solid state bounceback.pdf

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Sakshi Vora
IIT Roorkee

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1. Crystalline
2. Amorphous
Types of Solids

1. Particles are arranged in a regular pattern.
2. They have flat surfaces and sharp edges.
3. They have a fixed value of Melting Point.
Crystalline Solids

4. They are called true solids.
5. Sodium Chloride, Copper Sulfate, ICE, Fe, Na, Graphite, Diamond, etc
6. They are anisotropic ie, different properties along different directions.
Crystalline Solids

1. Particles are arranged in a random fashion.
2. No flat surfaces or sharp edges
3. They have diffused (not fixed) melting point.
Amorphous Solids

4. They are called Pseudo Solids or Super cooled Liquids
5. Glass, Plastics, Rubbers.
6. They are isotropic ie, they have same physical properties in all directions
over a macroscopic length ie, over a larger length
Amorphous Solids

1. Molecular Solids
2. Ionic Solids
3. Covalent or Network Solids
4. Metallic Solids
Classification of crystalline solids

[Main Jan. 09, 2020 (I)]
A.
B.
C.
D.
Zinc sulphide
Mercury
Silicon carbide
Carbon tetrachloride

The smallest part of a crystal which upon its continuous repetition in 3D,
again forms the crystal.
Physical properties of the crystal are directly proportional to that of the unit
cell.
Mass of crystal ∝ Mass of unit cell
Volume of crystal ∝ Volume of unit cell
Density of crystal = Volume of unit cell
Unit Cell

Classification of Unit Cell : On the basis of shape

In known solids, only 7 unique shapes of unit cells are observed so far.
Cubic
Tetragonal
Orthorhombic
Monoclinic
Triclinic
Rhombohedral
Hexagonal
Except HEXAGONAL, all shapes have 6 faces. Hence, for HEXAGONAL (8
faces), some properties are different as compared to other unit cells.

CTOMaTRiH
Cubic a=b=c 0
Tetragonal a=b≠c 1
Orthorhombic a≠b≠c 2
Monoclinic a≠b≠c 2
Triclinic a≠b≠c 2
Rhombohedral a=b=c 0
Hexagonal a=b≠c 1
𝞪 = 𝞫 = 𝛄 = 90 Primitive, Body, face
𝞪 = 𝞫 = 𝛄 = 90 Primitive, body
𝞪 = 𝞫 = 𝛄 = 90 Primitive, body, face, end
𝞪 = 𝛄 = 90 𝞫 ≠ 90 Primitive, end
𝞪 ≠ 𝞫 ≠ 𝛄 ≠ 90 primitive
𝞪 = 𝞫 = 𝛄 ≠ 90 primitive
𝞪 = 𝞫 = 90 𝛄 = 120 primitive

C Cubic a=b=c 0 𝞪 = 𝞫 = 𝛄 = 90 Primitive, Body, face 3
T Tetragonal a=b≠c 1 𝞪 = 𝞫 = 𝛄 = 90 Primitive, body 2
O Orthorhombic a≠b≠c 2 𝞪 = 𝞫 = 𝛄 = 90 Primitive, body,
face, end
4
Ma Monoclinic a≠b≠c 2 𝞪 = 𝛄 = 90 𝞫 ≠ 90 Primitive, end 2
T Triclinic a≠b≠c 2 𝞪 ≠ 𝞫 ≠ 𝛄 ≠ 90 primitive 1
Ri Rhombohedral a=b=c 0 𝞪 = 𝞫 = 𝛄 ≠ 90 primitive 1
H Hexagonal a=b≠c 1 𝞪 = 𝞫 = 90 𝛄 = 120 primitive 1

[Main Jan. 10, 2019 (I)]
A.
B.
C.
D.
Triclinic
Hexagonal
Monoclinic
Tetragonal

1. Simple/Primitive
2. Face centered
3. End centered
4. Body centered
Unit Cell : On the basis of position of particles

This classification is for all shapes except hexagonal unit cell.
Simple/Primitive Particles are at corners only
Face centered At all corners + at body centres
End centered At all corners + any 2 opposite faces
Body centered At all corners + all face centres
Unit Cell : On the basis of position of particles

