Solid State Chemistry Class 12 | Chapter 1 | CBSE | Term 1 |


Solid State Chemistry Class 12 | Chapter 1 |

Solid State

Solid State Chemistry Class 12 : Matter is anything that occupies space and has mass. Factors Deciding states of Matter; Intermolecular Force and Thermal Energy                                                 

  • Intermolecular Force ˃>˃ Thermal Energy     (For Solids)
  • Intermolecular Force ˃ Thermal Energy         (For Liquid)
  • Intermolecular Force ˂˂˂ Thermal Energy     (For Gases)

Ultimately, Properties of matter depend upon the nature of constituent particle and binding forces operating between them. 

Solids must have definite shape and volume.

  • The constituent particles in solids have fixed position and can only oscillate about their mean position (vibrational motion) that accounts for rigidity of solids .
  • They have strong attraction between constituent particles and have high density.
  • They are incompressible and diffusion is negligible.

Types of Solids 

  1. Crystalline Solids
  2. Amorphous Solids   
These are the solids in which constituent particle are arranged in regular ,3D orderly arrangement which gets repeated throughout. In Amorphous Solids , constituent particles are randomly arranged in 3D pattern of solids.
These solids are long range ordered solids. These solids are short range ordered solids (arrangement is fixed but not continuous for long distance.)
These solids have sharp (fixed) melting point. These solids have melting point in range.
On cutting with knife , they give sharp and smooth (even) edges. They form irregular edges when cut with sharp knife.
These solids are Anisotropic solids. These solids are isotropic solids.
Ex: Metals , Salts ( NaCl , KCl) , Quartz Ex : Rubber , Glass , Plastic.

ANISOTROPY: Physical Properties such as electrical resistance or refractive index, mechanical strength show different values when measured along  different directions in same crystal.  For e.g. : Crystalline Solids

ISOTROPY: Physical properties will have same value in all direction called Isotropy. For e.g. : Amorphous Solid      

Note : Amorphous silicon is one of the best material for conversion of sunlight into electricity.    

Important Questions of Solid State Chemistry Class 12

Question :  Why is glass considered a super-cooled liquid or pseudo solids ? 

Why does the window glass of old buildings become thicker at bottom?  

Answer: Glass is an amorphous solid of silica and it flows down very slowly making the bottom portion slightly thicker.   

Question : Why does window glass of old buildings look milky?

Answer: Due to heating during day and cooling at night (annealing) glass acquires some crystalline nature.   

Question : What makes a glass different from quartz ?

Answer: Glass is an amorphous solid of silica . In glass, SiO4 tetrahedral are randomly arranged which on heating softens and melts over a range of temperature. Whereas, Quartz is a crystalline form of silica in which SiO4 tetrahedral are joined regularly. If Quartz is melted and cooled rapidly, it changes into glass. 

Classification of Crystalline Solids

1) Molecular Solids     

2) Ionic Solids     

3) Metallic Solids         

4) Covalent or Network Solids

Molecular Solids

Solids having molecules as constituent particle are called Molecular Solids. On the basis of molecules, these solids are classified as:

Polar Molecular Solids Non Polar Molecular Solids Hydrogen Bonded Molecular Solids
Constituent Particles Polar Molecules Non Polar Molecules or Noble gas Atoms Molecules having F , O , N with H atoms
Nature of bonding Dipole-Dipole interaction Dispersion Force or London Force Hydrogen Bonding

1)These are soft.  

2)Non conductor    

3)Higher than non polar molecular 

4)Exist in Gaseous and Liquid state

1) very soft   

2)Non Conductor 

3)Low melting and boiling point 

4)Exist in gaseous and liquid at room temp.


2)Non conductor 

3)higher than polar and non polar molecular solids. 

4)exist as volatile liquid or soft solid at room temp.        

Example HCl , SO2 Ar , CCl4 , H2 , I2 H2O , NH3

Ionic Solids

Ions are the Constituent particles. Electrostatic force of attraction or columbic force between ions.


