Class 12 - Chemistry - The Solid State


Define the term 'amorphous'. Give a few examples of amorphous solids.


Amorphous solids are the solids whose constituent particles are of irregular shapes and have short range order.

These solids are isotropic in nature and melt over a range of temperature.

Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids.

They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces.

Examples of amorphous solids include glass, rubber, and plastic.




What makes a glass different from a solid such as quartz? Under what conditions quartz could be converted into glass?


The arrangement of the constituent particles makes glass different from quartz.

In glass, the constituent particles have short range order,

but in quartz, the constituent particles have both long range and short range orders.

Quartz can be converted into glass by heating and then cooling it rapidly.



Question 1.3.

Classify each of the following solids as ionic, metallic, molecular, network

(covalent) or amorphous.

(i) Tetra phosphorus decoxide (P4O10)

(ii) Ammonium phosphate (NH4)3PO4

(iii) SiC

(iv) I2

(v) P4

(vi) Plastic

(vii) Graphite

(viii) Brass

(ix) Rb

(x) LiBr

(xi) Si



(i) Tetra phosphorus decoxide (P4O10)   Molecular

(ii) Ammonium phosphate (NH4)3PO4    Ionic

(iii) SiC               Network (covalent)

(iv) I2                  Molecular

(v) P4                   Molecular

(vi) Plastic           Amorphous

(vii) Graphite      Network (covalent)

(viii) Brass            Metallic

(ix) Rb                  Metallic

(x) LiBr                    Ionic

(xi) Si                  Network (covalent)



Question 1.4.

(i) What is meant by the term “coordination number”?

(ii) What is the coordination number of atoms?

(a) In a cubic close-packed structure?

(b) In a body-centred cubic structure?


  • The coordination number of an atom, ion or molecule is the number of constituent particles
  • which touch that particular atom, atom or molecule.
  • The coordination number of atoms.
  • In cubic close packed structure coordination number is 12.
  • In body–centered cubic structure coordination number is 8.



Question 1.5.

How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.


Atomic mass of the metal can be determined easily if the density and the dimensions of unit cell of an unknown metal are known.

Let us assume a cubic crystal, its volume of unit cell will be = a3

No of atom in each cell = z

Also molar mass of the metal =M,

Mass of each atom = m

Also, mass of unit cell = (no of atom in each cell) x (Mass of each atom) --- (1)

Also Mass of each atom = (molar mass) / (Avogadro number)

Or = (M/NA) --- (2)

Putting the values from equation (2) to equation (1) we get,

Mass of unit cell = (Mz) / (NA) --- (3)

Now density of unit cell (d) = (mass of unit cell) / (volume of unit cell)

Therefore d = (Mz) / (a3 NA) ---- (4)

From 4th equation we can calculate the atomic mass of unknown metal.



Question 1.6.

'Stability of a crystal is reflected in the magnitude of its melting points'.

Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book.

What can you say about the intermolecular forces between these molecules?


Higher the melting point, greater are the forces holding the constituent particles together and hence greater is the stability.

Melting points of substances

Water = 273K

Ethyl alcohol = 155.7K

Diethyl ether = 156.8K

Methane = 90.5K.

Higher melting point of water than alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in water molecules.

The stability of crystals depends up on the packing of the atoms.

If there are strong force of attraction between the constituent particles then greater the stability.

Water – strong intermolecular hydrogen bonding

Ethyl alcohol – strong intermolecular hydrogen bonding.

Diethyl ether – it is a polar molecule so there is dipole –dipole attraction

Methane    – is a non–polar molecule so that there are weak van der Waal’s forces (London dispersion forces).



Question 1.7.

How will you distinguish between the following pairs of terms:-

(i) Hexagonal close-packing and cubic close-packing?

(ii) Crystal lattice and unit cell?

(iii) Tetrahedral void and octahedral void?


