a$
In $△AFD$,
$c^2=a^2+b^2=a^2+(\sqrt{2}a{)}^2=a^2+2a^2$
$c^2=3a^2$
$c=\sqrt{3}a$
As $△AFD$ is an equilateral triangle.
$\therefore \sqrt{3}a=4r$
$[\because C=3r+r+r]$
$\Rightarrow r=\frac{\sqrt{3}a}{4}$
Hence, $r_{Cs}+r_{Cl}{}^{-}=2r=2\times \frac{\sqrt{3}}{4}a=\frac{\sqrt{3}}{2}a$