  • They have high melting and boiling point.
  • They are soluble in polar solvent.
  • They are insulator in solid state but conduct electricity when present in molten state.                                         

For e.g.  NaCl , KCl , CaO , MgO         

Covalent Solids

Atoms (Especially Non metals) are the constituent particles.

Covalent bonding between the atoms.


  • They have high melting and boiling point and may decompose before melting.
  • They are hard and form a giant molecule due to network of covalent bond.
  • Exceptional behavior of Graphite: Carbon atom are arranged in different layer in which each atom is bonded to three neighbouring atom and fourth atom is free that makes graphite a good conductor.
  • Different layers slide over one another that makes graphite a soft solid and good solid lubricant.         

For e.g.  Silicon Carbide , Diamond , Graphite.

 Metallic Solids

Positive ions surrounded by free electrons are the constituent particles.                                               

They have electrical attraction between metal ion and mobile electron i.e metallic bonding .               


  • Free electron and mobile electron are responsible for high electrical and thermal conductivity of metals.
  • They are ductile , malleable ,  and show luster and colour.                                                                     

For e.g. Na , Mg , Ca, Zn and alloys.

Crystal Lattice and Unit Cells

Regular arrangement of constituent particle in a crystal in three dimensional space is called crystal lattice.

Lattice Sites or Points

The positions which are occupied by the constituent particle in a crystal lattice are defined as crystal point or crystal lattice. Lattice points are joined by straight lines to bring out the geometry of the lattice.

Unit cell is the smallest portion of crystal lattice which repeats in different direction and generates the entire lattice. Unit cell is characterized by the following factors:

  • Dimension along three edges : a , b and c    

Angle between edges i.e α ,β and Ƴ

Types of Crystal System on the basis of Parameters

Crystal System  Variations Edge Lengths Axial Angles Examples
Cubic Primitive ,BCC ,FCC      a=b=c α= β = Ƴ =90ᵒ NaCl , Cu
Tetragonal Primitive , BCC      a=b≠c α= β = Ƴ =90ᵒ TiO2 , CaSO4
Orthorhombic Primitive, BCC, FCC, End- Centered      a≠b≠c α= β = Ƴ =90ᵒ Rhombic Sulphur, KNO3 , BaSO4
Hexagonal Primitive      a=b≠c α= β=90ᵒ,Ƴ=120ᵒ ZnO , CdS
Rhombohedral Primitive      a=b=c α= β = Ƴ ≠90ᵒ Calcite, cinnabar
Monoclinic Primitive, End-centered     a≠b≠c α= Ƴ =90ᵒ, β≠90ᵒ Monoclinic Sulphur
Triclinic Primitive      a≠b≠c α≠ β ≠Ƴ ≠90ᵒ K2Cr2O7 , H3BO3

Types of Unit Cell

Primitive Unit Cell or Simple Cubic Cell  (SCC)

If all constituent particle are present only at corners is called primitive unit cell.

Non primitive Unit cell or Centered Unit Cell

If the constituent particles are present at position other than corners in addition to those at corners, it is called a centred unit cell.

Types of Centered Unit Cell 

a) Face Centered Unit cell (FCC) : When particles are present not only at corners but also at centre of each face of unit cell is known as FCC.

b) Body Centered Unit Cell (BCC) : Besides particles at corners , one constituent particle is at  body centre of such unit cell . 

c) End Centered Unit cell : when particles are present at corners as well as centre of any two opposite faces is called end centred unit cell .

d) Edge centered unit cell : When particles are present not only at corners but also at each edge centre , such a unit cell is called as edge centered unit cell. 

Number of Atoms in a Unit Cell

Points to be kept in mind:

  • An atom at corner is shared by 8 unit cells . Thus,  contribution of each atom present at corner = 1/8
  • An atom on face is shared by 2 unit cells. Hence, contribution of each atom present at face = 1/2
  • An atom present at edge is shared by 4 unit cells . Hence, contribution of each atom at face =1/4
  • An atom present within the body of unit cell is shared by no other unit cell. Hence , contribution of each atom within the body = 1

In simple cubic unit cell :

Contribution of each atom = 1/8

Contribution of 8 atoms , Z = 8 x 1/8 = 1

No. of atoms in SCC = 1

In Face Centered unit Cell :   

