MATTER: senses . E.g. solids , liquids, gases are forms of ...shsmoga.com/rwData/doc/23032015_122728States_of_matter_solids_.… · Anything which occupies space , has volume and

Class XII Chemistry states of matter [SOLID STATE]

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MATTER:

Anything which occupies space , has volume and can be perceived by our senses . E.g. solids , liquids, gases are forms of matter

Other states of matter are Plasma, BEC (Bose Einstein condensate).

PHYSICAL PROPERTIES OF SOLIDS LIQUIDS AND GASES: Sr.

No

.

PROPERTIES Related law

or formula

SOLIDS LIQUIDS GASES

1 Mass Units (kg) definite Definite Definite

2 Shape Area : Units (m2)

definite Acquires the shape of the

container

Acquires the shape of the

container

3 Volume Units (m3 ) definite Definite indefinite

4 Compressibility Factor

Z= PV nRT

Not possible Almost negligible

Highly compressible

5 Fluidity Viscocity (μ)

Not possible Can flow Can flow

6 Rigidity Strength Highly rigid Less rigid Not rigid

7 Diffusion r 1 √d

Slow Fast Very fast

8 No. of free

surfaces

Free sites Any no. of free

surfaces

Only one free

surface

None

9 Density Mass =

volume units kg/m3

High Slightly lower Very low

10 Packing of particles

Measure of

strength/den

sity

Most closely packed

Less closely packed

Least closely packed.

11 Inter particle force

Spacing

between

molecules.

Strongest: cohesive

forces> thermal energy

Slightly weaker than

in solids.

Negligible, thermal

energy >cohesive forces

12 Expansion on heating

vaporisation (KJ/mol)

Less More than solids

Most

13 Motion of constituent

particles

Energy Joule (J)

Oscillatory Translatory Translator

14 Kinetic

energy of particles

Energy (J)

= 1mu2

2

Least Large Very large

SOLID: Solid is a form of matter in which the constituting particles are arranged very closely. The constituent particles can be atoms, molecules or ions.

PROPERTIES OF SOLIDS: a. Definite mass, volume and shape.

b. Short Intermolecular distances , hence strong intermolecular forces (F 1/r2) .

c. Atoms/ ions occupies fixed positions and can only oscillate/vibrate about their mean

positions due to low thermal energy. Thermal energy temperature d. Closely packed, therefore low compressibility and rigid.


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CLASSIFICATION OF SOLIDS ON BASIS OF ARRANGEMENT OF ATOMS/IONS:

a. Crystalline solids: definite arrangement (regular) of atoms/ ions e.g. metals (iron,

copper), non metals (diamond, graphite) all elements, compounds, alum, sugar. Majority of solids are crystalline in nature.

b. Amorphous solids: no definite arrangement (irregular) of atoms / ions e.g. Glass, plastics, rubber, fused silica, tar, pitch, high molecular weight polymers.

PROPERTIES OF CRYSTALLINE SOLIDS:

a. Definite geometrical shape or regular geometry. b. A long range order, i.e. regular arrangement .

c. They have a sharp melting point. d. They are anisotropic in nature

i.e. their physical properties show different values when measured along different directions

in the same crystal. e. Clean cleavage i.e. when cut with a sharp edged tool,

they split into two pieces and the newly generated surfaces are plain and smooth.

f. Definite and characteristic heat of fusion g. Also called true solids.

Anisotropy: In case of crystalline substances, properties like electrical properties, refractive index, thermal expansion etc are having different values in different

directions.

POLYMORPHIC FORMS OR POLYMORPHS: The different crystalline forms of a substance are known as polymorphic forms or polymorphs. E.G. CaCO3 exists as Calcite and Aragonite.

NOTE: for elements it is called allotropy for example: graphite and diamond are

allotropes of Carbon.

ISOMORPHISM: When two or more crystalline solids have similar chemical composition that exists in

same crystalline form or structure e.g. Na3PO4, Na3AsO4

ANNEALING: regular heating and cooling a glass of glass make the glass milky.

CHARACTERISTICS OF AMORPHOUS SOLIDS:

a. An irregular shape or no regular arrangement of particles. b. A short range order i.e. regular arrangement of particle in small space. c. Do not have definite heat of fusion. Gradually soften/ melt/ fuses over a range of

temperature. (not fixed) d. Isotropic in nature i.e. their physical properties are the same in all directions.

e. Irregular cleavage, e.g. when cut with a sharp edged tool, they cut into two pieces with irregular surfaces. f. Also called as pseudo solids or super cooled liquids. This is because they have a

tendency to flow, though very slowly. h. Also called pseudo solids (highly super cooled liquid of very high viscocity).


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TYPES OF CRYSTALLINE SOLIDS:

Type of Solid

Constituent Particles

Bonding/ Attractive Forces

Examples Physical Nature (hardness / brittleness)

Electrical conductivity

Melting point

Solubility

1. Ionic solids

+ve and –ve Ions

Coulombic or electrostatic force of

attraction

NaCl, MgO, ZnS, CaF2

Hard but brittle

Insulators in solid state but conductors

in molten state and in aqueous solutions

High Soluble in polar and insoluble in non polar

2. Molecular solids

a. Non polar

Molecules Dispersion or London forces

Ar, CCl4, H2, I2, CO2

Soft Insulator Very low

Soluble as well as insoluble in both

b. Polar Molecules Dipole dipole Interactions

HCl, solid SO2, solid NH3

Soft Insulator Low Soluble in polar

c. Hydrogen

bonded

Molecules Hydrogen Bonding

H2O (ice)

Hard Insulator Low Soluble in polar

3. Covalent or network solids

Atoms/ molecule

Covalent Bonding

SiO2

(quartz), SiC, C (diamond), AlN C(graphite)

Hard Soft

Insulators Conductor (exception)

Very High

Insoluble in polar and usually soluble in non polar solvents

4. Metallic solids

Positive ions (+) in a sea of

delocalised electrons (-)

Metallic Bonding (electrostatic

attraction between cations and sea of electrons)

Fe, Cu, Ag, Mg

Hard but malleable and

ductile

Conductors in solid state as

well as in molten state

Fairly High

Insoluble in both

V.IMP. Q. DIFFERENCE BETWEEN ANY TWO CAN BE ASKED IN BOARD EXAMS.