Regular geometric pattern of similar type of particles in 3D space.
In known solids, only 14 unique lattices are observed : Bravais lattices
Bravais Lattice

These are the locations where similar type of particles can be present.
In a unit cell, there are four types of lattice points
1. Corners
2. Face centres
3. Centre of body
4. Edge Centre
1. Everything here is excluding hexagonal unit cell.
2. At the lattice points, since the centre of the particle is situated, hence
the particle may be shared in multiple unit cells.
Lattice Points

1. Each corner atom is shared by 8 unit cells.
2. Each body centre atom is shared by just 1 unit cell.
3. Each face centre atom is shared by 2 unit cells.
4. Each edge centre atom is equally shared by 4 unit cells.
Lattice Points

If particles are at corners only
8*(⅛) = 1
If particles are at all corners and all face centres
8*(⅛) + 6*(½) = 4
If particles are at all corners and at body centre
8*(⅛) + 1 = 2
If particles are at all corners and at two opposite edges.
8*(⅛) + 2*(½) = 2
Effective atoms per unit cell (Z)

[Main April 12, 2019 (II)]
A.
B.
C.
D.
8 : 1 : 6
1 : 2 : 4
4 : 2 : 1
4 : 2 : 3

The total number of surrounding atoms touching a particular atom.
If Coordination number is x for 1 atom of the system, that means
coordination number will be x for all the atoms of that system. (if nothing
mentioned)
Coordination Number (2D)

The total number of surrounding atoms touching a particular atom.
If Coordination number is x for 1 atom of the system, that means
coordination number will be x for all the atoms of that system.
Coordination Number (2D)

Considering a cube of edge length a
1. Corners = 8
2. Faces = 6
3. Edges = 12
4. Volume = a3
5. Face Diagonals = 12
6. Length of face diagonal = √2 a
7. Body diagonals = 4
8. Length of Body diagonal = √3a
Analysis of Cubic System

1. Similar types of particles are present at all the corners of the cube in
such a manner so that each corner atom touches all its adjacent
corner atoms.
Simple Cubic Lattice

Relation between r and edge length

1. a = 2r
2. Z = 1
3. CNo. = 6
Simple Cubic unit cell

Similar type particles are present at all corners and at body centre in
such a manner so that the body centre atom touches all the corner atoms
but corner atoms will not touch each other.
Body centred Cubic unit cell or BCC

Body centred Cubic unit cell or BCC

1. √3a= 4r
2. Z = 2
3. CNo. = 8
There are two types of atoms in BCC unit cell
Corners and body centre
and both these atoms have
CNo. =8
BCC

Similar type particles are present at all corners and at centre of all faces in
such a manner so that each face centre atom touches all the corner atoms
but corner atoms will not touch each other.
Face centred Cubic Unit cell

1. √2a= 4r
2. Z = 4
CNo. = 12 ( 4+ 4 +4 )
There are two types of atoms in FCC unit cell
Corners and face centres
1. Atom at the corner does not touch other corner atoms but touches all
the atoms at the face centres. So CNo. = 12
2. Atom at the face centre touches 4 other corner atoms, 4 face centres of
left cube and 4 face centres of right cube.
FCC

CNo 12 wrt atom at face centre

It is the measure of the space occupied by the atoms in a unit cell.
PF = Part occupied by the atoms
Total part of the cell
Packing Fraction

For 1D
PF = length occupied by the atoms
Total length of the cell
For 2D
PF =area occupied by the atoms
Total area of the cell
For 3D
PF = volume occupied by the atoms
Total volume of the cell
Packing Fraction

PF = Volume occupied by the atoms
Packing Fraction for Simple cubic unit cell

Packing Fraction for face centred unit cell

Packing Fraction for body centred unit cell

1- packing fraction
Void Fraction

Density of crystal = Density of unit cell
Density of unit cell = mass of unit cell
Volume of unit cell
= mass of 1 atom * total number of atoms
Volume of unit cell
= A*Z
No * volume of unit cell
Density of Crystalline Substance

Density of crystal = Density of unit cell
Density of unit cell = mass of unit cell
Volume of unit cell
= mass of 1 atom * total number of atoms
Volume of unit cell
= A*Z
No * a3
Density of Cubic unit cell

A.
B.
C.
D.
40 NA
8 NA
4 NA
2 NA
[Main Sep. 05, 2020 (I)]

[Main Jan. 11, 2019 (I)]
A.
B.
C.
D.
0.0432 kg mol-1
0.0216 kg mol-1
0.0305 kg mol-1
0.4320 kg mol-1

[Main Jan. 9, 2019 (II)]
A.
B.
C.
D.

