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

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.