 Atoms on corners and 6 atoms at each face

Contribution by 8 atoms on corners = 8 x 1/8 = 1

Contribution by 6 atoms on face = 6 x 1/2 = 3

No. of atoms in FCC, Z  = 1 + 3 = 4                                                               

In Body Centered Unit Cell :             

8 atoms at corners and 1 inside the cell

Contribution by 8 atoms on corners = 8 x 1/8 = 1

 Contribution by 1 atom inside the body = 1

No. of atoms in BCC , Z= 1 + 1 = 2                                                                        

 In End Centered Unit cell :

8 atoms at each corner and 2 at opposite face centres

Contribution by 8 atoms on corners = 8 x 1/8 = 1

Contribution by two atoms on opposite faces = 2 x 1/2 = 1

 No. of atoms in end centred cell , Z = 1 + 1 = 2                               

 In Edge Centered Unit Cell

8 atoms at each corner and 12 at each edge centre

Contribution  by 8 atoms on corners = 8 x 1/8 = 1

 Contribution by 12 atoms at edge centre =  12 x 1/4 = 3

No of atoms in edge centred cell, Z  = 1 + 3 = 4 

Note : Total no of particles in a unit cell or Z = 1/8 (occupied at centre) + 1/2 (occupied by face centre) + 1/4 (occupied by edge centre) + 1(occupied at body centre)


Numerical 1 : Calculate the number of atoms in a cubic based unit cell having one atom on each corner and two atoms on each body diagonal

Soln: Contribution by two atoms present on one body diagonal = 2

Contribution by two atoms on 4 body diagonal = 2 × 4 = 8 

Contribution by atoms present at corner = 1 

Total no of atoms present per unit cell = 8 + 1 = 9    

Numerical 2 : Calculate the no of unit cell present in 1 g of Au if it crystallizes in FCC lattice ?                           

 Soln : No of atom present per unit cell of FCC = 4                                                                                               

 No of Au atoms in 1g = No of moles × NA =   1/197 x 6.022 x 10 23 atoms                                                            

No of unit cell in 1g of Au = 3.056 x 10 21 / 4 = 7.64 x 10 20 unit cells     

Close Packing In Crystal       

The packing of crystal takes place in such a way that they occupy maximum available space and there is minimum empty space. This type of packing is called close-packing. The no of nearest neighbours of a particle is called its coordination number.                                                                                   

1) Close packing in one Dimension : there is only one way to arrange spheres in 1-D i.e spheres in a row touching each other Its coordination no is 2.

2) Close packing in two Dimension :

  • Square Close Packing : spheres of second row are exactly above those of first row are exactly above those of first row. Its coordination number is 4.                                                                                             
  • Hexagonal Close packing : spheres of second row are placed in depression of first row . Its coordination no is 6.

3) Close packing in three dimension :                                                                                                                 

  • 3-D closed packing from 2-D square closed packed layers : second layer is placed over first layer such that squares of upper layer are exactly above those of first layer .Thus , this lattice has AAA.. type pattern .                                                                                                                                                                                       
  • 3-D closed packing from 2-D hexagonal closed packed layers: Empty space present in layer are called voids . Spheres of second layer are in depression of first layer .                       

Tetrahedral Voids :  When one sphere is placed above other three spheres which are in same layer . These four spheres leaves a small space in between which is called Tetrahedral void.                                 

 Coordination no = 4   

Octahedral Voids : This void is surrounded by six spheres which lies at vertices of regular octahedron .

Coordination No = 6                                                                                                                                         

Note: Let no of atoms in unit cell = N                                                                                                                             

No of Octahedral voids = N                                                                                                                                     

No of tetrahedral voids = 2N

Numerical : In a compound the atoms of element Y form CCP lattice and those of element X occupy 2/3 rd of tetrahedral void. The formula of compound will be                                                                 

Soln : No of Atoms of Y in close packing =N                                                                                                     

No of tetrahedral voids = 2/3 x 2N = 4N / 3                                                                                                                                  

No of atoms of X = 4N/3 : N                                                                                                                                 

  X : Y =  = 4 : 3                                                                                                                                                     

Formula of compound = X4Y3

Packing Efficiency  : It is the percentage of total space filled by the particles.                                         