For a solid compound made of elements A,B and C, the empirical formula
may be written as
AZABZBCZC
ZA, ZB and ZC are the number of atoms per unit cell.
Formula of Solid Compound

Tell the formula of a compound which has A in BCC unit cell and B at all
edge centres
Practise Question

Tell the formula of a compound which has A in FCC unit cell and B is at
body centre and C is at half edge centres.
Practise Question

[Main Jan. 10, 2019 (II)]
A.
B.
C.
D.

The empty space between the closely packed similar type particles.
The type of void is given by the shape created by joining the centres of all
the atoms surrounding the void.
Voids

1- packing fraction
Voids in 2D

There are three types of voids
1. Cubic void
Formed in simple cubic lattice at the body centre position.
When the centres of all atoms surrounding the voids are joined, it forms a
cube.
1. Tetrahedral Voids
Whenever a triangular void is covered by an atom, a tetrahedral void is
formed.
1. Octahedral Voids (8 faces/ square bipyramidal voids)
Whenever a triangular void is covered by an another triangular void in
opposite orientation.
Voids in 3D

8 faces of octahedral void
Voids

If a lattice has Z as effective number of particles,
Then, Effectively per unit cell
Number of Octahedral Voids - Z
Number of Tetrahedral Voids = 2Z
Voids

In cubic lattices, only FCC has tetrahedral and Octahedral voids.
There are no tetrahedral and octahedral voids in Body centred cubic
unit cell.
Voids in cubic lattices

Tetrahedral voids:
There are 2 tetrahedral voids on each body diagonal at a distance of √3a/4
from each corner.
Location of voids in FCC unit cell

Tetrahedral voids:
When three face centred atoms touch each other, they form a triangular
void, the corner atom touches these 3 atoms thus forming a tetrahedral
void.

If all tetrahedral voids are joined, if forms a cube with the edge length a/2

Octahedral Voids:
Present at all edge centres and at the
body centre.
For the void at body centre, there are
two opposite triangles formed,
which on joining form octahedral void.

Octahedral Voids:
Present at all edge centres and at the
body centre.
For the void at edge centre,
two face centres and one corner will
form triangle, other two face centres
and corner will form another triangle,
once they overlap, they form
octahedral void.

At centre = 1
At edge centres = 12*¼ = 3
Total 1+3 = 4
Number of octahedral voids in FCC lattice

[Main Sep. 06, 2020 (II)]
A.
B.
C.
D.
+2, +4
+1, +3
+3, +1
+4, +2

Particles should be arranged in such a way that the packing fraction becomes
maximum.
1D Close Packing AAAA close packing/Linear close
packing
C No = 2
Close Packing in crystals

2D Close Packing
Formed by repetition of 1D layers in same or different alignment.
Square Close Packing/ AAA close packing
CNo. = 4 Void: Square planar

2D Close Packing
Formed by repetition of 1D layers in same or different alignment.
Hexagonal Close Packing/ ABAB close packing/
CNo. = 6 Void: TRIANGULAR

3D Close Packing
Three type of Close packing is present in the known solids.
1. Square Close packing (SCP)
2. Cubic Close Packing (CCP)
3. Hexagonal Close Packing (HCP)

Square Close Packing SCP in 3D/AAA Close
packing
CNo. 6 and Lattice: Primitive unit cell Void: Cubic void

CCP and HCP unit cell
Both are formed by repetition of 2D hexagonal layers in 3D

First Layer of atoms : Layer A
There are two types of triangular voids,
upper triangle c and lower triangle b

Second Layer of atoms : Layer B
● A second hexagonal layer this placed over Layer A in such a
manner that the c type (upper triangular) type of voids are covered
by the atoms of second layer.
● Since triangular void is covered by a 4th atom, this void turns out to
be a tetrahedral void.
● Because of this arrangement, the b (lower triangular) type voids of
layer A will be covered by the c (upper triangular) type voids of
layer B forming Octahedral Void.