V.IMP Q. DIFFERNENCE BETWEEN CRYSTALLINE AND AMORPHOUS SOLIDS? (2)


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PROPERTY CRYSTALLINE SOLID AMORPHOUS SOLIDS

Definition Definite arrangement

(regular) of atoms/ ions.

No definite arrangement

(irregular) of atoms / ions.

Shape They have definite

shape and regular geometrical form.

They do not have definite

shape and regular geometrical form.

Melting Point They have sharp (definite) melting point.

They melt over wide range of temperature.

Compressibility They are rigid and incompressible.

They too are usually rigid and cannot be compressed to any appreciable extent.

However graphite is soft because of its unusual

structure.

Cutting with a sharp

edged tool

They give clean

cleavage, i.e. they break into two pieces with plane surfaces.

They give irregular

cleavage, i.e. they break into two pieces with irregular surfaces.

Heat of fusion They have definite heat of fusion.

They do not have a definite heat of fusion.

Isotropic/ anisotropic

Anisotropic i.e. their mechanical and electrical

properties depend on the direction along which

they are measured.

Isotropic i.e. they have similar physical properties in

all directions because the constituents are arranged in

random manner.

Example All elements and

compounds Cu, Ni, NaCl etc

Rubber , glass etc.

NOTE:

Any material can be made amorphous or glassy either by rapidly cooling its melt or freezing its vapours. E.g. silica (SiO2) crystallises as quartz in which SiO4 tetrahedra are

linked in a regular manner but on melting and then rapid cooling, it gives glass in which SiO4 tetrahedra are randomly joined to each other. thus quartz is crystalline SiO2 whereas silica glass is amorphous solids SiO2.

CRYSTAL LATTICE/ SPACE LATTICE: Space lattice is a regular repeating arrangement of points in space.


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(a) Two dimensional lattices:

Regular arrangement of atoms in the plane of paper.

Sr. No

Lattice Unit cell

1 Square lattice Square

2 Rectangular lattice Rectangle

3 Parallelogram lattice Parallelogram

4 Rhombic lattice Rectangular with interior point

5 Hexagonal lattice Rhombus with an angle of 600

(b) 3dimensional crystal lattice: A regular orderly arrangement of constituent

particles (atoms/ ions/molecules) in three dimensional space. LATTICE POINTS OR LATTICE SITES:

The fixed positions on which the constituent particles (atoms/ ions / molecules) are present are called lattice points or lattice sites.

SCC: lattice points: 8 , BCC: 9, FCC = 14 CRYSTAL LATTICE:

A group of lattice points which when repeated over and over again in 3 dimensions give the complete crystal lattice.

UNIT CELL: Smallest repeating unit in space lattice which

when repeated over and over again generates the complete crystal lattice.

The crystal can consist of an infinite number of unit cells.

V.IMP Q. DEFINE UNIT CELL? CRYSTAL LATTICE? LATTICE POINTS, SPACE

LATTICE (1)?

PARAMETERS WHICH CHARACTERIZE A UNIT CELL:

a. Dimensions of the unit cell along the three edges, a, b and c: These edges may or may not be mutually

perpendicular (900). b. Inclination (angle) of the edges to each other:

This is denoted by the angle between the edges , ,

and respectively. is the angle between the edges b

and c, is the angle between the edges a and c, and is the angle between a and b.

TYPES OF CRYSTAL SYSTEMS:


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S. n

o

System Maximum

symmetry

elements

Axes and angles

Examples

1 Cubic or

regular (most symmetrical)

9 planes

13 axes 1 centre

= = = 90° ,

a = b = c

NaCl, Cu, KCl, CaF2, ZnS, Cu2O, Diamond,

Alums, Pb, Ag,, Au, Hg.

2 Tetragonal 5 planes 5 axes

= = = 90° ;

a = b c

SnO2,ZnO2,TiO2,CaSO4,NiSO4,ZrSiO4,PbWO4

,White Sn.

3 Hexagonal: 7 planes 7 axes

= = 90°,

=120°;

a = b c

ZnO, PbI2,CdS, AgI, cinnabar( HgS),

Graphite, Ice, Beryl, Mg, Zn, Cd

4 Rhombohedr

al or trigonal:

7 planes

7 axes = =

90°;

a = b = c

NaNO3, CaSO4, Calcite ,ICI, Quartz, As, Sb,

Bi

5 Orthorhombi

c Or Rhombic

3 plane

3 axis = = =

90°;

a b c

KNO3, K2SO4, PbCO3, BaSO4,CaCO3,

Rhombic Sulphur, MgSO4.7H2O

6 Monoclinic: 1 plane

1 axis == 90°,

90°;

a b c

Na2SO4.10H2O, Na2B4O7.10H2O,

CaSO4.2H2O, Monoclinic Sulphur

7 Triclinic (most un

symmetrical)

No plane No axis

90°

; a b c

CuSO4.5H2O, K2Cr2O7, H3BO3

TYPES OF UNIT CELLS:

a. Primitive or simple unit cells:


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Have constituent particles only at its corners.

b. Centred unit cells are those unit cells: in which one or more constituent particles

are present at positions in addition to those present at the corners.

