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Resonance Pre-foundation Career Care Programmes
(PCCP) Division
WORKSHOP TAPASYA
SHEET
CHEMISTRY
COURSE : KVPY (STAGE-) I
Subject : Chemistry
KVPY STAGE-I
S. No.
Topics
Page No.
1.
Organic Chemistry
1 - 13
2.
Mole Concept
14 - 25
3.
Study of Gas Laws
26 - 31
4.
Chemical Bonding
32 - 41
5.
Periodic Table
42-52
6.
Acids, Bases and Salts
53-60
7.
Nuclear Chemistry
61-71
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.13RPCCP
ORGANIC CHEMISTRY
INTRODUCTION
VERSATILE NATURE OF CARBON
The compounds like urea, sugars, fats, oils, dyes,
proteins, vitamins etc., which are isolated directly or
indirectly from living organisms such as animals and
plants are called organic compounds.The branch of
chemistry which deals with the study of these
compounds is called ORGANIC CHEMISTRY.
VITAL FORCE THEORY OR
BERZELIUS HYPOTHESIS
Organic compounds cannot be synthesized in the
laboratory because they require the presence of a
mysterious force (called vital force) which exists only
in living organisms.
WOHLER’S SYNTHESIS
About 3 million organic compounds are known today.
The main reasons for this huge number of organic
compounds are (i) Catenation : The property of self linking of carbon
atoms through covalent bonds to form long straight
or branched chains and rings of different sizes is
called catenation.Carbon shows maximum
catenation in the periodic table due to its small size,
electronic configuration and unique strength of carboncarbon bonds.
(ii) Electronegativity and strength of bonds : The
electronegativity of carbon (2.5) is close to a number
of other elements like H (2.1) , N(3.0) , P (2.1), Cl (3.0)
and O (3.5). So carbon forms strong covalent bonds
with these elements.
In 1828, Friedrich Wohler synthesized urea (a well
known organic compound) in the laboratory by heating
ammonium cyanate.
(iii) Tendency to form multiple bonds : Due to small
size of carbon it has a strong tendency to form multiple
bonds (double & triple bonds).
(NH 4)2 SO4 +
Ammonium
sulphate
(iv) Isomerism : It is a phenomenon by the virtue of
which two compounds have same molecular formula
but different physical and chemical properties.
2 KCNO
Potassium
cyanate
2NH 4CNO + K2SO 4
Ammonium Potassium
sulphate
cyanate
CLASSIFICATION OF ORGANIC COMPOUNDS

Note :
Urea is the first organic compound synthesized in
the laboratory.
The organic compounds are very large in number on
account of the self -linking property of carbon called
catenation. These compounds have been further
classified as open chain and cyclic compounds.
Organic compounds
MODERN DEFINITION OF ORGANIC CHEMISTRY
Organic compounds may be defined as hydrocarbons
and their derivatives and the branch of chemistry
which deals with the study of hydrocarbons and their
derivatives is called ORGANIC CHEMISTRY.
Open chain
compounds
Closed chain
compounds
(i) Organic compounds are large in number.
(ii) Organic compounds generally contain covalent
bond.
(iii) Organic compounds are soluble in non polar solvents.
(iv) Organic compounds have low melting and boiling
points.
Aromatic
compounds
Alicyclic
compounds
Organic chemistry is treated as a separate branch
because of following reasons-
(a) Open Chain Compounds :
These compounds contain an open chain of carbon
atoms which may be either straight chain or branched
chain in nature. Apart from that, they may also be
saturated or unsaturated based upon the nature of
bonding in the carbon atoms. For example.
(v) Organic compounds show isomerism .
(vi) Organic compounds exhibit homology.
,
,
PAGE # 1
1
1
,
e.g.
Benzene
Toluene
Phenol

n-Butane is a straight chain alkane while
2-Methylpropane is branched alkane.
Ethyl benzene
Aniline
Note :
Benzene is the parent compound of majority of
aromatic organic compounds.
HYDROCARBONS
(b) Closed Chain or Cyclic Compounds :
Apart from the open chains, the organic compounds
can have cyclic or ring structures. A minimum of three
atoms are needed to form a ring. These compounds
have been further classified into following types.
(i) Alicyclic compounds : Those carbocyclic
compounds which resemble to aliphatic compounds
in their properties are called alicyclic compounds .
The organic compounds containing only carbon and
hydrogen are called hydrocarbons. These are the
simplest organic compounds and are regarded as
parent organic compounds. All other compounds are
considered to be derived from them by the
replacement of one or more hydrogen atoms by other
atoms or groups of atoms. The major source of
hydrocarbons is petroleum.
Types of Hydrocarbons :
The hydrocarbons can be classified as :
e.g.
or
or
Cyclopropane
Cyclobutane
(i) Saturated hydrocarbons :
(A) Alkanes : Alkanes are saturated hydrocarbons
containing only carbon - carbon and carbon - hydrogen
single covalent bonds.
General formula- CnH2n + 2(n is the number of carbon atoms)
e.g.
or
Cyclopentane
CH4 ( Methane)
C2H6 (Ethane)
(ii) Unsaturated hydrocarbons :
(A) Alkenes : These are unsaturated hydrocarbons
which contain carbon - carbon double bond. They
contain two hydrogen less than the corresponding
alkanes.
General formula e.g.
or
Cyclohexane
CnH2n
C2H 4
C3H 6
(Ethene)
(Propene)
(ii) Aromatic compounds : Organic compounds which
(B) Alkynes : They are also unsaturated hydrocarbons
which contain carbon-carbon triple bond. They contain
four hydrogen atoms less than the corresponding
alkanes.
contain one or more fused or isolated benzene rings
are called aromatic compounds.
General formula e.g.
CnH 2n–2
C2H 2
(Ethyne)
C3H 4
(Propyne)
PAGE # 2
2
2
Examples :
NOMENCLATURE OF ORGANIC COMPOUNDS
a
Nomenclature means the assignment of names to
organic compounds . There are two main systems of
nomenclature of organic compounds (1) Trivial system
(2) IUPAC system (International Union of Pure and
Applied Chemistry)
(a) Basic rules of IUPAC nomenclature of
organic compounds :

Note :
The name of the compound, in general , is written in
For naming simple aliphatic compounds, the normal
saturated hydrocarbons have been considered as
the parent compounds and the other compounds as
their derivatives obtained by the replacement of one
or more hydrogen atoms with various functional
groups.
the following sequence(Position of substituents )-(prefixes ) (word root)-(p suffix).
(iii) Names of branched chain hydrocarbon : The
carbon atoms in branched chain hydrocarbons are
(i) Each systematic name has two or three of the
following parts-
present as side chain . These side chain carbon atoms
constitute the alkyl group or alkyl radicals. An alkyl group
(A) Word root : The basic unit of a series is word root
which indicate linear or continuous number of carbon
atoms.
is obtained from an alkane by removal of a hydrogen.
General formula of alkyl group = CnH2n+1
(B) Primary suffix : Primary suffixes are added to the
word root to show saturation or unsaturation in a
carbon chain.
(C) Secondary suffix : Suffixes added after the
primary suffix to indicate the presence of a particular
functional group in the carbon chain are known as
secondary suffixes.
M
E
P
B
E
Propene
Eth -
An alkyl group is represented by R.
e.g.
H
(A)
–H
H
H
Methyl
(ii) Names of straight chain hydrocarbons : The
name of straight chain hydrocarbon may be divided
into two parts(A) Word root
C
(B) Primary suffix
(A) Word roots for carbon chain lengths :
Chain
length
Word
root
Chain
length
Word
root
C1
C2
C3
C4
C5
MethEth Prop But Pent-
C6
C7
C8
C9
C10
HexHeptOctNonDec-
(B)
–H
H
H
H
C
C
H
H
Ethyl
(B) Primary suffix :
(C)
PAGE # 3
3
3
A branched chain hydrocarbon is named using the
e.g.
following general IUPAC rules :
Rule1: Longest chain rule : Select the longest possible
2–Methylpentane
4–Methylpentane
(Correct)
continuous chain of carbon atoms. If some multiple
(Wrong)
bond is present , the chain selected must contain the
e.g.
multiple bond.
(i) The number of carbon atoms in the selected chain
3–Methylbut–1– ene
determines the word root .
(ii) Saturation or unsaturation determines the primary
suffix (P. suffix).
(iii) Alkyl substituents are indicated by prefixes.
CH3
e.g.
Prefix : Methyl
CH3 – CH – CH2 – CH – CH3
Word root : Hept-
CH3
CH2 – CH2 – CH3
e.g. CH3 – CH2 – C – CH3
CH2
e.g. CH3 – CH 2– CH – CH2– CH 3
CH – CH3
e.g.
P. Suffix : -ane
Prefix : Methyl
Word root : ButP. Suffix : –ene
Prefixes : Ethyl, Methyl
Word root : PentP. Suffix : -ane
4
CH3
|
3
2
(Wrong)
1
1
3-Methylbut-1-yne
(Correct)
Prefix : Methyl
Word root : pentP. Suffix: - ane
e.g. CH3 – CH2 – CH – CH2 – CH3
2–Methylbut – 3 – ene
(Correct)
Rule 2 : Lowest number rule: The chain selected is
numbered in terms of arabic numerals and the
position of the alkyl groups are indicated by the number
3
4
2-Methylbut-3-yne
(Wrong)
Rule 3 : Use of prefixes di, tri etc. : If the compound
contains more than one similar alkyl groups,their
positions are indicated separately and an appropriate
numerical prefix di, tri etc. , is attached to the name
of the substituents. The positions of the substituents
are separated by commas.
CH3
5
4 3
2
1
CH3 – CH2– C – CH – CH3
e.g.
CH3 CH3
2,3 - Dimethylpentane
2,3,3 - Trimethylpentane
e.g.
2,3,5 -Trimethylhexane
CH3
CH3
|
2
2,2,4 - Trimethylpentane
Rule 4 : Alphabetical arrangement of prefixes: If there
are different alkyl substituents present in the
compound their names are written in the alphabetical
order. However, the numerical prefixes such as di,
tri etc. , are not considered for the alphabetical order.
of the carbon atom to which alkyl group is attached .
(i) The numbering is done in such a way that the
e.g.
substituent carbon atom has the lowest possible
number.
3-Ethyl - 2,3-dimethylpentane
(ii) If some multiple bond is present in the chain, the
carbon atoms involved in the multiple bond should
get lowest possible numbers.
Rule 5 : Naming of different alkyl substituents at the
equivalent positions :
Numbering of the chain is done in such a way that the
alkyl group which comes first in alphabetical order
gets the lower position.
e.g.
2–Methylbutane
(Correct)
3–Methylbutane
(Wrong)
e.g.
3-Ethyl-4-methylhexane
PAGE # 4
4
4
Rule - 6 : Lowest sum rule
According to this rule numbering of chain is done in
such a way that the sum of positions of different
substituents gets lower value.
e.g.
FUNCTIONAL GROUP
An atom or group of atoms in an organic compound
or molecule that is responsible for the compound’s
characteristic reactions and determines its properties
is known as functional group. An organic compound
generally consists of two parts (i) Hydrocarbon radical (ii) Functional group
(i)
e.g.
Hydrocarbon radical Functional group
• Functional group is the most reactive part of the
molecule.
• Functional group mainly determines the chemical
properties of an organic compound.
• Hydrocarbon radical mainly determines the physical
properties of the organic compound.
(a) Main Functional Groups :
Word root
: Hex Primary suffix : - ane
Substituent : two methyl & one ethyl groups
IUPAC name : 4-Ethyl - 2, 4 - dimethylhexane
(i) Hydroxyl group (– OH) : All organic compounds
containing - OH group are known as alcohols .
e.g. Methanol (CH3OH) , Ethanol (CH3 – CH2 – OH) etc .
(ii) Aldehyde group (–CHO) : All organic compounds
containing –CHO group are known as aldehydes.
e.g. Methanal (HCHO), Ethanal (CH3CHO) etc.
Some other example :
(i)
Word root
: Prop P. Suffix
: -ane
Substituent : two methyl groups
IUPAC name : 2, 2 - Dimethylpropane
(iii) Ketone group (–CO–) : All organic compounds
containing –CO– group are known as ketones.
e.g.
Propanone
(CH 3 COCH 3 ),
Butanone
(CH3COCH2CH3) etc.
(iv) Carboxyl group ( – COOH) : All organic compounds
containing carboxyl group are called carboxylic acids.
e.g. CH3COOH (Ethanoic acid)
CH3CH2COOH(Propanoic acid)
(v) Halogen group (X = F, Cl, Br, I) : All organic
compounds containing –X (F, Cl, Br or I) group are
known as halides.
e.g. Chloromethane (CH3Cl), Bromomethane (CH3Br)
etc .
(ii)
Word root
: But P. Suffix
: - ene
Substituent : two methyl groups
IUPAC name : 2, 3 - Dimethylbut - 1 - ene
(iii)
Word root
: Hex P. Suffix
: - yne
Substituent : one methyl group
IUPAC name : 4 - Methylhex - 2 - yne
(b) Nomenclature of Compounds Containing
Functional Group :
In case functional group (other than C = C and C  C)
is present, it is indicated by adding secondary suffix
after the primary suffix. The terminal ‘e’ of the primary
suffix is removed if it is followed by a suffix beginning
with ‘a’, ‘e’, ‘i’, ‘o’, ‘u’. Some groups like
–F, – Cl, – Br and –  are considered as substituents
and are indicated by the prefixes.
O
Some groups like – CHO, – C – , – COOH, and – OH
are considered as functional groups and are
indicated by suffixes.
PAGE # 5
5
5
Class
Functional
Group
General
Formula
Prefix
Carboxylic
acid
Carboxy
Suffix
- oic acid
IUPAC Name
Alkanoic acid
(R = CnH2n+1)
Carbalkoxy or
alkyl (R’) - oate
alkoxy carbonyl
Ester
Aldehyde
– CHO
Formyl
or oxo
R – CHO
Ketone
Alcohol
oxo
– OH
Alkenes
Alkynes
Halides
R – OH
Hydroxy
CnH2n
–C C–
–X
(X = F,Cl,Br,I)
–
CnH2n–2
–
R–X
Halo
- al
- one
Alkyl alkanoate
Alkanal
Alkanone
- ol
Alkanol
- ene
Alkene
- yne
Alkyne
–
Haloalkane
Steps of naming of an organic compound
Step 4 :
containing functional group :
The carbon atoms of the parent chain are numbered
in such a way so that the carbon atom of the functional
Step 1:
group gets the lowest possible number . In case the
Select the longest continuous chain of the carbon
functional group does not have the carbon atom, then
atoms as parent chain. The selected chain must
the carbon atom of the parent chain attached to the
include the carbon atoms involved in the functional
functional group should get the lowest possible
groups like – COOH, – CHO, – CN etc, or those which
number.
carry the functional groups like – OH, – NH2,– Cl,
Step 5 :
– NO2 etc.
The name of the compound is written as -
The number of carbon atoms in the parent chain
Prefixes - word root - primary suffix - secondary suffix
decides the word root.

Note :
Step 2 :
The presence of carbon - carbon multiple bond
decides the primary suffix.
The number of carbon atoms in the parent chain
decides the word root.
Step 3 :
The secondary suffix is decided by the functional
group.
PAGE # 6
6
6
S.No.
Compound
Common name Derived name
IUPAC Name
1
CH3 – OH
Methyl alcohol
or Wood spirit
Carbinol
Methanol
2
CH3 – CH2 – OH
Ethyl alcohol
Methyl carbinol
Ethanol
3
CH3 – CH2 – CH2 – OH
n-Propyl alcohol
Ethyl carbinol
1- Propanol
Structure
H
4
H
5
CH3 – CH2 – CH2 – CH2 – OH
n-Butyl alcohol
6
HCOOH
Formic acid
–
Methanoic acid
Acetic acid
–
Ethanoic acid
CH3COOH
7
n-Propyl carbinol
methyl acetic
acid
8
CH3 – CH2 – COOH
Propionic acid
9
CH3 – CH2 – CH2 – COOH
Butyric acid
ethyl acetic
acid
10
CH3 – CH2 – CH2 – CH2 – COOH
Valeric acid
n-Propyl acetic
acid
H
H – C – C – O – H
Isopropyl alcohol Dimethyl carbinol 2 - Propanol
C H3
1- Butanol
Propanoic acid
Butanoic acid
O
(iii) CH3 – CH2 – CH2 – NH2
Word root
: Prop Primary suffix
: - ane
Secondary suffix
: - amine
IUPAC name
: Propan - 1 - amine
Some more examples :
(i)
Word root
: HeptPrimary suffix : – ane
Functional group : – OH
Secondary suffix : – ol
IUPAC Name :
(iv)
2, 5-Dimethylheptan–1– ol
(ii)
Word root
: Pent Primary suffix
: – ene
Secondary suffix
: – oic acid
IUPAC name : Pent-2-en-1-oic acid/Pent-2-enoic acid
Pentanoic acid
Word root
Primary suffix
Substituent
IUPAC name
: Prop: - ane
: nitro(prefix)
: 1 - Nitropropane
Word root
Primary suffix
Prefix
IUPAC name
: But : – ane
: – chloro
: 2 - Chlorobutane
(v)
PAGE # 7
7
7
ISOMERISM
(vi)
Word root
Primary suffix
Secondary suffix
Prefix
IUPAC name
: But : – ane
: – one
: Methyl
: 3 - Methylbutan - 2- one
Such compounds which have same molecular
formula but different physical and chemical properties
are known as isomers and the phenomenon is
known as isomerism.
HOMOLOGOUS SERIES
Homologous series may be defined as a series of
similarly constituted compounds in which the members
possess similar chemical characteristics and the two
consecutive members differ in their molecular formula
by – CH2.
(a) Characteristics of Homologous Series :
(i) All the members of a series can be represented by
the same general formula.
e.g. General formula for alkane series is CnH2n+2 .
(ii) Any two consecutive members differ in their formula
by a common difference of – CH 2 and differ in
molecular mass by 14.
(a) Chain Isomerism :
The isomerism in which the isomers differ from each
other due to the presence of different carbon chain
skeletons is known as chain isomerism.
e.g.
(i) C4H10
,
(iii) Different members in a series have a common
functional group.
e.g. All the members of alcohol family have –OH group .
(iv) The members in any particular family have almost
identical chemical properties. Their physical
properties such as melting point, boiling point, density
etc, show a regular gradation with the increase in the
molecular mass.
2 - Methylpropane
(Isobutane)
(ii) C5H12
2 - Methylbutane
(Isopentane)
(v) The members of a particular series can be
prepared almost by the identical methods.
(b) Homologues :
The different members of a homologous series are
known as homologues.
2, 2 -Dimethylpropane
(neo - pentane)
e.g.
(i) Homologous series of alkanes
General formula : CnH2n+2
Value of n
n=1
n=2
n=3
(iii) C4H8
Molecular formula IUPAC name
CH 4
Methane
C2H 6
Ethane
C3H 8
Propane
(ii) Homologous series of alkenes
General formula :CnH2n
Value of n
n=2
n=3
n=4
Molecular
formula
C2H 4
C3H 6
C4H 8
IUPAC
name
Ethene
Propene
But-1-ene
Common
name
Ethylene
Propylene
 - Butylene
(iii) Homologous series of alkynes
General formula : CnH2n–2
Value of n
n=2
n=3
n=4
Molecular
formula
C2H 2
C3H 4
C4H 6
IUPAC
Common
name
name
Ethyne
Acetylene
Propyne Methyl acetylene
But -1-yne Ethyl acetylene
CH3 – CH2 – CH = CH2 ,
But - 1 - ene
Methylpropene
(b) Position Isomerism :
In this type of isomerism, isomers differ in the structure
due to difference in the position of the multiple bond
or functional group.
e.g.
(i) C4H8
CH3 – CH2 – CH = CH2 , CH3 – CH = CH – CH3
But -1 - ene
But -2 - ene
(ii) C3H8O
CH3 – CH2 – CH2 – OH ,
Propan-1-ol
CH3 – CH – CH3
OH
Propan-2-ol
PAGE # 8
8
8
only. These compounds are open chain compounds
which are also addressed as Acyclic compounds.
Alkanes have the general formula CnH2n+2 .The carbon
atoms in alkanes are in a state of sp3 hybridization,
i.e. the carbon atoms have a tetrahedral geometry.
(c) Functional Group Isomerism :
In this type of isomerism, isomers differ in the
structure due to the presence of different functional
groups.
e.g.
(a) Physical Properties :
(i) C3H8O
CH3 – CH2 – O – CH3
CH3 – CH2 – CH2 – OH
Methoxy ethane
Propan-1-ol
(i) Alkanes of no. of carbon atoms C1 to C4 are gases.
Carbon atoms C5 to C17 are liquids and C18 & onwards
are solids.
(ii) Alkanes are colourless and odourless.
(iii) They are non-polar in nature, hence they dissolve
only in non-polar solvents like benzene, carbon
tetrachloride etc.
(iv) Boiling point of alkanes increases as their
molecular weight increases.
(ii) C4H6
CH3 – CH2 – C  CH
But - 1- yne
CH2 = CH – CH = CH2
Buta - 1, 3 - diene
[or 1, 3 - Butadiene ]

ALKANES
Alkanes are aliphatic hydrocarbons having only
C – C single covalent bonds. These are also known
as saturated hydrocarbon as they contain single bond
Note :
Alkanes are unaffected by most chemical reagents
and hence are known as paraffins (parum-little, affinis
affinity).
SOME COMMON EXAMPLES OF ALKANES
M olcula r Form ula
Structure
CH 4
CH 4
C2H6
CH 3–CH 3
C3H8
CH 3–CH 2 –CH 3
C 4 H 10
Methane
C H 3 – C H 2 – CH 2 – C H 3
CH 3 – CH – CH 3
|
CH 3
C 5 H 12
CH
3
– CH
2
– CH
2
– CH
3
CH 3 – CH – C H 2 – C H 3
|
CH
3
CH
C 6 H 14
Trivia l Na m e
3
–
CH
| 3
C – CH
3
|
CH
3
CH 3 – CH 2 – CH 2 – CH 2 – CH 2 – CH 3
C H 3 – CH – CH 2 – CH 2 – CH 3
IUPAC Na m e
Methane
Ethane
Ethane
Propane
Propane
n-Butane
Isobutane
Butane
2–Methy lpropane
n-P entane
Pentane
Isopentane
2–Methy lbutane
Neopentane
2,2–Dimethylpropane
n-Hexane
Hexane
Isohexane
2–Methy lpentane
–
3–Methy lpentane
–
2,3–Dimethylbutane
Neohexane
2,2–Dimethylbutane
CH3
CH 3 – CH2 – CH – CH 2 – CH 3
CH 3
CH 3 – CH – CH – CH 3
CH 3 CH 3
CH 3
CH 3 – C – CH 2 – CH3
CH 3
PAGE # 9
9
9

Note :
The C – C bond distance in alkanes is 1.54 Å and the
bond energy is of the order of 80 Kcal per mole.
(vi) Laboratory Method : Methane is prepared in the
laboratory by heating a mixture of dry sodium acetate
and soda lime in a hard glass tube as shown in
figure. It is a decarboxylation reaction.
METHANE
It is a product of decomposition of organic matter in
absence of oxygen. It is found in coal mines (hence
the name damp fire), marshy places (hence the name
marsh gas) and the places where petroleum is found.

Note :
Hard glass tube
Delivery
tube
Sodium acetate
and soda lime
Gas jar
Cork
Bubbles of
methane gas
Burner
Methane is a major constituent of natural gas.
Trough
Beehive shelf
Water
Iron
stand
Gas
(a) Properties :
Methane is a colourless gas with practically no smell
and is almost insoluble in water. It melts at – 183º C
and boils at –162ºC. Methane has tetrahedral
geometry in which H–atoms are situated at four
corners of the regular tetrahedron. Bond angle is
109º28’. It has sp3 hybridisation.
Methane, so formed is collected by downward
displacment of water. This gas contains some
hydrogen, ethylene etc. as impurities which can be
removed by passing the impure gas through alkaline
potassium permanganate solution.
(b) Structure :
(d) General Reactions :
H
(i) Combustion :
(A) Methane burns with explosive violence in air
forming carbon dioxide and water.
CH4 + 2O2
CO2 + 2H2O + Heat
(B) In the presence of insufficient supply of oxygen.
2CH4 + 3O2
2CO + 4H2O + Heat
C
H
H
H
Preparation of methane gas
Tetrahedral
(ii) Halogenation :
(A) In direct sunlight
(c) Preparation of Methane :
(i) Direct synthesis :
CH4 + 2Cl2
h
C + 4HCl
(B) In diffused light
(ii) Sabatier and Senderens reductive method :
Methane can be prepared by passing carbon
monoxide or carbon dioxide and hydrogen over finely
powdered nickel catalyst at 300ºC.
CO + 3H2
CO2 + 4H2
Ni (powder)
300ºC
CH4 + H2O
Ni (powder)
300ºC
12H2O
Water
+
2H
CH4 + 2H2O
Methane
3CH4
Methane
+
4Al(OH)3
Aluminium
hydroxide
Zn–Cu Couple
CH4
H2O
Methane
+
H
Hydrogen
iodide
(v) Reduction of methanol or formaldehyde or formic
acid with H
CH3OH
+
2H 
Red P
HCHO
Methanal
CH4
+
+
H2O
Methane
Methanol
+
4H
Red P
Cl2
Cl2
CH2Cl2
Methylene
dichloride
Cl2
CHCl3
Chloroform
CCl4
Carbon
tetrachloride
Fluorine forms similar substitution products in the
presence of nitrogen which is used as a diluent
because of high reactivity of fluorine. Bromine vapours
react very sluggishly while iodine vapours do not react
at all.
CH4
(iv) Reduction of methyl iodide :
CH3 – 
Methyl
iodide
CH3Cl
Methyl
chloride
(iii) Nitration :
(iii) Hydrolysis of aluminium carbide :
Al4C3
+
Aluminium
carbide
Cl2
CH4
Methane
CH4
Methane
+
2+
H 2O
+
HO–NO2
Nitric acid
400ºC
10 atm.
CH3–NO2
+
H 2O
Nitromethane
(iv) Catalytic Air oxidation : This is a method for
commercial production of methanol.
When a mixture of methane and oxygen in a ratio of 9: 1
by volume is passed through a heated copper tube at
200ºC and at a pressure of 100 atmospheres,
methanol is formed.
CH4 +
1/2 O2
CH 3OH
Methane
Methanol
(e) Uses :
(i) Alkanes are used directly as fuels .
(ii) Certain alkanes, such as methane, are used as a
source of hydrogen.
(iii) The carbon obtained in decomposition of alkanes
is in very finely divided state and is known as carbon
black. This is used in making printer’s ink, paints, boot
polish and blackening of tyres.
(iv) Alkanes are used as starting materials for a
number of other organic compounds e.g.
methanol, methyl chloride, methylene dichloride etc.
10
10
PAGE # 10
(b) Uses :
ALKENES
Alkenes are the simplest unsaturated aliphatic
hydrocarbons with one carbon - carbon double bond.
Alkenes have general formula CnH2n. The carbon
atoms connected by the double bond are in a state of
sp2 hybridisation and this part of molecule is planar.
A double bond is composed of sigma () and a pi ()
bond. Alkenes are also called olefines (oil forming)
becuase they form oily products with halogens.
R – CH = CH2 + Br2
(i) Ethylene is mainly used in the manufacture of
ethanol, ethylene oxide and higher 1-alkenes.
Ethylene is used for ripening of fruits. It is also used
for preparation of mustard gas.
[Cl – CH2 – CH2 – S – CH2 – CH2 – Cl]
(ii) Polythene from ethylene, teflon from tetra
fluoroethylene and polystyrene from styrene are used
as plastic materials. Acrilon or orlon obtained from
vinyl cyanide is used for making synthetic fibres.
R – CH – CH2
Br Br
(Oily liquid)
(a) Properties :
(i) Alkenes of C2 to C4 are gases. Alkenes of carbon
atoms C5 to C14 are liquids and C14 and onwards are
solids.
(ii) Ethene is colourless gas with faint sweet smell.
All other alkenes are colourless and odourless.
(iii) Alkenes are insoluble in polar solvents like water,
but fairly soluble in non-polar solvents like benzene,
carbon tetrachloride etc.
ETHENE
Ethene occurs in natural gas, coal gas and wood
gas. It is also formed during the cracking of high boiling
petroleum fractions.
(a) Properties :
Ethene is a colourless gas (B.P. = –105ºC). It is very
sparingly soluble in water but dissolves in acetone,
alcohol etc. It burns with smoky flame. Ethene has
trigonal planar geometry. Bond angle is 120º. It has
sp2 hybridisation.
(b) Structure :
(iv) Boiling point of alkenes increases with increase
in molecular mass.
Bond length of C = C is 1.34 Å . The energy of the
double bond is 142 Kcal mol–1, which is less than
twice the energy of a single bond i.e. 80 Kcal mole-1.
This indicates that a pi () bond is weaker than a
sigma () bond.
(b) Some common examples of alkenes -
(c) Preparation of Ethene :
(i) By dehydration of alcohol (Lab. method) :
CH3 – CH2 – OH
Conc. H2SO4
165 – 170ºC
CH2 = CH2 + H2O
Ethene
Ethanol
(ii) By cracking of kerosene :
Cracking
CH3 – (CH2)4– CH3
CH3 – CH2 – CH2 – CH3 + CH2 = CH2
Butane
n-Hexane
Ethene
(iii) From alkyl halides (Dehydrohalogenation) :
CH2 – CH2 + KOH
(Alcoholic)
H
X
CH2 = CH2 + KX + H2O (Here X = Halogen)
Ethene
Ethyl halide
(d) General Reactions :
(i) Addition of halogens :
CH2 = CH2
Ethene
CH2 = CH2
Ethene
+
Cl2
Chlorine
CCl4
CH2– CH2
Cl Cl
1,2-Dichloroethane
(Ethylene dichloride)
CCl4
Br2
Bromine
(red-brown colour)
+
CH2– CH2
Br
Br
1,2-Dibromoethane
(colourless)
11
11
PAGE # 11

Note :
Addition of bromine on alkenes in presence of CCl4
ALKYNES
Alkynes are unsaturated aliphatic hydrocarbons
having a carbon-carbon triple bond. Alkynes have
general formula CnH2n–2. Thus, they have two hydrogen
atoms less than an alkene and four hydrogen atoms
less than an alkane with same number of carbon
atoms. A triple bond is composed of one sigma ()
and two pi () bonds. The carbon atoms connected
by a triple bond are in state of sp hybridisation.
is the test for unsaturation.
(ii) Addition of halogen acids (Hydrohalogenation) :
CH2 = CH2 + HCl
Ethene
CH2 – CH2
Cl
H
Chloroethane
(iii) Hydrogenation :
(a) Properties :
Ni or Pt
CH2 = CH2 + H2
CH3 – CH3
High T& P
Ethene
Ethane
(i) Alkynes of carbon atoms C2 to C4 are gases. Alkynes
of carbon atoms C5 to C12 are liquids.Alkynes of C13 &
onwards are solids.
(iv) Combustion :
C2H4 + 3O2
Ethene
2CO2
+
(ii) Alkynes are colourless and odourless, but ethyne
has characteristic odour.
2H2 O + Heat
(iii) Boiling point and solubilities in water are relatively
higher than those of alkanes and alkenes.
(v) Addition of oxygen :
(iv) Alkynes are weakly polar in nature.
(v) Alkynes are lighter than water and soluble in nonpolar solvents.
(vi) Boiling point of alkynes increases with the increase
in molecular mass.
(vi) Polymerisation :

nCH2 = CH2
High T
& High P
– (CH2– CH2 –)n
Polyethene
Ethene
Note :
The bond energy of a triple bond is 190.5 Kcal per
mole, which is less than thrice the energy of a single
() bond.
SOME COMMON EXAMPLES OF ALKYNES :
Molecular formula
Structure
C2H2
H–C
C3H4
CH3 – C
C4H6
C5H8
Derived Name
IUPAC name
Acetylene
Ethyne
Methyl acetylene
Propyne
Ethyl acetylene
1–Butyne
Dimethyl acetylene
2– Butyne
n-Propyl acetylene
1–Pentyne
Ethyl methyl acetylene
2-Pentyne
Isopropyl acetylene
3-Methyl- 1-butyne
C–H
C–H
CH3–CH2 – C  CH
CH3 – CH2 – CH2 – C  CH
C H 3 – CH – C  C H
C H3
ETHYNE
(a) Structure :
It is also known as acetylene. Acetylene is the first
member of alkyne series and has a linear geometry.
It has sp hybridisation.The carbon-carbon triple bond
distance and carbon-hydrogen bond distance have
been found to be 1.20 Å and 1.06 Å respectively. The
180º
H
C
C
H
carbon-carbon hydrogen bond angle is 180º.
Linear
12
12
PAGE # 12
(b) Properties :
It is a colourless gas which is slightly soluble in water.
Pure ethyne has ethereal odour. Acetylene burns with
luminous flame like aromatic compounds. This is a
highly exothermic reaction.