Hexagonal close packing:–

When the tetrahedral voids of the second layer are covered by the spheres of the third layer,

So that the spheres of the third layer are exactly aligned with those of the first layer,

we get a pattern of spheres which is repeated in alternate layers.

This pattern can be written in the form of ABAB ....... pattern.

This structure is called hexagonal close packed (hcp) structure.

Magnesium and zinc metals have this pattern


Cubic close packing: – When the third layer is placed above the second layer in a manner such that its spheres cover the octahedral voids.

And the spheres of the third layer are not aligned with first or the second layer, arrangement is called “C’ type.

This pattern of layers is often written as ABCABC........... This structure is called cubic close packed (ccp).


(ii)Crystal lattice:–

If the constituent particles are represented by dot (.)

The periodic and systematic arrangement of these dots in three dimensions is called Crystal lattice.


If each particle in three dimensional arrangements of constituent particles in a crystal is represented by a point,

the arrangement is called crystal lattice.

Thus, a regular three dimensional arrangement of constituent particles which represented by points in space is called a crystal lattice.


Unit cell: – the smallest repeating unit in space lattice is called unit cell.

When we repeat unit cell in all dimensions, we get a structure of the crystal.

Properties of unit cell are measured by length of edges and the angles between the edges.


(iii) Tetrahedral void:–

When a sphere of second layer is touching three sphere of first layer and the centers of all four spheres are at the corners of tetrahedron,

the empty space which has tetrahedral arrangement is called tetrahedral void.

Octahedral void:–

When the tetrahedral void is formed at the corners of six spheres, each octahedral void generates two set of

equilateral triangles which are opposite to each other are called octahedral voids.



Question 1.8.

How many lattice points are there in one unit cell of each of the following lattice?

(i) Face-centred cubic

(ii) Face-centred tetragonal

(iii) Body-centred


(i) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred cubic.

(ii) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred tetragonal.

(iii) There are 9 (1 from the centre + 8 from the corners) lattice points in body-centred cubic.



Question 1.9.


(i) The basis of similarities and differences between metallic and ionic crystals.

(ii) Ionic solids are hard and brittle.


(i) Both metallic & ionic crystals have strong forces of attraction between their atoms or ions.

In metallic crystals there is a strong metallic bond present between electrons & positively charged ions.

In ionic crystals there is strong ionic bond between anions & cations.

Both of them conduct electricity, but metallic crystals conduct electricity in all three states of matter while

ionic crystals conduct electricity only in molten state.

(ii) The ionic solids are hard & brittle because they have strong electrostatic forces of attraction between the anions & cations .

As a result the anions & cations are tightly held by these forces & due to

which they are unable to move & they are fixed at one position.

So when force is applied to break them, they appear to be hard.




Question 1.10.

Calculate the efficiency of packing in case of a metal crystal for

(i) Simple cubic

(ii) body-centred cubic

(iii) face-centred cubic (with the assumptions that atoms are touching each other).


(i) Simple cubic:

Suppose the edge length of the unit cell = a

Radius of the sphere = r


As the spheres are touching each other along the edge, therefore a = 2r

Now there are 8 spheres at the corners of the cube & each sphere at the corner is shared by

8 unit cells & the contribution per unit cell is (1/8) so that

Number of spheres per unit cell is (8) x (1/8) = 1

Volume of sphere = (4)/ (3πr3)

Volume of cube = a3 = (2r) 3 = 8r3

Now packing efficiency

= [(volume of one sphere / total volume of cubic unit cell)] x (100)


[(4/3) (πr3) / (8r3)] x (100) = 52.4%

Therefore the volume occupied in simple cubic arrangement = 52.4%

(ii) It can be observed from the above figure that the atom at the centre is in contact

with the other two atoms diagonally arranged.


From ΔFED, we have:

b2= (a2+ a2)

  • b2 = 2a2
  • b=√2a

Again, from ΔAFD, we have:

c2 = (a2+ b2)

  • c2= (a2 + 2a2) (because b2 = 2a2)
  • c2 = 3a2
  • c=√3a

Let the radius of the atom be r.

