Question

Aluminium metal has a density of $2.72{\text{ }}g{\text{ c}}{{\text{m}}^{ - 3}}$ and crystallizes in a cubic lattice with an edge of $404{\text{ pm}}$ . which is/are correct? This question has multiple correct options
A. It forms an fcc unit cell.
B. It forms a bcc unit cell.
C. Its coordination number is 8
D. Its coordination number is 12

Answer 1

You can use the following formula to calculate the number of aluminium atoms present in a one-unit cell.
$n = \dfrac{{\rho \times {a^3} \times {N_A}}}{M}$
From the number of aluminium atoms present in a one-unit cell you can determine the type of lattice as fcc or bcc. From the type of the lattice, you can obtain the coordination number.

Complete answer:
You can calculate the density of the unit cell by dividing its mass with volume. For this purpose, you can use the following formula.
$\rho = \dfrac{{n \times M}}{{{a^3} \times {N_A}}}$
Here, $\rho$ is the density of the unit cell, n is the number of aluminium atoms present in one unit cell, M is the atomic weight of aluminium, a is the edge length of the unit cell and ${N_A}$ is the Avogadro’s number.
Rearrange the above expression in terms of n
$\Rightarrow n = \dfrac{{\rho \times {a^3} \times {N_A}}}{M}$
Substitute $2.72{\text{ }}g{\text{ c}}{{\text{m}}^{ - 3}}$ as $\rho$, $404 \times {10^{ - 10}}{\text{cm}}$as a, ${\text{27 g/mol}}$as M and $6.023 \times {10^{23}}{\text{atoms/mol}}$ as ${N_A}$ in the above expression and calculate the value of n.

$$\Rightarrow n = \dfrac{{\rho \times {a^3} \times {N_A}}}{M} \\\ \Rightarrow n = \dfrac{{2.72{\text{ }}g{\text{ c}}{{\text{m}}^{ - 3}} \times {{\left( {404 \times {{10}^{ - 10}}{\text{cm}}} \right)}^3} \times 6.023 \times {{10}^{23}}{\text{atoms/mol}}}}{{{\text{27 g/mol}}}} \\\ \Rightarrow n = 4$$

For a bcc unit cell, the value of n is 2 and for a fcc unit cell, the value of n is 4. The coordination number in the fcc unit cell is 12.
Hence, aluminium metal forms the fcc unit cell and hac coordination number of 12.

**The correct options are the options A and D.

Note:**
The fcc unit cell is a face centred unit cell in which 8 atoms are present at eight corners of a cube. 6 atoms are present at 6 face centres.