TEST FOR ALKANES, ALKENES AND ALKYNES
(a) Alkanes :
(i) Bromine water test: It does not decolourise the
bromine water.
Note :
The temperature of oxyacetylene flame is about
3000ºC and is used for welding and cutting steel.
(ii) Baeyer’s test: It does not, react with Baeyer’s
reagent (alkaline solution of KMnO4).
(c) Preparation :
(b) Alkenes:
(i) From carbon and hydrogen (Direct synthesis ) : When
an electric arc is struck between carbon (graphite) rods
in an atmosphere of hydrogen, acetylene is formed.
(i) Bromine water test: It decolourises the orange
colour of Bromine water.
2C + H2
1200ºC
(iii) Silver nitrate Test: No reaction
H
H
C=C
C2H2
H
(ii) From calcium carbide (Lab. Method) :
H
Ethene
H
CCl
4
+ Br2
Bromine water
(red-brown colour)
H Br Br H
1,2-Dibromoethane
(Colourless)
Ca(OH)2 + C2H2
CaC2 + 2H2O
Calcium
carbide
H
C–C
(ii) Baeyer’s test: It decolourises the purple colour of
Baeyer’s reagent.
Calcium Ethyne
hydroxide
(iii) Dehydrohalogenation of dihaloalkanes :
(iii) Silver nitrate Test: No reaction
(d) Chemical Properties :
(c) Alkynes :
(i) Addition of halogens :
(i) Bromine water test : It decolourises the Br2 water.
H – C  C – H + Br2
Ethyne
(ii) Addition of Halogen acid :
Cl
HC  CH + HCl
H2C = CH
Ethyne
HCl
Cl
Chloroethene
(Vinyl chloride)
H2
Ni
Ethyne
HC  CH
Br
1,2-Dibromoethene
Br2
H
H
Br – C – C – Br
Br Br
1,1,2,2-Tetrabromoethane
(Colourless)
(ii) Baeyer’s test : It also decolourises the purple
colour of alkaline KMnO4 .
Cl
1,1-Dichloroethane
(Gem dihalide)
(iii) Silver nitrate Test : It gives white precipitate
H2C = CH2
Ethene
H2
H
C=C
H3C – CH
(iii) Hydrogenation :
HC  CH
Bromine
water
H
Br
H2 /Ni
CH3 – CH3
Ethane
H2C = CH2
Ethene
Pd/BaSO4
Ethyne
(iv) Combustion :
2C2H2 + 5O2
Ethyne
4CO2 + 2H2O + Heat
(v) Polymerisation :
H
3HC  CH
Ethyne
Fe

H
H
H
H
or (C6H6)
H
Benzene
13
13
PAGE # 13
MOLE CONCEPT
Symbol Derived from English Names
ATOMS
All the matter is made up of atoms. An atom is the
smallest particle of an element that can take part in a
chemical reaction. Atoms of most of the elements
are very reactive and do not exist in the free state (as
single atom).They exist in combination with the atoms
of the same element or another element.
Atoms are very, very small in size. The size of an atom
is indicated by its radius which is called "atomic
radius" (radius of an atom). Atomic radius is
measured in "nanometres"(nm). 1 metre = 10 9
nanometre or 1nm = 10-9 m.
Atoms are so small that we cannot see them under
the most powerful optical microscope.

Note :
Hydrogen atom is the smallest atom of all , having an
atomic radius 0.037nm.
(a) Symbols of Elements :
A symbol is a short hand notation of an element which
can be represented by a sketch or letter etc.
Dalton was the first to use symbols to represent
elements in a short way but Dalton's symbols for
element were difficult to draw and inconvenient to
use, so Dalton's symbols are only of historical
importance. They are not used at all.

English Name of
the Element
Symbol
Hydrogen
H
Helium
He
Lithium
Li
Boron
B
Carbon
C
Nitrogen
N
Oxygen
O
Fluorine
F
Neon
Ne
Magnesium
Mg
Aluminium
Al
Silicon
Si
Phosphorous
P
Sulphur
S
Chlorine
Cl
Argon
Ar
Calcium
Ca
Symbols Derived from Latin Names
It was J.J. Berzelius who proposed the modern
system of representing an element.
The symbol of an element is the "first letter" or the
"first letter and another letter" of the English name or
the Latin name of the element.
e.g. The symbol of Hydrogen is H.
The symbol of Oxygen is O.
There are some elements whose names begin with
the same letter. For example, the names of elements
Carbon, Calcium, Chlorine and Copper all begin with
the letter C. In such cases, one of the elements is
given a "one letter "symbol but all other elements are
given a "first letter and another letter" symbol of the
English or Latin name of the element. This is to be
noted that "another letter" may or may not be the
"second letter" of the name. Thus,
The symbol of Carbon is C.
The symbol of Calcium is Ca.
The symbol of Chlorine is Cl.
The symbol of Copper is Cu (from its Latin name
Cuprum)
It should be noted that in a "two letter" symbol, the
first letter is the "capital letter" but the second letter is
the small letter
English Name of
the Element
Symbol
Sodium
Na
Potassium
K
(b) Significance
Element :
of
Latin Name of
the Element
Natrium
Kalium
the
Symbol
of
an
(i) Symbol represents name of the element.
(ii) Symbol represents one atom of the element.
(iii) Symbol also represents one mole of the element.
That is, symbol also represent 6.023 × 1023 atoms of
the element.
(iv) Symbol represent a definite mass of the element
i.e. atomic mass of the element.
Example :
(i) Symbol H represents hydrogen element.
(ii) Symbol H also represents one atom of hydrogen
element.
(iii) Symbol H also represents one mole of hydrogen
atom.
(iv) Symbol H also represents one gram hydrogen
atom.
CLASS-XI_STREAM-SA_PAGE # 14
E.g. A nitrogen atom gains 3 electrons to form nitride
ion, so nitride ion bears 3 units of negative charge
and it is represented as N3-.
IONS
An ion is a positively or negatively charged atom or
group of atoms.
Every atom contains equal number of electrons
(negatively charged) and protons (positively charged).
Both charges balance each other, hence atom is
electrically neutral.
Note :
Size of a cation is always smaller and anion is always
greater than that of the corresponding neutral atom.

(c) Monoatomic ions and polyatomic ions :
(a) Cation :
(i) Monoatomic ions : Those ions which are formed
from single atoms are called monoatomic ions or
simple ions.
E.g. Na+, Mg2+ etc.
If an atom has less electrons than a neutral atom,
then it gets positively charged and a positively
charged ion is known as cation.
E.g. Sodium ion (Na+), Magnesium ion (Mg2+) etc.
A cation bears that much units of positive charge as
there are the number of electrons lost by the neutral
atom to form that cation.
(ii) Polyatomic ions : Those ions which are formed
from group of atoms joined together are called
polyatomic ions or compound ions.
E.g. Ammonium ion (NH4+) , hydroxide ion (OH–) etc.
which are formed by the joining of two types of atoms,
nitrogen and hydrogen in the first case and oxygen and
hydrogen in the second.
E.g. An aluminium atom loses 3 electrons to form
aluminium ion, so aluminium ion bears 3 units of
positive charge and it is represented as Al3+.
(b) Anion :
If an atom has more number of electrons than that of
neutral atom, then it gets negatively charged and a
negatively charged ion is known as anion.
E.g. Chloride ion (Cl¯), Oxide ion (O2-) etc.
(d) Valency of ions :
The valency of an ion is same as the charge present
on the ion.
If an ion has 1 unit of positive charge, its valency is 1
and it is known as a monovalent cation. If an ion has
2 units of negative charge, its valency is 2 and it is
known as a divalent anion.
An anion bears that much units of negative charge
as there are the number of electrons gained by the
neutral atom to form that anion.
LIST OF COMMON ELECTROVALENT POSITIVE RADICALS
Monovalent Electropositive Bivalent
Electropositive
1. Hydrogen
H+
1. Magnesium
Mg
2+
1. Aluminium
2. Ammonium
3. Sodium
NH4
Na +
2. Calcium
Ca
2+
2. Ferric [Iron (III)] Fe
Zn
2+
4. Potassium
K+
4. Plumbous [Lead (II)]
Pb
5. Cuprous [Copper (I)]
Cu
5. Cupric [(Copper) (II)]
Cu
6. Argentous [Silver (I)]
Ag +
6. Argentic [Silver(II)]
Ag
7. Stannous [Tin (II)]
8. Ferrous [Iron (II)]
Sn
2+
Fe
+
+
2+
7. Mercurous [Mercury(I)] Hg2
3. Zinc
3. Chromium
Al
3+
1. Stannic [Tin (IV)]
Sn
3+
2. Plumbic [Lead (IV)]
Pb
Cr
4+
4+
3+
2+
2+
2+
2+
9. Mercuric [Mercury (II)] Hg
10. Barium
Tetravalent
Electropositive
Trivalent
Electropositive
Ba
2+
2+
LIST OF COMMON ELECTROVALENT NEGATIVE RADICALS
Bivalent
Electronegative
Monovalent Electronegative
1. Fluoride
F–
2. Chloride
Cl
3. Bromide
Br
I
4. Iodide
5. Hydride
6. Hydroxide
7. Nitrite
8.Nitrate
12. Bisulphate or Hydrogen sulphate
13. Acetate
1. Nitride
N
2. Phosphide
P
3. Phosphite
PO 3
4. Phosphate
PO 4
SO 4
2. Sulphite
SO 3
–
3. Sulphide
S
4. Thiosulphate
S 2O3
–
H
22-
5. Zincate
ZnO2
6. Oxide
O
–
2
7. Peroxide
O2
8. Dichromate
Cr 2O7
9. Carbonate
CO 3
10. Silicate
SiO 3
NO
–
NO3
–
HSO3
–
HS
Tetravalent
Electronegative
1. Carbide
C
4-
333-
2-
–
OH
3-
2-
2-
1. Sulphate
–
9. Bicarbonate or Hydrogen carbonate HCO3–
10. Bisulphite or Hydrogen sulphite
11. Bisulphide or Hydrogen sulphide
Trivalent
Electronegative
222-
22-
–
HSO4
–
CH3COO
CLASS-XI_STREAM-SA_PAGE # 15

Atomic
Number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Note :
Cation contains less no. of electrons and anion
contains more no. of electrons than the no. of protons
present in them.
ATOMIC MASS UNIT
The atomic mass unit (amu) is equal to one-twelfth
(1/12) of the mass of an atom of carbon-12.The mass
of an atom of carbon-12 isotope was given the atomic
mass of 12 units, i.e. 12 amu or 12 u.
The atomic masses of all other elements are now
expressed in atomic mass units.
RELATIVE ATOMIC MASS
The relative atomic mass of an element is a relative
quantity and it is the mass of one atom of the element
relative to one -twelfth (1/12) of the mass of one
carbon-12 atom. Thus, Relative atomic mass
Element
Symbol
Hydrogen
H
Helium
He
Lithium
Li
Beryllium
Be
Boron
B
Carbon
C
Nitrogen
N
Oxygen
O
Fluorine
F
Neon
Ne
Sodium
Na
Magnesium
Mg
Aluminium
Al
Silicon
Si
Phosphorus
P
Sulphur
S
Chlorine
Cl
Argon
Ar
Potassium
K
Calcium
Ca
Atomic
mass
1
4
7
9
11
12
14
16
19
20
23
24
27
28
31
32
35.5
40
39
40
Mass of one atom of the element
1
=
12
RELATIVE MOLECULAR MASS
 mass of one C  12 atom
[1/12 the mass of one C-12 atom = 1 amu, 1 amu =
1.66 × 10–24 g = 1.66 × 10–27 kg.]

Note :
The relative molecular mass of a substance is the
mass of a molecule of the substance as compared
to one-twelfth of the mass of one carbon -12 atom
i.e.,
Relative molecular mass
One amu is also called one dalton (Da).
=
Mass of one molecule of the substance
1
GRAM-ATOMIC MASS
The atomic mass of an element expressed in grams
is called the Gram Atomic Mass of the element.
The number of gram -atoms
=
Mass of the element in grams
Gram atomic mass of the element
12
 mass of one C  12 atom
The molecular mass of a molecule, thus, represents
the number of times it is heavier than 1/12 the mass
of an atom of carbon-12 isotope.
GRAM MOLECULAR MASS
The molecular mass of a substance expressed in
grams is called the Gram Molecular Mass of the
substance . The number of gram molecules
e.g.
Calculate the gram atoms present in (i) 16g of oxygen
Mass of the subs tance in grams
=
Gram Molecular mass of the subs t ance
and (ii) 64g of sulphur.
e.g.
(i) The atomic mass of oxygen = 16.
 Gram-Atomic Mass of oxygen (O) = 16 g.
16
=1
16
(ii) The gram-atoms present in 64 grams of sulphur.
No. of Gram-Atoms =
64
64
= Gram Atomic Mass of sulphur =
=2
32
(i) Molecular mass of hydrogen (H2) = 2u.
 Gram Molecular Mass of hydrogen (H2) = 2 g .
(ii) Molecular mass of methane (CH4) = 16u
 Gram Molecular Mass of methane (CH4) = 16 g.
e.g. the number of gram molecules present in 64 g of
methane (CH4).
=
64
64
Gram molecular mass of CH4 = 16 = 4.
CLASS-XI_STREAM-SA_PAGE # 16
(a) Calculation of Molecular Mass :
The molecular mass of a substance is the sum of
the atomic masses of its constituent atoms present
in a molecule.
Ex.1 Calculate the molecular mass of water.
(Atomic masses : H = 1u, O = 16u).
Sol. The molecular formula of water is H2O.
Molecular mass of water = ( 2 × atomic mass of
H) + (1 × atomic mass of O)
= 2 × 1 + 1 × 16 = 18
i.e., molecular mass of water = 18 amu.
Ex.2 Find out the molecular mass of sulphuric acid.
(Atomic mass : H = 1u, O = 16u, S = 32u).
Sol. The molecular formula of sulphuric acid is H2SO4.
 Molecular mass of H2SO4
= (2 × atomic mass of H) + ( 1 × atomic mass of S)
+ ( 4 × atomic mass of O)
= (2 × 1) + (1× 32) + (4×16) = 2 + 32 + 64 = 98
i.e., Molecular mass of H2SO4= 98 amu.
FORMULA MASS
The term ‘formula mass’ is used for ionic compounds
and others where discrete molecules do not exist,
e.g., sodium chloride, which is best represented as
(Na+Cl–)n, but for reasons of simplicity as NaCl or
Na+Cl–. Here, formula mass means the sum of the
masses of all the species in the formula.
Thus, the formula mass of sodium chloride = (atomic
mass of sodium) + (atomic mass of chlorine)
= 23 + 35.5
= 58.5 amu
EQUIVALENT MASS
(a) Definition :
Equivalent mass of an element is the mass of the
element which combine with or displaces 1.008 parts
by mass of hydrogen or 8 parts by mass of oxygen or
35.5 parts by mass of chlorine.
(b) Formulae of Equivalent Masses of different
substances :
(i) Equivalent mass of an element =
Atomic wt. of the element
Valency of the element
(ii) Eq. mass an acid =
Mol. wt. of the acid
Basicity of the acid
Basicity is the number of replaceable H+ ions from
one molecule of the acid.
(iii) Eq. Mass of a base =
Mol. wt. of the base
Acidity of the base
Acidity is the number of replaceable OH– ions from
one molecule of the base
(iv) Eq. mass of a salt
Mol. wt. of the salt
= Number of metal atoms  valency of metal
(v) Eq. mass of an ion =
Formula wt. of the ion
Charge on the ion
(vi) Eq. mass of an oxidizing/reducing agent
=
Mol wt. or At. wt
No. of electrons lost or gained by one
molecule of the substance
Equivalent weight of some compounds are given in
the table :
S.No.
Compound
Equivalent
weight
1
HCl
36.5
2
H2SO4
49
3
HNO3
63
4
45
COOH
5
COOH
.2H2O
63
6
NaOH
40
7
KOH
56
8
CaCO3
50
9
NaCl
58.5
10
Na2CO3
53
In Latin, mole means heap or collection or pile. A
mole of atoms is a collection of atoms whose total
mass is the number of grams equal to the atomic
mass in magnitude. Since an equal number of moles
of different elements contain an equal number of
atoms, it becomes convenient to express the
amounts of the elements in terms of moles. A mole
represents a definite number of particles, viz, atoms,
molecules, ions or electrons. This definite number is
called the Avogadro Number (now called the Avogadro
constant) which is equal to 6.023 × 1023.
A mole is defined as the amount of a substance that
contains as many atoms, molecules, ions, electrons
or other elementary particles as there are atoms in
exactly 12 g of carbon -12 (12C).
CLASS-XI_STREAM-SA_PAGE # 17
(a) Moles of Atoms :
(i) 1 mole atoms of any element occupy a mass which
is equal to the Gram Atomic Mass of that element.
e.g. 1 Mole of oxygen atoms weigh equal to Gram
Atomic Mass of oxygen, i.e. 16 grams.
(iv) Number of moles of molecules
=
Mass of substance in grams
Gram Molecular Mass of substance
(v) Number of moles of molecules
No. of molecules of element
Ex.3 To calculate the number of moles in 16 grams of
Sulphur (Atomic mass of Sulphur = 32 u).
(b) Moles of Molecules :
Sol. 1 mole of atoms = Gram Atomic Mass.
(i) 1 mole molecules of any substance occupy a mass
which is equal to the Gram Molecular Mass of that
substance.
e.g. : 1 mole of water (H2O) molecules weigh equal to
Gram Molecular Mass of water (H2O), i.e. 18 grams.
(ii) The symbol of a compound represents 6.023 x
10 23 molecules (1 mole of molecules) of that
compound.
=
molecules and 2 H2O represents 2 moles of water
molecules.

22.4 litre
In term of
volume
23
6.023 × 10
(NA) Atoms
23
6.023 × 10
(NA) molecules
In terms of
particles
1 Mole
In terms of
mass
1 gram atom
of element
1 gram molecule
of substance
1 gram formula
mass of substance
Note :
The SI unit of the amount of a substance is Mole.
(c) Mole in Terms of Volume :
Volume occupied by 1 Gram Molecular Mass or 1
mole of a gas under standard conditions of
temperature and pressure (273 K and 1atm.
pressure) is called Gram Molecular Volume. Its value
is 22.4 litres for each gas.
Volume of 1 mole of gas = 22.4 litre (at STP)

NA
So, 1 mole of Sulphur atoms = Gram Atomic Mass of
Sulphur = 32 grams.
Now, 32 grams of Sulphur = 1 mole of Sulphur
So, 16 grams of Sulphur
= (1/32) x 16 = 0.5 moles
Thus, 16 grams of Sulphur constitute 0.5 mole of
Sulphur.
Note :
The symbol H2O does not represent 1 mole of H2
molecules and 1 mole of O atoms. Instead, it
represents 2 moles of hydrogen atoms and 1 mole
of oxygen atoms.

Avogadro number
e.g. : Symbol H 2O represents 1 mole of water

N
(ii) The symbol of an element represents 6.023 x 1023
atoms (1 mole of atoms) of that element.
e.g: Symbol N represents 1 mole of nitrogen atoms
and 2N represents 2 moles of nitrogen atoms.
PROBLEMS BASED ON THE MOLE CONCEPT
Ex.4 Calculate the number of moles in 5.75 g of
sodium. (Atomic mass of sodium = 23 u)
Sol. Number of moles
Mass of the element in grams
=
Gram Atomic Mass of element
Note :
The term mole was introduced by Ostwald in 1896.
=
5.75
= 0.25 mole
23
or
1 mole of sodium atoms = Gram Atomic mass of
SOME IMPORTANT RELATIONS AND FORMULAE
sodium = 23g.
23 g of sodium = 1 mole of sodium.
(i) 1 mole of atoms = Gram Atomic mass = mass of
6.023 × 1023 atoms
(ii) 1 mole of molecules = Gram Molecular Mass
= 6.023 x 1023 molecules
(iii) Number of moles of atoms
=
Mass of element in grams
Gram Atomic Mass of element
5.75 g of sodium =
5.75
23
mole of sodium = 0.25 mole
Ex.5 What is the mass in grams of a single atom of
chlorine ? (Atomic mass of chlorine = 35.5u)
Sol. Mass of 6.023 × 1023 atoms of Cl = Gram Atomic
Mass of Cl = 35.5 g.
 Mass of 1 atom of Cl =
35.5 g
6.023  10 23
= 5.9 × 10–23 g.
CLASS-XI_STREAM-SA_PAGE # 18
Ex.6 The density of mercury is 13.6 g cm–3. How many
moles of mercury are there in 1 litre of the metal ?
(Atomic mass of Hg = 200 u).
Sol. Mass of mercury (Hg) in grams = Density
(g cm–3)× Volume (cm3)
= 13.6 g cm–3 × 1000 cm3 = 13600 g.
 Number of moles of mercury
Mass of mercury in grams
13600
= Gram Atomic Mass of mercury =
200
= 68
Ex.7 The mass of a single atom of an element M is
3.15× 10–23 g . What is its atomic mass ? What
Ex.11 What mass in grams is represented by
(a) 0.40 mol of CO2,
(b) 3.00 mol of NH3,
(c) 5.14 mol of H5IO6
(Atomic masses : C = 12u,O = 16u, N = 14 u, H = 1u
andI = 127u)
Sol. Weight in grams = number of moles × molecular
mass.
Hence,
(a) mass of CO2 = 0.40 × 44 = 17.6 g
(b) mass of NH3 = 3.00 × 17 = 51.0 g
(c) mass of H5IO6 = 5.14 × 228 = 1171.92g
Ex.12 Calculate the volume in litres of 20 g of hydrogen
gas at STP.
Sol. Number of moles of hydrogen
could the element be ?
Mass of hydrogen in grams
Sol. Gram Atomic Mass = mass of 6.023 × 1023 atoms
= mass of 1 atom × 6.023 × 1023
=
Gram Molecular Mass of hydrogen
Molecular Volume.
= 10 ×22.4 = 224 litres.
= 3.15 × 6.023 g = 18.97 g.
 Atomic Mass of the element = 18.97u
Thus, the element is most likely to be fluorine.
Ex.13 The molecular mass of H 2 SO 4 is 98 amu.
Calculate the number of moles of each element
in 294 g of H2SO4.
Ex.8 An atom of neon has a mass of 3.35 × 10–23 g.
How many atoms of neon are there in 20 g of the
gas ?
Sol. Number of atoms
20
Total mass
=
= 5.97 × 1023
Mass of 1 atom
3.35  10 – 23
Ex.9 How many grams of sodium will have the same
number of atoms as atoms present in 6 g of
(Atomic masses : Na = 23u ; Mg =24u)
Sol. Number of gram -atom of Mg
1
6
Sol. Number of moles of H2SO4 =
= Gram Atomic Mass =
=
24
4
1
4
 1 Gram Atom of sodium = 23 g
gram atoms of sodium = 23 ×
1
= 5.75 g
4
Ex.10 How many moles of Cr are there in 85g of Cr2S3 ?
(Atomic masses : Cr = 52 u , S =32 u)
Sol. Molecular mass of Cr2S3 = 2 × 52 + 3 × 32 = 104
+ 96 = 200 u.
200g of Cr2S3 contains = 104 g of Cr.
 85 g of Cr2S3 contains =
104  85
200
Thus, number of moles of Cr =
=3.
1000
 Gram Atoms of sodium should be =
1
98
The formula H 2SO 4 indicates that 1 molecule of
H2SO4 contains 2 atoms of H, 1 atom of S and 4
atoms of O. Thus, 1 mole of H2SO4 will contain 2
moles of H,1 mole of S and 4 moles of O atoms
Therefore, in 3 moles of H2SO4 :
Number of moles of H = 2 × 3 = 6
Number of moles of S = 1 × 3 = 3
Number of moles of O = 4 × 3 = 12
=3×
4
294
Ex.14 Find the mass of oxygen contained in 1 kg of
potassium nitrate (KNO3).
Sol. Since 1 molecule of KNO 3 contains 3 atoms of
oxygen, 1 mol of KNO 3 contains 3 moles of
oxygen atoms.
 Moles of oxygen atoms = 3 × moles of KNO3
magnesium ?
Mass of Mg in grams
20
= 10
2
 Volume of hydrogen = number of moles × Gram
= (3.15 × 10–23g) × 6.023 × 1023
=
=
g of Cr = 44.2g
44.2
52
101
= 29.7
(Gram Molecular Mass of KNO3 = 101 g)
 Mass of oxygen = Number of moles × Atomic
mass
= 29.7 × 16 = 475.2 g
Ex.15 You are asked by your teacher to buy 10 moles of
distilled water from a shop where small bottles
each containing 20 g of such water are available.
How many bottles will you buy ?
Sol. Gram Molecular Mass of water (H2O) = 18 g
 10 mol of distilled water = 18 × 10 = 180 g.
Because 20 g distilled water is contained in 1
bottle,
180 g of distilled water is contained in =
bottles = 9 bottles.
= 0.85 .
 Number of bottles to be bought = 9
CLASS-XI_STREAM-SA_PAGE # 19
180
20
Ex.16 6.023 × 1023 molecules of oxygen (O2) is equal to
(b) Molarity :
Molarity of a solution is defined as the number of
moles of the solute dissolved per litre (or dm3) of
solution. It is denoted by ‘M’. Mathematically,
how many moles ?
Sol. No. of moles =
No. of molecules of oxygen
Avogadro' s no. of molecules
N

N
=
A
6.023  1023
6.023  1023
=1
Number of moles of solute
M=
PERCENTAGE COMPOSITION
Volume of the solution in litre
Mass of solute in gram/Gram Molecular Mass of solute
The percentage composition of elements in a
compound is calculated from the molecular formula
=
Volume of solution in litre
M can be calculated from the strength as given below :
of the compound.
The molecular mass of the compound is calculated
M=
from the atomic masses of the various elements
present in the compound. The percentage by mass
of each element is then computed with the help of the
Molecular mass of solute
If ‘w’ gram of the solute is present in V cm3 of a given
solution , then
1000
w
following relations.
M=
Percentage mass of the element in the compound
Total mass of the element
=
Strength in grams per litre
×
Molecular mass
V
e.g. a solution of sulphuric acid having 4.9 grams of it
dissolved in 500 cm3 of solution will have its molarity,
× 100
Molecular mass
w
M=
Ex.17 What is the percentage of calcium in calcium
carbonate (CaCO3) ?
Sol. Molecular mass of CaCO 3 = 40 + 12 + 3 × 16
= 100 amu.
M=
1000
Molecular mass
4.9
98
1000
×
500
×
V
= 0.1
(c) Formality :
Mass of calcium in 1 mol of CaCO3 = 40g.
Percentage of calcium =
40  100
100
= 40%
Ex.18 What is the percentage of sulphur in sulphuric
acid (H2SO4) ?
Sol. Molecular mass of H2SO4 = 1 × 2 + 32 + 16 × 4 = 98 amu.
 Percentage of sulphur =
32  100
98
= 32.65%
Mass of solute in gram/Formula Mass of solute
Ex.19 W hat are the percentage compositions of
hydrogen and oxygen in water (H2O) ?
(Atomic masses : H = 1 u, O = 16 u)
Sol. Molecular mass of water, H2O = 2 + 16 = 18 amu.
H2O has two atoms of hydrogen.
So, total mass of hydrogen in H2O = 2 amu.
Percentage of H =
2  100
18
In case of ionic compounds like NaCl, Na2CO3 etc.,
formality is used in place of molarity. The formality of
a solution is defined as the number of gram formula
masses of the solute dissolved per litre of the
solution. It is represented by the symbol ‘F’. The term
formula mass is used in place of molecular mass
because ionic compounds exist as ions and not as
molecules. Formula mass is the sum of the atomic
masses of the atoms in the formula of the compound.
= 11.11%
Volume of solution in litre
(d) Normality :
Normality of a solution is defined as the number of
gram equivalents of the solute dissolved per litre (dm3)
of given solution. It is denoted by ‘N’.
Mathematically,
Number of gram equivalents of solute
Similarly,
percentage of oxygen =
16  100
18
N =
= 88.89%
CONCENTRATION OF SOLUTIONS
(a) Strength in g/L :
The strength of a solution is defined as the amount of
the solute in grams present in one litre (or dm3) of the
solution, and hence is expressed in g/litre or g/dm3.
N=
Volume of the solution in litre
Weight of solute in gram / equivalent weight of solute
Volume of the solution in litre
N can be calculated from the strength as given below :
Strength in grams per litre
N=
Equivalent mass of solute
=
S
E
Mass of solute in gram
Strength in g/L =
Volume of solution in litre
CLASS-XI_STREAM-SA_PAGE # 20
If ‘w’ gram of the solute is present in V cm3 of a given
solution.

