Chemistry Book By Supun Ayeshmantha

hhheeemmmiiissstttrrryyy ®®®
Part - 1
CCC
S . A . W i j e b a n d a r a
ADVANCEDLEVELCHEMISTRY-2012

S . A . Wijebandara – Copyright Reserved Page 2
Content
Unit 1 –Atomic Structure 03
Unit 2 –Chemical Calculations 26
Unit 3 – State of matter 46
Unit 4 – Energetic 53
Unit 5 – S, P, D Blocks of Periodic table 62
Unit 6 – Organic Chemistry 91
Unit 7 – Hydrocarbons 103
Unit 8 – Alkyl Halides 114
Unit 9 – Oxygen containing organic compounds 117
Unit 10 –Nitrogen containing organic compounds 133

Unit 1 –Atomic Structure
Properties of Electrons, Neutrons and Protons
Properties Electron Proton Neutron
Mass of particle 9.107 x 10-13
1.6725 x 10-27
1.6725 x 10-27
Relative mass 1/1840 1 1
Charge c -1.602 x 10-19
1.602 x 10-19
0
Relative Charge -1 +1 0
Types of Rays
Cathode rays Canal rays
Million Discovered by Oil
drop experiment
Goldstyle used a perforated cathode
and found this ray
They have a mass. Consist positive particles
Travel in straight lines Mass depend on residual gas.
Negatively charged Mass is equal to A.M. of gas.
Made up of particles. E/M ratio is different in different
gases.
Generated from cathode.
E/M ratio is constant.
Radioactivity
-Discovered by Bekeral.
-Radioactivity is the spontaneous decompositional action of certain
element into a stable condition.
-This produces three types of rays
-Gamma
-Beta
-Alpha

Gold foil Experiment
-Gigan and Marsden under the guidance of Rutherford, directed a
Alpha ray to gold foil
-Majority travelled through
-1/8000 get deviated
-1/20000 Travelled backward
Atom
-Rutherford measured deflected angle and calculated the radius of
nucleus as 10-13
cm and 10-8
cm as average atomic radius.
-Rogan discovered, Atomic no = No of Protons = No of electrons
-Chadwick directed alpha particles to Berilium and produced
uncharged particles, after directed toward wax, it change into
charged particles. He named them as Neutrons
Isotope
-With same atomic no but different mass number.
-Using Mass spectrometer, we can detect
-No of isotopes.
-Relative isotopic mass.
-Relative abundance.
Spectrum
-When E is energy, h is plank constant and f is the frequency then
E =hf
-There are two types of spectrums,
-Continuous spectrum
-Discontinuous spectrum
1-Absorption
2-Emmision

Continuous spectrum
-When white light send through Sodium vapour, a dark region can
be seen. This is due to absorption of energy
-This spectrum is known as Atomic absorption spectrum.
Emission spectrum
-When white light send through a excited sodium vapour, only one
region can be seen.
-A combination of both the systems can form a complete spectrum.

Emission system of H2
Bohr Theory
-Every energy level has a definite energy.
-When energy is supplied to electrons it gets excited and go to higher
energy levels.
-To stay stable, electrons release energy and come to lower energy
levels.
-When rotating in an level no absorption or release of energy.
Four Quantum numbers
-Principle quantum no (n)
No of main energy level
-Azimuthal quantum no (l)
Shape of the orbital
n=1 S=0 0
n=2 S=0, P=1 0, 1
n=3 S=0, P=1, D=2 0, 1, 2

-Magnetic quantum no (m)
In which orbital the electron exist in a certain sub energy
level.
S 0(l) 0
P 1(l) -1, 0, 1
D 2(l) -2, -1, 0, 1 , 2
F 3(l) -3, -2, -1, 0, 1 , 2, 3
-Spin quantum no (ms)
The direction in which the orbital is
ms = ½ 0r - ⅟2