The centred unit cells are of three types:

i. Face centred unit cell: A face centred unit cell consists of one constituent particle present at the centre of each face in addition to those present at the corners.

ii. Body centred unit cell:

A body centred unit cell consists of a one constituent particle is present at its body centre in addition to those present at the corners.

iii. End centred unit cell: An end centred unit cells consists of one constituent particle present at the centre of any

two opposite faces in addition to those present at the corners.

CONTRIBUTION OF PARTICLES AT DIFFERENT LATTICE POSITIONS:

a. Corner: If an atom is present at any one corner, it is shared by eight unit cells. So, only one eighth of an atom actually belongs to the unit cell. CONTRIBUTION =1/8

b. Face centre: If an atom is present at the centre of the face, it is shared by two unit

cells. So, only half of the atom actually belongs to the unit cell. CONTRIBUTION =1/2

c. Body centre: If an atom is present at the body centre, it is not shared by any other unit cell. So, that one atom completely belongs to the same unit cell.

CONTRIBUTION =1 d. Edge centre: If an atom is present at the edge centre, it is shared by four unit cells. So, only one fourth of an atom belongs to the unit cell.

CONTRIBUTION =1/4 TOTAL NUMBER OF ATOMS IN DIFFERENT UNIT CELLS FOR ONE UNIT CELL:

a. Primitive unit cell (SCC): 1 atom (8 x 1/8) =1 b. Face centred unit cell (FCC): 4 atoms (8 x 1/8+ 6 x ½) =4

c. Body centred unit cell (BCC): 2 atoms (8 x 1/8+ 1) = 2


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Coordination number is the number of nearest neighbours of any particle.

CN for simple cubic (SCC) =6, BCC =8, FCC (hcp or ccp) = 12.

Q1. A compound formed by elements A and B has a cubic structure in which A

atoms are at the corners of the cube and B atoms are at the face centres. Derive the formula of the compound? Ans. As A atoms are present at the 8 corners of the cube, therefore, number no. of A

atoms in a unit cell = 1/8 x 8 =1 As B atoms are present at the face centres of the 6 face of the cube, therefore , no. of B

in the unit cell: 1/2x6 = 3 Therefore ratio of atoms A:B = 1:3 Hence the formula is AB3.

Q2. a cubic solid is made up of two elements X and Y. Atoms Y are present at

the corners of the cube and atoms x at the body centre. What is the formula of a compound? What are the coordination number of X and Y? Ans. As atoms Y are present at the 8 corners of the cube, therefore no of atoms of Y in

the unit cell = 1/8 x 8 =1 As atoms X are present at the body centre, therefore no of atoms of X, in the unit cell

=1, Therefore ratio of atoms X:Y ::1:1 Hence, the formula of the compound is XY. Coordination no of each of X and Y is 8

DO ASSIGNMENT NO:1 (GIVEN SEPARATELY)

CLOSE PACKING IN CRYSTAL STRUCTURES:

a. Close packing in one dimension: Each sphere is in contact with two of its neighbours. Coordination number is two.

b. Close packing in two dimensions: It is generated by stacking the rows of close

packed spheres in two ways:

i. Square close packing: AAA ----- TYPE When the spheres of the second row are placed exactly above those of the first row. This

way the spheres are aligned horizontally as well as vertically. The arrangement is AAA type. Coordination number is 4.


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ii. Hexagonal close packing: ABAB----type

When the spheres of the second row are placed above the first one in a staggered manner in such a way that its spheres fit in the depression of the first row. The

arrangement is ABAB type. Coordination number is 6. Space occupied 74% , empty space =26%

c. Close packing in three dimensions: They can be obtained by stacking the two dimensional layers one above the other. It can

be obtained in two ways: i. Three dimensional close packing from two dimensional square close packed

layers: AAAA --- pattern Here, The spheres of the upper layer are placed

exactly over the first layer such the spheres of the layers are perfectly aligned horizontally

and vertically. It has a AAAA.. type pattern. The lattice is simple cubic lattice.

ii. Three dimensional close packing from two dimensional hexagonal close

packed layers: There are two steps involved as: i. Placing the second layer over the first layer: if a two dimensional layer is considered as

A, the second layer which is placed above the first layer in such a way that the spheres of the second layer (considered as B) are placed in the depressions of the first layer.

This gives rise to two types of voids: tetrahedral voids and octahedral voids. ii. Placing the third layer over the second layer: There are two possibilities:

TYPES OF VOIDS: a. Tetrahedral voids: ABAB-- pattern

This type of void is formed at the centre when four spheres are joined in the form of a tetrahedron. Called as triangular void.

Radius of tetrahedral voids r = 0.225 R Where R is radius of sphere in close packing For cations in the voids and anions in the packing r+ = 0.225r-


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b. Octahedral void: ABCABC-- pattern This type of void is surrounded by six spheres. Double triangular void surrounded by six spheres is an octahedral voids.

Radius of octahedral voids r = 0.414 R where R is radius of sphere in close packing

For cations in the voids and anions in the packing r+ = 0.414r-

a. Covering the tetrahedral voids (hexagonal close packing): Here, tetrahedral voids of the second layer may be covered by the spheres of the third

layer. It gives rise to ABABAB… type pattern. The three dimensional structure is called hexagonal close packed structure. Coordination number is 12. Example: Mg, Zn


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b. Covering the octahedral voids (cubic close packing):

Here, octahedral voids of the second layer may be covered by the spheres of the third

layer. It gives rise to ABCABCABC… type pattern. The three dimensional structure is

called cubic close packed structure or face centred cubic structure.