1 milli equivalent of an acid neutralizes 1 milli
equivalent of a base.
1000
w
N = Equivalent mass of the solute ×
V
e.g. A solution of sulphuric acid having 0.49 gram of
it dissolved in 250 cm 3 of solution will have its
normality,
w
N = Equivalent mass of the solute ×
0.49
N=
49
1000
×
(e) Molality :
Molality of a solution is defined as the number of
moles of the solute dissolved in 1000 grams of the
solvent. It is denoted by ‘m’.
Mathematically,
1000
V
= 0.04
250
m=
(Eq. mass of sulphuric acid = 49).
Semi
normal
Solution
Deci
normal
‘m’ can be calculated from the strength as given below :
1
100
m=
(i) Milli equivalent of substance = N × V
where , N  normality of solution
V  Volume of solution in mL
m=
(ii) If weight of substance is given,
w  1000
E
(iii) S = N × E
S  Solubility in g/L
N  Normality of solution
E  Equivalent weight
m=

(iv) Calculation of normality of mixture :
N
10
N
HCl is mixed with 50 ml of
5
H2SO4 .
Find out the normality of the mixture.
Sol. Milli equivalent of HCl + milli equivalent H 2SO 4
= milli equivalent of mixture
N1 V1 + N2 V2 = N3 V3
{ where, V3 =V1 + V2 )
 1
 1

 100     50   N3 × 150

 10
 5

20
N3 =
150
2
=
15
Note :
Relationship Between Normality and Molarity of a
Solution :
Normality of an acid = Molarity × Basicity
Normality of base = Molarity × Acidity
Ex.22 Calculate the molarity and normality of a solution
containing 0.5 g of NaOH dissolved in 500 cm3 of
solvent.
Sol. Weight of NaOH dissolved = 0.5 g
Volume of the solution = 500 cm3
(i) Calculation of molarity :
Molecular weight of NaOH = 23 + 16 + 1 = 40
Molarity =
10
HCl is mixed with 25 ml of
=
N
5
NaOH. Find out the normality of the mixture.
Sol. Milli equivalent of HCl – milli equivalent of NaOH
= milli equivalent of mixture
N1 V1 – N2 V2 = N3 V3
{ where, V3 =V1 + V2 )
 1
 1

 100  –   25  = N3 × 125

10
5

 

1
N3 =
25
1.325 1000

= 0.05
106 250
= 0.133
N
Ex.21 100 ml of
1000
w
×
Mol. mass of the solute
W
e.g. A solution of anhydrous sodium carbonate
(molecular mass = 106) having 1.325 grams of it,
dissolved in 250 gram of water will have its molality -
Where, W  Weight of substance in gram
E  Equivalent weight of substance
Ex.20 100 ml of
Strength per 1000 gram of solvent
Molecular mass of solute
If ‘w’ gram of the solute is dissolved in ‘W’ gram of the
solvent then
Some Important Formulae :
milli equivalent (NV) =
Number of moles of the solute
× 1000
Weight of the solvent in gram
Centi
normal
1
10
1
2
Normality
Note :
Weight of solute/ molecular weight of solute
Volume of solution in litre
0.5/40
= 0.025
500/1000
(ii) Calculation of normality :
Normality
=
=
Weight of solute/ equivalent weight of solute
Volume of solution in litre
0.5/40
500/1000
= 0.025
CLASS-XI_STREAM-SA_PAGE # 21
Ex.23 Find the molarity and molality of a 15% solution
of H2SO4 (density of H2SO4 solution = 1.020 g/cm3)
(Atomic mass : H = 1u, O = 16u , S = 32 u)
Sol. 15% solution of H2SO4 means 15g of H2SO4 are
present in 100g of the solution i.e.
Wt. of H2SO4 dissolved = 15 g
Weight of the solution = 100 g
Density of the solution = 1.02 g/cm3 (Given)
15
98
urea solution.
weight of solute (urea) = 10 g
weight of solution = 100 g
weight of solvent (water) = 100 – 10 = 90g
Moles of solute
mole fraction of solute =
w/m
Calculation of molality :
Weight of solution = 100 g
Weight of H2SO4 = 15 g
Wt. of water (solvent) = 100 – 15 = 85 g
Molecular weight of H2SO4 = 98
 15 g H2SO4 =
Ex.24 Find out the mole fraction of solute in 10% (by weight)
=
10 / 60
w/m  W/M

Moles of solution
=
10 / 60  90 / 18
= 0.032
Note :
Sum of mole fraction of solute and solvent is always
equal to one.
= 0.153 moles
Thus ,85 g of the solvent contain 0.153 moles .
1000 g of the solvent contain=
STOICHIOMETRY
0.153
85
× 1000 = 1.8 mole
(a) Quantitative Relations in Chemical
Hence ,the molality of H2SO4 solution = 1.8 m
Reactions :
Calculation of molarity :
15 g of H2SO4 = 0.153 moles
Stoichiometry is the calculation of the quantities of
reactants and products involved in a chemical
Wt. of solution
Vol. of solution =
=
100
1.02
reaction.
Density of solution
It is based on the chemical equation and on the
= 98.04 cm3
relationship between mass and moles.
This 98.04 cm3 of solution contain H2SO4 = 0.153
moles
1000 cm3 of solution contain H2SO4
0.153
=
98.04
1 molecule N2 + 3 molecules H2  2 molecules
1 mol N2 + 3 mol H2  2 mol NH3
(Molar interpretation)
MOLE FRACTION
28 g N2 + 6 g H2  34 g NH3
The ratio between the moles of solute or solvent to
the total moles of solution is called mole fraction.
Moles of solute
n

Mole fraction of solute =
n

N
Moles of solution
w/m
(Mass interpretation)
1 volume N2 + 3 volume H2  2 volume NH3
(Volume interpretation)
Thus, calculations based on chemical equations are
divided into four types -
w/m  W/M
Moles of solvent
Mole fraction of solvent =
Moles of solution
W/M
=
A chemical equation can be interpreted as follows -
NH 3 (Molecular interpretation)
× 1000 = 1.56 moles
Hence the molarity of H2SO4 solution = 1.56 M
=
N2(g) + 3H2(g)  2NH3(g)
w/m  W/M
where,
n  number of moles of solute
N  number of moles of solvent
m  molecular weight of solute
M  molecular weight of solvent
w  weight of solute
W  weight of solvent

N
nN
(i) Calculations based on mole-mole relationship.
(ii) Calculations based on mass-mass relationship.
(iii) Calculations based on mass-volume relationship.
(iv) Calculations based on volume -volume
relationship.
(i) Calculations based on mole-mole relationship :
In such calculations, number of moles of reactants
are given and those of products are required.
Conversely, if number of moles of products are given,
then number of moles of reactants are required.
CLASS-XI_STREAM-SA_PAGE # 22
Ex.25 Oxygen is prepared by catalytic decomposition
of potassium chlorate (KClO 3). Decomposition
of potassium chlorate gives potassium chloride
(KCl) and oxygen (O2). How many moles and how
many grams of KClO 3 are required to produce
2.4 mole O2.
Ex.28 How many grams of oxygen are required to burn
completely 570 g of octane ?
Sol. Balanced equation
2C8H18 + 25O2
16CO2 + 18H2O
25 mole
25 × 32
2 mole
2 × 114
First method : For burning 2 × 114 g of the octane,
oxygen required = 25 × 32 g
Sol. Decomposition of KClO3 takes place as,
25  32
g
2  114
2KClO3(s)  2KCl(s) + 3O2(g)
For burning 1 g of octane, oxygen required =
2 mole KClO3  3 mole O2
Thus, for burning 570 g of octane, oxygen required
 3 mole O2 formed by 2 mole KClO3
2

 2.4 mole O 2 will be formed by   2.4 
3


mole KClO3 = 1.6 mole KClO3
Mass of KClO3 = Number of moles × molar mass
= 1.6 × 122.5 = 196 g
(ii) Calculations based on mass-mass relationship:
In making necessary calculation, following steps are
=
Mole Method : Number of moles of octane in 570
grams
570
= 5.0
114
For burning 2.0 moles of octane, oxygen required
= 25 mol = 25 × 32 g
For burning 5 moles of octane, oxygen required
followed =
(a) Write down the balanced chemical equation.
(b) Write down theoretical amount of reactants and
products involved in the reaction.
(c) The unknown amount of substance is calculated
using unitary method.
25  32
× 570 g = 2000 g
2  114
25  32
× 5.0 g = 2000 g
2 .0
Proportion Method : Let x g of oxygen be required for
burning 570 g of octane. It is known that 2 × 114 g of
the octane requires 25 × 32 g of oxygen; then, the
proportion.
x
25  32 g oxygen
=
570
g
oc
tan e
2  114 g oc tan e
Ex.26 Calculate the mass of CaO that can be prepared
by heating 200 kg of limestone (CaCO3) which is
25  32  570
= 2000 g
2  114
x=
95% pure.
95
 200 = 190 kg
Sol. Amount of pure CaCO 3 =
100
= 190000 g
CaCO3(s)  CaO(s) + CO2(g)
1 mole CaCO3  1 mole CaO
100 g CaCO3  56 g CaO
 100 g CaCO3 give 56 g CaO
56
 190000 g CaCO3 will give=
× 190000 g CaO
100
= 106400 g = 106.4 kg
dioxide
4FeS2 + 15O2 + 8H2O
2Fe2O3 + 8H2SO4
4 mole
4 × 120 g
8 mole
8 × 98 g
4 × 120 g of FeS2 yield H2SO4 = 8 × 98 g
1000 g of FeS2 will yield H2SO4 =
Ex.27 Chlorine is prepared in the laboratory by treating
manganese
Ex.29 How many kilograms of pure H 2SO 4 could be
obtained from 1 kg of iron pyrites (FeS2) according to
the following reactions ?
4FeS2 + 11O2  2Fe2O3 + 8SO2
2SO2 + O2  2SO3
SO3 + H2O  H2SO4
Sol. Final balanced equation,
(MnO 2 )
with
aqueous
8  98
× 1000
4  120
= 1633.3 g
(iii) Calculations involving mass-volume relationship :
hydrochloric acid according to the reaction -
In such calculations masses of reactants are given
MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O
and volume of the product is required and vice-versa.
How many grams of HCl will react with 5 g MnO2 ?
1 mole of a gas occupies 22.4 litre volume at STP.
Sol. 1 mole MnO2 reacts with 4 mole HCl
or 87 g MnO2 reacts with 146 g HCl
146
 5 g MnO2 will reacts with =
× 5 g HCl = 8.39 g HCl
87
Mass of a gas can be related to volume according to
the following gas equation PV = nRT
PV =
w
RT
m
CLASS-XI_STREAM-SA_PAGE # 23
Ex-30. What volume of NH3 can be obtained from 26.75 g
of NH4Cl at 27ºC and 1 atmosphere pressure
Sol. The balanced equation is -
NH4Cl(s)
NH3(g) + HCl(g)
1 mol
53.5 g
1 mol
 53.5 g NH4Cl give 1 mole NH3
1
 26.75 g NH4Cl will give
× 26.75 mole NH3
53.5
= 0.5 mole
PV = nRT
1 ×V = 0.5 × 0.0821 × 300
V = 12.315 litre
1 mol
79.5 g
Cu
H2
+ H2O
1 mol
22.4 litre at STP
79.5
× 2.80 g = 9.93 g
22.4
CO2
1 mol
= 22.4 litre at STP
100 g of CaCO3 evolve carbon dioxide = 22.4 litre
20 g CaCO3 will evolve carbon dioxide
22.4
=
× 20 = 4.48 litre
100
Ex.33 Calculate the volume of hydrogen liberated at 27ºC
and 760 mm pressure by heating 1.2 g of magnesium
with excess of hydrochloric acid.
Sol. The balanced equation is
Mg + 2HCl
Ex-35 What volume of air containing 21% oxygen by volume
is required to completely burn 1kg of carbon containing
100% combustible substance ?
Sol. Combustion of carbon may be given as,
1 mol
12 g
CO2(g)
1 mol
32 g
combustion
CaO +
1 mol
100 g
2 vol.
2(1000 – x)
 12 g carbon requires 1 mole O2 for complete
Ex-32 Calculate the volume of carbon dioxide at STP
evolved by strong heating of 20 g calcium carbonate.
Sol. The balanced equation is -
CaCO3
2CO(g)
1 vol.
(1000 –x)
C(s) + O2(g)
22.4 litre of hydrogen at STP reduce CuO = 79.5 g
2.80 litre of hydrogen at STP will reduce CuO
=
CO2(g) + C(s)
Total volume of the gas becomes = x + 2(1000 – x)
= x + 2000 – 2x = 1600
x = 400 mL
 volume of CO = 400 mL and volume of CO2 = 600 mL
Ex-31 What quantity of copper (II) oxide will react with
2.80 litre of hydrogen at STP ?
Sol. CuO +
Ex-34 One litre mixture of CO and CO2 is taken. This is
passed through a tube containing red hot charcoal.
The volume now becomes 1.6 litre. The volume are
measured under the same conditions. Find the
composition of mixture by volume.
Sol. Let there be x mL CO in the mixture , hence, there will
be (1000 – x) mL CO2. The reaction of CO2 with red
hot charcoal may be given as -
MgCl2
+
24 g
H2
1 mol
24 g of Mg liberate hydrogen = 1 mole
 1000 g carbon will require
1
 1000 mole O2 for
12
combustion, i.e. , 83.33 mole O2
Volume of O2 at STP = 83.33 × 22.4 litre
= 1866.60 litre
 21 litre O2 is present in 100 litre air
1866.60 litre O2 will be present in
100
× 1866.60 litre air
21
= 8888.57 litre or 8.89 × 103 litre
Ex-36 An impure sample of calcium carbonate contains
80% pure calcium carbonate 25 g of the impure
sample reacted with excess of hydrochloric acid.
Calculate the volume of carbon dioxide at STP
obtained from this sample.
Sol. 100 g of impure calcium carbonate contains = 80 g
pure calcium carbonate
25 g of impure calcium carbonate sample will contain
=
80
× 25 = 20 g pure calcium carbonate
100
The desired equation is 1.2 g Mg will liberate hydrogen = 0.05 mole
PV = nRT
1 × V = 0.05 × 0.0821 × 300
V = 1.2315 litre
(iv) Calculations based on volume volume
relationship :
These calculations are based on two laws :
(i) Avogadro’s law
(ii) Gay-Lussac’s Law
e.g.
N 2(g)
1 mol
1 × 22.4 L
+
3H 2(g)
3 mol
3 × 22.4 L
2NH 3(g) (Avogadro's law)
2 mol
2 × 22.4 L
(under similar conditions of temperature and
pressure, equal moles of gases occupy equal
volumes)
N2(g) +
3H 2(g)
2NH3(g) (Gay-lussac’s law)
1 vol.
3 vol.
2 vol.
under similar conditions, ratio of coefficients by mole
is equal to ratio of coefficient by volume.
CaCO3 + 2HCl
CaCl2 + CO2 + H2O
22.4 litre
at STP
1 mol
100 g
100 g pure CaCO3 liberate = 22.4 litre CO2.
20 g pure CaCO3 liberate =
22.4
 20
100
= 4.48 litre CO2
VOLUMETRIC CALCULATIONS
The quantitative analysis in chemistry is primarily
carried out by two methods, viz, volumetric analysis
and gravimetric analysis.In the first method the mass
of a chemical species is measured by measurement
of volume, whereas in the second method it is determined by taking the weight.
The strength of a solution in volumetric analysis is
generally expressed in terms of normality, i.e., number of equivalents per litre but since the volume in the
volumetric analysis is generally taken in millilitres
(mL), the normality is expressed by milliequivalents
per millilitre.
CLASS-XI_STREAM-SA_PAGE # 24
Ex.41 What is strength in gram/litre of a solution of H2SO4,
USEFUL FORMULAE FOR
VOLUMETRIC CALCULATIONS
12 cc of which neutralises 15 cc of
(i) milliequivalents = normality × volume in millilitres.
(ii) At the end point of titration, the two titrants, say 1
and 2, have the same number of milliequivalents,
i.e., N1V1 = N2V2, volume being in mL.
(iii) No. of equivalents =
m.e.
.
1000
1
× 15 = 1.5
10
Sol. m.e. of NaOH solution =
m.e. of 12 cc of H2SO4 = 1.5
1.5
12
Strength in grams/litre = normality × eq. wt.
=
Volume at STP
equivalent volume( vol. of 1eq. at STP)
1.5
× 49 grams/litre
12
= 6.125 grams/litre.
(v) Strength in grams per litre = normality × equivalent
weight.
(vi) (a) Normality = molarity × factor relating mol. wt.
and eq. wt.
(b) No. of equivalents = no. of moles × factor relat
ing mol. wt. and eq. wt.
Ex.37Calculate the number of milli equivalent of H2SO4
present in 10 mL of N/2 H2SO4 solution.
1
× 10 = 5.
2
Ex.38 Calculate the number of m.e. and equivalents of
NaOH present in 1 litre of N/10 NaOH solution.
Sol. Number of m.e. = normality × volume in mL
1
=
× 1000 = 100
10
Number of equivalents =
solution ?
 normality of H2SO4 =
(iv) No. of equivalents for a gas =
Sol. Number of m.e. = normality × volume in mL =
no. of m.e.
100
=
= 0.10
1000
1000
Ex.39 Calculate number of m.e. of the acids present in
(i) 100 mL of 0.5 M oxalic acid solution.
(ii) 50 mL of 0.1 M sulphuric acid solution.
Sol. Normality = molarity × basicity of acid
(i) Normality of oxalic acid = 0.5 × 2 = 1 N
m.e. of oxalic acid = normality × vol. in mL = 1 × 100
= 100.
(ii) Normality of sulphuric acid = 0.1 × 2 = 0.2 N
m.e. of sulphuric acid
= 0.2 × 50 = 10


molecular wt. 98
 eq. wt. of H2 SO 4 

 49 
basicity
2


Ex.42 What weight of KMnO4 will be required to prepare
250 mL of its
N
solution if eq. wt. of KMnO4 is 31.6 ?
10
Sol. Equivalent weight of KMnO4 = 31.6
Normality of solution (N) =
1
10
Volume of solution (V) = 250 ml
W
NEV
1000
W=
1 31.6  250

10
1000
31.6
 0.79 g
40
Ex.43 100 mL of 0.6 N H2SO4 and 200 mL of 0.3 N HCl
were mixed together. What will be the normality of the
resulting solution ?
Sol. m.e. of H2SO4 solution = 0.6 × 100 = 60
m.e. of HCl solution = 0.3 × 200 = 60
 m.e. of 300 mL (100 + 200) of acidic mixture
= 60 + 60 = 120.
m.e.
total vol.
120
2
=
=
N.
300
5
Normality of the resulting solution =
Ex.40 A 100 mL solution of KOH contains 10
milliequivalents of KOH. Calculate its strength in normality and grams/litre.
Sol. Normality =
no. of m.e.
10
volume in mL = 100  0.1 N
Ex.44 A sample of Na2CO3. H2O weighing 0.62 g is added
to 100 mL of 0.1 N H2SO4. Will the resulting solution
be acidic, basic or neutral ?
Sol. Equivalents of Na2CO3. H2O =
 strength of the solution = N/10
Again, strength in grams/litre = normality × eq. wt.
=
N
NaOH
10
1
 56 = 5.6 gram/litre.
10


molecular wt. 56
 eq. wt. of KOH 

 56 
acidity
1


0.62
= 0.01
62
124


 62 
 eq. wt. of Na 2CO 3 .H2 O 
2


m.e. of Na2CO3. H2O = 0.01 × 1000 = 10
m.e. of H2SO4 = 0.1 × 100 = 10
Since the m.e. of Na2CO3. H2O is equal to that of H2SO4,
the resulting solution will be neutral.
CLASS-XI_STREAM-SA_PAGE # 25
STUDY OF GAS LAWS
INTRODUCTION
Gas laws are the rules which the gases obey when
subjected to changes in volume, temperature or
pressure. Any change in one of the aforesaid variables
affects the other two variables.
For example, if the pressure of a gas undergoes some
significant change, its volume and temperature also
change. These variables are discussed below -
This behaviour was generalised and named as
Boyle's law as stated below "Temperature of an enclosed mass of dry gas
remaining constant, its volume is inversely
proportional to pressure.
" Let ‘V’ be the volume of an enclosed dry gas and ‘P’
is its pressure, such that temperature is constant.
(a) Volume :
According to Boyle's law : V 
Gases always occupy the complete volume of the
container on account of their high expansion. Thus,
the volume of a gas is always equal to the volume of
container.
V = K.
Units of Volume :
The volume of gases is measured in the following
units :
(i) 1 millilitre (1 ml) = 1 cm3 (1 cc)
(ii) 1 litre (1  ) = 1 cubic decimeter (dm3)
(iii) 1  = 1 dm3 = 1000 cc = 1000 ml.
(b) Temperature :
The temperature of a gas is the average kinetic energy
of its molecules. If the average kinetic energy of the
molecules of a gas increases, its temperature rises
and vice versa.
Units of temperature :
The temperature is measured in the following units (i) Celsius temperature is measured in degrees celsius
= °C
1
P
1
(at constant T)
P
[ K is constant of proportionality]
PV = K
If the temperature of a gas is kept constant, such that
its pressure changes to P 1 and then P 2 , when
corresponding volumes are V1 and V2 respectively,
then according to Boyle's law P1V1 = K ...(i)
P2V2 = K ...(ii)
Comparing (i) and (ii), we have
P1V1 = P2V2
The above relation is called Boyle's law equation.
From the above equation, the Boyle's law can be
defined as under "Temperature of an enclosed mass of dry gas
remaining constant, the product of its pressure and
volume is a constant quantity."
The table given below shows the experimental data
for the validity of Boyle's law. The data are also plotted
on graph to illustrate change in volume with the
change in pressure at constant temperature.
Pressure (P)
(in cm of Hg)
Volume (V)
(in litres)
P×V
(iii) Temperature in kelvin = 273 + temperature in °C
K = 273 + °C
10
2.0
20
20
1.0
20
(c) Pressure :
One of the fundamental properties of a gas is
pressure. Formally, pressure is defined as the force
per unit area.
30
0.67
20 (approx)
40
0.5
20
(ii) Kelvin temperature is measured in kelvin = K.
10 cm
of Hg
Units of pressure :
The pressure is measured in the following units (1) 1 atmosphere (atm) = 760 mm Hg
(2) N/m2 or pascals (Pa)
(3) 1 atm = 760 mm Hg = 760 torr = 101, 325 Pa
Sir Robert Boyle (1662) studied the relationship
between the volume of a fixed mass of an enclosed
gas at a constant temperature by increasing or
decreasing pressure on it. He found that on
increasing pressure the volume of the gas decreases
and vice versa.
30 cm
of Hg
40 cm
of Hg
30
40
2.0
Volume in Litres
BOYLE'S LAW
20 cm
of Hg
1.0
0.67
0.50
0
10
20
Pressure in cm of Hg
CLASS-XI_STREAM-SA_PAGE # 26
Conclusion : The product of pressure and volume (P
x V) is a constant quantity provided that the
temperature is constant.
T3
Pressure (p)
Pv

