Question

Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm. The edge length (in $pm$ ) of the unit cell and number of unit cells per cc of aluminium respectively are:
A.$354,2.25 \times {10^{22}}$
B.$648,3.3 \times {10^{23}}$
C.$298,25 \times {10^{22}}$
D.$436,6.3 \times {10^{22}}$

Answer 1

Hint:
In crystalline solid, there is a regular and repeating pattern of constituent particles. If we represent three dimensional arrangement of constituent particles diagrammatically, each particle is depicted as a point. Unit cell is the smallest part of such a crystal lattice.

Complete step by step answer:
In crystalline solids there is a regular and repeating pattern of constituent particles. If we represent the three dimensional arrangement of constituent particles diagrammatically and each particle is depicted as a point then such arrangement is called crystal lattice. Some features of the crystal lattice are-
Each point in the lattice is called a lattice point.
Each point in the crystal lattice represents a constituent particle which can be an atom, group of atoms or a molecule.
Unit cells are the smallest portion of such crystal lattice. When it is repeated in different directions we get the entire lattice.
The edge length of the unit cell is abbreviated as $a$ .
There are different types of unit cell such as primitive, body-centered, face-centered, end-centered.
These atoms, ions or molecules that form the crystal can be of various shapes and hence the mode of packing in such crystals will change according to their shapes. We consider these particles as spheres. The packing of spheres is done in such a way that the available space is used maximum.
There are two ways in which these spheres can be arranged-
Hexagonal close packing
Cubic close packing.
Let us understand the cubic close packing. Spheres form a first layer and we call it as layer A. now while forming the second layer the spheres are placed in unoccupied hollows of layer A. Now when the third layer is formed the atoms of layer C do not come exactly above layer A. this type of packing is called as ABC. Now the spheres of the fourth layer will be exactly above the layer A. this arrangement continues as ABCABCABC... and so on. When such arrangement continues indefinitely the system is found to have cubic symmetry and packing is called as cubic close packing (ccp).
If $r$ is the radius of sphere and edge length of unit cell is $a$ , then for cubic close packing the relation between the two can be given as-
$r = \dfrac{a}{{2\sqrt 2 }}$
In the given problem,
$r = 125pm$
To calculate $a$ ,
$a = 2\sqrt {2r} $
$a = 2\sqrt 2 \times 125 = 354pm$
The edge length is $354pm$ .
Now, the volume of one unit cell $ = {a^3}$
Now we need to convert the edge length into $cm$ .
$1pm = {10^{ - 10}}cm$
Volume of one unit cell $ = {(354pm)^3} = {(354 \times {10^{ - 10}})^3} = {(3.54 \times {10^{ - 8}})^3} = 4.4 \times {10^{ - 23}}c{m^3}$
Now we need to find the number of unit cell present in $1c{m^3}$ of aluminium-
Number of unit cell $ = \dfrac{\text{total volume}}{\text{volume of one unit cell}}$
Number of unit cells $ = \dfrac{1}{{4.4 \times {{10}^{ - 23}}}} = 2.26 \times {10^{22}}$ unit cells.
So the edge length of unit cell is $354pm$ and the number of unit cells in $1c{m^3}$ of aluminium is $2.26 \times {10^{22}}$ .
The correct option is A.

Note: In ccp structure, one unit cell has four atoms.
-Both the types of packing are highly efficient and $74% $ space in the crystal is filled.
-In hcp, the packing of atoms is of AB type.