Question

Aluminium (molecular weight\[ = 27\] ) crystallises in a cubic unit cell with the edge length \[\;a = 100pm\] with density $d = 180g/c{m^3}$, then type of unit cell is:
A) scc
B) bcc
C) fcc
D) hcp

Answer 1

Hint: As per question first we determine the number of unit cell in the crystal and then we will find the type of unit cell. There is a formula to calculate the number of unit cell are as -
\[\rho = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}\;\]

Complete Step by step answer:The unit cells which are all identical are defined in such a way that they fill space without overlapping.
This unit cell is the smallest recurring unit of the crystal lattice and the building block of a crystal.
The 3D arrangement of atoms, molecules or ions inside a crystal is called a crystal lattice. It is made up of numerous unit cells. A s per formula - \[\rho = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}\;\]
 \[\therefore \] where, \[\rho \] = Density of the crystal, \[M\] = Molecular weight
\[Z\] = Number of unit cells in an atom.
\[{N_A}\] = Avogadro number, \[a\] = Edge length of unit cell
in question given that Molecular weight of Aluminium = \[27g/mole\]
Edge length of unit cell = \[100{\text{ }}pm\] = \[100{\text{ }} \times \;{10^{ - 10}}Cm\] = \[{10^{ - 8\;}}Cm\] \[\therefore \]\[(1{\text{ }}pm = {10^{ - 10}})\]
Density = \[180\;g/c{m^3}\]
Avogadro's number = \[6.022 \times {10^{23}}\]
Putting these value in formula \[\rho = \dfrac{{Z \times M}}{{{N_A} \times {a^3}}}\;\]
\[Z = \;\dfrac{{\rho \times {N_A} \times {a^3}}}{M}\] = \[\dfrac{{180\left( {6.022 \times {{10}^{23}}{\text{ }}} \right){{10}^{ - 8\;}}}}{{27}}\]
\[\Rightarrow {\text{ }}Z{\text{ }} = {\text{ }}4.01466\; \approx 4\]
The number of atoms per unit cell is \[Z = 4.\]
Therefore, the number of atoms per unit cell is \[4\] and the type of unit cell is FCC (Face centered cubic).

So the option (C) is correct.

Note: A face-centred cubic unit cell comprises of atoms placed at all the corners and at the centre of all the faces of the cube.In face-centered cubic (fcc) has a coordination number of \[12\] and contains \[4\] atoms per unit cell. The body-centered cubic (bcc) has a coordination number of \[\;8\] and contains \[2\] atoms per unit cell. The simple cubic has a coordination number of \[6\] and contains \[1\] atoms per unit cell.