0
was found 800 cm3, when pressure was 760 mm of
mercury. If the pressure increases by 25% , find the
new volume of gas.
T2
Pv
Ex.4 At constant temperature the volume of a certain gas
T1
Isotherms
Sol. Initial volume of gas (V1) = 800 cm3
Final volume of gas (V2) = ?
Initial pressure of gas (P1) = 760 mm of Hg
T3 > T 2 > T 1
Volume(1/V)
Increase in pressure = 760 ×
25
= 190 mm of Hg
100
Note :
A curve plotted between P and V at constant
temperature is known as isotherm.
 Final pressure of gas (P2) = 760 + 190 = 950 mm
The size of weather balloon keeps on becoming
larger as it rises to higher altitude because at higher
altitude the external pressure (i.e., atmospheric
pressure) on balloon goes on decreasing and thus,
size of balloon increases.
By Boyle’s law : P1V1 = P2V2
Ex.1 A gas occupies 1500 cm3 at pressure of 720 mm of
mercury. Find at what pressure its volume is 1000
cm 3 . Assume temperature remains constant
throughout the experiment.
Sol. Initial volume of gas (V1) = 1500 cm3
Initial pressure of gas (P1) = 720 mm of Hg
Final volume of gas (V2) = 1000 cm3
Final Pressure of gas (P2) = ?
By Boyle's law : P1V1 = P2V2
P1V1
720  1500
 P2 = V =
= 1080 mm of Hg.
1000
2
Ex.2. A gas occupies a volume of 800 cm3 at a pressure P.
If the pressure is altered to 2.5 atm, the volume of
gas found 900 cm3. Calculate the value of P.
Sol. Given that - P1 = P , P2 = 2.5 atm , V1 = 800 cm3 ,
V2 = 900 cm3
By Boyle's law : P1V1 = P2V2
2.5  900
 P × 800 = 2.5 x 900 P =
= 2.81 atm
800
Ex.3. At constant temperature, a gas is at a pressure of
1080 mm of mercury. At what pressure its volume will
decrease by 40% ?
Sol. Let initial volume of gas (V1) =V
40 V
 40% of initial volume =
= 0.4 V
100
 Final volume of gas (V2) = V - 0.4 V = 0.6 V
Initial pressure of gas (P1) = 1080 mm of Hg
Final pressure of gas (P2) = ?
By Boyle's law : P1V1 = P2V2
P1V1 1080  V
 P2 = V = 0 .6 V = 1800 mm of Hg.
2
of Hg
P1V1
760  800
 V2 = P =
= 640 cm3.
950
2
Ex.5. A vessel of capacity 12 dm3, contains nitrogen gas at
a pressure of 152 cm of Hg. If this vessel is connected
to another evacuated vessel of 6 dm3 capacity, what
will be the pressure of nitrogen in both vessels.
(Assume that temperature remains constant).
Sol. Given that - V1 = 12 dm3 , V2 = 12 + 6 = 18 dm3,
P1 = 152 cm of Hg , P2 = ?
By Boyle’s law : P1V1 = P2V2
 P2 =
P1V1 152  12
= 101.33 cm of Hg.
V2 =
18
CHARLES' LAW
In 1787, Jacques Charles experimentally studied the
relationship between the volume of a fixed mass of
an enclosed dry gas and the temperature, when the
pressure of the gas was kept constant, throughout
the experiment. He found out ''The volume of the fixed mass of an enclosed dry gas
increases by 1 / 273th part of its initial volume at 0 ºC
for every 1 ºC rise in temperature and vice versa,
provided the pressure remains same throughout the
experiment.”
For example, if the initial volume of an enclosed dry
gas is 273 cc, at 0 ºC, then its volume for 1 ºC rise in
temperature will be -
1
× 273 = 274 cc
273
1
Conversely, volume at -1°C = 273 –
x 273 = 272 cc.
273
Volume at 1 °C = 273 +
CLASS-XI_STREAM-SA_PAGE # 27
(a) Concept of Absolute Zero (Kelvin Zero)
Temperature :
(e) Definition of Charles’ Law Based on
Kelvin Scale :
Lord Kelvin, by applying Charles' experimental
deductions, theoretically tried to calculate the volume
of fixed mass of an enclosed gas at constant
pressure. According to him, if initial volume of dry
enclosed gas is 273 cc at 0 °C, then :
Pressure of an enclosed mass of dry gas remaining
constant, the volume of the gas is directly proportional
to its kelvin temperature (absolute temperature).
Thus, if V is the volume of an enclosed mass of dry
gas and T is the kelvin temperature, then
V T
1
x 273 x –1 = 272 cc.
273
1
Volume of gas at –100 °C = 273 +
x 273 x – 100 = 173 cc.
273
1
Volume of gas at – 273 °C = 273 +
x 273 x – 273 = 0 cc.
273
Volume of gas at –1 °C = 273 +
or
Now, if we consider V1 as the volume of gas at T1 (K)
and V2 as the volume of gas at T2 (K), such that
pressure of the given mass of dry gas remains
constant, then V1  T1
;
V2  T2
Thus, according to Lord Kelvin, if the temperature of
an enclosed gas at 0 °C is lowered to –273°C, its
volume becomes zero. However, this is not possible.
It is because gas is one of the states of matter and
hence, must have some definite mass and volume.
The other alternative can be that Charles'
experimental deductions were wrong. However, this
is not possible, as the experimental observations can
be verified independently.
Lord Kelvin offered the solution to the above riddle.
He
suggested
that
in
all
probability
- 273°C is the last limit of temperature, which cannot
be reached and hence, an enclosed gas will never
have zero volume.
V1
T1 = K ;
V1 V2
 T = T
1
2
V2
T2 = K
or
[at constant pressure]
The above equation is called Charle’s law equation.
A graph is shown between the volume and kelvin
temperature, when a fixed mass of dry gas is heated
at a constant pressure.
(i) If the gas is initially at 100 K, its volume increases
to four times when heated to 400 K.
(ii) If the temperature is lowered below 100 K, the
gas liquefies. Thus, the experimental points cannot
be plotted as shown by a dotted line.
Note :
The last limit of temperature was named absolute
zero by Lord Kelvin. However, in order to honour Lord
Kelvin, the absolute zero was renamed as Kelvin zero.
P
(b) Definition of Absolute Zero (Kelvin Zero) :
P
2.00
It is the last limit of the lowest temperature, where the
volume of a given mass of dry enclosed gas at
constant pressure becomes zero. Its theoretical value
is -273 °C.
1.50
or it is defined as theoretical temperature, when the
molecules of an enclosed dry gas at constant
pressure have zero kinetic energy, i.e., they stop
vibrating.
P
1.00
P
0.50
(c) Concept of Absolute Scale (Kelvin
Scale) of Temperature :
The new temperature scale with its zero at -273°C,
such that each degree on it is equal to 1 degree on
the celsius scale is called absolute scale or kelvin
scale.
(d) Characteristics
(Kelvin Scale)
of
Absolute
Scale
(i) The temperature scale, with kelvin zero as starting
point, is called kelvin scale.
(ii) All temperature on kelvin scale are positive.
(iii) Temperature on kelvin scale = 273 + temperature
in ºC.
K = 273 + oC
(iv) Temperature on kelvin scale is not expressed in
degrees. For example, 273 K is the correct
temperature and not 273 oK.
0

300
100
200
Temperature in Kelvin
400
Note :
A curve plotted between V and T at constant pressure
is known as isobar.
P1
P2
P3
Volume

V
= K [K is the constant of proportionality]
T
P4
Isobar
P4 > P3 > P2 > P1
–300 –200–100 0 100
Temperature (ºC)
CLASS-XI_STREAM-SA_PAGE # 28
CONCEPT OF STANDARD
TEMPERATURE AND PRESSURE
From the Boyle's law and Charles' law it is very clear
that the volume of a given mass of dry enclosed gas
depends upon the
Ex.8. A gas is enclosed in a vessel, at standard
temperature. At what temperature the volume of
enclosed gas is 1/8 of its initial volume, pressure
remaining constant ?
Sol. Let initial volume of gas (V1) = x
Initial temperature of gas (T1) = 0 ºC = 273 K
(i) pressure of the gas
(ii) temperature of the gas in kelvin.
Thus, we cannot correctly express the volume of a
given mass of dry enclosed gas, unless we specify
or standardise the temperature and pressure. Thus,
to compare the mass or the density of two or more
gases having same volume, we must standardise
the temperature and pressure at which the volume of
the gases is measured.
This standard temperature and standard pressure
for all gases is called standard temperature and
pressure whose short form is written as S.T.P.
Standard temperature is taken as 0 oC or 273 K.
Standard pressure is taken as 76 cm or 760 mm of
Hg or one atmosphere.
Final volume (V2) =
x
8
Final temperature of gas (T2) = ?
By Charle’s law :

V1 V2

T1 T2
x
x
=
8  T2
273
T2 =
273  x
= 34.125 K
8x
= 34.125 – 273 = – 238.875 ºC.
CONCEPT OF GAS EQUATION
PRESSURE-TEMPERATURE LAW
OR GAY-LUSSAC'S LAW
According to this law the pressure of a given mass of
gas is directly proportional to absolute temperature
at constant volume.
P  T (at constant V and for a fixed amount of gas)
or
P1 P2
P
= constant or T  T
1
2
T
Ex.6. 279 cm 3 of gas at 87°C is cooled to standard
temperature, at constant pressure.
Calculate the volume of gas at standard temperature.
Sol. Initial volume of gas (V1) = 279 cm3
Initial temperature of gas (T1) = 87°C = (273 + 87)
K = 360 K
Final temperature of gas (T2) = 0 °C = (0 + 273) = 273 K
Final volume of gas (V2) = ?
By Charle’s law :
V1 V2
T1 = T2
V2
279
279  273
=
V2 =
= 211.57 cm3
360
360
273
Ex.7. A dry gas occupies a volume of 1054 dm3, at a
temperature of –73 ºC . At what temperature its
volume is 4216 dm 3 when the pressure remains
constant throughout experiment ?
Sol. Initial volume of gas (V1) = 1054 dm3
Initial temperature of gas ( T1) = – 73º C = (–73 + 273)
K = 200 K
Final volume of gas (V2) = 4216 dm3
Final temperature of gas (T2) = ?
By Charle’s law :

V1 V2
T1 = T2
4216
1054
4216  200
=
= 800 K
T2  T2 =
200
1054
 Temperature in ºC = (800 – 273) = 527 ºC.
It has been found that in all practical situations , the
volume, the pressure and the temperature of an
enclosed gas change simultaneously, when any of
the above variables are altered. Thus, there is a need
to have a mathematical equation which connects
these variables.
(a) Definition :
“A mathematical equation used in calculating the
change in volume when the initial temperature and
pressure of an enclosed gas simultaneously change
is called gas equation.”
The perfect gas equation can be derived by combining
Boyle’s law and Charles’ law .
Consider an enclosed dry gas of volume V at pressure
P and temperature T kelvin.
Applying Boyle’s law - V 
1
P
(At constant temperature)
-- (i)
Applying Charles’ law - V  T
(At constant pressure)
Combining (i) and (ii)
V
1
×T
P
or
V=K×
or
PV
=K
T
T
P
---- (ii)
(K = constant)
If the initial volume of an enclosed dry gas is V1, when
its pressure is P1 and temperature is T1(K), such that,
its volume changes to V2, when its pressure is P2 and
temperature is T2(K), then -
CLASS-XI_STREAM-SA_PAGE # 29
P1V1
T1 = K (constant)
------ (iii)
P2 V2
T2 = K (constant)
Combining (iii) and (iv)
------ (iv)
P1V1
P2 V2
T1 = T2
[Gas equation]
Initial pressure Initial volume
or Initial temperature in kelvin
=
AVOGADRO’S LAW
According to this law equal volumes of gases under
similar conditions of pressure and temperature
possess equal number of moles or molecules.
V  NA
(at constant P and T)
Vn
where, N and n are number of molecules and moles
of gas taken respectively.
IDEAL GAS EQUATION
Final pressure  Final volume
Final temperature in kelvin
According to Boyle’s law : V  (1/P)
According to Charle’s law: V  T
According to Avogadro’s law : V  n
Thus,
V  (nT/P)
PV  nT or PV = nRT ----- (i)
Equation (i) is known as ideal gas equation. R is
known as universal gas constant or molar gas
constant. The term ideal gas refers to one which
obeys equation (i) in all temperature and pressure
ranges. However, since none of the gases present in
universe obeys this equation rigidly and thus, gases
are named as real or non-ideal gases.
Ex.9 1.57 dm3 of dry hydrogen gas is at a pressure of 750
mm of mercury when the temperature is 37.5 ºC. Find
the volume of gas at S.T.P.
Sol. P1 = 750 mm of Hg
P2 = 760 mm of Hg
V1 = 1.57 dm3
V2 = ?
T1 = (273 + 37.5) K = 310.5 K T2 = 273 K
By gas equation :
V2 =
P1V1 P2 V2
T1 = T2
Numerical values of R :
P1V1 T2
750  1.57  273
×
=
= 1.36 dm3
T1
P2
310.5  760
Ex.10. Sulphur dioxide occupies a volume of 512 cm3 at
S.T.P. Find its volume at 27 ºC and at a pressure of
720 mm of mercury.
Sol. P1 = 760 mm of Hg
P2 = 720 mm of Hg
V1 = 512 cm3
V2 = ?
T1 = 273 K
T2 = 273 + 27 = 300 K
By gas equation :
P1V1 P2 V2
T1 = T2
(a) In litre atmosphere : R = 0.0821 litre atm K–1 mol 1
(b) In C.G.S. System : R = 8.314 × 107 erg K–1 mol–1
(c) In S.I. system : R = 8.314 JK–1 mol–1

Note :
Real gases behave almost ideally at low P and high T.
GRAHAM’S LAW OF DIFFUSION
According to Graham, the rate of diffusion of a gas at
constant P and T is inversely proportional to square
root of its molecular weight.
P1V1 T2 760  512 300

 V2 = T × P =
= 593.89 cm3.
273
720
1
2
Ex.11. A gas occupies a volume of 100 cm3 at 0 ºC and
760 mm Hg pressure. If the kelvin temperature of the
gas is increased by one fifth and its pressure is
increased one half times, calculate the final volume
of gas.
760 

 = 1140 mm of Hg
Sol. P1 = 760 mm of Hg P2 =  760 
2 

V1 = 100 cm
V2 = ?
3

T1 = 0 ºC = 273 K T2 =  273 

By gas equation :
 V2 =
273 
 K = 327.6 K
5 
P1V1 P2 V2
T1 = T2
P1V1 T2
760  100  327.6
×
=
= 80 cm3.
T1
P2
273  1140
1
M
r1
M2

r2
M1
r

(at constant P and T)
(at constant P and T)
Since, molecular wt. of a gas = 2 × vapour density ()

r1

r2
2
1
(at consant P and T )
The rate of diffusion r for two gases under different
pressure can be given by r1   M 2  P1 


r2 =   M1  P2  (at constant T)
Further rate of diffusion (r) can be expressed in terms
of Volume diffused ( V )
Moles diffused (n)
r = Time taken in diffusion = Time taken in diffusion
r=
Distance travelled in a narrow tube (d)
Time taken
CLASS-XI_STREAM-SA_PAGE # 30
Therefore, according to Graham’s law of diffusion at
constant P and T -
V1 t 2
M2


 2 
t1 V2
1
M1
A mixture of gases that do not react with one another
where V1, V2 are volumes diffused in time t1 and t2
n1 t 2
M2


 2 
t1 n2
1
M1
where n1, n2 are moles diffused in time t1 and t2.
d1 t 2
M2


 2 
t1 d2
1
M1
where d1, d2 are distances travelled by molecules in
narrow tube in time t1 and t2
w

M
Also, we have
n= 
 By equation
n1 t 2


t1 n2
or
M2
M1
M2
M1
w 1 M2 t 2

M1t1
w2 =
w1 t 2


t1 w 2
M1
M2
where w1, w2 are weights diffused in time t1 and t2.
Ex.12. 20 cm3 of SO2 diffuses through a porous plug in 60
seconds. What volume of O2 will diffuse under similar
conditions in 30 seconds ?
Sol. V1 = 20 cm3
t1 = 60 sec.
V2 = V
t2 = 30 sec.
M1 = 64
V1 t 2


t1 V2

M2 = 32
M2
M1
20 30

=
60 V
32
64
 V = 14.14 cm3
Ex.13. One mole of nitrogen gas at 0.8 atm takes 38
seconds to diffuse through a pinhole, whereas one
mole of an unknown compound of xenon with fluorine
at 1.6 atm takes 57 seconds to diffuse through the
same hole. Calculate molecular mass of the
compound.
Sol.
r1
M2 P1


r2
M1 P2
n1 t 2


or
t1 n2
or
1 57


38 1
 M = 252
DALTON’S LAW OF PARTIAL PRESSURES
behaves like a single pure gas. For example, we can
treat air as a single gas when we want to use the
ideal gas laws to predict its properties. The total
pressure exerted by a gaseous mixture is equal to
the sum of partial pressures of each component (gas)
present in mixture.
Thus,
PM = P1 + P2 + P3 + ............ (i)
Partial pressures P1, P2 , ............ are defined as the
pressure exerted by that component if same amount
of gas is filled in the same container at the same
temperature.
Let n1, n2, n3 , ......... are moles of gases 1,2,3, .......
which are filled in a container of volume V at
temperature T, then -
n1RT
........ (ii)
(R is gas constant)
V
n 2RT
P2 =
......... (iii)
V
n1RT n 2RT

 PM = P1 + P2 + P3 + ......... =
+ .........
V
V
RT
or PM = (n1 + n2 + ...... )
---------- (iv)
V
P1 =
By Equations (ii) and (iv)
P1
n1

PM (n1  n 2  ....)


n1

 mole fraction 


n
n
....
1
2


 P1= PM × mole fraction of 1
i.e. Partial pressure of any constituent gas
= Total pressure × its mole fraction
Ex.14 Two gases A and B having molecular weights 60
and 45 respectively are enclosed in a vessel. The
weight of A is 0.50g and that of B is 0.20 g . The total
pressure of the mixture is 750 mm. Calculate partial
pressure of the two gases.
Sol. Given that weight of gas A = 0.50 g and weight of gas
B = 0.2 g.
Molecular weight of A = 60 and mol. weight of B = 45
Pmixture = 750 mm, pA = Partial pressure of A and pB =
Partial pressure of B
From pA = PM × mole fraction of A
M2 P1

M1 P2
0.5 / 60
pA = 750 ×  0.5  0.2  = 489.13 mm
 60 45 
M 0.8

28 1.6
 pB = PM – pA = 750 – 489.13
= 260.87 mm
CLASS-XI_STREAM-SA_PAGE # 31
CHEMICAL BONDING
(ii) Octet rule does not explain the formation of electron
deficient molecules such as BeCl2, BF3 etc. in which
the central atom has less than eight electrons in its
valence shell.
ELECTRON DOT REPRESENTATION
(LEWIS SYMBOLS)
(i) A useful short hand notation (given by G.N. Lewis )
known as electron dot structure which is used to
indicate the manner in which the constituent atoms
of a molecule are bonded.
(ii) To draw the Lewis symbol for an element, we write
its chemical symbol surrounded by a number of dots
or crosses, which represents the valence electrons
of the atom.
e.g.
•
H ,
(Z=1)

• C••• ,
(Z=6)
••N•••
(Z=7)
(iii) Octet rule does not explain the existence of
molecules like PCl5, SF6, IF7 which have 10, 12, 14
electrons around central atom P, S and  respectively.
etc.
(2,4)
(2,5)
Note :
The number of dots in the Lewis symbols represent
the number of valence electrons in the atom of the
element.
VARIABLE VALENCY
A few elements like manganese (Mn), chromium (Cr),
phosphorus (P), tin (Sn), iron (Fe), copper (Cu) etc.,
do not exhibit single valency but they exhibit more
than one type of valency i.e., variable valency.
Variable valency of a few elements are as follows-
ELECTRONIC THEORY OF VALENCY
The theory was put forward by Lewis and Kossel to
explain the nature of bonds between the atoms within
organic and inorganic molecules. The theory is based
upon the electronic configuration of atoms and, that
is why, called electronic theory of valency.
Element
According to this theory helium (He) and those
elements whose atoms have eight electrons in their
valence shell (inert gas configuration) are chemically
inert i.e., they do not react with any other element. On
the other hand, elements which do not have inert gas
configuration have a tendency to acquire it by either
transferring electron(s) or by sharing of
electron(s).This results in bond formation between
the atoms.
EXCEPTIONS TO THE OCTET RULE
Valency Element
Valency
Iron (Fe)
2,3
Tin (Sn)
2,4
Copper (Cu)
1,2
Arsenic (As)
3,5
Mercury (Hg)
1,2
Lead (Pb)
2,4
For the lower number valency suffix -ous is used and
for valency of higher number suffix -ic is used.
e.g. :
Cuprous (Cu+),
Ferrous (Fe+2),

Cupric (Cu+2)
Ferric (Fe+3) etc.
Note :
Usually transition elements show variable valency.
Octet rule was given by G.N. Lewis and W. Kossel in
1916.
CHEMICAL BOND
According to octet rule “an atom whose outermost
shell contains 8 electrons (octet) is stable.”
This rule, however, does not hold good in case of
certain small atoms like helium (He) in which
presence of 2 electrons (duplet) in the outermost
When atoms of elements combine to form molecules,
a force of attraction is developed between the atoms
which holds them together. This force of attraction is
called a chemical bond.
e.g.
In a chlorine molecule (Cl2) two chlorine atoms are
held together by a chemical bond.
shell is considered to be the condition of stability.
(i) Hydrogen has only one electron in its valence shell
and one more electron is needed to fulfil the valence
shell. The completed shell has electronic configuration
1s 2 like helium & hence is stable. In this case,
therefore, octet rule is not needed to achieve a stable
noble gas configuration.

Note :
A chemical bond is accompanied by release of energy,
thus the resulting molecule has less energy and is
more stable.
CLASS-XI_STREAM-SA_PAGE # 32
Atoms combine with one another to achieve the inert
gas electron arrangement and become stable. Atoms
form chemical bonds to achieve stability by acquiring
the inert gas configuration or by completing their octet
or duplet ( in case of small atoms) in outermost shell.
An atom can achieve the inert gas electron
arrangement in three ways -
(iii) High lattice energy (L.E) : The amount of energy
released when isolated ions form 1 mole of a crystal
lattice is called lattice energy. The value of lattice
energy depends on the charges present on the two
ions and the distance between them. It shall be
high if charges are high and ionic radii are small.
(iv) The summation of the three energies should be

(i) by losing one or more electrons.
(ii) by gaining one or more electrons.
(iii) by sharing one or more electrons.
negative i.e., energy should be released.
I.E. +
E.A.
+
L.E.
= -ve
Note :
Noble gases do not usually form bonds with other
elements, because they are stable. So, atoms of
elements have the tendency to combine with one another
to achieve the inert gas configuration.
(b) Born - Haber Cycle :
IONIC OR ELECTROVALENT BOND
This bond is formed by the atoms of electropositive
and electronegative elements. Electropositive
elements lose electrons in chemical reaction and
electronegative elements gain electrons in chemical
reaction. When an atom of electropositive element
come in contact with that of an electronegative element
then the electropositive atom loses electron &
becomes positively charged, while the electronegative
atom gains the electron to become negatively
charged. Electrostatic force of attraction works
between the positively and negatively charged ions
due to which both ions are bonded with each other.
As a result, a chemical bond is produced between
the ions, forming Ionic or Electrovalent compound.

Note :
Number of electrons donated or accepted by any
element is called electrovalency.
In an ionic compound every cation is surrounded by a
fixed number of anions and every anion is surrounded
by a fixed number of cations and they are bonded in a
fixed geometry in a three dimensional structure which
is called lattice.
(a) Necessary Conditions for the Formation
of Ionic Bond
:
(i) Low ionisation energy (I.E) : Ionic bond is favoured
by the low I.E. of the element that forms the cation.
The element should be a metal i.e. electropositive in
nature.
(ii) High electron affinity (E.A) : Ionic bond is
favoured by the high E.A. of the element that forms
the anion. The element should be a non - metal i.e.
electronegative in nature.
The formation of an ionic compound and the
determination of lattice energy can be explained by
Born - Haber cycle. This cycle is based on Hess’s
law i.e., the formation of an ionic crystal may occur
either by the direct combination of elements or by an
alternate process in which various steps are involved.
The energy involved in both the cases is same.
e.g. Formation of ionic compound Potassium Fluoride
(KF) :
Direct combination :
K(s) +
1
 KF(s)
F (g) 
2 2
H or heat of formation = –562.6 kJ mol–1
Steps :
(i) Conversion of solid potassium into gaseous state.
 K(g)
K(s) 
HA (heat of atomisation) or HS heat of sublimation = +
89.6 kJ mol–1
(ii) Formation of a cation.
 K+(g) + e–
K(g) 
I.E. or first ionisation energy =+ 419.0 kJ mol–1
(iii) Conversion of molecular fluorine into gaseous
atomic fluorine.
1
 F(g)
F (g) 
2 2
1
1
H or
(heat of dissociation)
2 D
2
1
= +
(158.2) = 79.1 kJ mol–1
2
(iv) Formation of anion.
 F–(g)
F (g) + e– 
E.A. or first electron affinity = – 332.6 kJ mol–1
(v) Combination of K+(g) and F– (g) to form KF (s).
 KF(s)
K+(g) + F–(g) 
L.E. or lattice energy = – U kJ mol –1
on the basis of Hess’s law
– 562.6 = 89.6 + 419.0 + 79.1 - 332.6 – U
U = 817.7 kJ mol–1,
i.e. lattice energy of KF = 817.7 kJ mol – 1
CLASS-XI_STREAM-SA_PAGE # 33
Born - Haber cycle for formation of KF(s) may be
represented as :
(v) Solubility : Ionic compounds are generally soluble
in polar solvents like water and insoluble in non - polar
solvents like carbon tetrachloride, benzene, ether,
K(s)
1
F (g)
2 2
+HA or HS
+
K(g)
1
HD
2
(vi) Brittle nature : Ionic compounds on applying
F(g)
+I.P.
external force or pressure are broken into small
pieces, such substances are known as brittle and
–E.A.
–
F
KF
alcohol etc.
+
K
–U
( b ) Properties of Ionic Compounds :
(i) Ionic compounds consist of ions : All ionic
compounds consist of positively and negatively
charged ions and not molecules. For example,
sodium chloride consists of Na + and Cl – ions,
magnesium fluoride consists of Mg2+ and F– ions and
so on.
this property is known as brittleness. When external
force is applied on the ionic compound, layers of ions
slide over one another and particles of the same
charge come near to each other as a result due to the
strong repulsion force, crystals of compounds are
broken.
(ii) Physical nature : Ionic compounds are solid and
relatively hard due to strong electrostatic force of
attraction between the ions of ionic compound.
(iii) Crystal structure : X-ray studies have shown that
ionic compounds do not exist as simple single ion
pair as Na+Cl–. This is due to the fact that the forces of
attraction are not restricted to single unit such as Na+
and Cl– but due to uniform electric field around an
ion, each ion is attracted to a large number of other
ions. For example, one Na+ ion will not attract only
one Cl– ion but it can attract as many negative charges
as it can. Similarly, the Cl– ion will attract several Na+
ions. As a result, there is a regular arrangement of
these ions in three dimensions as shown in diagram.
Such a regular arrangements is called crystal lattice.
+
–
+
–
+
+
–
+
–
+
–
+
–
+
–
–
+
–
+
–
+
–
+
–
+
+
–
+
–
+
–
+
–
+
–
–
+
–
+
–
Brittle nature of ionic compounds
(vii) Electrical conductivity : Electrical conductivity in
any substance is due to the movement of free
electrons or ions. In metals electrical conductivity is
due to the free movement of valency electrons. An
ionic compound exhibits electrical conductivity due to
the movement of ions either in the fused state or in
the soluble state in the polar solvent. But in the solid
state due to strong electrostatic force of attraction free
ions are absent so they are insulator in the solid state.
W hen two atoms of
electronegative elements
approach each other then they share electrons. Bonds
formed in this manner are called covalent bonds and
compounds are termed as covalent compound.
Number of electrons taking part in the process of
sharing is called covalency of that element.
e.g.,
••
••
F •• F••
••••
••
F – F or F2 molecule
••
H• •O • •H
••
(iv) Melting point and boiling point : Strong
electrostatic force of attraction is present between
ions of opposite charges. To break the crystal lattice
more energy is required so their melting points and
boiling points are high.
H – O – H or HO
molecule
2
The electrons that participate in bond formation are
called bond pairs of electrons while those which do
not take part in bond formation are called lone pairs
of electrons.
CLASS-XI_STREAM-SA_PAGE # 34
(a) Types of Covalent Bonds :

Note :
A polar covalent bond is formed when two atoms
having different electronegativities combine or in other
words a polar covalent bond is formed between
(i) Depending upon the number of electrons shared
between the two bonded atoms, covalent bonds are
of following two types :
different types of atoms.
(A) Single covalent bond
(B) Multiple covalent bond
e.g. HCl molecule
(A) Single covalent bond : A single covalent bond
is formed by sharing of one pair of electrons
between two atoms. It is represented by a single
short line ( –) between the two atoms.
H•
+
Hydrogen
atom
••
•Cl ••
••
Chlorine
atom
••
H • • Cl••
••
Hydrogen
chloride
molecule
or H
Cl
In this example the electronegativity of Cl atom is more
than that of H atom & thus the shared pair of electrons
remains more near or attracted to Cl atom, making
the H – Cl bond polar.
• The shifting of shared electrons from one atom to
another is indicated by putting an arrow-head in the
centre of the line representing the bond between the
atoms.
e.g.
Cl – Cl or Cl2 molecule
Other examples : H – H, F – F, H – Cl etc.
(B) Multiple covalent bond : It is of following two types :
• Double covalent bond : A double covalent bond is
formed by sharing of 2 pairs of electrons between the
atoms. It is represented by two lines (=) between the
atoms.
• The arrow head points towards the more
electronegative element (like Cl in this example).
H
Cl
• Since the electrons are attracted more towards the
e.g.
Cl atom it acquires a slight negative charge (–) while
H acquires a slight positive charge (+) as electrons
move away from it. This can be represented as -
O = O or O2 molecule
Other examples : CO2, C2H4 etc.
+
–
Cl
H
(+ is pronounced as delta plus)
• Triple covalent bond : A triple covalent bond is
formed by sharing of 3 pairs of electrons and is
atoms.
eg.
(B) Non - Polar covalent bond :
• A covalent bond in which there is an equal attraction
for the shared pair(s) electrons between the two
combining atoms is called non-polar covalent bond.
N  N or N2 molecule