Length of the body diagonal, c = 4π

=>√3a = 4r

=> a = (4r)/ (√3)


r= (√3a/4)

Volume of the cube,

a3 = (4r)/ (√3)3

A body-centred cubic lattice contains 2 atoms.

So, volume of the occupied cubic lattice

= (2 π) (4/3) r3

= (8/3) πr3

Therefore Packing efficiency

 = (volume occupied by two spheres in the unit cell)/ (total volume of the unit cell) x 100%

= [(8/3) πr3)/ (4r/√3)3] x 100%

= [(8/3) πr3)/ (64/3√3) r3] x 100%

= 68%

(iii)    Face-centred cubic

Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.


From ΔABC, we have:

AC2 = AB2 + BC2

  • b2 = (a2+ a2)
  • b2 = 2a2
  • b = √(2a)

Let r be the radius of the atom.

Now, from the figure, it can be observed that:


  • √(2a) = 4r
  • Or a =2√2r

Now, volume of the cube,

a3 = (2√2r) 3

We know that the number of atoms per unit cell is 4.

So, volume of the occupied unit cell

= (4π) [(4r3)/ (3)]

Therefore Packing efficiency

 = (volume occupied by two spheres in the unit cell)/ (total volume of the unit cell) x 100%

= [(4π) [(4r3)/ (3)]/ (2√2r) 3] x 100%

= [(16/3) π r3] / (16√2r3) x 100%

= 74%



Question 1.11.

Silver crystallises in FCC lattice. If edge length of the cell is 4.07 × 10–8cm and density is 10.5 g cm–3,

calculate the atomic mass of silver


It is given that the edge length, a = 4.07 × 10-8 cm and density d = 10.5 g cm-3

As the lattice is FCC type the number of atoms per unit cell z = 4

We also know that, NA = 6.022 × 1023 mol-1

Using the relation:

d= (zM)/ (a3 NA)

  • M = (d a3 NA)/(z)
  • =(10.5 gcm3 x (4.077 x 108 cm)3 x 6.022 x 1023 mol-1)/(4)
  • =107.13 gmol-1

Therefore, atomic mass of silver = 107.13u



Question 1.12.

A cubic solid is made of two elements P and Q.

Atoms of Q are at the corners of the cube and P at the body-centre.

What is the formula of the compound? What are the coordination numbers of P and Q?



It is given that the atoms of Q are present at the corners of the cube.

Therefore, number of atoms of Q in one unit cell = 8 x 1/8 = 1

It is also given that the atoms of P are present at the body-centre.

Therefore, number of atoms of P in one unit cell = 1

This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1

Hence, the formula of the compound is PQ.

The coordination number of both P and Q is 8.



Question 1.13.

Niobium crystallises in body-centred cubic structure. If density is 8.55g cm–3,

calculate atomic radius of niobium using its atomic mass 93 u.


It is given that the density of niobium, d = 8.55 g cm−3

Atomic mass, M = 93 gmol−1

As the lattice is bcc type, the number of atoms per unit cell, z = 2

We also know that, NA = 6.022 × 1023 mol−1

Applying the relation:


d= (zM)/ (a3NA)

  • a3=(zM)/(d NA)
  • =( 2 x 93 gmol-1)/(8.55 gcm-3 x 6.022 x 1023 mol-1)

= 3.612 × 10−23 cm3

So, a = 3.306 × 10−8 cm

For body-centred cubic unit cell:

r= (√3a/4)

= (√3/4) x 3.306 x 10-8 cm

= 1.432 × 10−8 cm

= 14.32 × 10−9 cm

= 14.32 nm



Question 1.14.

If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.


A sphere fitting into an octahedral void is shown by shaded circle. ABC is a right angle triangle.