Arrangement of Electrons
-Hund’s rule
Electrons will fill orbital by keeping their spins parallel.
-Pauli’s exclusive principle
Two electrons do not have similar quantum numbers.
-Aufban’s Principle
When arranging electrons, they go to the lowest energy level.
The electrons arrange according to the following diagram.
Special Configurations
Half filled sub energy level is more stable than normal configuration.
Variation Ionization Energies
-Ionization energy across the period increases, but Group (ii) and (iv)
elements have higher IE.
-Group (i) Minimum IE
-Group (viii) Maximum IE
- Group (v) > Group (iv) (Partially filled P orbital)

Electro negativity
-The capacity or tendency of an atom to attract the shared pair of
electrons is known as EN.
-Variation
En increases from left to right.
En decreases down the group.

Chemistry of Bonds
Chemical bonds
H + H H2 Exothermic Reaction
H2 H + H Endothermic Reaction
Ionic bond
- Ionic bonding is the attraction of oppositely charged ions (cations
and anions) in
Large numbers to form a solid. Such a solid compound is called an
ionic solid.
-When electronegativity difference is higher than 2.1 ,it shows ionic
characters.
- When electronegativity difference is equal to 2.1, then it is 50%
ionic and 50% covalent.
-In nature they form giant 3d lattice like NaCl, MgCl
Physical properties
-High melting and boiling points.
-Do not conduct electricity and heat.(Because the electrons are
immobile)
-Solid at room temperature.
Elements that can form ionic bonds
-(i) and (ii) groups form Anions.
-(vii) group forms Cations
Covalent Bond
- A covalent bond is formed when two atoms share one or more pairs
of electrons.
-Covalent bonding occurs when the electro negativity difference,
(EN), between
Elements (atoms) is zero or relatively small.

- Ionic bonding cannot result from a reaction between two
nonmetals, because their electro negativity difference is not great
enough for electron transfer to take place. Instead,
Reactions between two nonmetals result in covalent bonding.
Comparison between Ionic and Covalent Compounds
Sigma bond Pie bond
Axial overlapping Lateral overlapping
Symmetrical overlapping P-P overlapping
Overlap is greater Overlap is lesser
Bond is strong Bond is weaker
Electron density is high
between nuclei
Less electron density
One atom relative to other can
rotate
No movement

Dative bond
-In covalent bond the two atomic orbitals should have unpaired
electrons
-One should contain vacant orbital while other should have lone pair
of electrons.
Polar Covalent bond
-If covalent molecule is formed between similar atoms electrons
spread symmetrically.
-If not electron doesn’t spread symmetrically.
Due to that one atom get slight negative charge while other
get slight positive charge.
H+β
- Cl-β
-
The attractive forces between them are
Static attraction
Covalent bond force
-This type of bonds are more stronger than non polar bonds.
-Physical properties
They exist in simple molecules.
Do not conduct electricity and heat.
Melting and boiling points are very low.
Metallic bonds.
-The valency electrons of metallic elements release and form common
stream of electrons and positive ions.
-In metallic bonds, lattices are formed due to that.
-As a result of delocalized electrons it conducts electricity.
-The strength of a metallic bond depend on
No of available electrons for bond.
Charge density of Cation
Polarizability
-When an ideal ionic compound is considered, the anions and cations
which are its components are regarded as existing in the form of
regular solid spheres.
-But depending on the nature of the cations and the anion which are
the constituents of the ionic compound, the electron cloud of the
anion gets attracted towards the electron cloud of the cations and as a
result distortions occur in the electron cloud of the anion.

Factors which increase the polarizability
1) Cations : Smaller in size. Highly charged
2) Anion : Larger in size. Highly charged
Dot cross Structures
The writing of Lewis formulas is an electron bookkeeping method
that is useful as a first approximation to suggest bonding schemes. It
is important to remember that Lewis dot formulas only show the
number of valence electrons, the number and kinds of bonds, and the
order in which the atoms are connected. They are not intended to
show the three-dimensional shapes of molecules and polyatomic ions.