Other examples : C2H2 etc.
(ii) Depending on whether the two bonded atoms differ
in their electronegativities or not i.e. on bond polarity
covalent bonds are of two types :
.
Other examples : Hydrogen fluoride (HF),water (H2O),
ammonia (NH3) etc.
represented by three lines () between the bonded
Note :
A non-polar covalent bond is formed when two atoms
having equal electronegativities combine or when
same type of atoms combine.
e.g. Chlorine molecule
(A) Polar covalent bond
(B) Non - polar covalent bond
(A) Polar covalent bond :
• A covalent bond in which there is an unequal
attraction for the shared electrons between the two
combining atoms, resulting in one end of the
molecule becoming slightly positive and the other
end becoming slightly negative is called a polar
covalent bond.
••
••
• + • Cl••
•• Cl
••
••
Chlorine Chlorine
atom
atom
••
••
• • Cl••
••Cl
••
••
or Cl – Cl
In the above example the two chlorine atoms have
equal electronegativities & so the shared pair of
electron remains at equal distance from both the
chlorine atoms making the bond non-polar.
Other examples :
Fluorine molecule (F 2), hydrogen molecule (H 2),
oxygen molecule (O2) etc.
CLASS-XI_STREAM-SA_PAGE # 35
(b) Characteristic Properties of Covalent
Compounds :
The important characteristic properties of covalent
compounds are :
(iii) Crystal structure : Covalent compounds exhibit
(i) Covalent compounds consist of molecules : The
covalent compounds consist of molecules. They do
not have ions. For example - water, hydrogen chloride,
methane consist of H 2 O, HCl, CH 4 molecules
respectively.
break the crystal is less due to the presence of weak
(ii) Physical state : Weak Vanderwaal’s forces are
present between the molecules of covalent
compounds. So, covalent compounds are in solid,
gaseous or liquid state at normal temperature and
pressure.
bad conductors of electricity due to the absence of
For example : Hydrogen chloride , methane are gases
while carbon tetrachloride, ethyl alcohol, ether etc.
are liquids. Glucose, sugar, urea etc. are some solid
covalent compounds.
both crystalline and non crystalline structure.
(iv) Melting point and boiling point : Energy required to
Vanderwaal’s force, so their melting and boiling points
are less.
(v) Electrical conductivity - Covalent compounds are
free electrons or free ions.
(vi) Solubility : Due to the non - polar nature of covalent
compounds they are soluble in non - polar solvents
like benzene, carbon tetrachloride etc. and insoluble
in polar solvents like water etc.
DIFFERENCES BETWEEN IONIC & COVALENT COMPOUNDS
S.No.
1
2
3
4
5
Ionic compounds
Covalent compounds
Thes e are usually crystalline solids.
These are usually liquids or gas es. Only some of them are solids.
Ionic com pounds have high m elting
& boiling points i.e., they are non-volatile.
Ionic com pounds conduct electricity
when dissolved in water or m elted i.e., they are
electrolytes .
Ionic com pounds are us ually soluble in polar
solvents like water.
Covalent com pounds have usually low m elting & boiling points
i.e., they are volatile.
Ionic com pounds are us ually ins oluble
in organic solvents like alcohol, ether etc.
Covalent com pounds do not conduct electricity i.e., they are nonelectrolytes.
Most of the covalent Com pounds are usually insoluble in polar
s olvent like water. (except s om e like glucose, s ugar, urea etc.)
Covalent com pounds are s oluble in organic solvents.
MODERN CONCEPT OF COVALENT BOND
The modern concept of covalent bond can be
explained on the basis of Valence Bond Theory.
The valence bond theory : This theory was presented
by Heitler and London, in 1927, to explain how a covalent bond is formed.
The main points of the theory are (i) A covalent bond is formed by overlapping of atomic
orbitals of valence shell.
(ii) Only half filled atomic orbitals, i.e., orbitals singly
occupied can enter into overlapping process.
(v) Greater the overlapping, higher is the strength of
the chemical bond.
(vi) Electrons which are already paired in valence shell
can enter into bond formation if they can be unpaired
first and shifted to vacant orbitals of slightly higher
energy of the same main energy shell (valence shell).
This point explains the trivalency of boron, tetravalency
of carbon etc.
(vii) Two types of bonds are formed on account of
overlapping. These are (a) sigma () and (b) Pi ()
bonds.
(a) Sigma bond :
(iii) The atoms with half filled orbitals must come
closer to one another with their axes in proper directions for overlapping.
(iv) As a result of overlapping there is maximum electron density somewhere between the two atoms. A
large part of binding force comes into existence from
the electrostatic attraction between the nuclei and the
accumulated electron cloud between them.
A bond formed between two atoms by the overlapping of singly occupied orbitals along their axes (end
to end overlap) is called sigma () bond. In such a
bond formation, maximum overlapping is possible
between electron clouds and hence, it is a strong
bond.
Sigma bond can, thus, be defined as “Bond orbital which is symmetrical about the line joining the two nuclei is known as sigma bond”.
CLASS-XI_STREAM-SA_PAGE # 36

Note :
(b) Pi () Bond :
Electron cloud of sigma bond is symmetrical about
the line joining the nuclei of the two atoms.
- bonds are formed by the sidewise or lateral overlapping of p-orbitals. The overlapping takes place at
the side of the two lobes and hence, the extent of
Sigma bond can be formed by the overlapping of s-s
orbitals, s-p orbitals or head to head overlapping of
p-p orbitals as shown :
•
•
s-s overlapping
•
overlapping is relatively smaller. Thus, -bond is a
weaker bond in comparison to sigma bond.
•
•
p
p
p-p overlapping
s-p overlapping
Electron cloud of a pi bond is oriented above and
below the plane containing nuclear axis.
p-p overlapping
COMPARISON OF SIGMA BOND AND pi BOND
Dipole moment is a vector quantity, i.e., it has both
magnitude as well as direction. The overall value of
the dipole moment of a polar molecule depends on
the geometry and shape of the constituent bonds. A
symmetrical molecule is non-polar even though it
contains polar bonds. For example , CO2, BF3, CH4,
CCl4 being symmetrical molecules have zero resultant dipole moments and hence are non-polar since
dipole moments summation of all the bonds present
in the molecule cancel each other.
H
F
DIPOLE MOMENT

A covalent bond, in which electrons are shared unequally and the bonded atoms acquire a partial positive and negative charge, is called a polar covalent
bond or a covalent bond between two dissimilar atoms is a polar covalent bond.
Difference in polarities of bonds is expressed on a
numerical scale. The polarity of a molecule is indicated in terms of dipole moment.
The dipole moment is defined as the product of the
distance separating charges of equal magnitude and
opposite sign, with the magnitude of the charge. The
distance between the positive and negative centres
is called the bond length. Thus,
µ = electric charge × bond length = q × d
Note :
Dipole moment is measured in ‘Debye’ unit (D).1 D =
3.33 × 10–30 coulomb metre.
Generally, as electronegativity difference increases
in diatomic molecules, the value of dipole moment
increases. Greater the value of dipole moment of a
molecule, greater is the polarity of the bond between
the atoms.
O
C
F

C
B
O
F
H
H
H
Note :
Dipole moment is usually indicated by an arrow having + on the tail (
, above the polar bond
and pointing towards the negative end.
Unsymmetrical non-linear polyatomic molecules always have net value of dipole moment, thus such
molecules are polar in nature. H2O, CH3Cl, NH3, etc ,
are polar molecules as they have some positive values of dipole moments.
CLASS-XI_STREAM-SA_PAGE # 37
orbitals of an atom which differ in energy slightly may
S
O
mix with each other to form new orbitals called hybrid
orbitals. The process of mixing or amalgamation of
O
O
Sulphur dioxide
µ = 1.60 D
H
H
Water µ =1.84D
atomic orbitals of nearly same energy to produce a
set of entirely new orbitals of equivalent energy is
known as hybridization.
Cl

N
C
H
H
H
H
The following are the rules of hybridization -
H
Ammonia
µ = 1.46 D
H
Methyl chloride
µ =1.86 D
Note :
This is a hypothetical concept and has been introduced by Pauling and Slater.
(i) Only orbitals (atomic) of nearly same energy belonging to same atom or ion can take part in hybridization.
The valence bond theory (overlapping concept)
(ii) Number of the hybrid orbitals formed is always
equal to the number of atomic orbitals which have
taken part in the process of hybridization.
explains satisfactorily the formation of various
molecules but it fails to account for the geometry and
(iii) The hybrid orbitals are similar. They may differ
from one another in orientation in space.
HYBRIDIZATION
shapes of various molecules.
In order to explain these , the valence bond theory
has been supplemented by the concept of hybridization. According to this concept any number of atomic
(iv) Hybrid orbitals form only sigma bond.
The following table shows the type of hybridisation
and the geometry of the molecules -
Number of
hybrid
orbitals
Type of hybridisation of
central atom
No. of lone
pairs of electron
Geometry/Shape
Examples
2
sp
[W hen one s and one p
orbitals intermix]
0
Linear
BeF 2 , CO 2 , CS 2 , NO 2 +
0
Trigonal Planar
B F 3 , AlCl3 , S O 3 , NO 3 – , CO 3 2–
3
sp 2
[W hen one s and two p
orbitals intermix]
1
Bent
0
Tetrahedral
CH 4 , SiCl4 , SO 4 –2 , ClO 4 –
1
Pyramidal
NH 3 , NF 3 , PCl3 , PH 3 , H 3 O +
2
Bent
H 2 O, NH 2 –
0
Trigonal bipyramidal
P Cl 5 , PF 5 , SbCl 5
1
See saw
S F 4 , TeF 4
2
T-shape
ClF 3 , BrF 3
3
Linear
XeF 2 , I3 – , Br3 –
0
1
Octahedral
Square pyramidal
SF 6 , SeF 6
IF 5 , BrF 5
2
Square planar
XeF 4 .
0
P entagonal bipyramidal
IF 7
4
5
6
sp3
[W hen one s and three p
orbitals intermix]
sp 3 d
[W hen one s three p and one
d orbitals intermix]
sp 3 d2
[W hen one s, three p and two
d orbitals intermix]
SO 2 , PbCl 2 , SnCl 2 , NO 2 –
3 3
7

sp d
[W hen one s, three p and
three d orbitals intermix]
Note :
Bond angle is 180º, 120º and 109º 281 respectively in the case of sp, sp2 and sp3 hybridization without any lone pair
of electrons.
CLASS-XI_STREAM-SA_PAGE # 38
(b)
Characteristics
compounds :
COORDINATE BOND
may be defined as “a covalent bond in which both
electrons of the shared pair are contributed by one of
(i) Physical state : These exist as gases, liquids and
solids under ordinary conditions.
the two atoms”.
Note :
(ii) Melting and boiling points : Their melting and
boiling points are higher than purely covalent
compounds and lower than purely ionic compounds.
Coordinate bond is also called dative bond.
A coordinate or a dative bond is established between
two such species, one of which has a complete octet
and possesses a pair of valency electrons while the
(iii) Solubility : These are sparingly soluble in polar
solvents like water but readily soluble in non-polar
(organic) solvents.
other is short of a pair of electrons.
(iv) Stability : These are as stable as the covalent
compounds. The addition compounds are, however,
not very stable. The bond present in coordination
compounds is strong because the shared electrons
cannot be separated easily.
This bond is represented by an arrow ()

coordinate
The properties of coordinate compounds are
intermediate between the properties of electrovalent
compounds and covalent compounds. The main
properties are described below :
It is a special type of covalent bond in which both the
shared electrons are contributed by one atom only. It

of
Note :
(v) Conductivity : Like covalent compounds, these
are also bad conductors of electricity. The solutions
or fused mass do not allow the passage of electricity.
The atom which contributes electron pair in a
coordinate bond is called the donor while the atom
which accepts it is called acceptor.
The compound consisting of the coordinate bond is
HYDROGEN BONDING
termed coordinate compound.
In a compound containing hydrogen, when hydrogen
is bonded to highly electronegative atoms (such as
F,O, N) by a covalent bond, the electron pair is attracted towards electronegative atom so strongly that,
one end carries a slight positive charge (H– end) and
other end carries a slight negative charge (X-end).
(a) Examples of Coordinate Bond :
(i) Combination of ammonia and boron trifluoride :
Although the nitrogen atom has completed its octet
in ammonia, it still has a lone pair of electrons in the
valence shell which it can donate. The boron atom in
boron trifluoride is short of two electrons which it
accepts and completes its octet.
H
F
H
F

Note :
In 1920, Latimer and Rodebush introduced the idea
of “hydrogen bond.”
–
H – N ••
H
Donor
+
B–F
F
H–N
H
B–F
F
Acceptor
(ii) Formation of ammonium ion : Hydrogen ion (H+)
has no electron and thus accepts a lone pair donated
by nitrogen of ammonia.
X
H
Electronegative atom
If a number of such molecules are brought nearer to
each other, the positive end of one molecule and negative end of the other molecule will attract each other
and weak electrostatic force will develop. Thus, these
molecules will associate together to form a cluster of
molecules.
The attractive force that binds hydrogen atom of one
molecule with electronegative atom of the same or
different molecule is known as hydrogen bond. This
bond is represented by a dotted line.
CLASS-XI_STREAM-SA_PAGE # 39
e.g. : In oxalic acid (C2O4H2 ) ratio of carbon, oxygen
and hydrogen atoms is 1 : 2 : 1 so its empirical formula
is CO2H.
HYDROGEN BONDING IS OF TWO TYPES
(a) Intermolecular Hydrogen Bonding :
This type of bonding results between the positive and
the negative ends of different molecules of the same
or different substances.
e.g. Hydrogen bonding in ammonia
H
N
H
H
H
N
N
H
H
(B) Molecular formula : Formula which represents
the actual number of atoms of each element present
in one molecule of the substance is called molecular
formula.
H
H
N
H
N
e.g. : In oxalic acid the actual number of atoms of C, H
and O in one molecule are 2, 2, 4 respectively.
So, molecular formula would be C2H2O4 or C2O4H2.
H
H
H
H
H
H
e.g. CH2O is the empirical formula of formaldehyde
(HCHO), acetic acid (CH 3COOH) and glucose
(C6H12O6).
Significance of molecular formula :
Hydrogen bonding in water
H
O

H
H
H
O
H
H
O
H
(i) Name of substance : Molecular formula represents
the name of substance. For example, name of NaCl
is sodium chloride.
O
H
(ii) Knowledge of constituent elements : From the
molecular formula we come to know about the constituent
elements.
e.g. : In NaCl, constituent elements are sodium (Na)
and chlorine (Cl) whereas in H2O constituent elements
are hydrogen (H) and oxygen (O).
Note :
This type of hydrogen bonding increases the boiling
point of the compound and also its solubility in water.
(b) Intramolecular Hydrogen Bonding :
This type of bonding results between hydrogen and
an electronegative element both present in the same
molecule. This type of bonding is generally present
in organic compound. Examples are o -chlorophenol, o-nitrophenol, o-hydroxy benzoic acid , etc.
(iii) Number of atoms : With the help of the molecular
formula actual number of various atoms in the
molecule of any element or compound can be
determined.
e.g. : In one mole of H2O, two mole atoms of hydrogen
and one mole atoms of oxygen are present.
O
(iv) Molecular weight : From the weight of atoms
present in the molecule of an element or a compound
molecular weight can be determined.
e.g. :Molecular weight of water is - H2O=2(1)+16=2+16= 18
H
Cl
o - chlorophenol

Note :
This type of bonding decreases the boiling point and
the solubility of the compound in water.
(v) Quantity as per weight : From molecular formula
we get information pertaining to weight of element
present in the compound.
e.g. :In water as per weight 2 parts hydrogen and 16
parts oxygen are present.
CHEMICAL FORMULA
Smallest particle of elements or compounds which
can exist independently is known as molecule.
Formula of a substance is a group of symbols of the
elements which represents one molecule of the
substance.
(vi) Number of molecules : If any number is written
before the molecular formula then it represents the
number of molecules.
e.g. : In CuSO4. 5H2O five molecules of water are
present.
e.g. - Water molecule (H2O), Chlorine molecule (Cl2),
Hydrogen molecule (H2) etc.
(C) Structural formula : From the molecular formula
of element or compound we get no information
regarding the bonds between the atoms.
(a) Types of Formulae :
Formula representing the bonding of atoms in the
molecule of element or compound is known as
structural formula.
There are three types of formulae :
(A) Empirical formula
e.g. : In water molecule (H2O) one atom of oxygen is
linked with the two atoms of hydrogen.
H–O–H
(B) Molecular formula
(C) Structural formula
(A) Empirical Formula : Formula which represents
the simplest relative whole number ratio of atoms of
each element present in one molecule of that
substance is known as empirical formula.

Note :
Many compounds may have the same empirical
formula but their molecular and structural formula
may be different.
CLASS-XI_STREAM-SA_PAGE # 40
(b) Determination of
Molecular Formulae :
Empirical
and
2.
The following steps are involved in determining the
empirical formula of a compound :
(i) The percentage composition of each element is
divided by its atomic mass. It gives atomic ratio of the
elements present in the compound.
(ii) The atomic ratio of each element is divided by the
minimum value of atomic ratio as to get the simplest
ratio of the atoms of elements present in the
compound.
(iii) If the simplest ratio is fractional, then values of
simplest ratio of each element is multiplied by
smallest integer to get the simplest whole number
for each of the element.
(iv) To get the empirical formula, symbols of various
elements present are written side by side with their
respective whole number ratio as a subscript to the
lower right hand corner of the symbol.
(v) The molecular formula of a substance may be
determined from the empirical formula if the
molecular mass of the substance is known. The
molecular formula is always a simple multiple of
empirical formula and the value of simple multiple
(n) is obtained by dividing molecular mass with
empirical formula mass.
n
=
Molecular Mass
Empirical Formula Mass
Sol.
A compound on analysis, was found to have the
following composition :
(i) Sodium = 14.31%, (ii) Sulphur = 9.97%, (iii) Oxygen
= 69.50%, (iv) Hydrogen = 6.22%. Calculate the
molecular formula of the compound assuming that
whole of hydrogen in the compound is present as
water of crystallisation. Molecular mass of the
compound is 322.
Element Percentage
Atomic
mass
Relative Number
of atoms
Simplest
ratio
Sodium
14.31
23
14 .31

23
0.622
0.622
2
0.311
Sulphur
9.97
32
9 .97

32
0.311
0.311
1
0.311
Hydrogen
6.22
1

6.22
6.22
 20
0.311
Oxygen
69.50
16
69 .50

16
4.34
4.34
 14
0.311
6 . 22
1
The empirical formula = Na2SH20O14
Empirical formula mass
= (2 × 23) + 32 + (20 × 1) + (14 × 16)
= 322
Molecular mass
= 322
Molecular formula = Na2SH20O14
Whole of the hydrogen is present in the form of water
of crystallisation. Thus, 10 water molecules are
present in the molecule.
So, molecular formula = Na2SO4. 10H2O
(c) Sample Problems :
1.
Sol.
A compound of carbon, hydrogen and nitrogen
contains these elements in the ratio of 9:1:3.5
respectively. Calculate the empirical formula. If its
molecular mass is 108, what is the molecular formula ?
S.No. Element
Element Atomic Relative Number
Ratio
Mass
of Atoms
Simplest
Ratio
1
C arbon
9
12
9
 0.75
12
0.75
3
0.25
2
Hydrogen
1
1
1
1
1
1
 4
0.25
3
Nitrogen
3.5
14
3.5
 0.25
14
0.25
1
0.25
Empirical ratio = C3H4N
Empirical formula mass = (3 × 12)+(4× 1) + 14 = 54
n=
Molecular Mass
108
2
Empirical Formula Mass = 54
Thus, molecular formula of the compound
= (Empirical formula)2
= (C3H4N)2
= C6H8N2
CLASS-XI_STREAM-SA_PAGE # 41
PERIODIC TABLE & PERIODICITY IN PROPERTIES
(ii) Limitations of Dobereiner’s Classification :
DEFINITION
A periodic table may be defined as the table giving
the arrangement of all the known elements according
to their properties so that elements with similar
properties fall within the same vertical column and
elements with dissimilar properties are separated.
(A) Atomic mass of the three elements of some triads
are almost same.
e.g. Fe, Co, Ni
and
Ru, Rh, Pd
(B) It was restricted to few elements, therefore
discarded.
(c) Newlands’ Law of Octaves :
In 1866, an English chemist, John Newlands,
EARLY ATTEMPTS TO CLASSIFY ELEMENTS
(a) Metals and Non-Metals :
Among the earlier classifications, Lavoisier classified
the elements as metals and non-metals. However,
this classification proved to be inadequate. In 1803,
John Dalton published a table of relative atomic
weights (now called atomic masses). This formed
proposed a new system of grouping elements with
similar properties. He tried to correlate the properties
of elements with their atomic masses. He arranged
the then known elements in the order of increasing
atomic masses. He started with the element having
the lowest atomic mass (hydrogen) and ended at
thorium which was the 56th element. He observed
that every eighth element had properties similar to
an important basis of classification of elements.
that of the first.
(b) Dobereiner’s Triads:
Newlands called this relation as a law of octaves due
(i) In 1817, J.W. Dobereiner a German Chemist gave
to the similarity with the musical scale.
this arrangement of elements.
(A) He arranged elements with similar properties in
the groups of three called triads.
(B) According to Dobereiner the atomic mass of the
central element was merely the arithmetic mean of
atomic masses of the other two elements.
(i) Newlands’ arrangement of elements into ‘Octaves’:
Note s of
Music
Elements
sa (do)
re (re ) ga (m i)
m a (fa )
pa (so) dha (la ) ni (ti)
H
Li
Be
B
C
N
F
Na
Mg
Al
Si
P
O
S
Cl
K
Ca
Cr
Ti
Mn
Fe
Co and Ni
Cu
Zn
Y
In
As
Se
Br
Rb
Sr
Ce and La
Zr
–
–
e.g.
Elements of
the triad
Symbol Atomic mass
Lithium
Li
7
Sodium
Na
23
Potassium
K
39
Atomic mass of sodium =
Atomic mass of lithium  Atomic mass of potassium
2
=
7  39
= 23
2
Some examples of triads are given in the table :
(ii) Limitations of law of octaves : The law of octaves
has the following limitations :
(A) The law of octaves was found to be applicable
only upto calcium. It was not applicable to elements
of higher atomic masses.
(B) Position of hydrogen along with fluorine and
chlorine was not justified on the basis of chemical
properties.
(C) Newlands’ placed two elements in the same slot
to fit elements in the table. He also placed some
unlike elements under the same slot. For example,
cobalt and nickel are placed in the same slot and in
the column of fluorine, chlorine and bromine. But
cobalt and nickel have properties quite different from
fluorine, chlorine and bromine. Similarly, iron which
has resemblances with cobalt and nickel in its
properties has been placed far away from these
elements.
Thus, it was realized that Newlands’ law of octaves
worked well only with lighter elements. Therefore, this
classification was rejected.
CLASS-XI_STREAM-SA_PAGE # 42
( d ) Lother Meyer’s Classification :
In 1869, Lother Meyer studied the physical properties
like volume, melting point, boiling point etc. of different
elements.
He plotted a graph between atomic masses against
their respective atomic volumes for a number of
elements. He found the following observations (i) Elements with similar properties occupied similar
positions on the curve.
(ii) Alkali metals (Li, Na, K, Rb, Cs etc.) having larger
atomic volumes occupied the crests .
(iii) Transition elements (V, Fe, Co, Cu etc.) occupied
the troughs.
(iv) The halogens (F, Cl, Br,  etc.) occupied the
ascending portions of the curve before the inert
gases.
(v) Alkaline earth metals (Mg, Ca, Sr, Ba etc.) occupied
positions at about the mid points of descending
portions of the curve.
Among chemical properties, Mendeleev concentrated
mainly on the compound formed by elements with
oxygen and hydrogen. He selected these two
elements because these are very reactive and formed
compound with most of the elements known at that
time. The formulae of the compounds formed with
these elements (i.e. oxides and hydrides) were
regarded as one of the basic properties of an element
for its classification.
(i) Mendeleev’s periodic law : This law states that
the physical and chemical properties of the elements
are the periodic function of their atomic masses.
This means that when the elements are arranged in
the order of their increasing atomic masses, the
elements with similar properties recur at regular
intervals. Such orderly recurring properties in a cyclic
fashion are said to be occurring periodically. This is
responsible for the name periodic law or periodic
table.
(ii) Merits of Mendeleev’s periodic table : Mendeleev’s
periodic table was one of the greatest achievements in
the development of chemistry. Some of the important
contributions of his periodic table are given below :
Cs
A tom ic Vo lum e (cm 3 per m ole of atom s)
70
(A) Systematic study of elements : He arranged
known elements in order of their increasing atomic
masses considering the fact that elements with
similar properties should fall in the same vertical
column.
•
60
Rb
•
50
K
40
•
Li
30
20
10
•
• • Sr
•
• Ba
Na
I
Br •
•
Ca
• • •
•• Cl • •
Mg
• •Be •• • •• VFe Co ••••Zn • • ••
••• ••••• •••Cu •• •
•
••
•••• ••
•••
F
0
20
40
60 80
100 120
(B) Correction of atomic masses : The Mendeleev’s
periodic table could predict errors in the atomic
masses of elements based on their positions in the
table. Therefore atomic masses of certain elements
were corrected. For example, atomic mass of
beryllium was corrected from 13.5 to 9. Similarly, with
the help of this table, atomic masses of indium, gold,
platinum etc. were corrected.
140
Atomic mass
Change of Atomic Volume with Atomic Mass.
Drawback of Lother Meyer’s classification : This
was a hypothetical classification and it was difficult to
remember the positions of different elements.
(e) Mendeleev’s Periodic Table :
The major credit for a systematic classification of
elements goes to Mendeleev. He tried to group the
elements on the basis of some fundamental property
of the atoms. When Mendeleev started his work, only
63 elements were known. He examined the
relationship between atomic masses of the elements
and their physical and chemical properties.
(C) Mendeleev predicted the properties of those
missing elements from the known properties of the
other elements in the same group. Eka -aluminium
and eka -silicon names were given for gallium and
germanium (not discovered at the time of Mendeleev ).
(D) Position of noble gases : Noble gases like helium
(He), neon (Ne) and argon (Ar) were mentioned in
many studies. However, these gases were discovered
very late because they are very inert and are present
in extremely low concentrations. One of the
achievements of Mendeleev’s periodic table was that
when these gases were discovered, they could be
placed in a new group without disturbing the existing
order.
(iii) Limitations of Mendeleev’s periodic table : Inspite
of many advantages, the Mendeleev’s periodic table
has certain defects also. Some of these are given
below -
CLASS-XI_STREAM-SA_PAGE # 43
For example :
(A)Position of hydrogen : Position of hydrogen in the
periodic table is uncertain. It has been placed in 1A
group with alkali metals, but certain properties of
hydrogen resemble those of halogens. So, it may be
placed in the group of halogens as well as in the
group of alkali metals.
(B) Position of isotopes : Isotopes are the atoms of
the same element having different atomic masses.
Therefore, according to Mendeleev’s classification
these should be placed at different places depending
upon their atomic masses. For example, hydrogen
isotopes with atomic masses 1, 2 and 3 should be
placed at three places. However, isotopes have not
been given separate places in the periodic table
because of their similar properties.
• The atomic mass of argon is 39.9 and that of
potassium 39.1. But argon is placed before
potassium in the periodic table.
• The positions of cobalt and nickel are not in proper
order. Cobalt (at. mass = 58.9) is placed before nickel
( at. mass = 58.7).
• Tellurium (at. mass = 127.6) is placed before iodine
(at. mass = 126.9).
(D) Some similar elements are separated, in the
periodic table. For example copper (Cu) and mercury
(Hg). On the other hand, some dissimilar elements
have been placed together in the same group.
e.g. : Copper (Cu), silver (Ag) and gold (Au) have been
placed in group 1 along with alkali metals. Similarly,
manganese (Mn) is placed in the group of
halogens.
(C) Anomalous pairs of elements : In certain pairs of
elements, the increasing order of atomic masses
was not obeyed. In these, Mendeleev placed
elements according to similarities in their properties
and not in increasing order of their atomic masses.
Zr = 90
51
128
(E) Cause of periodicity : Mendeleev could not explain
the cause of periodicity among the elements.
CLASS-XI_STREAM-SA_PAGE # 44
CLASS-XI_STREAM-SA_PAGE # 45
Li
Hydrogen 2
6.941 4
K
4
7
6
Mg
87.62
Sr
85.468 38
Rb
Calcium
4
Ti
V
Cr
8
Mn
54.938 26
7 VIIB
51.996 25
6 VIB
50.942 24
5 VB
47.867 23
IVB
B
Boron
Fe
55.845 27
9
10
Co
58.933 28
VIII
Ni
58.963 29
11
12
Cu
63.546 30
IB
Zn
65.39
IIB
Ra
Radium
(226)
Barium
Fr
88
Ba
Francium
Caesium
87 (223)
Cs
N
O
F
18.998 10
17 VIIA
15.999 9
16 VIA
8
Ne
20.180
Helium
He
4.0026
18 XVIII
2
Si
Ga
Ge
Aluminium
Silicon
31 69.723 32 72.64
Al
33
As
S
Se
78.96
Sulphur
74.922 34
Phosphorus
P
Ar
Br
Kr
Argon
Chlorine
35 79.904 36 83.80
Cl
Oxygen
Fluorine
Boron
Nitrogen
Carbon
Neon
13 26.982 14 28.086 15 30.974 16 32.065 17 35.453 18 39.948
C
B
14.007
15 VA
12.011 7
14 IVA
10.811 6
13 IIIA
5
p–Block Elements
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Ac
Rf
Rutherfordium
ACTINOIDES
LANTHANOIDES
Actinium
104 (261)
Hafnium
(227)
Lanthanum
89
Hf
La
Sg
Rhenium
Bh
Nd
Protactinium
Hs
Ds
110 (281)
Platinum
Pt
Rg
Gold
111 (272)
Au
Meitnerium Darmstadtium Roentgenium
Mt
109 (268)
Iridium
Ir
Mercury
Hg
Thallium
Ti
Lead
Pb
83
208.98 84
(145)
Neptunium
Np
(237)
Promethium
Pm
(244)
Pu
Plutonium
94
(243)
Am
Americium
95
(247)
Cm
Curium
96
(247)
Bk
Berkelium
97
(251)
Cf
Californium
98
(252)
Es
Einsteinium
99
Er
Fermium
Fm
100 (257)
Erbium
164.93 68
Ho
Holmium
162.50 67
Dy
Dysprosium
158.93 66
Tb
Terbium
157.25 65
Gd
Gadolinium
151.96 64
Eu
Europium
150.36 63
Sm
Samarium
62
Te
Mendelevium
Md
101 (258)
Thulium
Tm
I
At
(210)
Iodine
86
Nobelium
No
102 (259)
Ytterbium
Yb
Xe
Lu
174.97
Radon
Rn
(222)
Xenon
Lawrencium
Lr
103 (262)
Lutetium
173.04 71
Astatine
85
168.93 70
Polonium
Po
(209)
Tellurium
197.26 69
Bismuth
Bi
Inner - Transition Metals (f-Block Elements)
Hassium
238.03 93
U
Uranium
231.04 92
Pa
Osmium
Os
108 (277)
144.24 61
Bohrium
Praseodymium Neodymium
Pr
Re
107 (264)
140.91 60
Seaborgium
232.04 91
Th
Thorium
90
Tungsten
W
106 (266)
140.12 59
Ce
Cerium
58
Dubnium
Db
105 (262)
Tantalum
Ta
Antimony
Sb
Iron
Selenium
Scandium
Titanium
Vanadium
Chromium Manganese
Copper
Zinc
Gallium Germanium
Arsenic
Bromine
Cobalt
Nickel
Krypton
39 88.906 40 91.224 41 92.906 42 95.94 43 (98)
44 101.07 45 102.91 46 106.42 47 107.87 48 112.41 49 114.82 50
118.71 51 121.76 52
127.60 53
126.90 54
131.29
Sc
44.956 22
IIIB
IIIA
10.811
Element name
5
13
Transition Metals (d –Block Elements)
Symbol
Atomic number
Group IUPAC
Zirconium
Niobium
Yttrium
Rhodium
Palladium
Cadmium
Strontium
Molybdenum Technetium Ruthenium
Silver
Indium
Tin
Rubidium
55 132.91 56 137.33 57 138.91 72 178.49 73 180.95 74 183.84 75 186.21 76 190.23 77 192.22 78 195.08 79 196.97 80 200.59 81 204.38 82 207.2
37
Potassium
Ca
Magnesium 3
Sodium
19 39.098 20 40.078 21
Na
5
Be
9.0122
IIA
Beryllium
Lithium
11 22.990 12 24.305
3
H
1.0079
1
3
2
Period 1
IA
Elements
1
Group
s–Block
Relative atomic mass
PERIODIC TABLE OF THE ELEMENTS
(f) Modern Periodic Table :
(iv) Demerits of Modern Periodic Table :
(i) Introduction :
In 1913, an English physicist, Henry Moseley showed
that the physical and chemical properties of the atoms
of the elements are determined by their atomic
number and not by their atomic masses.
Consequently, the periodic law was modified.
Following are the demerits of modern periodic table -
(ii) Modern Periodic Law (Moseley’s Periodic Law) :
uncertain in the periodic table.
“Physical and chemical properties of an element are
the periodic function of its atomic number’’.The
atomic number gives us the number of protons in
the nucleus of an atom and this number increases
by one in going from one element to the next in a row.
Elements, when arranged in the order of increasing
atomic number Z, lead us to the classification known
as the Modern Periodic Table. Prediction of properties
of elements could be made with more precision when
elements were arranged on the basis of increasing
atomic number.
(A) Position of hydrogen : Position of hydrogen was
uncertain in the periodic table.
(B) Position of lanthanides and actinides : The
positions of lanthanides and actinides were also
CLASSIFICATION OF THE ELEMENTS
It is based on the type of subshells which receives
the differentiating electron (i.e. last electron).
(a) s- Block Elements :
W hen last electron enters the s- orbital of the
outermost (nth) shell, the elements of this class are
called s- block elements.
Characteristics :
(i) Group 1 & 2 elements constitute the s - block.
(ii) General electronic configuration is ns1–2 .
(iii) Merits of Modern Periodic Table :
(A) Anomalous pairs : The original periodic law
based on atomic masses is violated in case of four
pairs of elements in order to give them positions on
the basis of their properties. The elements having
higher atomic masses have been assigned position
before the elements having lower atomic masses at
four places as shown below -
(iii) s - block elements lie on the extreme left of the
periodic table.
(iv) This block includes metals only.