Applying Pythagoras theorem,



(2R) 2=(R+r) 2+(R+r) 2

⇒2(R+r) 2

Or (2R) 2/ (2) =(R+r) 2

√ [(2R) 2/ (2)] =(R+r)

Or √ [(4R2)/ (2)] =(R+r)

√ (2R) 2 =(R+r)


r= (√2R−R)

r=R (√2–1)

=R (1.414−1)

r=R (0.414)



Question 1.15.

Copper crystallises into a fcc lattice with edge length 3.61 × 10–8 cm.

Show that the calculated density is in agreement with its measured value of 8.92 g cm–3.


Edge length, a = 3.61 x 10−8 cm

As the lattice is FCC type, the number of atoms per unit cell, z = 4

Atomic mass, M = 63.5 g mol−1

We also know that NA = 6.022 x 1023 mol−1

Applying the relation:

d= (zM)/ (a3 NA)

= (4 x 63.5 gmol-1)/ (3.61 x 10-8 cm) 3 x 6.022 x 1023 mol-1)

= 8.97 g cm−3

The measured value of density is given as 8.92 g cm−3

Hence, the calculated density 8.97 g cm−3 is in same as the measured value.



Question 1.16.

Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?


The formula of nickel oxide is Ni0.98O1.00.

Therefore, the ratio of the number of Ni atoms to the number of O atoms,

Ni: O = (0.98: 1.00) = (98: 100)

Now, total charge on 100 O2− ions = 100 × (−2)

= −200

Let the number of Ni2+ ions be x.

So, the number of Ni3+ ions is (98 – x).

Now, total charge on Ni2+ ions = x (+2)

= +2x

And, total charge on Ni3+ ions = (98 − x) (+3)

= (294 − 3x)

Since, the compound is neutral, we can write:

2x + (294 − 3x) + (−200) = 0

⇒ −x + 94 = 0


⇒ x = 94

Therefore, number of Ni2+ ions = 94

And, number of Ni3+ ions = (98 – 94) = 4

Hence, fraction of nickel that exists as Ni2+

= (94/98)

= 0.959

And, fraction of nickel that exists as

Ni3+ = (4/98)

= 0.041

Alternatively, fraction of nickel that exists as Ni3+ = (1 − 0.959)

= 0.041



Question 1.17.

What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.


Semiconductors are substances having conductance in the intermediate range of 10-6 to 104 ohm−1m−1.

The two main types of semiconductors are:

(i) n-type semiconductor (ii) p-type semiconductor

n-type semiconductor: The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor.

When the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As, an n-type semiconductor is generated.

Si and Ge have four valence electrons each. In their crystals, each atom forms four covalent bonds.

On the other hand, P and As contain five valence electrons each.

When Si or Ge is doped with P or As, the latter occupies some of the lattice sites in the crystal.

Four out of five electrons are used in the formation of four covalent bonds with four neighbouring Si or Ge atoms.

The remaining fifth electron becomes delocalised and increases the conductivity of the doped Si or Ge.

p-type semiconductor: The semiconductor whose increased in conductivity is a result of electron hole is called a p-type semiconductor.

When a crystal of group 14 elements such as Si or Ge is doped with a group 13 element such as B, Al, or Ga

(which contains only three valence electrons), a p-type of semiconductor is generated.

When a crystal of Si is doped with B, the three electrons of B are used in the formation of three covalent bonds and an electron hole is created.

An electron from the neighbouring atom can come and fill this electron hole, but in doing so, it would leave an electron hole at its original position.

The process appears as if the electron hole has moved in the direction opposite to that of the electron that filled it.

Therefore, when an electric field is applied, electrons will move toward the positively-charged plate through electron holes.

However, it will appear as if the electron holes are positively-charged and are moving toward the negatively- charged plate.




Question 1.18.

Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1.

Can you account for the fact that this substance is a p-type semiconductor?