Lewis Structures
Chemical bonding usually involves only the outermost electrons of
atoms, also called valence electrons. In Lewis dot representations,
only the electrons in the outermost occupied s and p orbital’s are
shown as dots. Paired and unpaired electrons are also indicated.
All elements in a given group have the same outer-shell electron
configuration. It is somewhat arbitrary on which side of the atom
symbol we write the electron dots. We do, however, represent an
Electron pair as a pair of dots and an unpaired electron as a single dot

Octet Rule
Representative elements usually attain stable noble gas electron
configurations when they share electrons. In the water molecule eight
electrons are in the outer shell of the O atom, and it has the neon
electron configuration; two electrons are in the valence shell of each
H atom, and each has the helium electron configuration. Likewise,
the C and O of CO2
In most of their compounds, the representative elements achieve
noble gas configurations.
VALENCE SHELL ELECTRON PAIR REPULSION
(VSEPR) THEORY
Molecule
(Example)
Total pairs Bonded
pairs
Lone
pairs
Shape Name
BeCl2 2 2 0 Linear
BeCl3 3 3 0 Trigonal plainer
O3 3 2 1 Angular
CF4 4 4 0 Tetrahedron
NH4 4 3 1 Trigonal
Pyramidal
H2O 4 2 2 Angular
PH5 5 5 0 Trigonal by
pyramidal
SF4 5 4 1 See-saw
5 3 2 T-shape
Xecl2 5 2 3 Linear
SF6 6 6 0 Octahedron
6 5 1 Square pyramid
XeCl4 6 4 2 Square Plainer

-There are three types of repulsions in-between atoms
Lone-Lone pair repulsion
Lone-Bond pair repulsion
Bond-Bond pair repulsion
Hybridization
-The electron configuration of carbon is 1s2
, 2s2
, 2p2
-
In the last shell of carbon there’s only 4 electrons and only 2
unpaired
-Therefore carbon can form only 2 bonds.
-But carbon can form 4 bonds using all valency electrons.
-Using hybridization we can explain this.
SP3
Hybridization
- Four sigma bonds are formed with 4 covalent bonds.
- One S orbital get the energy of P orbital.
- Tetrahedron in shape.

SP2
Hybridization
-One S orbital and 2 P orbitals are involved.
-A sigma and a pie bond is formed.
SP1
Hybridization
-One sigma and two pie bonds are formed.

Resonance Structures.
-A molecule or polyatomic ion for which two or more Lewis formulas
with the same
-Arrangements of atoms can be drawn to describe the bonding is said
to exhibit resonance.
- The three structures above are resonance structures of the carbonate
ion. The relationship among them is indicated by the double-headed
arrows. This symbol does not mean that the ion flips back and forth
among these three structures. The true structure can be described as
an average, or hybrid, of the three.
- The dashed lines include that some of the electrons shared between
C and O atoms are delocalized among all four atoms; that is, the four
pairs of shared electrons are equally distributed among three C-O
bonds.
-Example, Draw resonance structure of SO2

Formal Charge.
-Formal charge is the hypothetical charge on an atom in a molecule
or polyatomic ion; to find the formal charge, we count bonding
electrons as though they were equally shared between the two
bonded atoms. The concept of formal charges helps us to write
correct Lewis formulas in most cases. The most energetically
favorable formula for a molecule is usually one in which the formal
charge on each atom is zero or as near zero as possible.
- The formal charge, abbreviated FC, on an atom in a Lewis formula
is given by
the relationship
FORMAL CHARGE = (GROUP NO)-(NO OF BONDS – NO OF UNSHARED ELECTRONS)
Rules for applying formal charge.
-Molecules FC = 0
-Poly Atomic ions FC = Charge it carries
Dipole Moments.
It is convenient to express bond polarities on a numerical scale. We
indicate the polarity of a molecule by its dipole moment, which
measures the separation of charge within the molecule. The dipole
moment, µ, is defined as the product of the distance, d, separating
charges of equal magnitude and opposite sign, and the magnitude of
the charge, q. A dipole moment is measured by placing a sample of
the substance between two plates and applying a voltage. This causes
a small shift in electron density of any molecule, so the applied
voltage is diminished very slightly. Diatomic molecules that contain
polar bonds, however, such as HF, HCl, and CO, tend to orient
themselves in the electric field . This causes the measured voltage
between the plates to decrease more markedly forthese substances,
and we say that these molecules are polar.

