Question

Aluminium (molecular weight = $ 27 $ ) crystallizes in a cubic unit cell with edge length a = $ \text{100 pm} $ , with density d = $ \text{180 g/c}{{\text{m}}^{\text{3}}} $ , then the type of unit cell is:
(A) SCC
(B) BCC
(C) FCC
(D) HCP

Answer 1

Hint: The unit cell in a crystal lattice is defined as the smallest repeating unit having the full symmetry of the crystal structure. There can be different unit cells depending on the number and the position of the lattice points that constitute the atoms. The lattice points can be atoms, molecules, or ions. We shall find the number of atoms in the unit cell which will tell us the type of arrangement.

Formula Used: Density of the unit cell, $ \text{ }\rho\text{ }=\dfrac{\text{Z }\times\text{ M}}{{{\text{N}}_{\text{A}}}\text{ }\times\text{ }{{\text{a}}^{\text{3}}}} $
Where Z is the number of the unit cells, M is the mass of the compound or element, $ {{\text{N}}_{\text{A}}} $ is the Avogadro, and “a” is the length of unit cell.

Complete step by step solution:
Given that, Z = unknown, M = 27, a = $ \text{100 pm} $ = $ 100\times {{10}^{-12}}=1\times {{10}^{-10}} $ m = $ 11\times {{10}^{-8}} $ m
 d = $ \text{180 g/c}{{\text{m}}^{\text{3}}} $ , $ {{\text{N}}_{\text{A}}}=6.023\times {{10}^{23}} $
Putting the values of the given quantities in the equation we get,
 $ \text{Z}=\dfrac{\text{ }\rho\text{ }\times {{\text{N}}_{\text{A}}}\text{ }\times\text{ }{{\text{a}}^{\text{3}}}}{\text{M}}=\dfrac{180\times \left( 6.023\times {{10}^{23}} \right)\text{ }\times\text{ }{{\left( {{10}^{-8}} \right)}^{\text{3}}}}{27} $ = $ \text{Z}=\dfrac{180\times \left( 6.023\times {{10}^{-1}} \right)\text{ }}{27}=4.014 $ which is equal to 4 atoms of aluminium which the number of atoms present in an FCC unit cell.
Hence, the correct answer is option C.

Note:
Depending on the number of the lattice points there can be different unit cells. If there are eight lattice points at the eight corners of the cube and as each cube share that lattice points with seven other unit cells, then the total lattice points in the unit cell = $ 8\times \dfrac{1}{8}=1 $ . This the simple cubic unit cell.
If there is one more lattice points at the centre of the unit cell along with eight at the corners then it is called the body centred cubic cell and the number of lattice points = $ \left( 8\times \dfrac{1}{8} \right)+1=2 $ .
In the face centred cubic unit cell there are six lattice points at the six faces of the cubes along with the eight lattice points at the eight corners of the unit cell. These lattice points are shared by another unit cell on each face. Hence the total number of lattice points here are: $ \left( 8\times \dfrac{1}{8} \right)+\left( 6\times \dfrac{1}{2} \right)=3+1=4 $