Packing Efficiency = {z x Volume of sphere / Vol of cubic cell } x 100                                                               

1) Packing Efficiency in Simple unit cell :                                                                                                               

Let edge of unit cell = a                                                                                                                                         

Radius of unit cell = r                                                                                                                                    .       

                   a = 2r                                                                                                                                               

No of particles in SCC =  1                                                                                                                                     

Vol. of sphere = 4/3 Π r3                                                                                                                                       

Vol. of cube = a3 = (2r)3  = 8r3                                                                                                                             

Packing Efficiency = 1 x 4/3 Π r3 / 8r3             

 P.E OF SCC  = 52.4 %     


2) Packing Efficiency in Body Centered Cell                                                                                                           

Let edge length be a and radius of sphere be r 

                                                                                                                                                                                                   FD = √ 2 a                                                                                           

By Pythagoras Theorem :

(AF)2 = a2 + (√ 2 a )2                                                                                                                      

AF = √3a                          ‘

also,  AF=4r                                                                                           

4r =  √3a          a = 4r / √3                                                                                                               

Volume of cube = a3  = ( 4r / √3) 3 = 64 /3√3 r3                                                                                                     

No of atoms per unit cell = 2                                                                                                                               

Packing Efficiency = (2 x 4/3 Π r3 ) / (64 /3√3 r3 ) x 100 = 68%

Packing Efficiency of BCC = 68%           


3) Packing Efficiency in Face Centered Cubic Unit Cell 

For FCC;  4r = √2a                                                                                                                                                 

a = 4r / √2                                                                                                                                                               

Volume of cube = a3 = (4r / √2  )3 = 32 / √2 r3                                                                                                     

No of atoms per unit cell = 4                                                                                                                                 

Packing Efficiency = ( 4 x 4/3 Π r3 ) / (32 / √2 r3 ) = 74% 

Packing Efficiency of FCC = 74% 


Note :

1) Distance of neighbouring atoms or closest distance (d)= 2r

2) Interatomic  distance or distance b/w cation and anion :                                                                                               

In FCC :     ( rcation + ranion ) = a /2                                                                                                                 

In BCC :        (rcation + ranion) = √3a /2

Packing of Ions in Ionic Crystal

NaCl  Type Cl ions form FCC Na+ ion in octahedral voids Na+ = 6

Cl = 6

             4 NaCl , KCl , LiCl, AgBr
CsCl   Type Cl ions form SCC Cs+ at centre of cube Cs+ = 8

Cl = 8

             1 CsCN , CaS
Zinc Blende S2- ions form CCP . Zn2+ at alternate tetrahedral voids Zn2+ = 4

S2- = 4

            4 AgI , CuCl
Wurtzite S2- form HCP Zn2+ ions at half of tetrahedral voids. Zn2+ = 4

S2- = 4

           6 ZnO ,  SiC
Fluorite (CaF2) F ions at all tetrahedral voids Ca2+ ions form FCC. Ca2+ = 8

F = 4

            4 BaF2 , BaCl2
Antifluorite Type (Na2O) O2-  form CCP Na+  ions at all tetrahedral voids. Na+  = 4

O2- = 8

              4 K2O , Li2O


Calculation of Density of Cubic Crystal     

Suppose edge of unit cell be a cm                                                                                                                       

No of atom present per unit cell be Z                                                                                                             

Volume of unit cell = ( a cm)3 = a3cm3                                                                                                              

Density of Unit cell = mass of unit cell / volume of unit cell                                                                                 

 Mass of unit cell = no of atom in unit cell × mass of each atom = Z × m                                                                                                 

 (mass of each atom , m = Atomic mass (M) / Avogadro number ) )                                                               

Thus, density of unit cell (ρ or d) = Z x M / ax N 

Z=1 (for SCC)             Z = 2(for BCC)                       Z= 4 (for FCC)                                                                    

Relationship between Radius (r) of void and radius of atom (R)