Dipole – dipole Interaction.
-DDI causes between covalent molecules because of +ve and –ve
attractions.
-DDI effective only for short distances.
-Approximately the DDI energy is 4 KJ/Mol
-This is weaker than covalent and ionic bonds.
-When temperature increases this become less important
*Percentage of Ionic character = µ(Observed) x 100 /µ (Calculated)
Dispersion Interaction.
-DI are weaker attractive forces only over extremely short distances
because they vary as 1/d7
.
-Present in-between all types of molecules in condensed phase, but
weaker for small molecules.
-DI are the only forces in-between symmetrical non-polar
(noble gases) substances such as SO2 ,CO2 ,O2 and mono atomic
molecules.
Dipole – Induced Interaction.
-A polar molecule may polarize a neutral molecule which lies in it’s
vicinity and thus induces dipolarity molecule there.
-In that molecule magnetic dipolarity occurs.
-This induced dipole interact with first atom and thereby the 2
molecules are attracted each other.

-Without DI these cannot form liquids and solids.
-DI increases when the size of the molecule increases.
Hydrogen Bond
- Hydrogen bond is a special electrostatic interaction formed
by a hydrogen atom covalently bonded to an electro negative atom (x)
, with an electronegative atom (Y) containing one or more lone pairs
of electrons.
- Hydrogen bonds become more stronger when X and Y are
more electronegative atoms such as F, O, N and Cl. Strength of
hydrogen bonds decreases as F>O>N>Cl. This is named as FONCl
rule. In some special cases, when the positions of X and Y are taken
by atoms other than the above ones, strong hydrogen bonds are
formed
- Importance of hydrogen bonding
Without HB water is a gas
HB exist in many tissues, organs, skin and bones of animals.
Play important role in determination of structure of protein.
-There are two types of hydrogen bondings.
Intermolecular hydrogen bond
Intramolecular hydrogen bond
Intermolecular hydrogen bonding.
-When HB take place between two difference molecules of same
compound such as in HF , H2O it’s called IHB
-Homointermolecular hydrogen bonding
-This type of H bonding is formed in-between same
molecule.
-Heterointermolecular hydrogen bonding
-This bonding take place in-between different compounds.
Intermolecular hydrogen bonding.
-If HB takes place in-between Hydrogen and the electronegative
atom presence within the same molecule.

Lattices
-A formation where the building units are attached to one another in
an orderly pattern can be described as a lattice.
- The presence of a formal pattern and the formation from a repetitive
basic unit is a common feature of the lattices.
- The substances with lattice arrangements can be classified according
to their building units as follows.
Homoatomic lattices
Heteroatomic lattices
Non – polar molecular lattices
Polar molecular lattices
Ionic lattice
- The bonding formed during the formation of the lattices are
different, depending on the nature of the building unit of the lattice
substance.
- The nature of the bonding formed during the formation of the
lattice affects the physical properties of the lattice.
- Diamond and graphite lattices which are formed from
homogeneous atoms are examples for homoatomic lattices.
- Silicon dioxide which is formed from heterogeneous atoms is an
example for heteroatomic lattices.
- Iodine crystals which are formed from nonpolar iodine molecules
are examples for non- polarized molecular lattices.
-Ice which is formed from polar molecules is an example for polar
molecular lattice.
-Sodium chloride which consists of sodium ions and chloride ions is
an example for ionic lattices.
- Homogeneous and heterogeneous atomic lattices are formed by
atoms covalently bonding with one another.
-Non polar molecular lattices are built by the bonding of non polar
molecules to one another by attractive forces formed between
induced dipoles.
-Polar molecular lattices are formed by the bonding together of polar
molecules to one another by attractive forces developed between
permanent dipoles.
-Ionic lattices are formed by the bonding together of ions by strong
electrostatic attractions developed between positively charged and
negatively charged ions.
- The substances composed of homogeneous and heterogeneous
atomic lattices have a high hardness and also higher melting points/
boiling points because such lattices are formed by strong covalent
bonds.