Since Cu2O is non-stoichiometric oxide, it contains Cu in two oxidation states, +1 and +2. Cu2+ provides an excess of positive charge.

As a result an electron from a neighboring Cu+ is transferred to Cu2+. The transfer of electron leaves behind a hole,

which carries an extra positive charge and a negative hole is created.

It appears that the positive hole moves through the lattice, hence it appears as P-type semiconductor.



Question 1.19.

Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions.

Derive the formula of the ferric oxide.


Let the number of oxide (O2−) ions be x.

So, number of octahedral voids = x

It is given that two out of every three octahedral holes are occupied by ferric ions.

So, number of ferric (Fe3+) ions = (2/3) x

Therefore, ratio of the number of Fe3+ ions to the number of O2− ions,

(Fe3+): (O2-) = (2/3x): (x)

= (2/3): (1)

= 2: 3

Hence, the formula of the ferric oxide is Fe2O3.



Question 1.20.

Classify each of the following as being either a p-type or a n-type semiconductor:

(i) Ge doped with In (ii) B doped with Si.


(i)Ge belongs to group 14 & In belongs to group 13, In has 3 valence electrons while Ge has 4 valence electron.

On replacement an electron deficient hole was created, therefore it is a p-type semiconductor.


2) B belongs to group 13 with 3 valence electrons & Si belongs to group 14 with 4 valence electrons.

Therefore on replacement there will be a free electron & it is n-type semiconductor.



Question 1.21.

Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell.

What is the length of a side of the cell?


For a face-centred unit cell:

a= (2√2) r

It is given that the atomic radius, r = 0.144 nm


a= (2 x 0.144) nm

= 0.407 nm

Hence, length of a side of the cell = 0.407 nm




Question 1.22.

In terms of band theory, what is the difference:-

(i) Between a conductor and an insulator

(ii) Between a conductor and a semiconductor?



  • The valence band of a conductor is partially-filled or it overlaps with a higher energy, unoccupied conduction band.



On the other hand, in the case of an insulator, the valence band is fully- filled and there is a

large gap between the valence band and the conduction band.


  • In the case of a conductor, the valence band is partially-filled or it overlaps with a higher energy,
  • unoccupied conduction band. So, the electrons can flow easily under an applied electric field.


On the other hand, the valence band of a semiconductor is filled and there is a small gap

between the valence band and the next higher conduction band.

Therefore, some electrons can jump from the valence band to the conduction band and conduct electricity.



Question 1.23.

Explain the following terms with suitable examples:

(i) Schottky defect (ii) Frenkel defect (iii) Interstitials and (iv) F-centres.


  • Schottky defect: Schottky defect is basically a vacancy defect shown by ionic solids.
  • In this defect, an equal number of cations and anions are missing to maintain electrical neutrality.
  • It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids.
  • For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature.
  • Ionic substances containing similar sized cations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc.


  • Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of defect.
  • When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created.
  • It creates a vacancy defect as well as an interstitial defect.
  • Frenkel defect is also known as dislocation defect.
  • Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.


  • Interstitials: Interstitial defect is shown by non-ionic solids.
  • This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial site of the crystal.
  • The density of a substance increases because of this defect.


  • F-centres: When the anionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called F-centres.
  • These unpaired electrons impart colour to the crystals.
  • For example, when crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal.
  • The Cl ions diffuse from the crystal to its surface and combine with Na atoms, forming NaCl.
  • During this process, the Na atoms on the surface of the crystal lose electrons.
  • These released electrons diffuse into the crystal and occupy the vacant anionic sites, creating F-centres.



Question 1.24.

Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.

(i) What is the length of the side of the unit cell?

(ii) How many unit cells are there in 1.00 cm3 of aluminium?


(i) For cubic close-packed structure:

a= (2√2) r

= (2 x √2) x 125 pm

=353.55 pm

=354 pm (approx.)

