Question

In the cubic crystal of $CsCl\left(d=3.97gc{m}^{-3}\right)$ the eight corners are occupied by $C{l}^{-}$with a $C{s}^{+}$at the centre and vice-versa. What is the radius ratio of the two ions? (Atomic mass of $Cs=132.91$ and $C{l}^{-}=35.45)$

• A. 0.16
• B. 0.24
• C. 0.73
• D. 0.88

Answer 1

Answer: (C)

In a unit cell, $n=1$ for cubic crystal
$\therefore$ Density $=\frac{n\times \text{ Formula mass }}{V\times \text{ At. no. }}$
$=\frac{n\times \text{ Formula mass }}{a^3\times At\text{. no. }}$
$\therefore 3.97=\frac{1\times 168.36}{a^3\times 6.023\times 10^{23}}$
$a=4.13\times 10^{-8}cm$
$a=4.13\mathring{A}$
For a cube of side length $4.13\mathring{A}$,
Diagonal $=\sqrt{3}\times 4.13=7.15\mathring{A}$
As it is a bcc with $C{s}^{+}$at centre (radius $r^{+}$) and $C{l}^{-}$at corners (radius $r^{-}$) so,
$2r^{+}+2r^{-}=7.15$ or $r^{+}+r^{-}=3.57\mathring{A}$
i.e., distance between neighbouring $C{s}^{+}$and $C{l}^{-}=3.57\mathring{A}$
Now, assum two $C{l}^{-}$ion touch with each other so,
Length of unit cell $=2r^{-}=4.13\mathring{A}$
$\therefore r^{-}=2.06\mathring{A}$
$\therefore r^{+}=3.57-2.06=1.51$
$\therefore r^{+}/r^{-}=1.51/2.06=0.73$