-There is no tendency to go into solution because the covalent bonds
present in atomic lattices are very strong.
- Electricity is not conducted through atomic lattices as they lack
mobile electrons.
-The substances with nonpolar molecular lattices consisting of
molecules bonded to one another by weak Van der Waals forces have
a low hardness and their melting points/ boiling points are also low
relative to substances formed by other lattices.
-The substances formed by nonpolar molecular lattices dissolve in
nonpolar solvents because they consist of nonpolar molecules and
they do not conduct electricity as they do not possess mobile
electrons.
-The substances formed by polar molecular lattices consisting of
molecules bonded by permanent dipole – permanent dipole attractive
forces (or hydrogen bonds) have a high hardness and also higher
melting points / boiling points compared to substances formed by
nonpolar molecular lattices.
-The substances formed by polar molecular lattices show a high
tendency to dissolve in polar solvents but do not conduct electricity
due to the lack of mobile electrons.
- The substances formed by ionic lattices bonded by strong
electrostatic attractions show higher melting points/ boiling points
and also a high hardness.
-The substances with ionic lattices show a tendency to dissolve in
polar solvents.
-The substances consisting of ionic lattices do not conduct electricity
in the solid state due to the lack of mobile electrons.
-The substances with ionic lattices conduct electricity in the molten
state or in solution because of the presence of mobile ions.

Atomic lattices
The lattices formed by one type of atoms are known as homoatomic
lattices. Diamond and graphite which are allotropic forms of carbon
are examples for these. The lattices formed by the combination of
atoms which are different from one another are called heteroatomic
lattices and silica (SiO2) is an example of this.
Diamond
- Diamond is formed by covalently bonding together of carbon atoms
which are subjected to sp3 hybridization. This lattice is arranged so
that each carbon atom is bonded by single covalent bonds to four
other carbon atoms.
Graphite
-Graphite is formed by covalently bonding together of carbon atoms
which are subjected to sp2 hybridization.
-The lattice is arranged so that each carbon atom is attached to three
other carbon atoms by single covalent bonds.
-Within the lattice structures, each carbon atom is surrounded by
three other carbon atoms which are arranged so that they occupy the
vertices of a planar triangle.
- This is a two dimensional giant structure.
-In graphite each carbons atom has a p– orbital which is not
hybridized and also there is an unpaired electron in it.
-The unhybrid p– orbital is situated perpendicular to the plane of the
two dimensional lattice.
-Therefore the p- orbitals containing these unpaired electrons help to
form attractions between the two dimensional carbon layers.
Molecular Lattice
-Some of the molecular lattices are formed by homo atomic molecules
while some others are formed by heteroatomic molecules.
-Considering the polarity of the molecule which forms the lattice,
they can be classified as nonpolar molecular lattices and polar
molecular lattices.
-Iodine crystal is an example for a non polar molecular lattice. Iodine
molecule is a large molecule.

-Therefore due to various reasons iodine molecules develop induced
dipoles temporarily.
-Iodine lattice is formed by attractions made between such induced
dipole - induced dipole attractions.
- Water molecule is polar and ice is a good example for polar
molecular lattice. Due to the permanent dipole in -OH bond water
molecules are bonded by hydrogen bonds in ice.
Ionic Lattice
-Many ionic compounds exist naturally as solid substances with a
crystalline structure.
-This crystal lattice is formed by the packing of oppositely charged
ions of the ionic compound in a definite pattern is space.
-Electrostatic attractions exist between these oppositely charged ions.
-Sodium chloride is a good example for an ionic lattice.

