Question

$CsCl$ has cubic structure of ions in which $C{s}^{+}$ ions is present in the body-centre of the cube. If density is $3.99gc{m}^{-3}$.
(a) Calculate the length of the edge of a unit cell.
(b) What is the distance between $C{s}^{+}$ and $C{l}^{-}$ ions?
(c) What is the radius of $C{s}^{+}$ ion if the radius of $C{l}^{-}$ ion is $180pm$?

Answer 1

We know that,
(a) $Z=\frac{a^3\times \rho \times N}{M}$
$Z=1$

$M=168.5$
$1=\frac{a^3\times 3.99\times 6.023\times {10}^{23}}{168.5}$
$a=4.123\times {10}^{-8}cm=412pm$

(b) Distance between $C{s}^{+}$ and $C{l}^{-}$ ions
$=\frac{a\sqrt{3}}{2}=\frac{412\times \sqrt{3}}{2}=356.8pm$

(c) $r_{C{s}^{+}}+r_{C{l}^{-}}=356.8$
$r_{C{s}^{+}}+180=356.8$

$r_{C{s}^{+}}=176.8pm$