Note :
The total number of elements in s-block is 13 (including hydrogen).
(b) p-Block Elements :
(a)
(b)
Ar
K
Co
Ni
At. Mass
40
39
60 58.6
At. No.
18
19
27
28
(c)
(d)
Te
I
Th
Pa
127.5
127
232
231
52
53
90
91
When differentiating electron enters the p - orbital of
the nth orbit, elements of this class are called p - block
elements.
Characteristics :
(i) Elements of group 13 to 18 constitute the p - block.
(ii) General electronic configuration is ns2np0-2 .
(iii) p - block elements lie on the extreme right of the
periodic table.
The discrepancy disappears, if the elements are
arranged in order of increasing atomic numbers.
(B) Position of isotopes : Isotopes are atoms of the
same element having different atomic masses, but
same atomic number. All the isotopes of an element
will be given different positions, if atomic mass is
taken as a basis. This shall disturb the symmetry of
the table. In modern table, one position is fixed for
one atomic number and since all the isotopes of an
element have the same atomic number, these are
assigned only one position.
(iv) This block includes some metals, all non-metals
and metalloids.

Note :
The total number of elements in p-block is 31.
(c) d - Block Elements:
When differentiating electron enters the (n–1)d orbital,
then elements of this class are called d - block
elements.
Characteristics :
(C) Elements with similar properties were placed
together and elements with dissimilar properties were
separated in modern periodic table.
(i) Elements of group 3 to 12 constitute the d - block.
(D) Cause of periodicity : Modern periodic table
explains the cause of periodicity among the elements.
(iv) All the transition elements are metals and most
of them form coloured complexes or ions.
(ii) General electronic configuration is (n – 1)d1–10 ns0- 2.
(iii) d - block elements lie in the centre of the periodic table.

Note :
The total number of elements in d-block is 39.
CLASS-XI_STREAM-SA_PAGE # 46
(d) f - block elements :
When last electron enters into f - orbital of (n – 2)th
shell, elements of this class are called f - block
elements.
Characteristics :
(i) All f - block elements belong to 3rd group.
(ii) General electronic configuration is (n – 2)f
(n– 1)d0-1 ns2.
1 – 14
(iii) f-block elements are present in two separate rows
below the periodic table.
(iv) All f-block elements are metals only.

Note :
Zn, Cd and Hg are d-block elements , but not transition
elements, because they do not contain partially filled
d-orbitals.
(iv) Inner transition elements :- They contain three
incomplete outermost shell and were also referred
to as rare earth elements, since their oxides were
rare in earlier days.
(v) Diagonal relationship : Some elements of 2nd and
3rd period show diagonal relationship among them.
They represent the same properties of two periods.
This relation is known as diagonal relation.
The elements of f- block have been classified into
two series :
(A) Lanthanides : 14 elements present after element
lanthanum (57La) are called lanthanides.1 st inner
transition series of metals or 4f - series, contains 14
elements i.e. 58Ce to 71Lu .
(B) Actinides : 14 elements present after element
actinium (89Ac) are called actinides. 2nd inner transition
series of metals or 5f- series, also contains 14 elements
i.e. 90Th to 103Lr.
Note :
The total number of elements in f-block is 28.
DIVISION OF THE PERIODIC TABLE
INTO VARIOUS BLOCKS
PERIODICITY IN PROPERTIES
Periodicity :
The repetition of elements with similar properties after
certain regular intervals, when the elements are
arranged in order of increasing atomic number, is
called periodicity.
(a) Atomic Volume :
Atomic volume increases in a group from top to
bottom. The increase is due to the increase in the
number of energy levels.
In a period from left to right, atomic volume varies
cyclically, i.e. it decreases at first for some elements,
becomes minimum in the middle and then increases.
The following two factors explain this trend (i) atomic radii decrease due to increase of nuclear
charge.
(ii) the number of valence electrons increases in a
period.
(i) Noble gases : The elements belonging to group
18 are called noble gases or aerogenous. They are
also known as inert gases because their outermost
orbitals are completely filled. Helium (He) is an
exception. It has only two electrons which are present
in s-orbital.
As to accommodate all the valence electrons, the volume
increases. These two factors oppose each other. The
effect of first factor is more on the left hand side and that
of the second factor is more on the right hand side in the
periodic table . The volumes are in cm3.
(ii) Representative elements : Elements in which
atoms have all shells complete except outermost shell
which is incomplete. Except 18th group, all s - block
and p - block elements are collectively called normal
or representative elements.
Group/
Period
(iii) Transition elements : Those elements which have
partially filled d - orbitals in neutral state or in any
stable oxidation state are called transition elements.
2
3
4
1
2
13 14 15 16 17 18
Li
Be
B
C
N
O
F
Ne
13
5
5
5
14
11
15
17
Na Mg Al
Si
P
S
Cl
Ar
24
14
12
17
16
19
24
K
Ca Ga Ge As Se Br
Kr
46
26
33
10
12
13
16
16
23
CLASS-XI_STREAM-SA_PAGE # 47
Increases

(vi) Transuranium elements : The elements with
atomic number greater than 92 (Z > 92) are known as
transuranium elements.

Note :
The maximum value of atomic volume (87 cm3) is
observed in the case of francium.
(d) Atomic Radius :
(i) Covalent radius : It may be defined as one - half of
the distance between the centres of the nuclei of two
similar atoms bonded by a single covalent bond.
(b) Density :
The density of the elements in solid state varies
periodically with their atomic numbers . At first, the
density increases gradually in a period and becomes
maximum somewhere for the central members and
then starts decreasing afterwards gradually. The value
of densities in the table are in g/cc.
Group/
Period
2
2
13
14
15
16
17
Li
Be
B
C
2.35
2.26
O
–
S
F
–
Cl
–
Br
Na Mg
Al
Si
N
–
P
0.97 1.74
2.70
2.34
1.82
2.1
Se
0.54 1.85
3
Ca
Ga
Ge
As
0.86 1.55
5.90
5.32
5.77 4.19 3.19
In
Sn
Sb
7.31
7.27
6.70 6.25 4.94
Tl
Pb
K
4
Rb
5
Sr
1.53 2.63
Cs
6

1
Ba
Bi
Te
Po
1.90 3.62 11.85 11.34 9.81 9.14
e.g. The internuclear distance between two hydrogen
atoms in H 2 molecule is 74 pm. Therefore, the
covalent radius of hydrogen atom is 37 pm.

Note :
Covalent radius is generally used for non - metals.
(ii) Vander Waal’s radius : It may be defined as half
of the internuclear distance between two adjacent
atoms of the same element belonging to two nearest
neighbouring molecules of the same substance.
I
At
–
Note :
In solids, Iridium (Ir) has the highest density (22.61 g/
cc) and in liquids, mercury (Hg) has the highest
density (13.6 g/cc).
(c) Melting and Boiling Points :
Characteristics :
The melting points of the elements exhibit some
periodicity with rise of atomic number. It is observed
that elements with low values of atomic volumes have
high melting points, while elements with high values
of atomic volumes have low melting points. In general,
melting points of elements in any period at first
increase and become maximum somewhere in the
centre and thereafter begin to decrease. Values in
the table are in ºC.
Group/
Period
2
3
1
2
13
14
15
16
17
(A) Covalent radius of the elements is shorter than
its Vander Waal’s radius.
(B) The formation of covalent bond involves overlapping
of atomic orbitals, as a result of this, the internuclear
distance between the covalently bonded atoms is less
than the internuclear distance between the non bonded
atoms.
e.g. : Vander Waals radius of helium is 1.20 Å
(iii) Metallic radius (Crystal radius) : Metallic radius
may be defined as half of the internuclear distance
between two adjacent atoms in a metallic lattice.
18
Li
Be
B
C
N
O
F
Ne
181
1287
2180
3727
–210
–219
–220
–249
Na
Mg
Al
Si
P
S
Cl
Ar
98
650
660
1420
44
119
–101
–189
Tungsten has the maximum melting point (3410ºC)
amongst metals and carbon has the maximum
melting point (3727ºC) amongst non-metal. Helium
has the minimum melting point (–270ºC) amongst
all elements . The metals Cs (m.p. = 28.5ºC), Ga
(m.p. = 30ºC) and Hg (m.p. = –39ºC) are known in
liquid state at 30ºC.
The boiling points of the elements also show similar
trends, however, the regularities are not striking as
noted in the case of melting points.
• The metallic radius of an atom is always larger than
its covalent radius.
Metallic radius >
Covalent radius
e.g.
K
231pm
203 pm
Na
186 pm
154 pm

Note :
The order of different radius is - r Vander Waals > rMetallic > rCovalent
CLASS-XI_STREAM-SA_PAGE # 48
(iv) Variation of atomic radii in a period :
As we move from left to right across a period, there is
a regular decrease in atomic radii of the
representative elements. This is due to the fact that
number of energy shells remains the same in a
period, but nuclear charge increases gradually as
the atomic number increases. This increases the
force of attraction towards nucleus which brings
contraction in size.
This can also be explained on the basis of effective
nuclear charge which increases gradually in a period
i.e. electron cloud is attracted more strongly towards
nucleus as the effective nuclear charge becomes
more and more as we move in a period. The increased
force of attraction brings contraction in size.
(v) Variation of atomic radii in a group :
Atomic radii in a group increase as the atomic
number increases. The increase in size is due to
extra energy shells which outweigh the effect of
increased nuclear charge. The following table
illustrates the periodicity in atomic radii (covalent radii)
of representative elements. The radii given in the
table are in angstrom (Å).
1
1
H
5
6
2
13
14
15
16
remains almost constant.
However, in vertical columns of transition elements,
there is an increase in size from first member to
second member as expected, but from second
member to third member, there is very small change
in size and sometimes sizes are same. This is due
to Lanthanide contraction (in the lanthanide elements
differentiating electrons enter into 4f-levels). Since
these electrons do not effectively screen the valence
electrons from the increased nuclear charge, the size
gradually decreases. This decrease is termed
lanthanide contraction.
(e) Ionisation Energy (IE) :
Ionisation Energy (IE) of an element is defined as the
amount of energy required to remove an electron from
an isolated gaseous atom of that element resulting
17
(A) The energy required to remove the outermost
electron from an atom is called first ionisation energy
0.32
Li
Be
B
C
N
O
F
1.23
0.89
0.80
0.77
0.75
0.73
0.72
Na
Mg
Al
Si
P
S
Cl
1.54
1.36
1.20
1.17
1.10
1.04
0.99
K
Ca
Ga
Ge
As
Se
Br
2.03
1.74
1.26
1.22
1.20
1.16
1.14
Rb
Sr
In
Sn
Sb
Te
I
2.16
1.91
1.44
1.41
1.40
1.36
1.33
Cs
Ba
Tl
Pb
Bi
Po
2.35
1.98
1.48
1.47
1.46
1.46
(IE)1.
Increases
4
levels effectively screen much of increased nuclear
charge on the outer ns electrons and therefore, size
(i) Characteristics :
Group/
Period
3
small since the differentiating electrons enter into
inner ‘d’ levels. The additional electrons into (n–1)d
in the formation of a positive ion.
Periodicity in atomic radii (covalent radii)
2
The decrease in the size of transition elements is
After removal of one electron, the atom changes into
monovalent positive ion.
M(g) + IE1  M+(g) + e–
(B) The minimum amount of energy required to
remove an electron from monovalent positive ion of
the element is known as second ionisation energy
Decreases
(IE)2.
M+(g) + IE2  M2+(g) + e–
Atoms of zero group elements do not form chemical
bonds among themselves. Hence for them Vander
Waals radii are considered.
(C) The first, second etc. ionisation energies are
collectively known as successive ionisation energies.
M2+(g) + IE3  M3+(g) + e–
In general (IE)1 < (IE)2 < (IE)3 so on, because, as the
Element
Vander Waals
radii (in Aº )
He
Ne
Ar
Kr
Xe
1.20 1.60 1.91 2.00 2.20
The sudden increase in atomic radii in comparison
to the halogens (the elements of 7th group) in case
of inert gases, is due to the fact that, Vander Waals
radii are considered which always possess higher
values than covalent radii.
number of electrons decreases, the attraction
between the nucleus and the remaining electrons
increases considerably and hence subsequent
ionisation energies increase.
(D) Units : Ionisation energy is expressed either in
terms of electron volts per atom (eV/atom) or Kilojoules
per mole of atoms (KJ mol – 1) or K cal mol – 1.
1 eV/atom = 96.49 KJ/mol = 23.06 Kcal/mol = 1.602 × 10–19
J/atom
CLASS-XI_STREAM-SA_PAGE # 49
(ii) Factors influencing ionisation energy :
(A) Size of the atom : Ionisation energy decreases
with increase in atomic size. As the distance between
the outermost electrons and the nucleus increases,
the force of attraction between the valence shell
electrons and the nucleus decreases. As a result,
outermost electrons are held less firmly and lesser
amount of energy is required to knock them out.
For example, ionisation energy decreases in a group
from top to bottom with increase in atomic size.
(B) Nuclear charge : The ionisation energy increases
with increase in the nuclear charge. This is due to the
fact that with increase in the nuclear charge, the
electrons of the outermost shell are more firmly held
by the nucleus and thus greater amount of energy is
required to pull out an electron from the atom. For
example, ionisation energy increases as we move from
left to right along a period due to increase in nuclear
charge.
(C) Shielding effect : The electrons in the inner shells
act as a screen or shield between the nucleus and
the electrons in the outermost shell. This is called
shielding effect or screening effect. Larger the number
of electrons in the inner shells, greater is the
screening effect and smaller the force of attraction
and thus ionisation energy decreases.
(E) Electronic Configuration : If an atom has exactly
half-filled or completely filled orbitals, then such an
arrangement has extra stability.The removal of an
electron from such an atom requires more energy
than expected. For example,
E1 of Be > E1 of B
1s 2 , 2s 2



Be (Z = 4) Completely filled
orbital (more stable)
B (Z = 5)
1s 2 , 2s 2 , 2 p1


Partially filled
orbital (less stable )
As noble gases have completely filled electronic
configurations, they have highest ionisation energies
in their respective periods.
(iii Variation of ionisation energy in a period :In
general, the value of ionisation energy increases with
increase in atomic number across a period. This can
be explained on the basis of the fact that on moving
across the period from left to right(A) nuclear charge increases regularly.
(B) addition of electrons occurs in the same shell.
(C) atomic size decreases.
(iv) Variation of ionisation energy in a group : In
general, the value of ionisation energy decreases
while moving from top to bottom in a group.This is
because -
These electrons
shield the outer
electrons from the
nucleus

This electron does
not feel the full inward pull
of the positive charge of
the nucleus
(A) effective nuclear charge decreases regularly.
(B) addition of electrons occurs in a new shell.
(D) Penetration effect of the electrons : The
ionisation energy increases as the penetration effect
of the electrons increases. It is a well known fact that
the electrons of the s-orbital have the maximum
probability of being found near the nucleus and this
probability goes on decreasing in case of p, d and f
orbitals of the same energy level.
(C) atomic size increases.
Greater the penetration effect of electrons more firmly
the electrons will be held by the nucleus and thus
higher will be the ionisation energy of the atom.
For example, ionisation energy of aluminium is
comparatively less than magnesium as outermost
electron is to be removed from p-orbital (having less
penetration effect) in aluminium, whereas in
magnesium it will be removed from s-orbital (having
larger penetration effect) of the same energy level.
Cl(g) + e–  Cl– (g) + 349 KJ/mol
Note :
(EA -) is exothermic whereas, (EA-) is endothermic.
Within the same energy level,the penetration effect
decreases in the order s > p > d > f
(i) Units : Kilo joules per mole (KJ/mol) of atoms or
electron volts per atom (eV/atom).
(f) Electron Affinity (EA) :
Electron affinity is defined as the energy released in
the process of adding an electron to a neutral atom in
the gaseous state to form a negative ion.
X(g) + e–  X– (g) + Energy (E.A.)
The electron affinity of chlorine is 349 KJ/mol.
The addition of second electron to an anion is
opposed by electrostatic repulsion and hence the
energy has to be supplied for the addition of second
electron.
O(g) + e–  O– (g) + Energy (EA -)
O–(g) + e–  O2– (g) – Energy (EA-)
CLASS-XI_STREAM-SA_PAGE # 50
(ii) Factors affecting electron affinity:
(i) Factors influencing electronegativity :
(A) Nuclear charge : Greater the magnitude of
nuclear charge greater will be the attraction for the
incoming electron and as a result, larger will be the
value of electron affinity.
(A) The magnitude of electronegativity of an element
depends upon its ionisation potential & electron
affinity. Higher ionisation potential & electron affinity
values indicate higher electronegativity value.
Electron affinity  Nuclear charge.
(B) With increase in atomic size the distance between
nucleus and valence shell electrons increases,
therefore the force of attraction between the nucleus
and the valence shell electrons decreases and hence
the electronegativity values also decrease.
(B) Atomic size : Larger the size of an atom is, more
will be the distance between the nucleus and the
incoming electron and smaller will be the value of
electron affinity.


1
 E.A. 

Atomic size 

(C) Electronic configuration : Stable the electronic
configuration of an atom lesser will be its tendency to
accept the electron and lower will be the value of its
electron affinity.
(iii) Variation of electron affinity in a period : On
moving across the period the atomic size decreases
and nuclear charge increases. Both these factors
result into greater attraction for the incoming electron,
therefore electron affinity in general increases in a
period from left to right.
(iv) Variation of electron affinity in a group : On
moving down a group, the atomic size as well as
nuclear charge increase, but the effect of increase in
atomic size is much more pronounced than that of
nuclear charge and thus, the incoming electron feels
less attraction consequently, electron affinity
decreases on going down the group.
(v) Some irregularities observed in general trend:
(A) Halogens have the highest electron affinities in
their respective periods. This is due to the small size
and high effective nuclear charge of halogens.
Halogens have seven electrons in their valence shell.
By accepting one more electron they can attain stable
electronic configuration of the nearest noble gas. Thus
they have maximum tendency to accept an additional
electron.
(B) Due to stable electronic configuration of noble
gases electron affinities are zero.
(C) Be, Mg, N and P also have exceptionally low values
of electron affinities due to their stable electronic
configurations.
2
2
Be = 1s , 2s
Mg = 1s2, 2s2 , 2p6, 3s2
2
2
3
N = 1s , 2s , 2p
P = 1s2, 2s2, 2p6, 3s2, 3p3
(g) Electronegativity :
Electronegativity is a measure of the tendency of an
element to attract electrons towards itself in a
covalently bonded molecule .
(C) In higher oxidation state, the element has higher
magnitude of positive charge. Thus, due to more
positive charge on element, it has higher polarising
power. Thus, with increase in the oxidation state of
element, its electronegativity also increases.

Note :
Polarising power is the power of an ion (cation) to
distort the other ion.
(D) With increase in nuclear charge, force of attraction
between nucleus and the valence shell electrons
increases and, therefore electronegativity value
increases.
(E) The electronegativity of the same element
increases as the s-character in the hybrid orbitals
increases.
Hybrid orbital
sp 3
sp 2
sp
s-character
25%
33%
50%
Electronegativity increases
(ii) Variation of Electronegativity in a group : On
moving down the group atomic number increases,
so nuclear charge also increases. Number of shells
also increases, so atomic radius increases. But the
increase in size outweigh the effect of increased
nuclear charge. Therefore electronegativity
decreases on moving down the group.
(iii) Variation of Electronegativity in a period : While
moving across a period left to right atomic number,
nuclear charge increase & atomic radius decreases.
Therefore electronegativity increases along a period.
(h) Valency :
(i) The valency of an element may be defined as the
combining capacity of the element.
(ii) The electrons present in the outermost shell are
called valence electrons and these electrons
determine the valency of the atom.
Valency of an element is determined by the number
of valence electrons in an atom of the element.
CLASS-XI_STREAM-SA_PAGE # 51
The valency of an element = number of valence
electrons (when number of valence electrons are
from 1 to 4 ) The valency of an element = 8–number of
valence electrons. (when number of valence electrons
are more than 4)
(iii) Variation of valency across a period : The number
of valence electrons increases from 1 to 8 on moving
across a period. The valency of the elements with
respect to hydrogen and chlorine increases from 1 to
4 and then decreases from 4 to zero. With respect to
oxygen valency increases from 1 to 7.
Variation of valency of elements of second and third
periods :
Elements of second period :
Li
Be
B
C
N
O
F
Valency with respect to H
L iH BeH2 BH3
CH 4
(1)
(2) (3)
(4)
H 2O
(2)
HF
(1)
NH 3
(3)
Valency with respect to Cl
L iCl BeCl2 BCl3 CCl4 NCl3 Cl2O ClF
(1)
(2)
(3)
(4)
(3)
(2) (1)