Unit 2 –Chemical Calculations
Significant Figures
-There is a difference between counting and measuring. Counting
can be done with certainty while measuring is only stating the nearest
value. The number of meaningful figures included in the numerical
value of the statement of a certain physical quantity is known as the
significant figures.
-Example :
The length of a pen can be measured by different
measuring instruments and stated as follows.
-Measurement by a ruler marked with centimeters 18 cm
-Measurement by a meters ruler 18.1 cm
-Measurement by a vernier caliper 18.12 cm
Therefore, the stated value after measurement by a measuring
instrument is an approximate value and not a definite value.
Here the first measurement is stated with 2 significant figures. The
second measurement is stated with 3 significant figures and the third
measurement with 4 significant figures. It is necessary to manipulate
significant figures correctly in addition, subtraction, multiplication
and division of numbers
Rules
-Nonzero digits are always significant.
-Zero’s at beginning are not significant.
-Zero’s between nonzero numbers are significant.
-Zero’s at the end of a number with decimals are significant
Relative Atomic Unit (AMU)
This defined as exactly the 1/12 of the mass of an atom of particular
kind carbon atom.
AMU = Mass of atom x 12 / Mass of carbon

Percentage composition
Examples: 158.04 g of KMnO4 contains 39.10 g of K, 59.94 g of Mn
and 64.00 g of O.
Empirical formula
The formula that shows the simplest whole number ratio between the
number of atoms in agreement with the composition of a compound
is its empirical formula
Example: Empirical formula of benzene is CH
Molecular formula
The formula that shows the exact number of atoms in the molecule of
a compound is its molecular formula. The molecular formula of water
is H2O. The ratio between the empirical formula and the molecular
formula of a compound is a whole number.

Composition of a compound
-Composition stated as the number of parts per million of parts
(ppm) = fraction x 106
-Composition stated as the number of parts per billion of parts
(ppb) = fraction x 10 9

Avogadro Constant (L)
-Avogadro constant is given as L= N/n, Where
N = number of particles
n = amount of substance
Avogadro constant (L) = 6.022 x 1023
mol -1
Gas Constant (R)
-Gas constant (R) is encountered during the study of ideal gases.
-Consider a sample of gas. When pressure of the gas is P, volume of
the gas is V amount of the gas is n mole and absolute temperature of
the gas is T
PV = nRT
R = 8.314 J K-1 mol -1
Faraday constant (F)
-Faraday constant is defined as the molar charge of the proton.
F = eL
L – Avogadro constant
e – Charge of the electron
Faraday constant (F) = 96485 C mol-1

Balancing Chemical Equation.
-There are three ways of balancing chemical equations.
Inspection method.
Mathematical method.
By using half reactions.
Inspection method
Balancing an equation by balancing the number of atoms of each
kind in the reactants and products is called inspection balancing.
In this method, the elements in the reactants and products are
separately balanced. Elements with the least number of atoms are first
considered and molecules composed of atoms of same element.
Mathematical method
Balancing equations when the reactants and the products are known
by comparing the coefficients is called the mathematical method.

Balancing equations using oxidation / reduction half reaction

Oxidation number
-This represents the no of electrons lost or gain to change into
compound from free state.
-This gets positive when electrons are given.
-This gets negative when electrons are gained.
-In covalent bond it’s a imagination.
Rule
-Uncombined or free state elements oxidation number is zero.
-F always -1 -Metals always positive.
-Oxygen -2 , when in peroxides -1
-Hydrogen +1 in ionic hydrides -1
-Alkali metals +1
-Alkaline earth metals +2
-Halogen halides -1
-Sulphur in sulphides -2
-The algebraic sum of oxidation number is equals to the net charge.
-Oxidation number may be a fraction.
-Maximum oxidation number = Group no (Except O,F)
-Minimum oxidation number = Group no -8 (Except metals)