Coordination number is 12. Example: Cu, Ag

Cubic close packing ccp: ABCABCABC type space occupied 74% empty 26%

Body centred cubic packing BCC space occupied 68% empty 32% Scc packing space occupied 52.4%

In hcp (ABAB pattern) or ccp (ABCABC pattern) arrangement, octahedral and tetrahedral voids are present. The number of octahedral voids present in

a lattice is equal to the number of close packed particles. The number of tetrahedral voids is twice the number of octahedral voids.

Example: If the number of close packed particles = n Number of particles present in octahedral voids = n

Number of particles present in tetrahedral voids = 2n

NO OF VOIDS FILLED AND FORMULA OF A COMPOUND:

No .of octahedral voids = n, where n is no. of particles present in close packing

Effective no of octahedral voids in ccp structure

One octahedral void is present at the body centre, 12 octahedral voids are present on the centre of 12 edges of the cube = 1+ 1/12 x ¼ =1+3 =4 Note: Each edge centre is shared by 4 unit cell therefore contribution to one is ¼ for edge centre voids.


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No .of tetrahedral voids = 2 x n where n is no. of octahedral voids In ccp, total no. of voids per unit cell = 4 (octahedral) + 8 (tetrahedral) =12 In hcp, total no. of voids per unit cell = 6 (octahedral) + 12 (tetrahedral) =12 Assuming n particles of B are present in the packing and 1/3rd of octahedral voids are occupied by particles A, then ratio of A:B = n/3 : n :: 1/3: 1 Hence formula is A3B.

SIZE OF VOIDS: Octahedral void: For an atom to occupy an octahedral void, its radius must be 0.414

times the radius of the sphere. r = 0.414 R Tetrahedral void: For an atom to occupy a tetrahedral void, its radius must be 0.225

times the radius of the sphere. (no. of tetrahedral voids= double the no. of spheres)

r = 0.225R TRIGONAL VOIDS: Radius of trigonal void is r = 0.155R where r is radius of spherical trigonal site, R is

radius of closely packed sphere.

CUBIC VOIDS: r= 0.732 R

Derivation of radius of voids as per class lecture.

Imp. DIFFERENCE BETWEEN TETRAHEDRAL VOIDS AND OCTAHEDRAL VOIDS:

Sr. No Tetrahedral voids Octahedral voids

1 r = 0.225R r = 0.414 R

2 Much smaller than the size of the sphere in the packing

Smaller than the size of the sphere in the packing but larger than the

tetrahedral voids.

3 Coordination no.4 Coordination no.6


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4 In hcp, ccp packing, each sphere is in contact with 3 spheres in the layer below and three above it,

thus forming one tetrahedral void. Hence there are 2 tetrahedral

voids per sphere. i.e. no of tetrahedral voids is double the no.

of spheres in the packing.

As octahedral voids is a combination of two voids of the two layer, no of octahedral voids is

equal to half the no of tetrahedral voids and hence equal to no of

spheres in the packing.

No .of tetrahedral voids = 2 x no. of octahedral voids

No .of octahedral voids = no. of particles present in close packing

Coordination number is the number of nearest neighbours of any particle.

CN for simple cubic =6, Bcc =8, fcc (hcp, ccp) = 12.

PACKING EFFICIENCY: It is the percentage of total space occupied by constituent

particles (atoms, molecules or ions)

Packing efficiency= Volume occupied by spheres x 100% Total volume of unit cell

a. Packing efficiency for face centred unit cell = 74% b. Packing efficiency for body centred cubic unit cell = 68% c. Packing efficiency for simple cubic unit cell = 52.4%


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Derivations: As per class lecture. Q3. Formula of a Compound and Number of Voids Filled A compound is formed

by two elements X and Y. Atoms of the element Y (as anions) make ccp and those of the element X (as cations) occupy all the octahedral voids. What is the

formula of the compound? Ans The ccp lattice is formed by the element Y. The number of octahedral voids

generated would be equal to the number of atoms of Y present in it. Since all the octahedral voids are occupied by the atoms of X, their number would also be equal to

that of the element Y. Thus, the atoms of elements X and Y are present in equal numbers or 1:1 ratio. Therefore, the formula of the compound is XY.

Q4. Atoms of element B form hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the

elements A and B?

Ans The number of tetrahedral voids formed is equal to twice the number of atoms of element B and only 2/3rd of these are occupied by the atoms of element A. Hence the ratio of the number of atoms of A and B is 2 × (2/3):1

or 4:3 and the formula of the compound is A4B3.

Q5. In a crystalline solid, anions B are arranged in a cubic close packing. Cations A are equally distributed between octahedral voids and

tetrahedral voids. If all the octahedral voids are occupied, what is the formula of the solid?

Ans. Suppose the no of anions B =n

Then no. of octahedral voids = n No of tetrahedral voids =2n

As octahedral voids and tetrahedral voids are equally occupied by cations A and

all the octahedral voids are occupied (given), therefore n cations A are present in octahedral voids and n cations A are present in tetrahedral voids. In other words

, corresponding to n anions B, there are n+n= 2n cations A. Thus cations A and anions B are in ratio of 2n : n :: 2:1

Thus formula A2B Q6. In the mineral, spinel, having the formula MgAl2O4, oxide ions are

arranged in the cubic close packing, Mg+2 ions occupy the tetrahedral voids while Al+3 ions occupy the octahedral voids.

(i) what %age of tetrahedral voids is occupied by Mg+2 ions? (ii) what %age of octahedral voids is occupied by Al+3 ions?

Ans According to the formula MgAl2O4:

If there are 4 oxide ions, there will be 1 Mg+2 ions and 2Al+3 ions. But if thw 4O2- ions are in ccp arrangement , there will be 4 octahedral voids and 8 tetrahedral

voids.