Note :
Valency of elements changes in a period. Elements
of third period :
Na
Mg
Al
Si P S Cl
Valency with respect to H
NaH MgH2 AlH3 SiH4 PH3 H2S
(1)
(2)
(3)
(4)
Valency with respect to ‘O’
Na2O MgO Al2O3 SiO2
(1)
(2)
(3)
(4)
(3)
(2)
P2O5
(5)
HCl
(1)
SO3
(6)
Cl2O7
(7)
(iv)Variation of valency along a group : On moving
down a group, the number of valence electrons
remains the same and, therefore, all the elements in
a group exhibit the same valency.
e.g. All the elements of group 1 have valency equal to
1 and those of group 2 have valency equal to 2.
CLASS-XI_STREAM-SA_PAGE # 52
ACIDS, BASES AND SALTS
ACIDS
USES OF ACIDS
An acid may be defined as a substance which
releases one or more H+ ions in aqueous solution.
A list of commonly used acids along with their
chemical formula and typical uses, is given below
Name
Type
Chemical Formula
Where found or used
Carbonic acid
Mineral acid
H2CO3
In soft drinks and lends fizz.
Nitric acid
Mineral acid
HNO3
Used in the manufacture of explosives (TNT,
Nitroglycerine) and fertilizers (Ammonium
nitrate, Calcium nitrate, Purification of Au, Ag)
HCl
In purification of common salt, in textile industry
as bleaching agent, to make aqua regia, in
stomach as gastric juice, used in tanning
industry
Mineral acid
H2SO4
Commonly used in car batteries, in
the manufacture of fertilizers (Ammonium
sulphate, super phosphate) detergents etc, in
paints, plastics, drugs, in manufacture of
artificial silk, in petroleum refining.
Phosphoric acid Mineral acid
H3PO4
Used in antirust paints and in fertilizers.
Hydrochloric acid Mineral acid
Sulphuric acid
Formic acid
Organic acid
HCOOH(CH2O2)
Acetic acid
Organic acid
CH3COOH(C2H4O2)
Lactic acid
Organic acid CH3CH(OH)COOH (C3H6O3)
Found in the stings of ants and bees, used in
tanning leather, in medicines for treating gout.
Found in vinegar, used as solvent in the
manufacture of dyes and perfumes.
Responsible for souring of milk in curd.
Benzoic acid
Organic acid
C6H5COOH
Used as a food preservative.
Citric acid
Organic acid
C6H8O7
Present in lemons, oranges and citrus fruits.
Tartaric acid
Organic acid
C4H6O6
Present in tamarind.
TYPES OF ACIDS
e.g.
(a) On the basis of source :
Carbonic acid (H2CO3), phosphoric acid (H3PO4),
formic acid (HCOOH), acetic acid (CH3COOH) are
(i) Mineral Acids : Acids which are obtained from
rocks and minerals are called mineral acids.
weak acids.
CH3COOH + Water  CH3COO– (aq) + H+ (aq)
(ii) Organic Acids : Acids which are present in animals
and plants are known as organic acids.
(b) On the basis of strength :
(i) Strong acid : Acids which are completely ionised
in water are known as strong acids.
e.g.
Hydrochloric acid (HCl), sulphuric acid (H2SO4), nitric
acid (HNO3) etc. are all strong acids.
–
HCl + Water  H+(aq) + Cl (aq)
Chemical Properties of Acids :
1.
Action with metals :
Dilute acids like dilute HCl and dilute H2SO4 react
with certain active metals to evolve hydrogen gas.
2Na(s) + 2HCl (dilute)  2NaCl(aq) + H2(g)
H2SO4 + Water  2H+(aq) + SO42–(aq)
Mg(s) + H2SO4 (dilute)  MgSO4(aq) + H2(g)
Metals which can displace hydrogen from dilute acids
are known as active metals. e.g. Na, K, Zn, Fe, Ca, Mg
etc.
(ii) Weak acids : Acids which are partially ionised in
water are known as weak acids.
Zn(s) + H2SO4 (dilute)  ZnSO4(aq) + H2(g)
CLASS-XI_STREAM-SA_PAGE # 53
e.g.
The active metals which lie above hydrogen in the
activity series are electropositive and more reactive
in nature. Their atoms lose electrons to form positive
ions and these electrons are accepted by H+ ions of
the acid. As a result, H2 is evolved.
e.g.
Zn(s)  Zn2+ (aq) + 2e–
CaCO3(s)+ 2HCl(aq)
Zn(s) + 2H+(aq)  Zn++(aq) + H2(g)
Action with bases :
Acids react with bases to give salt and water.
HCl + NaOH  NaCl + H2O
A base may be defined as a substance capable of
releasing one or more OH¯ ions in aqueous solution.
Alkalies :
Some bases like sodium hydroxide and potassium
hydroxide are water soluble. These are known as
alkalies. Therefore water soluble bases are known
as alkalies eg. KOH, NaOH.
Bases like Cu(OH)2, Fe(OH)3 and Al(OH)3 these are
not alkalies.
MgO(s) + H2SO4(aq)  MgSO4(aq) + H2O()
CuO(s) + 2HCl(aq.)  CuCl2(aq) + H2O()
3.
Sodium
sulphate
BASE
ZnO(s) + 2HCl (aq)  ZnCl2(aq) + H2O()
(Black)
 Na2SO4(aq) + 2H2O(aq) + 2CO2(g)
Sodium
bicarbonate
4.
Action with metal oxides :
Acids react with metal oxides to form salt and water.
These reactions are mostly carried out upon heating.
e.g.
Calcium
chloride
2NaHCO3(s) + H2SO4(aq)
2H+(aq) + SO42– (aq) + 2e–  H2(g) + SO42–(aq)
2.
 CaCl2(aq) + H2O()+ CO2(g)
Calcium
carbonate
(Bluish green)
Action with metal carbonates and metal
bicarbonates :
Both metal carbonates and bicarbonates react with
USES OF BASES
A list of a few typical bases along with their chemical
formulae and uses is given below-
acids to evolve CO2 gas and form salts.
Name
Commercial Chemical
Name
Formula
Sodium
hydroxide
Caustic
soda
NaOH
Potassium
hydroxide
Caustic
potash
KOH
Calcium
hydroxide
Magnesium
hydroxide
Aluminium
hydroxide
Ammonium
hydroxide
Slaked
lime
Milk of
magnesia
Ca(OH)2
Mg(OH)2
Uses
In manufacture of soap,
paper, pulp, rayon, refining of
petroleum etc.
In alkaline storage batteries,
manufacture of soap, absorbing
CO2 gas etc.
In manufacture of bleaching powder,
softening of hard water etc.
As an antacid to remove acidity
from stomach.
–
Al(OH)3
As foaming agent in fire extinguishers.
–
NH4OH
In removing grease stains from
clothes and in cleaning window panes.
Both NaOH and KOH are deliquescent in nature which
means that they absorb moisture from air and get
liquefied.
TYPES OF BASES
(a) On the basis of strength
(i) Strong base : A base contains one or more hydroxyl
(OH–) groups which it releases in aqueous solution
upon ionisation. Bases which are almost completely
ionised in water, are known as strong bases.
e.g.
Sodium hydroxide (NaOH), potassium hydroxide
(KOH), barium hydroxide Ba(OH)2 are all strong
bases.
NaOH(s) + Water  Na+(aq) + OH–(aq)
KOH(s) + Water  K+(aq) + OH–(aq)
(ii) Weak bases : Bases that are feebly ionised on
dissolving in water and produce less concentration
of hydroxyl ions are called weak bases.
e.g. Ca(OH)2, NH4OH
Chemical Properties :
1.
Action with metals :
Metals like zinc, tin and aluminium react with strong
alkalies like NaOH (caustic soda), KOH (caustic
potash) to evolve hydrogen gas.
CLASS-XI_STREAM-SA_PAGE # 54
Zn(s) + 2NaOH(aq)  Na2ZnO2(aq) + H2(g)
Sodium zincate
B+ + OH¯
Base
Sn(s) + 2NaOH(aq)  Na2SnO2(aq) + H2(g)
Sodium stannite
2.
Water
BOH
NaOH
2Al(s)+ 2NaOH + 2H2O  2NaAlO2(aq) + 3H2(g)
sSodium meta
aluminate
Base
Action with non-metallic oxides :
Acids react with metal oxides, but bases react with
oxides of non-metals to form salt and water.
e.g.
2NaOH(aq) + CO2(g)  Na2CO3(aq) + H2O()
Ca(OH)2(s) + SO2(g)  CaSO3(aq) + H2O()
Ca(OH)2(s) + CO2(g)  CaCO3(s) + H2O()
Base
NH4OH
Water
Water
Na+ + OH¯
NH4+ + OH¯
(A) Limitations of Arrhenius theory :
• It is applicable only to aqueous solutions. For the
acidic or basic properties, the presence of water is
absolutely necessary.
• The concept does not explain the acidic or basic
properties of acids or bases in non - aqueous solvents.
THEORIES OF ACIDS AND BASES
• It fails to explain the basic nature of compounds like
NH3, Na 2CO3 etc., which do not have OH in their
molecules to furnish OH– ions.
(a) Arrhenius Theory :
This concept was given in 1884 .
• It fails to explain the acidic nature of non - protic
compounds like SO2, P2O5, CO2, NO2 etc., which do
not have hydrogen in their molecules to furnish H+
ions.
According to this theory all substances which give H+
ions when dissolved in water are called acids, while
those which ionise in water to give OH– ions are called
bases.
• It fails to explain the acidic nature of certain salts like
AlCl3 etc., in aqueous solutions.
The main points of this theory are (i) An acid or base when dissolved in water, splits into
ions. This is known as ionisation.
(b)
(ii) Upon dilution, the ions get separated from each other.
This is known as dissociation of ions.
Lowry :
This theory was given by Bronsted, a Danish chemist
and Lowry, an English chemist independently in 1923.
According to it, an acid is a substance, molecule or
ion which has a tendency to release the proton
(protogenic) and similarly a base has a tendency to
accept the proton (protophilic).
(iii) The fraction of the acid or base which dissociates
into ions is called its degree of dissociation and is
denoted by alpha  which can be calculated by the
following formula :
=
Acid Base Concept of Bronsted and
No. of molecules dissociate d at equilibriu m
total no. of molecules
e.g.
(iv) The degree of dissociation depends upon the
nature of acid or base. Strong acids and bases are
highly dissociated, while weak acids and bases are
dissociated to lesser extent.
(v) The electric current is carried by the movement of
ions. Greater the ionic mobility more will be the conductivity of the acid or base.
(vi) The H+ ions do not exist as such and exist in
combination with molecules of H2O as H3O+ ions
(known as hydronium ion).
H+ + H2O
HCl + H2O
e.g.
HA + H2O
Acid
H2SO4 + 2H2O
Acid
H3O+ + Cl–
HCl + H2O
In this reaction, HCl acts as an acid because it donates a proton to the water molecule. Water, on the
other hand, behaves as a base by accepting a
proton.

Note :
Bronsted and Lowry theory is also known as proton
donor and proton acceptor theory.
Other examples :
(i) CH3COOH + H2O
H3O+
H3O+ + Cl–
H3O+ + CH3COO–
(ii) NH4+ + H2O
H3O+ + NH3
(iii) NH3 + H2O
NH4+ + OH–
+
H3O + A¯
2H3O+ + SO4–2
In the reactions (i) and (ii) water is acting as a base,
while in reaction (iii) it is acting as an acid.Thus water
can donate as well as accept H+ and hence can act
as both acid and base.
CLASS-XI_STREAM-SA_PAGE # 55

Note :
The species like H2O, NH3, CH3COOH which can act
(B) Molecules in which the central atom has empty
d-orbitals : The central atom of the halides such as
TiCl4, SnCl4, PCl3, PF5, SF4, TeCl4. etc., have vacant dorbitals. These can, therefore, accept an electron pair
and act as Lewis acids.
as both acid and base are called amphiprotic.
Moreover according to theory, an acid on losing a proton becomes a base, called conjugate base, while
the base by accepting proton changes to acid called
conjugate acid.
••
SiF4 + 2 ••F ••
••
Lewis
+
accepts H
acid
–
(C) Simple cations : Cations are expected to act as
Lewis acid, since they are electron deficient in nature.
••
+
+
+
loses H
Here CH3COO– ion is conjugate base of CH3COOH,
while H3O+ ion is conjugate acid of H2O.
Ag + 2NH3
(i) Merits :
(B) This theory states that the terms acid and base
are comparative. A substance may act as an acid in
one solvent, while as a base in another solvent.
(A) Many acid - base reactions proceed without H+
transfer.
+ SO4
(B) Negatively charged species or simple anions :
For example chloride (Cl–), cyanide (CN–), hydroxide
(OH–) ions etc. act as Lewis bases.
2-
(C) Multiple bonded compounds : The compounds
such as CO, NO, ethylene, acetylene etc. can act as
Lewis bases.
(c) Lewis theory :
The theory was given by G.N. Lewis in 1938. According to it, an acid is a species which can accept a pair
of electrons, while the base is one which can donate
a pair of electrons.


Note :
It is also known as electron pair donor and electron
(A) Lewis theory fails to explain the relative strength
of acids and bases.
INDICATORS
,
(ii) NH3 is a Lewis base as it has a pair of electrons
which can be easily donated.
Lewis acids :- CH3+, H+, BF3, AlCl3, FeCl3 etc.
Lewis base :- NH3, H2O, R–O–R, R – OH, CN¯, OH¯
etc.
(i) Characteristics of species which can act as
Lewis acids :
(A) Molecules in which the central atom has incomplete octet : Lewis acids are electron deficient molecules such as BF3, AlCl3, GaCl3 etc.
Note :
It may be noted that all Bronsted bases are also Lewis
bases, but all Lewis acids are not Bronsted acids.
(iv) Limitations of Lewis theory :
pair acceptor theory.
e.g.
(i) FeCl3 and AlCl3 are Lewis acids, because the central atoms have only six electrons after sharing and
need two more electrons.
NH3]
(A) Neutral species having at least one lone pair of
electrons : For example, ammonia, amines, alcohols
etc, act as Lewis bases as they contain a pair of electrons.
(ii) Demerits :
SO
Ag
(ii) Characteristics of species which can act as
Lewis bases :
e.g. Acetic acid acts as an acid in water while as a
base in HF.
e.g. SO2 + SO3
[H3N
(D) Molecules having a multiple bond between atoms of dissimilar electronegativity : Typical
examples of molecules belonging to this class of
Lewis acids are CO2, SO2 and SO3.
(A) Besides water any other solvent, which has the
tendency to accept or lose a proton may decide the
acidic or basic behaviour of the dissolved substance.
2+
Complex
+
CH3COO + H3O
Base
Acid
CH3COOH + H2O
Base
Acid
Lewis
base
2–
[SiF6]
An indicator indicates the nature of a particular
solution whether acidic, basic or neutral. Apart from
this, indicator also represents the change in nature
of the solution from acidic to basic and vice versa.
Indicators are basically coloured organic substances
extracted from different plants. A few common acid base
indicators are(a) Litmus :
Litmus is a purple dye which is extracted from ‘lichen’
a plant belonging to variety Thallophyta. It can also be
applied on paper in the form of strips and is available
as blue and red strips. A blue litmus strip, when dipped
in an acid solution acquires red colour. Similarly a
red strip when dipped in a base solution becomes
CLASS-XI_STREAM-SA_PAGE # 56
i.e., their pH do not change on dilution or on standing
for long. Such solutions are called buffer solutions.
blue.
(b) Phenolphthalein :
It is also an organic dye and acidic in nature. In neutral
or acidic solution, it remains colourless while in the
basic solution, the colour of indicator changes to pink.
(c) Methyl Orange :
Methyl orange is an orange or yellow coloured dye
and basic in nature. In the acidic medium the colour
of indicator becomes red and in the basic or neutral
medium, its colour remains unchanged.
(i) Manufacture : Sodium carbonate is prepared on
large scale by solvay process.
It is purple in colour in neutral medium and turns red
or pink in the acidic medium. In the basic or alkaline
medium, its colour changes to green.
• Sodium chloride (NaCl)
• Ammonia (NH3)
• Lime stone (CaCO3)
(e) Turmeric Juice :
(A) In this process, a cold and concentrated solution
of sodium chloride (called ‘brine’) is saturated with
ammonia in the saturation tank.
Note :
Litmus is obtained from LICHEN plant.
SCALE
According to Arrhenius theory, an acid releases H+
ion in aqueous solution. The concentration of these
ions is expressed by enclosing H+ in square bracket
i.e. as [H+]. Thus, greater the [H+] ions, stronger will
be the acid. However, according to pH scale, lesser
the pH value, stronger will be the acid. From the above
discussion, we can conclude that pH value and H+
ion concentration are inversely proportional to each
other.
The relation between them can also be expressed
aspH = – log [H+]
So, negative logarithm of hydrogen ion concentration
is known as pH.
e.g.
Let the [H+] of an acid solution be 10–3 M. Its pH can be
calculated as pH = – log [H+]
= – log [10–3]
= – (–3) log 10
= 3log10
( log 10 = 1)
=3
Note :
Just as the [H+] of a solution can be expressed in
terms of pH value, the [OH–] can be expressed as
pOH.
Mathematically , pOH = – log [OH–] = log
(B) The ammoniacal brine is fed from the top of a
tower, known as carbonating tower. Carbonating tower
is fitted with perforated plates. CO2 is introduced from
the base of the tower.
(C) As the ammoniacal brine trickles down the tower,
it meets the upcoming CO2 and produces sparingly
soluble sodium hydrogen carbonate (NaHCO3 ).
NaCl
+ NH 3 + H2O + CO 2
Sodium Ammonia Water
chloride
Carbon-dioxide
NH 4Cl + NaHCO 3
+ NH4HCO3
Ammonium Sodium
chloride bicarbonate
(Traces)
(D) The CO2 used in the above reaction is obtained by
heating limestone (CaCO3) in a lime kiln.
CaCO 3
Lime stone
heat
CaO + CO2
Quick lime
(E) Quick lime (CaO) obtained in the above reaction
is changed into slaked lime by dissolving in water.
CaO
+
Quick lime
Ca(OH)2
Slaked lime
H2 O
Water
(F) The slaked lime is then boiled with ammonium
chloride to liberate ammonia, which is recycled for
further use.
 1 
= log   
H 
Ca(OH)2 + 2NH4Cl
Ammonium
Slaked
lime
chloride
Thus, if pH value of solution is known, its pOH value
can be calculated.
Note :
There are some solutions which have definite pH
CaCl2 + 2NH3 + 2H2O
Calcium Ammonia Water
chloride
(G) Most of the ammonia (NH3) can be recovered. So,
sodium chloride and limestone are only consumed
in the process. CaCl2 is produced as a by-product.
(H) Ammonium hydrogen carbonate formed in traces
is heated
NH4HCO3

NH3 + H2O + CO2


(I) Sodium hydrogen carbonate formed in the step (C),
is sparingly soluble in water. It is separated by filtration.
Sodium carbonate is then obtained by heating sodium
bicarbonate.
1
[OH – ]
Moreover, pH + pOH = 14

Chemical name : Sodium carbonate decahydrate
Chemical formula : Na2CO3. 10H2O
The raw materials used in the manufacture of
sodium carbonate are :
PH

(a) Washing soda :
(d) Red Cabbage Juice :
It is yellow in colour and remains as such in the neutral
and acidic medium. In the basic medium its colour
becomes reddish or deep brown.

SOME IMPORTANT CHEMICAL COMPOUNDS
2NaHCO3
Sodium
bicarbonate
heat
Na2CO3 + H2O + CO2
Sodium
carbonate
(J) Sodium carbonate is recrystallized by dissolving
in water to get washing soda.
CLASS-XI_STREAM-SA_PAGE # 57
(H) Sodium carbonate when reacts with silica gives
sodium silicate.
Na2CO3.10H2O
Washing soda
Na2CO3 + 10H2O
Sodium
carbonate
Na2SiO3 (s) + CO2(g)
Na2CO3(s) + SiO2 (s)
NH3 + CO2( Traces)
(iii) Uses :
Filter
Lime
kiln
CO 2
Filter
CO2 CaO
+
H2O
Ammonia recovery
tower
(A) It is used as cleansing agent for domestic purposes.
Slaked
lime
Saturating
tank
Brine
Cooling Pipes Ammoniacal brine
Carbonating tower
NH4Cl + a little NH4HCO3
(B) It is used in softening hard water and controlling
the pH of water.
(C) It is used in manufacture of glass.
(D) Due to its detergent properties, it is used as a
constituent of several dry soap powders.
(E) It also finds use in photography, textile and paper industries etc.
(F) It is used in the manufacture of borax (Na2B4O7. 10H2O).
(ii) Properties :
(b) Bleaching Powder :
(A) Washing soda is a translucent, efflorescent, crystalline solid, having formula Na2CO3.10H2O.
(B) When exposed to air, crystals of sodium carbonate
decahydrate lose 9 molecules of water of crystallization to the atmosphere. This process is called efflorescence.
Na2CO3 . 10H2O
Sodium carbonate
decahydrate
Na2CO3. H2O + 9H2O
Sodium carbonate
monohydrate
(C) When heated above 373 K, it first becomes anhydrous and then melts, but does not decompose.
Na2CO3 + 10H2O
Sodium carbonate
(Anhydrous)
Soda ash
(D) It is soluble in water and aqueous solution is
alkaline. The alkalinity is due to hydrolysis.
Na2CO3.10H2O
Sodium carbonate
decahydrate
(i) Manufacture : Bleaching powder is prepared by
passing chlorine over slaked lime at 313 K.
Ca(OH)2 (aq) + Cl2 (g)
Slaked lime
313 K
Ca(OCl)Cl (s) + H2 O (g)
Bleaching powder
(A) Slaked lime is spread over the floor of a chamber
that is provided with wooden stirrers. Dry chlorine
gas is now allowed to pass into the chamber. Chlorine gas is slowly absorbed by lime. The temperature
is not allowed to rise above 313 K. The bleaching powder formed is left as such and is removed after about
24 hours.
(F) When aqueous solution of sodium carbonate is
treated with excess of CO2 gas, sodium bicarbonate
is produced.
(B) On a large scale, bleaching powder is prepared
in Hasenclever plant. It consists of cylinders of cast
iron horizontally placed one above the other. Each of
the cylinders is provided with a screw that keeps revolving slowly. Dry slaked lime is poured through a
hopper at the top and it passes through all the cylinders, one by one. A current of chlorine gas is led into
the plant at the bottom. Slaked lime comes in contact
with the current of chlorine gas travelling in the opposite direction. Bleaching powder is formed and falls
into the casks below.
Na2CO3(s) + H 2O (l) + CO2(g)
Ca(OH)2(aq)+Cl2 (g)
(E) When treated with a mineral acid, it produces effervescence due to evolution of CO2 gas.
Na2CO3 (s) + 2HCl (aq)  2NaCl (aq) + CO2 (g) + H2O ()
Na2CO3(s) + H2SO4(aq)  Na2SO4(aq) + CO2(g) + H2O()
2NaHCO3 (s)
Ca(OCl)Cl(s) + H 2O (g)
(G) When a solution of sodium carbonate is treated
with SO2 gas, sodium sulphite and sodium bisulphite
are formed and CO2 is released.
Na2CO3(aq) + SO2 (g)
Na2SO3 (aq) + CO2(g)
Sodium
sulphite
Na2CO 3 (aq) + H2O ( ) + 2SO2 (g)
2NaHSO3 (aq) + CO2 (g)
Sodium bisulphite
CLASS-XI_STREAM-SA_PAGE # 58
Ca(OH)2
(c) Baking soda
Waste gases
Hopper
Chlorine
Ca(OCl)Cl
•••••••••
••••••••
•••••••
Hasenclever plant for the manufacture of
bleaching powder
(ii) Properties :
(A) It is a white, amorphous powder that smells
strongly of chlorine.
(B) When exposed to air, it deteriorates giving off
chlorine.
CaOCl2 + CO2  CaCO3 + Cl2
(C) Action of water - When treated with water, it
decomposes into calcium chloride and calcium
hypochlorite.
2CaOCl2  CaCl2 + Ca(OCl)2
(D) Action of acid - With excess dilute acids, chlorine
is liberated.
CaOCl2 (s) + 2HCl (aq)  CaCl2 (aq) + H2O () + Cl2 (g)
CaOCl2 (s) + H2SO4 (aq)  CaSO4 (s) + H2O () + Cl2 (g)
The amount of chlorine so produced is known as
available chlorine.
Baking soda is sodium hydrogen carbonate or sodium bicarbonate (NaHCO3).
(i) Preparation :
It is obtained as an intermediate product in the preparation of sodium carbonate by Solvay process. In this
process, a saturated solution of sodium chloride in
water is saturated with ammonia and then carbon
dioxide gas is passed into the solution. Sodium chloride is converted into sodium bicarbonate which, being less soluble, separates out from the solution.
2NH3 (g) + H2O () + CO2 (g)  (NH4)2CO3(aq)
(NH4)2CO3 (aq) + 2NaCl (aq)  Na2CO3 (aq) + 2NH4Cl (aq)
Na2CO3 (aq) + H2O () + CO2 (g)  2NaHCO3 (s)
(ii) Properties :
(i) It is a white, crystalline substance that forms an
alkaline solution with water. The aqueous solution of
sodium bicarbonate is neutral to methyl orange but
gives pink colour with phenolphthalein.
(Phenolphthalein and methyl orange are dyes used
as acid-base indicators.)
(ii) When heated above 543 K, it is converted into
sodium carbonate.
2NaHCO3 (s)
 543 K


 Na2CO3 (s) + CO2 (g) + H2O ()
(iii) Uses :
(A) It is used in the manufacture of baking powder.
Baking powder is a mixture of potassium hydrogen
tartarate and sodium bicarbonate. During the preparation of bread the evolution of carbon dioxide causes
bread to rise (swell).
(E) In the presence of a very small amount of dilute
acid, it gives off oxygen.
CH(OH)COOK
2CaOCl2 (s) +H2SO4 (aq)  CaCl2 (aq) + CaSO4 (s) + 2HOCl (aq)
CH(OH)COOH
2HOCl (aq)  2HCl (aq) + 2[O]
[O] + [O]  O2 (g)
(B) It is largely used in the treatment of acid spillage
and in medicine as soda bicarb, which acts as an
antacid.
Due to the evolution of nascent oxygen, it behaves as
a bleaching agent.
(C) It is an important chemical in the textile, tanning,
paper and ceramic industries.
(D) It is also used in a particular type of fire extinguishers. The following diagram shows a fire extinguisher
that uses NaHCO3 and H2SO4 to produce CO2 gas.
The extinguisher consists of a conical metallic container (A) with a nozzle (Z) at one end. A strong solution of NaHCO3 is kept in the container. A glass ampoule (P) containing H2SO4 is attached to a knob (K)
and placed inside the NaHCO3 solution. The ampoule
can be broken by hittin g the knob. As soon as the
acid comes in contact with the NaHCO3 solution, CO2
gas is formed. When enough pressure in built up
inside the container, CO2 gas rushes out through the
nozzle (Z). Since CO2 does not support combustion,
a small fire can be put out by pointing the nozzle
towards the fire. The gas is produced according to
the following reaction.
(F) It liberates iodine from an acidified aqueous solution of potassium iodide.
CaOCl2 (aq) + 2KI (aq) + 2HCl (aq)  CaCl2 (aq) +
H2O () + 2KCl (aq) + 2 (aq)
(iii) Uses :
(A) It is commonly used as a bleaching agent in paper and textile industries.
(B) It is also used for disinfecting water to make it free
from germs.
(C) It is used to prepare chloroform.
(D) It is also used to make wool shrink-proof.
CH(OH)COOK
+ NaHCO3
+ CO2 + H2O
CH(OH)COONa
2NaHCO3 (aq) + H2SO4 (aq)  Na2SO4 (aq) + 2H2O() + 2CO2(g)
CLASS-XI_STREAM-SA_PAGE # 59
CO2
(Z)
NaHCO3
(A)
(P)
•
•••
•••••••
••••••
•••••
H2SO4
Knob (K)
Fire Extinguisher
CLASS-XI_STREAM-SA_PAGE # 60
NUCLEAR CHEMISTRY
(b) A nucleus which has lower n/p ratio , is placed
below the belt of stability either emits positrons or
undergoes electron capture. Both modes of decay
decrease the number of protons and increase the
number of neutrons in the nucleus and thus, positron
emission or electron capture results in an increase
in n/p ratio.
e.g.
INTRODUCTION
Atoms have three fundamental particles that are
electrons, protons and neutrons. Protons and
neutrons are present inside the nucleus and electrons
are present in the extranuclear region. Changes
occurring in the nucleus which are a source of
tremendous energy are called nuclear reactions. The
branch of science which deals with the study of atomic
nucleus and nuclear changes is called nuclear
chemistry.
11
11
6C  5
(Positron emission)
1
1p
STABILITY OF NUCLEUS
Neutrons help to hold protons together within the
nucleus. The number of neutrons necessary to create
a stable nucleus increases rapidly as the number of
protons increases. The number of neutron to proton
ratio (n/p) of stable nuclei increases with increasing
atomic number . The area within which all stable nuclei
are found is known as the belt of stability. Radioactive
nuclei occur outside this belt.
Unstable region
Unstable
Region
110
100
Stable
nuclei
=1
)
=N
um
be
ro
fp
ro
to
ns
(n
/p
80
70
60
50
ro
ns
40
30
20
10
0 10
20 30 40 50 60 70 80 90 100
Number of protons
The type of radioactive decay that a particular radio
isotope will undergo depends to a large extent on its
neutrons to protons ratio compared to those of nearby
nuclei that are within the belt of stability.
(a) A nucleus whose high n/p ratio places it above the
belt of stability emits a  -particle in order to lower n/p
ratio and move towards the belt of stability.
1
0n
11 p  –10 e  –
( – is Anti neutrino)

0
–1
e  
1
0
n + X-rays
(Electron capture)

Note :
A positron has same mass as electron but carries
opposite charge. The positron has a very short life
because it is annihilated when it collides with
an electron, producing gamma rays. This
phenomenon is known as pair production.
an -particle decreases both the number of protons
and neutrons and thereby decreases n/p ratio.
Thus,
Nu
m
be
ro
fN
eu
t
Number of neutrons
90

(c) The nuclei with atomic number > 83, outside the
belt of stability, undergo -emissions. Emission of
140
120
B  01 e
Note :
Antineutrino is the antiparticle of neutrino, which is
neutral particle produced in nuclear beta decay.
n
 1.0 , unstable nuclei.
Z
n
 1.0 , stable nuclei.
Z
EVEN-ODD NATURE OF THE NUMBER
OF PROTONS AND NEUTRONS
(i) The number of stable nuclides is maximum when
both Z and n are even numbers. About 60% of stable
nuclides have both Z and n even.
(ii) The number of stable nuclides in which either the
Z or n is odd is about one third of those, where both
are even.
MAGIC NUMBERS
Just as certain numbers of electrons (2,8,18,36,54
and 86) correspond to stable closed shell electron
configuration, certain number of nucleons leads to
closed shell in nuclei. The protons and neutrons can
achieve closed shell. Nuclei with 2, 8, 20, 28, 50 or 82
protons or 2,8,20, 28, 50, 82, or 126 neutrons
correspond to closed nuclear shell. Closed shell
nuclei are more stable than those that do not have
closed shells. These numbers of nucleons that
correspond to closed nuclear shells are called magic
numbers.
CLASS-XI_STREAM-SA_PAGE # 61
NUCLEAR EXCHANGE FORCE-MESON THEORY
If only the short range charge independent attractive
forces were to operate between the nucleons, the
attractive force would grow limitlessly till the nucleus
ultimately collapsed. There is a need, therefore, for
some repulsive forces to lead to the saturation of
attractive forces.
The nucleus stability has been explained in terms of
existence of nuclear exchange force. An exchange
interaction is possible between two particles which can
exist in a state capable of sharing some common
property, thereby, the total energy level gets lowered
and the system becomes more stable. A similar
situation is proposed for the atomic nucleus by
assuming a ceaseless exchange of common property
(may be charge, spin or position) between neighb
ouring nucleons in the same state of motion.
Yukawa’s prediction (1935) of the existence of mesons
(particles with mass intermediate between an electron
and a nucleon) and very soon, the - meson or pions
(m= 237 me) were recognized as exchange of common
property.
n
p + – ;
p
n + +
p
p + 0 ;
n
n + 0.
where –, + 0 are pions.
The range of pions being of the same order as the
nuclear radius and thus, emission of pion by a nucleon
are reabsorption by another nucleon goes on
incessantly.
Ex-1
14
6C
nuclide undergoes  -decay. Which stable nuclide
RADIOACTIVITY
Radioactivity is a process in which nuclei of certain
elements undergo spontaneous disintegration
without excitation by any external means.
All heavy elements from bismuth (Bi) to uranium
and a few of lighter elements have naturally
occurring isotopes which possess the property of
radioactivity. All those substances which have the
tendency to emit these radiations are termed as
radioactive materials. Radioactivity is a nuclear
phenomenon i.e., the kind of intensity of the radiation
emitted by any radioactive substance is absolutely
the same whether the element is present as such
or in any one of its compounds.
E.g. Elements like uranium (U) , thorium
(Th) , polonium (Po), radium (Ra) etc. are radioactive
in nature.
(a) History of the Discovery of Radioactivity :
In 1895, Henri Becquerel was studying the effect of
sunlight on various phosphorescent minerals, one
of the substance being studied was uranium ore.
He accidently left a crystal of uranium sample ;
Potassium uranyl sulphate [K2UO2 (SO4)2. 2H2O] in
a drawer along with some photographic plate
wrapped in black paper. Much to his surprise, he
discovered that the photographic plate had been
fogged by exposure to some invisible radiations
from uranium. He called this mysterious property of
the ore as ‘radioactivity’ (Radioactivity means rayemitting activity). A year later, in 1896, Marie Curie
found that besides uranium and its compounds,
thorium was another element which possessed the
property of radioactivity. 1898 Marie Curie and her
husband Pierrie Curie isolated two new radioactive
elements polonium and radium.
is formed ? Give equation.
(b) Natural & Artificial Radioactivity :
Sol.
14
6C

