Chemistry - Solid State Question with Solution | TestHub

The density of a unit cell is given by the formula:

$d=\frac{Z\times M}{N_A\times a^3}$

Rearranging the formula to solve for the number of atoms per unit cell $Z$:

$Z=\frac{d\times N_A\times a^3}{M}$

Given values are:
$d=6.25{\text{ g/cm}}^3$
$a=400\text{ pm}=4\times 10^{-8}\text{ cm}$
$M=120\text{ g/mol}$
$N_A=6.02\times 10^{23}{\text{ mol}}^{-1}$

Calculating the volume of the unit cell $a^3$:

$a^3=(4\times 10^{-8}{)}^3=64\times 10^{-24}{\text{ cm}}^3$

Substituting the values into the expression for $Z$:

$Z=\frac{6.25\times 6.02\times 10^{23}\times 64\times 10^{-24}}{120}$

$Z=\frac{400\times 6.02\times 10^{-1}}{120}$

$Z=\frac{240.8}{120}\approx 2$

Since the number of atoms per unit cell is $2$, the unit cell is body-centered cubic (bcc).

Answer: bcc