Thus 1Mg+2 ions is present in one of the 8 tetrahedral voids. Therefore %age tetrahedral voids occupied by Mg+2 = 1/8 x 100= 12.5%

Similarly , 2Al+3 ions are present in two octahedral voids out of 4 available.


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Therefore %age of octahedral voids occupied by Al+3 ions = 2/4 x100= 50%

SUMMARY OF STRUCTURE AND PACKING OF SOLIDS:

Property Hexagonal

close packing

Cubic close

packing

Body centred

cubic (bcc)

Arrangement of

packing

Close packed Closed packed Not close packed

Type of packing ABABAB.. ABCABCA... AB AB AB A

Available space

occupied

74% 74% 68%

Coordination number

12 12 8

Malleable and ductility

Less malleable , hard and brittle

Malleable and ductile

Examples Be, Mg, Ca, Cr,

Mo, V, Zn

Cu, Ag, Au, Pt Alkali metals, Fe

RADIUS RATIO:

Ratio of radius of cation to that of anion is called radius ratio.

Radius ratio = Radius of cation (r+) Radius of anion (r-) Larger is the radius ratio, larger is the size of cation and hence greater is its coordination

no.

Coordination

number

Radius

ratio

Geometrical shape i.e.

position of anions around cations

Examples

3 0.155-0.225 Planar triangle B2O3

4 0.225- 0.414 Tetrahedral ZnS (sphalerite), CuCl, CuI, BaS, HgS

6 0.414- 0.732 Octahedral NaCl (rock salt ) NaBr, KBr, MgO, MnO, CaO, CaS, NH4Br

8 0.732-1.00 Body centred cube CsCl (caesium chloride), CsBr, TlBr, TlCl

Density of a unit cell is same as the density of the substance. RADIUS RATIO in an octahedral void:

For an atom to occupy an octahedral void, its radius must be 0.414 times the radius of the sphere. r = 0.414 R

Radius ratio for tetrahedral void: For an atom to occupy a tetrahedral void, its radius must be 0.225 times the radius of

the sphere. (no. of tetrahedral voids= double the no. of spheres) r = 0.225R TRIGONAL VOIDS: radius ratio of trigonal void is r = 0.155R r is radius of spherical trigonal site, R is radius of closely packed sphere.

CUBIC VOIDS: r= 0.732 R

DECREASING ORDER OF SIZE OF VARIOUS VOIDS IS: Cubic> octahedral> tetrahedral > trigonal


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Q7. A solid A+B- has NaCl type close packed structure. If the anion has a radius

of 241.5pm, what should be the ideal radius of the cation? can a cation C+ having a radius of 50pm e fitted into the tetrahedral hole of the crystal A+B-?

Solution : As A+B- has NaCl structure, A+ ions will be present in the octahedral voids. Ideal radius of the cation will be equal to the radius of octahedral void because in that

case, it will touch the anions and the arrangement will be close packed. Hence Radius of octahedral void = rA

+ = 0.414rB- = 0.414x 241.5pm = 100pm

Radius of tetrahedral void = 0.225x rB- = 0.225 x241.5pm = 54.3pm

As the radius of the cation C+ (50pm) is smaller than the size of the tetrahedral hole, it

can be placed into the tetrahedral void (but not exactly fit into it).

Q8. the two ions A+ and B- ions have radii 88 and 200pm respectively.

In the close packed structure of compound AB, predict the coordination no of A+ ?

Ans.

r+ = r(A+) = 88pm = 0.44 r- r(A-) 200pm

it lies in the range of 0.414-0.732 hence coordination no of A+ is 6 RELATIONSHIP BETWEEN RADIUS OF CONSTITUENT PARTICLE (R) AND EDGE LENGTH (A) AND NEAREST NEIGHBOUR DISTANCE(d):

a. Simple cubic unit cell: a= 2r (or r= a/2) d =2r

b. Face centred unit cell: a 2 √2 r (or r= a/2√2) d = 2r c. Body centred unit cell: a = 4r/√3 (or r= √3a/4) d = 2r Volume of a unit cell = (edge length)3

= a3

a. Simple cubic unit cell: Volume= (2r)3

b. Face centred unit cell: Volume(2√2 r)3

c. Body centred unit cell: Volume=( 4r√3 )3

Number of atoms in a unit cell (z):

a. Simple cubic unit cell: z = 1 (8 x 1/8)

b. Face centred unit cell: z = 4 (8 x 1/8+ 6 x ½)

c. Body centred unit cell: z = 2 (8 x 1/8+ 1)


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SUMMARY OF UNIT CELLS AND THEIR PROPERTIES:

Property SCC BCC FCC

No of atoms per

unit cell (Z)

1 2 4

Packing efficiency

52.4% 68% 74%

Relation between r and

(a) edge length

r= a/2 r= √3a/4 r= a 2√2

Nearest neighbour

distance d=2r

d=2r= a d=2r= 2 x √3a 4

d=2r = 2 r= 2xa 2√2

Volume (a3) (2r)3

(2√2 r)3

( 4r√3 )3

Coordination

number

6 8 12 (hcp, ccp)

examples Mn BCC (alkali

metals)

NaCl ionic

compound,ZnS DENSITY OF UNIT CELL:

Density of unit cell ()g/cm3 = Z M

a3Na Z no. of formula units in the unit cell, (1 for sc, 4 for fcc, and 2 for bcc)

M is atomic mass of the element and a is length of the unit cell in pm

for cubic crystal of ionic compound , Z= No. of formula units per unit cell,

M = formula mass Remember, for ionic compounds, A+B- having fcc structure

(e.g. Na+Cl-) Edge (a) = 2 x distance between A+ and B- ions. X RAY DIFFRACTION AND CRYSTAL STRUCURE:

Brag equation : nλ = 2dsinθ where n =1,2,3,4,------

d/λ = n/ 2sinθ

Where: λ is wavelength, d is distance between the successive atomic planes and the angle of reflection (θ).