Second Layer of atoms : Layer B

Third Layer of atoms (Layer B or Layer C)
If the third layer is Layer C : CCP
● A third 2D hexagonal layer is placed over the second layer in such a
manner that all the octahedral voids formed by layers A and B are
covered by the atoms of this third layer

CCP Unit cell

CCP Unit cell or ABCABC Close packing

Third Layer of atoms (Layer B or Layer C)
ABCABC close packing : CCP
1. Coordination number = 12
2. Voids: Octahedral and tetrahedral
3. Unit cell: Face centred Cubic Unit cell

Looking at this figure from two angles
Side view: ABC ABC arrangement
Top view: FCC unit cell

Third Layer of atoms (Layer B)
ABABAB close packing : HCP
1. A similar 2D layer is placed over layer B in same form as the layer A

HCP Unit cell
1. A similar 2D layer is placed over layer B in same form as the layer A
2. The type of packing is ABABAB close packing
3. Both Octahedral and tetrahedral voids are present
4. C.No. will be 12
5. Type of lattice will be HEXAGONAL LATTICE

1. No previous definitions are valid here.
2. Similar type of particles are present at all corners, at all hexagonal
faces and three particles within the body.
Hexagonal Lattice

Effective number of atoms per unit cell: 6
12*(⅙) + 2*(½) + 3 = 6
Octahedral Voids = 6
Tetrahedral voids = 12
Hexagonal Lattice

Relation between a and r
a = 2r
Height of unit cell
2a√(⅔)
Volume of hexagonal unit cell
Volume = height * area of hexagonal face
= 2a√(⅔)* 6*(area of eq triangle)
= 2a√(⅔)* 6*(√(¾)𝜋a2
= 3√2𝜋a3
Hexagonal Lattice

Packing fraction of hexagonal unit cell
Packing Fraction = Volume occupied/ volume available
= 6*volume of 1 atom/ 3√2a3
= 0.74
= 74%
Hexagonal Lattice

Hexagonal Lattice
In a compound atoms of element Y form ccp lattice and those of element
X occupy 2/3rd of tetrahedral voids. The formula of the compound would
be :
A.
B.
C.
D.
X2Y3
X2Y
X3Y4
X4Y3

The type of void occupied by the smaller ion will be decided by Radius
Ratio.
RR = Radius of smaller ion
Radius of larger ion
RR can vary from 0 to 1
Radius Ratio

RR Type of void Coordination Number
RR<0.155 Linear Void 2
0.155≤RR<0.225 Triangular Void 3
0.225≤RR<0.414 Tetrahedral 4
0.414≤RR<0.732 Square Planar (2D) 4
Octahedral (3D) 6
0.732≤RR<1 Cubic 8
Radius Ratio

The situation is called IDEAL SITUATION.
Here the cation will touch the anions and also anions will touch each others.
Case 1: If RR is 0.155: Perfect Triangular void

Here the cation will touch the anions and but anions will not touch each
other.
● This case is valid for both the situations:
● the size of anion is kept the same but the size of cation increases.
● The size of cation is kept the same but the size of anion decreases then also
the anions won't touch each other.
Case 2: If RR is 0.20: Imperfect Triangular void

Whatever the case may be, ideal or non ideal / perfect or imperfect
Remember,
The cation will always touch the anions
Most important

If no other data is given in the question, we need to consider the
limiting values for the voids:
Radius of tetrahedral void = 0.225*radius of lattice
Radius of octahedral void = 0.414*radius of lattice
Limiting cases

