Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

Chemistry Class 12

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Class 12 : Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

The document Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev is a part of the Class 12 Course Chemistry Class 12.
All you need of Class 12 at this link: Class 12

What is a crystal:

A crystalline solid consists of a large number of small units, called crystals, each of which possesses a definite geometric shape bounded by plane faces. The crystals of a given substance produced under a definite set of conditions are always of the same shape.

Unit Cells

In this topic we would be studying certain properties of a solid which depend only on the constituents of the solid and the pattern of arrangement of these constituents. 

Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev
Fig: A unit cell (Primitive unit cell)

To study these properties, it would be convenient to take up an as small amount of the solid as possible because this would ensure that we deal with only the minimum number of atoms or ions. 

This smallest amount of the solid whose properties resemble the properties of the entire solid irrespective of the amount taken is called a unit cell

It is the smallest repeating unit of the solid. Any amount of the solid can be constructed by simply putting as many unit cells as required.


Bravais Lattices

Bravais (1848) showed from geometrical considerations that there are only seven shapes in which unit cells can exist. These are: 

(i) Cubic 

(ii) Orthorhombic 

(iii) Rhombohedral 

(iv) Hexagonal 

(v) Tetragonal 

(vi) Monoclinic  

(vii) Triclinic 

Moreover, he also showed that there are basically four types of unit cells depending on the manner in which they are arranged in a given shape.

These are: Primitive, Body-Centered, Face Centered and End Centered. He also went on to postulate that out of the possible twenty-eight unit cells (i.e. seven shapes, four types in each shape = 28 possible unit cells), only fourteen actually would exist. These he postulated based only on symmetry considerations. These fourteen unit cells that actually exist are called Bravais Lattices.

Primitive Cubic Unit Cell 

In a primitive cubic unit cell the same type of atoms are present at all the corners of the unit cell and are not present anywhere else. It can be seen that each atom at the corner of the unit cell is shared by eight unit cells (four on one layer, as shown, and four on top of these). 

Therefore, the volume occupied by a sphere in a unit cell is just one-eighth of its total volume. Since there are eight such spheres, the total volume occupied by the spheres is one full volume of a sphere. Therefore, a primitive cubic unit cell has effectively one atom. 

A primitive cubic unit cell is created in the manner as shown in the figure. Four atoms are present in such a way that the adjacent atoms touch each other.
Therefore, if the length of the unit cell is 'a', then it is equal to 2r, where r is the radius of the sphere. Four more spheres are placed on top of these such that they eclipse these spheres. The packing efficiency of a unit cell can be understood by calculating the packing fraction

It is defined as ratio of the volume occupied by the spheres in a unit cell to the volume of the unit cell and void fraction is given as 1 Packing fraction. Therefore, PF = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev. (This implies that 52 % of the volume of a unit cell is occupied by spheres). VF ≈ 0.48


Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev
Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

Fig: Primitive cubic unit cell


Body-Centered Cubic Unit Cell 

A Body Centred unit cell is a unit cell in which the same atoms are present at all the corners and also at the center of the unit cell and are not present anywhere else.This unit cell is created by placing four atoms which are not touching each other. 

Then we place an atom on top of these four. Again, four spheres eclipsing the first layer are placed on top of this. The effective number of atoms in a Body Centered Cubic Unit Cell is 2 (One from all the corners and one at the center of the unit cell). 

Moreover, since in BCC the body-centered atom touches the top four and the bottom four atoms, the length of the body diagonal (√3a) is equal to 4r. The packing fraction in this case is  = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev , VF ≈ 0.32.


Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRevFig: Body centered cubic unit cell

Face Centered Cubic Unit Cell 

In an fcc unit cell, the same atoms are present at all the corners of the cube and are also present at the center of each square face and are not present anywhere else. The effective number of atoms in fcc is 4 (one from all the corners, 3 from all the face centers since each face-centered atom is shared by two cubes). 

Since, here each face-centered atom touches the four corner atoms, the face diagonal of the cube (Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev) is equal to 4r.

Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev. VF ≈ 0.26.

Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRevFig: Face centered cubic unit cell


Hexagonal Primitive Unit Cell

Each corner atom would be common to 6 other unit cells, therefore their contribution to one unit cell would be 1/6. Therefore, the total number of atoms present per unit cell effectively is 6.

Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

Figure 6(b) ABCD is the base of the hexagonal unit cell AD=AB=a. The sphere in the next layer has its center F vertically above E and it touches the three spheres whose centers are A,B and D.




 
Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRevFig: Hexagonal close packing


Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

Hence FE = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

The height of unit cell (h) = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

The area of the base is equal to the area of six equilateral triangles, = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev  . The volume of the unit cell = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev.

Therefore

PF = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev= 0.74,  VF ≈ 0.26 


Density of crystal lattice 

The density of crystal lattice is same as the density of the unit cell which is calculated as

ρ = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

ρ = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

Seven crystal systems: The seven crystal systems are given below.

Cubic:
Bravais Lattices: Primitive, face Centered, Body Centered = 3
Parameters of Unit Cell:

  1. Intercepts: a=b=c
  2. Crystal angle: α = β = γ = 90o
    Example: Pb,Hg,Ag,Au Diamond, NaCl, ZnS

Orthorhombic:
Bravais Lattices: Primitive, Face Centered, Body Centered, End Centered = 4
Parameters of Unit Cell:

  1. Intercepts: a ≠ b≠ c
  2. Crystal angle: α = β = γ = 90o
    Example: KNO2, K2SO4

Tetragonal:
Bravais Lattices: Primitive, Body Centered =2
Parameters of Unit Cell:

  1. Intercepts: a = b≠ c
  2. Crystal angle: α = β = γ = 90o
    Example: TiO2,SnO2

Monoclinic:
Bravais Lattices: Primitive, End Centered =2
Parameters of Unit Cell:

  1. Intercepts: a ≠ b≠ c
  2. Crystal angle: α = β = γ = 90o
    Example: CaSO4,2H2O

Triclinic:
Bravais Lattices : Primitive = 1
Parameters of Unit Cell:

  1. Intercepts: a ≠ b≠ c
  2. Crystal angle: α = β = γ =
    Example:  K2Cr2O7, CaSO5H2O

Hexagonal:
Bravais Lattices: Primitive = 1
Parameters of Unit Cell:

  1. Intercepts: a = b≠ c
  2. Crystal angle: α = β = 90γ =1200
    Example:  Mg, SiO2, Zn, Cd

Rhombohedral:
Bravais Lattices: Primitive = 1
Parameters of Unit Cell:

  1. Intercepts: a = b= c
  2. Crystal angle: α = γ = 90o, β≠ 90o
    Example:  As, Sb, Bi, CaCO3

Illustration 1: Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a = 6.8Å,  b = 4.4Å and C = 7.2Å. If the molar mass is 21.76g. Calculate density of crystal. 


Solution: 
Since

Density, ρ = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev

Here n = 4, Av. No = 6.023 x 1023, &

Volume = V = a x b x c

= 6.8 x 108 x 4.4 x 108 x 7.2 x 108 cm= 2.154 x 1022 cm3

density,  = Doc: Crystal Lattice and Unit cells Class 12 Notes | EduRev= 0.6708 gm/cm3

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