SUMMARY OF STRUCTURE OF METALS

Crystal structure Arrangement of ions Coordination no

Formula units

AB type Rock salt

Cl- = fcc arrangement Na+ ions = edge centres

and body centre

Na+ = 4

Cl- = 4

4

Caesium chloride Bcc arrangement Cl- = corners

Cs+ = body centre

Cs+ = 8 S2- = 4

1


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Zinc blende (ZnS) type

Ccp arrangement S2- = fcc ,

Zn+2=in tetrahedral voids

Zn+2 = 4 S2- = 4

4

AB2 type

(i) Fluorite (CaF2) type

Ccp arrangement

Ca+2 =fcc, F- = all tetrahedral voids

Ca+2 =8

F- = 4

4

(ii) Antifluorite

(A2B)

B2- = ccp , A+ = half

tetrahedral voids

Na+ = 4

O2- = 8

4

EFFECT OF TEMPERATURE ON CRYSTAL STRUCTURE: Increase of pressure increase coordination no

While increase of temperature decreases coordination no.

NaCl Structure High pressure CsCl (6:6 coordination) 760K (8: 8 coordination) Q9. An element has a body-centred cubic (bcc) structure with a cell edge of 288

pm. The density of the element is 7.2 g/cm3. How many atoms are present in 208 g of the element? Solution: Volume of the unit cell = (288 pm)3

= (288×10-12 m)3 = (288×10-10 cm)3 = 2.39×10-23 cm3

Volume of 208 g of the element = mass = 208 g = 28.88 cm3 density 7.2 g/cm3 Number of unit cells in this volume = 28.88 cm3 = 12.08×1023 unit cells

2.39 x10-23 cm3/ unit cell Since each bcc cubic unit cell contains 2 atoms,

therefore, the total number of atoms in 208 g = 2 (atoms/unit cell) × 12.08 × 1023 unit cells = 24.16×1023 atoms

Q10. X-ray diffraction studies show that copper crystallises in an fcc unit cell with cell edge of 3.608×10-8 cm. In a separate experiment, copper is

determined to have a density of 8.92 g/cm3, calculate the atomic mass of copper. Solution : In case of fcc lattice, number of atoms per unit cell, z = 4 atoms

Therefore, M = dNA a3

Z

= 8.92g/cm3x 6.022x1023 atoms/mol x (3.608x10-8 cm)3 4atoms

= 63.1 g/mol Atomic mass of copper = 63.1u Q11. Silver forms ccp lattice and X-ray studies of its crystals show that the

edge length of its unit cell is 408.6 pm. Calculate the density of silver (Atomic mass = 107.9 u).

Solution : Since the lattice is ccp, the number of silver atoms per unit cell = z = 4 Molar mass of silver = 107.9 g mol–1 = 107.9×10-3 kg mol–1 Edge length of unit cell = a = 408.6 pm = 408.6×10–12 m

Density, d = z.M a3 .NA

= 4 x107.9x10-3 kgmol-1 (408.6 x10-12m)3 x 6.022x 1023 mol-1 = 10.5×10-3 kg m–3 = 10.5 g cm-3


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IMPERFECTONS OR DEFECTS IN SOLIDS: Crystal defects: Irregularities or departure from perfectly ordered arrangement of constituent particles.

Defects are of two types: a. Point defects - Point defects are the irregularities or deviations from ideal

arrangement around a point or an atom in a crystalline substance. b. Line defects - Line defects are the irregularities or deviations from ideal arrangement in entire rows of lattice points.

Point defects are of three types:

a. Stoichiometric or intrinsic or thermodynamic defects: These are the point defects that do not disturb the stoichiometry of the solid.

b. Non–stoichiometric defects: These are the point defects that disturb the stoichiometry of the solid. c. Impurity defects: These are the defects in ionic solids due to the presence of

impurities present in them. Stoichiometric defects for non- ionic solids are of two types:

VACANCY DEFECT INTERSTITIAL DEFECT

A crystal is said to have vacancy defect

when some of the lattice sites are vacant.

A crystal is said to have interstitial defect

when some constituent particles (atoms or molecules) occupy an interstitial

site.

This results in decrease in density of the

substance.

This results in increase in density of the

substance.

Stoichiometric defects for ionic solids are of two types:

SCHOTTKY DEFECT FRENKEL OR DISLOCATION DEFECT

In this defect equal number of cations and anions are missing

In this defect, the smaller ion (usually cation) is dislocated from

its normal site to an interstitial site.

It is basically a vacancy defect in ionic

solids.

It creates a vacancy defect at its original

site and an interstitial defect at its new location.

It decreases the density of a solid It does not change the density of the solid

Schottky defect is shown by

(a) ionic substances in which the cation and anion are of almost similar sizes.

(b) Occurs where coordination no. is high.

Frenkel defect is shown by

(a) ionic substance in which there is a large difference in the size of ions (cations and

anions). (b) Occurs where coordination No is low.