1) For Triangular Voids


Cos 30 o =  R / R+ r                                                        

√3 /2 =  R / R+ r 

2 / √3 = 1 + r / R

 r / R   = 0.155      

; rtriangular void  = 0.155 R  

2) For Tetrahedral Voids 


AB = 2R = √2 a ….(i)               

AD = 2R + 2r = √3a                                                                                             

2r = √3a – 2R  →      

2r = √3a – √2a…(ii)                                                                                             

dividing (ii) by (i) , we get

2r / 2R = (√3a – √2a) / √2a    →  r/R = 0.225                  

 rtetrahedral  = 0.225 Rsphere                                                         

3) For Octahedral Voids 

(2R + 2r)2 = (2R)2 + (2R)2     → 

2R + 2r = √8R       →  2r = √8R – 2R

2r / R = √8 – 2      →     r / R = 0.414  

              roct. void = o.414R  

Imperfection or Defects in Solids

Defects are basically irregularities in arrangement of constituent particle.

Point defect is irregularity around a point or atom in crystalline solid. 

Types of Point Defects

Stoichiometric Defect : These are the defect that do not disturb the stoichiometry of crystalline solid i.e constituent in a particular solid are present in the  same ratio as predicted by their formula. 

In Non-Ionic Solids 

a) Vacancy Defect : When some sites in the crystal lattice are vacant , crystal is said to have defect . It results in decrease in density of substance. 

b) Interstitial Defect : When some constituent particle occupy interstitial sites, crystal is said to have interstitial defect. This defect increases the density of substance.

In Ionic Solids 

 Schottky Defect  Frenkel  Defect (Dislocation)
These defect arises when equal no of cations and anions are missing from their crystal lattice and vacancies or holes are created at their position. It arises when cations  leaves their lattice site and occupy position elsewhere in the crystal lattice.
This decreases the density of crystal. It has no effect on density of crystal.
In this , size of cation and anion are similar. Anions are bigger in size than cation.
Ionic compound having high coordination number favours it. Ionic compound having low coordination no favours it.
Ex: NaCl , KCl , AgBr Ex : ZnS , AgCl , AgBr

Non Stoichiometric Defect: This defect arises when ratio of cation to anion becomes different to that indicated by the chemical formula.

Metal Excess Defect

a) By Anion vacancies : A –ve ion may be missing from lattice site leaving a vacancy which is occupy by electrons thereby maintaining the electrical neutrality.                                                                                       

 F-Centre : Anion vacancy which is occupied by an electron.                                                                           

NaCl impart yellow colour due to presence of metal excess defect.                                                                   

b) By presence of extra cation on interstitial sites : It arises due to presence of an extra cation occupying interstitial site. Electrical neutrality is maintained by one electron in another interstitial site.                                                    

Metal Deficiency Defect

This defect arises when metal shows variable valency  i.e transition metals . It occurs due to missing of the cation from lattice site and presence of cation having higher charge in adjacent site.

Impurity Defect

Defect introduced in crystal due to presence of certain impurity.

Electrical Properties

Classification of Solids on basis of Electrical Conductivity                                                                    

               TYPES OF SOLID      RANGE OF CONDUCTIVITY  (ohm-1m-1)
                CONDUCTORS                          104 to 107
               INSULATORS                         10-20 to 10-10
              SEMICONDUCTORS                        10-6   to 104

Conductivity of electricity of semi-conductors: With increase in temperature , electron gets enough energy to jump to conduction band and hence its conductivity increases. Example of intrinsic semiconductor are  Silicon and Germanium.


The process of adding desired impurities to an intrinsic semiconductor to increase its conductivity is called doping.

1) Electron rich impurities : When a pentavalent impurity or group 15 element  is added to an intrinsic semiconductor then an extra electron is added to semiconductor and become delocalized. It is also called n-type semiconductor. 

2) Electron Deficit Semiconductor : When a trivalent impurity or group 13 element is added to an intrinsic semiconductor then a hole is generated in the crystal . This type of semiconductors are called as p-type semiconductors.

# Solid State Chemistry Class 12 

# Solid State Chemistry Class 12 Notes

# Solid State Chemistry Class 12 Questions Answers

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