[Adv.2013]
A.
B.
C.
D.
104 pm
183 pm
125 pm
57 pm

PYQ JEE Adv 2020
The cubic unit cell structure of a compound containing cation M and anion X
is shown below. When compared to the anion, the cation has smaller ionic
radius. Choose the correct statement (s)
A.
B.
C.
D.
The empirical formula of the compound is MX
The cation M and anion X have different
coordination geometries
The ratio of M-X bond length to the cubic unit cell
edge length is 0.866.
The ratio of the ionic radii of cation M to anion X is 0.414.
JEE Adv 2020

1. For a crystalline substance at 0K, entropy = 0
2. There are no defects in a crystalline substance at 0K
3. As the temperature increases, the defects arise in a crystal.
1. Point Defects: Imperfections at few random lattice points in
the crystal
Line Defects: Imperfections in a row of large number of lattice
points. Ie. when so many point defects are seen together, it is
seen as a line defect.
Plane defect: Imperfection in a plane of a large number of
lattice points. I.e, so many line defects are seen together
Defects or imperfection in Crystals

They are of 2 types
1. Stoichiometric Defect: The formula of compound does not change
2. Non Stoichiometric defect: The formula of the compound changes.
Point Defect

1. Pairs of cations and anions are missing from the crystal.
Density of crystal decreases.
2. Generally shown by compounds having high Coordination
Number.
3. Schottky defect = Theoretical density - observed density
4. For example, NaCl, KCl, CsCl and AgBr.
Schottky Defect: Stoichiometric defect

1. Some ions will leave their original positions and will be present at
different positions in the crystal.
2. The formula of the compound remains same.
3. The density of the compound remains same.
4. Shown by compounds having large difference in ionic radii.
5. for example, ZnS, AgCl, AgBr and AgI due to small size of Zn2+
and Ag+ ions
Frenkel Defect: Stoichiometric defect

1. Some anions will be missing from the crystal creating anion
vacancies.
2. These vacant sites will be occupied by electrons called F centres,
which are responsible for the colour of the system
3. Formula of the compound changes but electrical neutrality is
same.
4. Density decreases
5. Good conductors of heat and electricity.
Non Stoichiometric defect: Metal Excess defect by anion
vacancy

1. Alkali halides like NaCl and KCl show this type of defect.
2. When crystals of NaCl are heated in an atmosphere of sodium
vapour, the sodium atoms are deposited on the surface of the
crystal.
3. The Cl– ions diffuse to the surface of the crystal and combine with
Na atoms to give NaCl. This happens by loss of electron by sodium
atoms to form Na+ ions
vacancy

1. Some extra cations are present in the interstitial sites of the crystal.
2. Electrons are present in the other sites to maintain electrical neutrality.
3. Zinc oxide is white in colour at room temperature. On heating it loses
oxygen and turns yellow
Now there is excess of zinc in the crystal and its formula becomes Zn1+xO.
The excess Zn2+ ions move to interstitial sites and the electrons to
neighbouring interstitial sites.
vacancy

1. There are many solids which are difficult to prepare in the
stoichiometric composition and contain less amount of the metal as
compared to the stoichiometric proportion.
2. A typical example of this type is FeO which is mostly found with a
composition of Fe0.95O.
3. It may actually range from Fe0.93O to Fe0.96O.
4. In crystals of FeO some Fe2+ cations are missing and the loss of
positive charge is made up by the presence of required number of
Fe3+ ions
Non Stoichiometric defect: Metal Deficiency defect by
cation vacancy

1. It is due to doping of a small amount of substance to a crystalline
substance.
2. If molten NaCl containing a little amount of SrCl2(cations now should
be of greeter charge) is crystallised, some of the sites of Na+ ions are
occupied by Sr2+
3. Each Sr2+ replaces two Na+ ions. It occupies the site of one ion and
the other site remains vacant. The cationic vacancies thus produced
are equal in number to that of Sr2+ ions.
4. Another similar example is the solid solution of CdCl2 and AgCl.
Impurity defect

[Main Jan. 08, 2020 (II)]
A.
B.
C.
D.
AgBr
CsCl
KBr
ZnS

[2018
]
A.
B.
C.
D.
Schottky defect
Vacancy defect
Frenkel defect
Metal deficiency defect

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