For example: NaCl, KCl, CsCl and AgBr For example: ZnS, AgCl, AgBr and AgI

Substance having both schottky defects and frenkel defects : AgBr


20

Non-stoichiometric defects are of two types:

a. Metal excess – This type of defect is due to excess of metal cations. These may be due to:

vimp i. Anionic vacancies: A compound may have an extra metal ion if the negative ion is absent from its

lattice site. This empty lattice site is called a hole. To maintain electrical neutrality this site is occupied

by an electron. The hole occupied by an electron is called f-centre or Farbenzenter centre . F- centre is responsible for the colour of the compound.

e.g. on heating NaCl in presence of Na ions, some anions (Cl-) leave lattice sites which

are occupied by electrons called F centres. ii. Presence of extra cations: A compound is said to have extra cations if a cation is present in the interstitial site. An electron is present in the interstitial site to maintain

the electrical neutrality. E.g. ZnO, O2 gas is lost, Zn+2 ions and e- occupy interstitial sites.

b. Metal deficiency: This defect arises because of absence of metal ions

from its lattice sites. The electrical neutrality is maintained by an adjacent ion having a higher

positive charge. E.g 2A+ in lattice sites may be replaced by A2+ in one lattice site and one lattice site remains vacant. Fe2O ideal composition

composition (actual Fe0.95O)

CLASSIFICATION OF SOLIDS BASED ON THEIR ELECTRICAL CONDUCTIVITIES:

A. CONDUCTORS-The solids with conductivities () ranging between 104to 107

ohm–1m–

1 are called conductors.

B. INSULATORS - These are the solids with very low conductivities () ranging between 10–20

to 10–10 ohm–1m–1.

C. SEMI-CONDUCTORS - These are the solids with conductivities () in the intermediate range from 10–6

to 104 ohm–1m–1.

Band theory – A metal is characterized by a band structure. The highest filled band is called valence band and the lowest unoccupied band is called conduction band. The gap

between the two bands is called forbidden band.


21

a. In case of conductors, the valence band and conduction band overlap

b. In case of insulators, the forbidden gap is very large and the electrons are unable to excite to the conduction band.

c. In case of semiconductors, forbidden gap is small. Therefore, some electrons may jump to conduction band and show some conductivity. Electrical conductivity of

semiconductors increases with rise in temperature, since more electrons can jump to the conduction band.

TYPES OF SEMICONDUCTORS:

a. Intrinsic: These are those semiconductors in which the forbidden gap is small. Only some electrons may jump to conduction band and show some conductivity. They have very

low electrical conductivity. Example: Silicon, germanium b. Extrinsic:

When an appropriate impurity is added to an intrinsic semiconductor. Their electrical conductivity is high.

DIFFERENCE BETWEEN INTRINSIC SEMICONDUCTORS AND EXTRINSIC SEMI CONDUCTIORS:

INTRINSIC SEMI CONDUCTORS EXTRINSIC SEMI CONDUCTORS

It is a pure semi conductor and no impurity

is added to it.

It is prepared by doping small quantity of

impurity atoms to the pure semi conducting materials.

2. examples are crystalline forms of pure Si and germanium.

e.g.si and ge crystals doped with impurity as As, Sb, P etc.

3. no of free electrons in the conduction band and no. of holes in the valence band

is exactly equal and very small indeed.

The no. of holes and valence electrons are never equal.

4. low electrical conductivity. High electrical conductivity.

5. electrical conductivity is a function of temperature alone.

Electrical conductivity depends upon the temperature as well as the quantity of

impurity atoms doped in the structure.

DOPING: The process of adding an appropriate amount of suitable impurity to increase

the conductivity of semiconductors. TYPES OF EXTRINSIC SEMI CONDUCTORS:


22

N-TYPE SEMICONDUCTORS P –TYPE SEMICONDUCTORS

1. It is obtained by doping the impurity atoms of group 15th (Si) . to pure semi

conductor e.g. As, Sb, P.

It is obtained by doping the impurity atoms of group 13th . to pure semi

conductor of group 15 (Si).

2. The increase in conductivity is

due to the negatively charged electrons.

The increase in conductivity is

due to the positively charged holes.

3. the electrons are majority carrier and holes are minority carriers. i.e. ne

>>nh

Holes are majority carrier and electrons are minority carriers.

i.e. nh >>ne

DIODE: It is a combination of n-type and p-type semiconductors and is used as a rectifier. e.g. LED light emitting diode, Photo diode used in solar cell

TRANSISTORS:

They are made by sandwiching a layer of one type of semiconductor between two layers of the other type of semiconductor. npn and pnp type of transistors are used to detect or amplify radio or audio signals.

12- 16 COMPOUNDS – These compounds are formed by the combination of group 12

and group 16 compounds. They possess an average valency of 4. Example: ZnS, CdS, CdSe and HgTe

13- 15 COMPOUNDS – These compounds are formed by the combination of group 13 and group 15 compounds. They possess an average valency of 4. Example: InSb, AlP

and GaAs Every substance has some magnetic properties associated with it. The origin of these

properties lies in the electrons. Each electron in an atom behaves like a tiny magnet. Its magnetic moment originates from two types of motions (i) its orbital motion around the

nucleus and (ii) its spin around its own axis.

VARIATION OF RESISTANCE OF SOME ELECTRICAL MATERIAL AND TEMPERATURE:

Electric resistance (R) – is the ratio of potential difference (V) across the ends of the conductor to the current (I) flowing through it, i.e.

R =V Ohm (Ω) I Units of R is ohm (Ω)


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RESISTANCE :

Property of a substance by virtue of which it opposes the flow of current through it.

VARIATION OF RESISTANCE: R= ρL where R is called resistance of the material, ρ is the resistivity of the A material, A is the area of cross-sectional of wire, L is the length of wire.

R also depends on Material of the conductor.

Rt = R0 (1 + t) Where Rt = Resistance at temperature toC Ro = Resistance at temperature 0oC

(i) METALS: for metals Temperature coefficient of resistance ()>0. Therefore resistance increases wrt Temperature. Physical explanation: collision frequency of free electron with immobile +ve ions

increases.

(ii) SOLID NON METALS: For these =0, so resistance is indepenedent with temperature. Physical explanation: complete absence of free electrons.