 (ii)    Volume of one unit cell = (354 pm) 3

= 4.4 × 107 pm3

= 4.4 × 107 × 10−30 cm3

= 4.4 × 10−23 cm3

Therefore, number of unit cells in 1.00 cm3  

= (1.00 cm3)/ (4.4 x 10-23) cm

= 2.27 × 1022



Question 1.25.

If NaCl is doped with 10–3 mol % of SrCl2, what is the concentration of cation vacancies?


NaCl is doped with 10-3 mole % of SrCl2.This means, 100 moles of NaCl are doped with 10-3 moles of SrCl2.

Therefore, 1mole of NaCl is doped with (10-3) / (100) = 10-5 moles of SrCl2 or

10-5 moles of Sr2+


One Sr2+ ion create one cation vacancy.

Therefore, no. of cation vacancies created by 10-5 moles Sr2+

= (10-5 moles/mole of NaCl)

= (10-5 x 6.022 x 1023 mol-1) of NaCl

= 6.022 x 1018 mol-1 of NaCl




Question 1.26.

Explain the following with suitable examples:

(i) Ferromagnetism

(ii) Paramagnetism

(iii) Ferrimagnetism

(iv)  antiferromagnetism

(v) 12-16 and 13-15 group compounds.


(i)      Ferromagnetism: The substances that are strongly attracted by a magnetic field are called ferromagnetic substances.

Ferromagnetic substances can be permanently magnetised even in the absence of a magnetic field.

Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium, and CrO­2.

In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains and each domain acts as a tiny magnet.

In an un-magnetised piece of a ferromagnetic substance, the domains are randomly-oriented and so, their magnetic moments get cancelled.

However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field.

As a result, a strong magnetic effect is produced. This ordering of domains persists even after the removal of the magnetic field.

Thus, the ferromagnetic substance becomes a permanent magnet.


Schematic alignment of magnetic moments in ferromagnetic

(ii)     Paramagnetism: The substances that are attracted by a magnetic field are called paramagnetic substances.

Some examples of paramagnetic substances are O2, Cu2t, Fe3t, and Cr3t.

Paramagnetic substances get magnetised in a magnetic field in the same direction, but lose magnetism when the magnetic field is removed.

To undergo paramagnetism, a substance must have one or more unpaired electrons.

This is because the unpaired electrons are attracted by a magnetic field, thereby causing paramagnetism.

(iii)    Ferrimagnetism: The substances in which the magnetic moments of the domains are aligned in parallel and anti-parallel directions,

in unequal numbers, are said to have ferrimagnetism. Examples include Fe3O4 (magnetite), ferrites such as MgFe2O4 and ZnFe2O4.

Ferrimagnetic substances are weakly attracted by a magnetic field as compared to ferromagnetic substances.

On heating, these substances become paramagnetic.


Schematic alignment of magnetic moments in ferrimagnetism

(iv)    Antiferromagnetism: Antiferromagnetic substances have domain structures similar to ferromagnetic substances,

but are oppositely-oriented. The oppositely-oriented domains cancel out each other’s magnetic moments.

Schematic alignment of magnetic moments in antiferromagnetic


 (v)    12-16 and 13-15 group compounds: The 12-16 group compounds are prepared by combining group 12 and group 16

elements and the 13-15 group compounds are prepared by combining group 13 and group15 elements.

These compounds are prepared to stimulate average valence of four as in Ge or Si. Indium (III) antimonide (IrSb), aluminium phosphide (AlP),

and gallium arsenide (GaAS) are typical compounds of groups 13-15.

GaAs semiconductors have a very fast response time and have revolutionised the designing of semiconductor devices.

Examples of group 12-16 compounds include zinc sulphide (ZnS), cadmium sulphide (CdS), cadmium selenide (CdSe), and mercury (II) telluride (HgTe).

The bonds in these compounds are not perfectly covalent. The ionic character of the bonds depends on the electronegativities of the two elements.

Share this with your friends  

Download PDF

You can check our 5-step learning process