Question

$CsCl$ crystals are formed by bcc crystal lattice where $C{l}^{-}$ occupies the corners and $C{s}^{+}$ occupies the centre of the cube.

The density of crystalline $CsCl$ is $4g/cc$. The effective volume occupied by a single $CsCl$ ion pair in the crystal is
[Molar mass of $CsCl=168{\text{g mol}}^{-1}]$

• A. $17\times {10}^{-23}cc$
• B. $10\times {10}^{-23}cc$
• C. $7\times {10}^{-23}cc$
• D. $9\times {10}^{-22}cc$

Answer 1

The correct option is C $7\times {10}^{-23}cc$

In bcc, the ions are present in the corners and body centre of the cube.
$\text{Number of}C{l}^{-}\text{per unit cell}=\frac{1}{8}\times 8=1$
$\text{Number of}C{s}^{+}\text{per unit cell}=1$
Hence, an unit cell of $CsCl$ comprise of one ion pair of $CsCl$
Thus, volume of the unit cell gives the effective volume occupied by a single $CsCl$ ion pair.

$\text{Density of unit cell,}\rho =\frac{ZM}{V{N}_A}$
where,
Z is the number of atoms in unit cell
M is the molar mass
$N_A$ is the avagadro number.
V is the volume of unit cell.
Since, one ion pair of $CsCl$ is present in the unit cell, the value of Z is 1

$d=\frac{1\times 168}{6.023\times {10}^{23}\times a^3}=4g/cc$
$a^3=\frac{1\times 168}{6.023\times 4\times {10}^{23}}=7\times {10}^{-23}cc.$