(iii) SEMI CONDUCTORS: For semi conductors <0, i.e. resistance decreases with the temperature rise.

Physical explanation: covalent bonds breaks, liberating more free electron and conduction increases.

(IV) ELECTROLYTE: for electrolyte <0, i.e. resistance decreases with temperature rise.

Physical explanation: the degree of ionisation increases and solution and solution becomes less viscous and more conducting.

(V) IONISED GAS: for ionised gas <0, i.e. resistance decreases with temperature rise.

Physical explanation: degree of ionisation increases.

(VI) ALLOYS: for alloys α has a small +ve value so with rise of temperature resistance of alloys is almost constant. Further alloy resistance are slightly higher than the pure metal resistance.

Alloys are used to made standard resistances, wires of resistance box, potentiometer wire, meter bridge wire etc.

Commonly used alloys are : constantan, maganin, Nichrome etc. (VII) SUPER CONDUCTORS: At low temperature , the resistance of certain substances becomes exactly zero. (e.g. Hg below 4.2K or Pb below 7.2K)

These substance are called super conductors and phenomenon is super conductivity. The temperature at which resistance becomes zero is called critical temperature and depends

upon the nature of the substance. USES: (i) Helps in producing super computers.

(ii) Superconductor cables for electricity transmission without any loss. (iii) To create strong magnetic field with small electrical power.

Therefore electrical conductivity of metals decreases with temperature (due to vibtations of +ve ions which hinders the flow of electrons).


24

Semi conductors increases with temperature (electron jump from valence band to conduction band)

Some transition metal oxides and their conductivities: (i) TiO, CrO2, ReO3 are metallic

(ii) MnO, FeO and CuO are insulators (iii) VO, VO2, VO3 and TiO3 changes from metallic to insulators at a certain temperature.

MAGNETIC PROPERTIES OF SOLIDS: CLASSIFICATION OF SOLIDS ON THE BASIS OF MAGNETIC PROPERTIES:

A. DIAMAGNETIC: Diamagnetic substances are weakly repelled by a magnetic field. Diamagnetism is shown by those substances in which all the electrons are paired and there are no unpaired electrons. E.g N2, TiO2, H2O, Zn, Cd, NaCl,Cu+, benzene etc.

B. PARAMAGNETIC: These are those substances which are weakly attracted by the magnetic field. It is due to presence of one or more unpaired electrons and hence

magnetic moment. E.g. O2,NO, metal ions (Cu+2, Fe+3, Cr+3 ), metals (Cr, Mn, Ni, Co, Fe) , metal ocide (CuO, VO2)etc. C. FERROMAGNETIC: These are those substances which are attracted very strongly by

a magnetic field even in the absence of magnetic field. E.G. Fe, Ni, Co, Gd,CrO2 (magnetic tapes, recorders)

D. ANTIFERROMAGNETIC: Which are expected to possess paramagnetism or ferromagnetism but actually have zero net magnetic moment. They have equal number of parallel and anti parallel magnetic dipoles (domains) resulting in a zero net dipole

moment. E.g. MnO, Mn2O3, MnO2 E. FERRIMAGNETIC: which are expected to possess large magnetism but actually have

small net magnetic moment, e.g. magnetite (Fe3O4), ferrites (M2+Fe2O4). Because of unequal number of parallel and anti parallel magnetic dipoles resulting in a net dipole moment.

CURIE TEMPERATURE: It is the temperature above which a ferromagnetic substance shows no ferromagnetism.

Some magnetic properties of common substance: (a) TiO, VO, CuO are paramagnetic whereas MnO, CoO, NiO are antiferromagentic. (b) TiO2 is diamagnetic, VO2 is paramagnetic, CrO2 is ferromagnetic whereas MnO2 is

antiferromagnetic.


25

PIEZOELECTRICITY: When mechanical stress is applied on such crystals, electricity is

produced due to displacement of ions which is known as piezoelectricity. Piezoelectric crystals: titanates of barium and lead, lead zirconate (PbZrO3), ammonium

dihydrogen phosphate (NH4H2PO4) PYROELECTROCITY: Some piezoelectric crystals when heated produce a small electric

current. Electricity thus produced is called pyroelectricity (pyro means heat) FERROELECTROCITY: In some piezoelectric crystals, the dipoles are permanently polarised even in the absence of electric field. However on applying electric field,

direction of polarisation changes. This phenomenon is called ferro-electricity. e.g. barium titanate (BaTiO3) , sodium potassium tartarate (Rochelle salt) and potassium

di hydrogen phosphate (KH2PO4). It may be pointed out here that all ferroelectric solids are piezoelectric but the reverse is not true. ANTIFERROELECTRICITY: In some crystals, the dipoles align themselves in such a

way that alternatively, they point up and down so that the crystals does not possess any net dipole moment. Such crystals are said to be anti ferroelectric. E.g. lead zirconate.

Q12. analysis shows that a metal oxide has the empirical formula of M0.96O1.00 . calculate the percentage of M2+ and M3+ ions in this crystal?

Ans. Formula of metal oxide = M0.96O1.00 means M = 0.96 = 96

O 1.00 100 Thus if there are 100 O atoms M atoms would be 96 Charge on 100 O2- ions =100 x (-2) = -200

Suppose M atoms as M2+ = x and M3+ = 96-x

total charge on M2+ and M3+ =(+2) x + (+3)x (96-x) = 288-x =200 or x= 88 therefore %age of M as M2+ = 88 x 100 = 91.7%

96 %age of M as M3+ = 100 - 91.7% = 8.3%

Curie temperature: It is the temperature above which a ferromagnetic substance shows no ferromagnetism.

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