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

Hint: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.