14
0
7 N + –1 e
Ex-2 Calculate number of  and  - particles emitted when
238
92 U
changes into radioactive
Sol. No. of -particles =
=
206
82 Pb
.
Change in mass number
4
238 – 206
32
=
=8
4
4
No. of  -particles = 2 × -particles - (ZA – ZB)
If a substance emits radiations by itself it possesses
natural radioactivity but if a substance does not
possess radioactivity and starts emitting radiations
on exposure to rays from a natural radioactive
substance, it is called induced or artificial
radioactivity.
e.g.
When aluminium is bombarded with - particles , a
radioactive isotope of phosphorus is formed which
disintegrate spontaneously with the emission of
positrons (which are positively charged electron, +1e0).
Here ZA and ZB are atomic no. of parent and daugther
nuclei respectively.
= 2 × 8 – (92 – 82)
16 – 10
=6
Thus, no. of -particles emitted out = 8
No. of  -particles emitted out = 6
CLASS-XI_STREAM-SA_PAGE # 62

-rays
-rays
ys
-ra
Note :
Natural radioactivity was discovered by Becquerel
while artificial radioactivity was discovered by Irene
Curie and Joliot.
ys
-ra
(c) Analysis of Radioactive Radiations :
In 1904, Rutherford and his co-workers observed that
when radioactive radiations were subjected to a
+
+
+
+
+
+
+
+
+
-rays
-rays
Magnetic
field
Radioactive
substances
magnetic field or a strong electric field, these were
split into three types, as shown in the figure. The rays
which are attracted towards the negative plate, are
positively charged , are called alpha () rays. The
rays which are deflected towards the positive plate
are negatively charged and are called beta ( ) rays.
The third type of rays which are not deflected on any
side but move straight are known as gamma () rays.
(A)
(B)
Figure :
(A) Emission of radioactive rays and deflection of
radioactive rays in electric field.
(B) Deflection in a magnetic field. (The direction of
magnetic field is inward perpendicular to the page).
(i) The important properties- rays ,  rays and -rays are as follows :
(d) Units of Radioactivity :
(i) SI unit is Becquerrel (Bq) which is defined as
disintegration per sec (dps).
(ii) Earlier radioactivity was given in terms of Curie
(Ci).
1 Ci refers to the activity of Radium.
1 Ci = 3.7 × 1010 dps = 3.7 × 1010 Bq.
1 Milli Ci = 4.7 × 107 Bq.
1 Micro Ci = 4.7 × 104 Bq.
These laws were given by Soddy,Fajans and Russel
(1911_1913).The element emitting the  or  particle
is called parent element and the new element formed
is called daughter element.
(i) When an _ particle is emitted, the new element
formed is displaced two positions to the left in the
periodic table than that of parent element (because
the atomic number decreases by 2).
e.g.
(iii) Another unit is Rutherford (Rd).
1 Rd = 106 dps
238
92 U
 90 Th234  2 He 4
(ii) When a  _ particle is emitted the new element
formed is displaced one position to the right in the
periodic table than that of parent element (because
the atomic number increases by 1).
CLASS-XI_STREAM-SA_PAGE # 63
e.g.
The emission of

 _ particle by 6C 14 may be
The emission of
represented as follows:
6
C
14

14
7
N +–1e
Note :
_decay and  _ decay.
0
(a) Explanation :
The results of the group displacement laws may be
explained as follows:
Since an _ particle is simply a helium nucleus
(containing two neutrons and two protons) therefore,
loss of _ particle means loss of two neutrons and
two protons . Thus, the new element formed has
atomic number less by 2 unit and mass number less
by 4 unit.
The  _ particle is simply an electron and there are no
electrons present in nucleus .However , the loss of  _
particle is also found to be a nuclear phenomenon
because the change in external conditions
(temperature etc.) has no effect on the rate of the
emission of  _ particle. It is therefore, believed that
for emission of  _ particle to occur, a neutron changes
to a proton and an electron i.e.
Ex.3 90Th234 disintegrates to give 82Pb206 as final product .
How many alpha and beta particles are emitted during
this process ?
Sol. Suppose the no. of  particles emitted = x and no. of
 _ particles emitted = y.
Then
90
Th234
No. of -particles =
4
=
C14 
7
 90 Th234  2 He 4
Hence, rate  [N0 – N]/t because rate continuously
decreases with time. Let dN be the change in no. of
atoms in an infinitesimal small time dt, then rate of
decay can be written as -
144
90
54
–
dN
 [N]1 = N. The negative sign indicates the
dt
decreasing trend of N with increasing time.
where  is the proportionality constant.
N14 +-1e0
Integration of this equation finally gives -
isotope of parent element.
e.g.
or  =

234
234


 92U234
U238 
 90Th 
 91Pa 
=7
4
Radioactive disintegration is an example of first order
reaction, i.e., the rate of decay is directly proportional
to the no. of atoms (amount) of the element present
at the particular time.
A Decay product
No. of atoms at t = 0  N0
No. of atoms left after t = t N
(iv) Emission of 1 and 2 particles produces an
92
=
RATE OF RADIOACTIVE DECAY
(iii)  _ decay produces isobars i.e.parent and the
daughter nuclides have different atomic numbers but
same mass number .
E.g.
6
4
28
Ans.7 and 6 particles will be emitted.
(ii) _ decay produces isodiaphers i.e.parent and the
daughter nuclides have same isotopic mass (which
is the difference between number of neutrons and
protons) .
E.g.
146
92
54
234 – 206
No. of -particle=( 2 × no. of -particles) – (ZA – ZB)
where , ZA = Atomic number of reactant
ZB = Atomic number of product
= (2 × 7) – (90 – 82) = 14 – 8 = 6
Note :
No. of neutrons:
No. of protons:
Difference
Pb206 + x 2He4 + y–1e0
Difference in atomic mass of reactants and products
n1  1P1 + –1e0
As a result ,the number of protons in the nucleus
increases by 1 and so does the atomic number.
238
92 U
82
Alternative Method:
0
(i) Increase or decrease in the number of protons in
the nucleus (due to loss of  particle or  _ particle)
is accompanied simultaneously by the loss or gain
of electrons in the extranuclear part (from the
surroundings) so that the electrical neutrality is
maintained in the new atom formed.

Equating the mass number on both sides ,we get
234 = 206 + 4x + 0y
or 4x = 28 or
x=7
Equating the atomic number on both sides ,we get
90=82+2x-y
y=6
Ans. 7 and 6 particles will be emitted.
Neutron  Proton + Electron ( _ particle)

 and  _ particles is also known as

2.303
N0
log10
t
N
Note :  is also known as decay or disintegration or
radioactive constant.
CLASS-XI_STREAM-SA_PAGE # 64
CHARACTERISTICS OF RATE OF DISINTEGRATION
(i) Rate of disintegration continuously decreases with
time.
EX.5 The half -life period of a radioactive element is 27.96
days . Calculate the time taken by a given sample to
reduce to 1/8th of its activity.
Sol. The amount of substance left after ‘n’ number of half
lives can be given as :-
(ii) Rate of disintegration as well as  are independent
of P and T.
 1n

2
N = No 
(iii) (a) Unit of rate of decay : disintegration per time
(b) Unit of decay constant : time–1
(iv) Time required to complete a definite fraction is
independent of initial no. of atoms (amount) of
radioactive species.
According to question , N =
or
or
The time required for the decay of radioactive element
to half of the original amount is called half-life period.
1
(2)
3
=
 n = 3
n
(2)
(a) Characteristics of Half-Life Period :
So, time taken by the sample to reduce to 1/8th of its
reactivity will be -
• It is denoted by t1/2 .
T= n×t½
• Each radioactive element has a characteristic halflife period .
T= 3×27.96
• Half-life period for an element is a constant.
0.693

• t1/2 =

 1n
1
N0 = No  
8
2
1
HALF-LIFE PERIOD
1
N
8 0
= 83.88 days
EX.6 Half- life period of a radioactive element is 100
seconds. Calculate the disintegration constant.
Where  is a constant known as disintegration
constant or decay constant. It is the characteristic of
the nature of the radioactive element.
Sol. t1/2 = 100 Seconds,
Note :
Half-life period does not depend upon initial amount
of element.
t
(b) Significance of Half-Life Period :
0.693
 =
 1n

2
Where ;
N = Amount of the substance left after ‘n’ half-lives.
No = Initial amount of the substance.
N=
N=
Average life ()=
t½
0.693
= 1.44 t1/2.
(i) Atoms having the same difference of neutrons and
proton.
(ii) Nuclide and its decay product after -emission
are called isodiaphers.
N0
2n
(iii) e.g.,
1
m
Z
A
m–4
Z–2
p=Z
n=m–Z
n – p = m – 2Z
23
1
%=
× 100 = 12.5%
8
1


ISODIAPHERS
180
Total time ( t )
=
= 3 days
60
Halflife period ( t ½ )
1
8
1/2
Evidently , the whole of the radioactive element can
never disintegrate or in other words
, the time required for the disintegration of the whole of
a radioactive element will be infinity.
Thus, it is meaningless to talk of the total life of a
radioactive element . However, sometimes another
term is used ,called average life () which is the
reciprocal of the disintegration constant () i.e.
N = No 
Applying the formula N=
= 0.00693 s–1
AVERAGE LIFE
(ii) The amount of substance left after ‘n’ number of
half lives can be given as :-
n=
100
= 6.93 × 10–3 sec–1
(i) Stability of nuclei : The value of half-life period can
give an idea about relative stability of radio isotopes.
All isotopes with longer t1/2 are more stable.
Ex.4 The half-life period of 53I125 is 60 days .What percent
of the original radioactivity would be present after 180
days ?
Sol. t½ = 60 days, t = 180 days
0.693
=

B
p=Z–2
n=m–Z–2
n – p = m – 2Z
Note :
Isotopic no. n – p = m – 2Z
CLASS-XI_STREAM-SA_PAGE # 65
ISOSTERS
(i) Molecules having same no. of atoms and same
no. of electrons are called isosters.
e.g., CO2 and N2O (There are three atoms and 22
electrons in both the molecules.)
During fission, there is always loss of mass, known
as mass defect ,which is converted into energy
according to Einstein equation i.e.
E = mc2.
e.g.
235
92 U
NUCLEAR ISOMERS
(i) Nuclides having identical atomic no. and mass
no. but differing in radioactive properties are known
as nuclear isomers.
(ii) Nuclear isomers differ in their energy state and
spins.

235
118
.009
.
1


236.127amu
140
93
56 Ba + 36Kr +
310 n +energy


143
.881
89
2.018


.947



235.846
 m = 236.127 – 235.846
= 0.281 amu
E(in MeV) = 0.281 × 931.5 = 261.75 MeV
Br
Energy released in one fission is equal to 261.75
MeV.
The symbol m with mass no. represents the
metastable state of parent element.
(i) Chain reaction : Whatever are the primary products
e.g.,
etc.
60m
60
CO and
Co
60m
Isomeric
Transition
CO,
60
69
Zn and
69m
Zn,
80
Br and
80m
Co + -rays
(iii) Nuclear isomers, thus have different rate of decay,
decay constant, half life, average life and binding
energy.

1
0n
+
Note :
In a metastable state, a system is in equilibrium (not
changing with time),but is susceptible to fall into lower
energy states with only slight interaction.
of fission of uranium, it is certain that neutrons are
set free.If the conditions are so arranged that each of
these neutrons can, in turn, bring about the fission,
the number of neutrons will increase at a continuously
accelerating rate until whole of the material is
exhausted. Such type of reaction is called chain
reaction. It takes very small time and is uncontrolled.
It ends in terrible explosion due to release of
enormous amount of energy.
235
1
92 U + 0 n
141
92
1
  56 Ba  36 Kr 3 0 n + Energy
NUCLEAR REACTIONS
The reactions in which nuclei of atoms interact with
other nuclei or elementary particles such as alpha
particle, proton, deutron, neutron etc. resulting in the
formation of a new nucleus and one or more
elementary particles are called nuclear reactions.
Nuclear reactions are expressed in the same fashion
as chemical reactions. In a nuclear reaction ,atomic
number and mass number are conserved.
e.g. the nuclear reaction :
The chain reaction is shown in the figure .
Ba
U
n
235
92
Ba
n
n
E
U
n
235
Kr
239
Uranium
(ii)
239
–
–
10 n 
 93 Np 
 94 Pu
235
92 U
Neptunium
Plutonium
captures slow neutron and splits up into
fragments.
235
1
92 U + 0 n

236
92 U
Unstable
144
90
1
 56 Ba 36 Kr 2 0 n +Energy
Kr
Ba
n
n
n
E Kr
Ba
n
E
The process of artificial transmutation in which heavy
nucleus is broken down into two lighter nuclei of nearly
comparable masses with release of large amount of
energy is termed as nuclear fission.
e.g.
238
92 U
235
U
92
Kr
92
(a) Nuclear Fission :
(i)
n
E
n
14
4
 178 O  11H
7 N  2He 
Ba
n
n
235
92 U
n
n
235
U
U
235
92
Ba
n
n
n
n
92
E
n
E
Kr
Kr
Ba
n
n
235
92 U
n
E
Kr
(ii) Critical mass : The minimum mass which the
fissionable material must have so that one of the
neutrons released in every fission hits another
nucleus and causes fission so that the chain reaction
continues at a constant rate is called critical mass .If
the mass is less than the critical mass , it is called
sub-critical mass . If the mass is more than critical
mass, it is called super-critical mass.
CLASS-XI_STREAM-SA_PAGE # 66
(iii) Applications of Nuclear Fission : Three practical
applications of nuclear fission are as follows (A) Atomic bomb
(B) Nuclear reactor
(C) Nuclear power plants
(A) Atomic Bomb :
• The basic principle of atomic bomb is uncontrolled
nuclear fission reaction (chain reaction).
• It requires several small samples of U-235 or Pu-239.
• An explosive like TNT (Tri Nitro Toluene) is placed
behind the samples which explodes to initiate the
reaction which causes the small samples to join and
form large mass.
• Neutron from Ra-Ba source (s) initiate the reaction
which starts the chain reaction finally leading to
explosion and release of large amount of energy.
• The rapid release of energy raises the temperature
enormously and generates a very high pressure front
in the atmosphere.
When a nuclear reactor is used for the production of
electricity it is termed as a nuclear power plant.The
heat produced during a nuclear reaction is utilized in
generating steam which runs the steam turbines. The
electric generator is connected to the turbine. The
electric power is then obtained from the generator.
Thus, a nuclear power plant consists essentially of
the following four parts:
1. Reactor core
2. Heat exchanger
3. Steam turbine
4. Steam condensing system
Reactor core is the main part of nuclear reactor. It
consists of the following parts :
Fuel rod :
The fissionable material used in the reactor is called
fuel. The fuel used is enriched uranium -235 . This is
obtained from the naturally occurring U-235
(containing about 0.7% of U-235 ) by raising the
percentage of U-235 to about 2-3%.
• Control rods : Cadmium or boron rods are used to
raise or lower and control the fission process.
Because they can absorb neutrons.
Atomic Bomb

(C ) Nuclear Power Plants :
Note :
• Moderator : The material used to slow down the
The first atomic bomb dropped over Hiroshima city
during the second world war in 1945 utilized 235U and
the second atomic bomb dropped on Nagasaki
made use of 239Pu. India exploded her first atomic
bomb at Rajasthan in May 1974,and used 239Pu as
the fissionable material.
neutrons (without absorbing them so that they can be
easily captured by the fuel, is known as moderator.
Heavy water (D2O) or graphite is used as moderator
material in nuclear power plant.
(B) Nuclear reactor :
fission, a liquid is used. This liquid is known as coolant.
Usually heavy water is used as coolant so that it also
acts as a moderator.
• An equipment in which nuclear chain reaction is
carried out in a controlled manner is called a nuclear
reactor.
• The energy thus liberated can be used for
constructive purposes like generation of steam to
run turbines and produce electricity.
• In a nuclear reactor, fission is controlled by controlling
the number of neutrons released.
• Coolant : To carry away the heat produced during
• Shield : To prevent the losses of heat and to protect
the persons operating the reactor from the radiation
and heat, the entire reactor core is enclosed, in a
heavy steel or concrete dome, called the shield.
Steam
• In a nuclear reactor, fission is based on the fact that
cadmium and boron can absorb neutrons thus
forming corresponding isotopes which are not
radioactive.
113
1
 114
48 Cd  0n 
48 Cd 
10
1
 11
5 B  0n 
5B 

  rays
Reactor
Primary
coolant
Heat
exchanger
  rays
D2O
Condenser
Note :
The first nuclear reactor was assembled by Fermi
and his coworkers at the University of Chicago in the
United states of America, in 1942. In India, the first
nuclear reactor was put into operation at Trombay
(Mumbai) in 1956.
Electricity
Generator
Turbine
D2O
Pump
Heavy water
Nuclear Reactor
CLASS-XI_STREAM-SA_PAGE # 67
(b) Nuclear Fusion :
than or around 107 K, fusion takes place dominantly
A nuclear reaction in which two lighter nuclei are fused
by proton-proton cycle as follows -
together to form a heavier nucleus is called nuclear
fusion. A fusion reaction is difficult to occur because
1
positively charged nuclei repel each other. At very high
1
temperature of the order of 106 to 107 K, the nuclei
2
may have sufficient energy to overcome the repulsive
forces and fuse.Therefore, fusion reactions are also
1
3
2
1
2
H + 1H
1
1
3
H + 1H
2
4
2
1
He +1 H
4 1H
highly exothermic in nature because loss of mass
0
He ++1
4
1
called thermonuclear reactions. Fusion reaction are
0
H ++1
2
0
He + 2 +1e + 24.7 MeV
occurs when heavier nucleus is formed from the two
lighter nuclei.
note that the first two reactions should occur twice to
4
2
2
2 He  23.85 MeV
1 H 1 H 
4
3
3
 2 He  210 n  11.3 MeV
1 H 1 H 
4
1
3
 2 He  20.0 MeV
1H 1 H 
4
2
3
 2 He 10 n  17 .6 MeV
1 H 1 H 
produce two 32 He nuclei and initiate the third reaction.
Hydrogen bomb is based on fusion reaction. Energy
released is so enormous that it is about 1000 times
As a result of this cycle, effectively, four hydrogen nuclei
combine to form a helium nucleus. About 26.7 MeV
energy is released in the cycle. Thus, hydrogen is the
fuel which ‘burns’ into helium to release energy. The
sun is estimated to have been radiating energy for
the last 3.5 × 109 years and will continue to do so till
all the hydrogen in it is used up. It is estimated that
that of an atomic bomb.
It is believed that the high temperatures of stars
including the sun is due to fusion reactions.
(i) Applications of Nuclear Fusion :
the present store of hydrogen in the sun is sufficient
for the next 5 × 109 years.
In hotter stars where the temperature is  10 8K,
another cycle known as CNO (Carbon-nitrogenoxygen cycle) cycle takes places.
(A) Hydrogen bomb :
• Its principle is nuclear fusion.
• It consists of an arrangement of nuclear fission in
the centre surrounded by a mixture of deuterium (12 H)
and lithium isotopes ( 36 Li) .
12
6
• The nuclear fission provides heat and neutrons.
• Neutrons convert
6
3 Li
3
1 H and the heat
between 12 H & 13 H .
to tritium
liberated is used for fusion
13
6
7
Fission (in the centre)  heat +
6
3 Li
+ 10 n 
3
1H
1
0n
+ 24 He + 3.78 MeV
2
1H
+ 13 H 
4
2He
+ 10n + 17.6 MeV
2
1H
+ 12 H 
3
2 He
+ 10n + 3.2 MeV
3
1H
+ 13 H 
4
2 He
+ 2 10n + 13.14 MeV
13
13
7
6
N
1
15
15
15
8
7
1
N + 1H
1
4 1H
0
C + +1e
14
O
N
7
1
N + 1H
15
7
13
C + 1H
14
• The reactions occurring are :
1
C + 1H
7
8
12
6
4
2
0
N + 1e
4
C +2
0
He + 2 +1e + 24.7 MeV
The end result of this cycle is again the fusion of four
hydrogen nuclei into a helium nucleus. Carbon
nucleus acts only as a catalyst.
(B) Fusion in sun : Among the celestial bodies in which
energy is produced, the sun is relatively cooler. There
are stars with temperature around 108 K inside. In
sun and other stars, where the temperature is less
CLASS-XI_STREAM-SA_PAGE # 68
DIFFERENCES BETWEEN NUCLEAR FISSION AND NUCLEAR FUSION
S.No.
Nuclear fission
Nuclear fusion
1
This process occurs in heavy nuclei.
This process occurs in lighter nuclei.
2
The heavy nucleus splits into lighter nuclei
of comparable masses.
The lighter nuclei fuse together to form a
heavy nucleus.
3
The reaction occurs at ordinary temperature. This occurs at very high temperature.
4
The energy liberated in one fission is about
200 MeV.
The energy liberated in one fusion is about
24 MeV.
5
This can be controlled.
This cannot be controlled.
6
Products of fission are usually unstable
and radioactive in nature.
Products of fusion are usually stable and
non-radioactive in nature.
7
The links of fission reactions are neutrons.
The links of fusion reactions are protons.
DIFFERENCES BETWEEN NUCLEAR REACTIONS AND CHEMICAL REACTIONS
Some of the characteristics that differentiate between nuclear reactions and ordinary chemical reactions are
summarized ahead :
Nuclear reactions
Chem ical reactions
Involve convers ion of one nuclide
into another.
Involve rearrangem ent of atom s and not
change in the nucleus .
Particles within the nucleus are
involved .
Only outerm os t electrons participate.
Often accom panied by releas e
of trem endous am ount of energy.
Accom panied by releas e or abs orption of
relatively s m all am ount of energy.
Rate of reaction is independent of
Rate of reaction is influenced by external
external factors s uch as
factors .
tem perature, pres s ure and catalys t.
No breaking or m aking of bonds
involved.
Involves breaking or m aking of bonds .
Irrevers ible.
Can be revers ible or irrevers ible.
APPLICATIONS OF RADIOACTIVITY
AND RADIOISOTOPES
(a) In Medicine :
Radioisotopes are used to diagnose many diseases.
E.g. arsenic - 74 tracer is used to detect the presence
of tumour, sodium -24 tracer is used to detect the
presence of blood clots , iodine-131 tracer is used to
study the activity of the thyroid glands and cobalt-60
is used in the treatment of cancer . It should be noted
that the radioactive isotopes used in medicine have
very short half life periods.
(b) In Agriculture :
The use of radioactive phosphorus 32P in fertilizers has
revealed how phosphorus is absorbed by plants. This
study has led to an improvement in the preparation of
fertilizers.
14
C is used to study the kinetics of
photosynthesis.
CLASS-XI_STREAM-SA_PAGE # 69
(ii) To determine the age of animals or objects of
vegetable origin such as wood, charcoal and textiles
by radio carbon dating technique.
(c) In Industry :
(i) The thickness of a material (e.g. cigarettes, metal
plates etc.) can be determined by placing a radioactive
source on one side of the material and a counting
device on the other. From the amount of radiation
reaching the counter, the thickness of the material
can be calculated.
(ii) When a single pipe line is used to transfer more
than one petroleum derivative, a small amount of
radioactive isotope is placed in last portion of one
substance to signal its end and the start of another.
(d) In Geological Dating :
The age of the earth and rocks can be predicted by
geological dating. Age of a rock sample can be
calculated by finding out the amounts of the parent
radioactive element and the isotope of lead (e.g.
92 U
238
and
82 Pb
206
) in rock sample.
DATING
(i) The determination of age of minerals and rocks,
an important part of geological studies involves
determination of either a species formed during a
radioactive decay or the residual activity of an isotope
which is undergoing decay. For example
a decay (t1/2 =4.5 × 109 years) series
206
82 Pb
forming a stable isotope
and He. Helium
obtained as a result of decay of 238
has almost
92 U
certainly been formed from -particles. Thus, if
238
and
92 U
He contents are known in a rock we can
determine the age of rock sample (1g of 238
in
92 U
equilibrium with its decay products produces about
10–7 g He in a year). Also by assuming that initially
rock does not contain
206
82 Pb
and it is present in rocks
due to decay of 238
, we can calculate the age of
92 U
rocks and minerals by measuring the ratio of
238
and 206
82 Pb
92 U
.The amount of
to be obtained by decay of
238
92 U

Mole of
Mole of
206
82 Pb
238
+ 8 24 He + 6 –01e
U left = N at time t i.e., Nt
206
Pb formed = N’ at time t
238
U = N + N’ (at time 0) i.e., (N0)
Thus, time t can be evaluated by-
N0
2.303
log
Nt

is supposed
U . Thus,
238
 Initial mole of
t=
206
82 Pb
Note :
Radio carbon dating technique was given by W.F.
Libby and was awarded Nobel Prize.
Carbon-14 has been used to determine the age of
organic material. The procedure is based on the
formation of 14C by neutron capture in the upper
atmosphere.
14
1
7 N + 0n

14
1
6 C + 1H
This reaction provides a small, but reasonably
constant source of
(e) In Radio Carbon Dating :
The age of a fallen tree or dead animal can be
predicted by measuring the amount of C-14 in dead
plants or animals.
238
undergoes
92 U

14
C . The
14
C isotope is
radioactive, undergoing  - decay with a half life of
5730 years.
14
6C

14
0
7 N + –1 e
In using radio carbon dating, we generally assume
that the ratio of 14C to 12C in the atmosphere has been
constant for at least 50,000 years. The 14 C is
incorporated into CO2, which is in turn incorporated
through photosynthesis, into more complex carbon
containing molecules within plants. When these
plants are eaten by animals, the 14C becomes
incorporated within them. Because a living plant or
animal has a constant intake of carbon compounds,
it also has to maintain 14C to 12C ratio that is identical
with that of atmosphere . However, once the organism
dies, it no longer ingests carbon compounds to
compensate 14C which is lost through radioactive
decay. The ratio of 14 C to 12 C therefore,
decreases.Thus, by knowing the equilibrium
concentration of 14C in a living matter as well as in a
dead piece of matter at a particular time, the age of
material can be determined.
HAZARDS OF RADIATIONS
(i) Radioactive radiations cause atmospheric
pollution.
(ii) When living organisms are exposed to radiations,
the complex organic molecules get ionized, break up
and disrupt the normal functioning of the organisms.
(iii) Effects of radiations :
(A) Pathological damage : i.e. permanent damage
to living body which causes death and development
of diseases e.g. cancers or leukemia etc.
(B) Genetic damage : i.e. effect on chromosomes
causing mutations.
CLASS-XI_STREAM-SA_PAGE # 70
RADIOACTIVE POLLUTION
Radioactive pollution is a special form of physical
pollution, relating to all systems air, water and soil.
This type of pollution is not only harmful for the present
generation but also for future generations. The
radioactive substances with long half-life are usually
the main sources of environmental concern.
Neutrons released during nuclear tests make other
materials radioactive in the surrounding. These
materials include 90Sr, 137Cr and 131. The radioactive
materials are converted into gases. These gases and
fine particles are thrown high up into the air and carried
away by wind to distant areas. They ultimately settle
down and cause pollution to water and soil. From
soil the radioactive substances enter in the food chain
and thus affect all forms of life including man. Cosmic
radiations and explosion of a hydrogen bomb produce
14
C in air.
Nuclear power plants and reprocessing plants
discharge 90Sr, 137Cs, 131, 140 Ba, 140La, 144Rh, etc. Coal
based thermal power stations release radioactive
gases such as 85Kr, 133Xe and particulates such as
137
, 60Co, 54Mn and 137 Cs through chimney.
Nuclear dumping within land or in ocean leads to
radiation pollution.
(a) Effects of Radioactive Pollution :
(i) Radiations induce mutations and breaks in
chromosomes particularly at the time of cell division.
(ii) Higher doses of radiations can cause cancer,
leukaemia, anaemia and sterility. Excessive use of
X-rays causes death of tissues.
(iii) Radiations induce mutations in plants also.
Morphological deformities occur.
(b) Control of Radiation Pollution :
(i) Manufacture and use of nuclear weapons should
be stopped.
(ii) Nuclear tests and further development should be
suspended.
(iii) Ocean dumping of nuclear wastes should be
suspended.
(iii) Proper handling of radio isotopes during their
use in various fields should be done.
CLASS-XI_STREAM-SA_PAGE # 71