Chemical Calculations
-There are various ways in which the composition can be expressed. -
-They are
Mass fraction
Volume fraction
Mole fraction
Mass/volume
Moles /mass
-When the composition is expressed in terms of moles / volume it is
called as concentration.
-Composition of substances present in very small amounts is
commonly expressed in terms of parts per millions (ppm) and parts
per billions (ppb).
-ppm and ppb can be expressed as ppm = mass fraction x 106 and
ppb = mass fraction x 109. ppm and ppb can be also be expressed as
ppm = volume fraction x 106 and ppb = volume fraction x109.
-Density of water is 1000 g dm-3
. Density of a dilute aqueous solution
can be considered as approximately equals to the density of water.
Therefore,
Mass of 1 dm3 of a solution = 1 kg = 1000 g = 1000000 mg
-For such instances, mass/ volume ratio can also be expressed in
ppm. As a mass fraction, 1 ppm means that 100000 mg of the mixture
contains 1 mg of the particular substance for dissolve solutions. For
dilute solutions,

CALCULATIONS BASED ON CHEMICAL EQUATIONS
- Let us again consider the combustion of methane in excess oxygen.
The balanced chemical equation for that reaction is

-A chemical equation also indicates the relative
amounts of each reactant and productin a given
chemical reaction. We showed earlier that
formulas can represent moles of substances.
Suppose Avogadro’s number of CH4 molecules, rather than just one
CH4 molecule, undergo this reaction. Then the equation can be
written

THE LIMITING REACTANT CONCEPT
-In the problems we have worked thus far, the presence of an excess
of one reactant wasstated or implied. The calculations were based on
the substance that was used up first,called the limiting reactant.

PERCENT YIELDS FROM CHEMICAL REACTIONS
- The theoretical yield from a chemical reaction is the yield calculated
by assuming that the reaction goes to completion. In practice we
often do not obtain as much product froma reaction mixture as is
theoretically possible.
-The actual yield is the amount of a specified pure product actu-ally
obtained from a given reaction.
-The term percent yield is used to indicate how much of a desired
product is obtainedfrom a reaction.

SEQUENTIAL REACTIONS
-Often more than one reaction is required to change starting materials
into the desired product. This is true for many reactions that we carry
out in the laboratory and for many industrial processes. These are
called sequential reactions. The amount of desired product from each
reaction is taken as the starting material for the next reaction.

CONCENTRATIONS OF SOLUTIONS
- Concentrations of solutions are expressed in terms of either the
amount of solute present in a given mass or volume of solution, or the
amount of solute dissolved in a given mass or volume of solvent.
- Percent by Mass
Concentrations of solutions may be expressed in terms of
percent by mass of solute,which gives the mass of solute per 100 mass
units of solution. The gram is the usual mass
unit.
Molarity
-Molarity (M), or molar concentration, is a common unit for
expressing the concentrations of solutions. Molarity is defined as the
number of moles of solute per liter of solution:

DILUTION OF SOLUTIONS
-Recall that the definition of molarity is the number of moles of solute
divided by the volume of the solution in liters:
USING SOLUTIONS IN CHEMICAL REACTIONS

Unit 3 – State of matter
Behaviour of matter
Gases
Ideal gases
-The temperature, pressure, volume and the amount (moles) of
substance of a gas are the factors that affect the behaviour of a gas.
-The ideal gas equation can be described as a relationship of the
above four variables regarding a gas.
Derivation of Boyle law from the ideal gas equation

Derivation of Charles law from the ideal gas equation
Derivation of Avogadro law from the ideal gas equation
Pressure
-Total pressure = PA + PB + PC +……..+ PN
PA = XA . PTotal
Where XA = Mole fraction

Kinetic molecular theory
Assumptions of the molecular kinetic theory
-The molecules of a gas are in a state of continuous random motion.
- When molecules collide with one another and bounce off the total
kinetic energy of the system remains the same.
-The pressure exerted by a gas is the result of collisions of the
molecules on the walls of the container.
-Molecular kinetic equation is
PV =⅟3mNC-2
Where m -Mass
N -Moles
C-2
- Mean square velocity
Escape velocity
-The minimum velocity at which any object should be projected
from the surface of the Earth to enter into space completely escaping
from the gravitational attraction of the Earth is called its escape
velocity.
-When the mean velocity of a gas molecule exceeds its escape velocity
it leaves the atmosphere of the Earth.
-The mean velocity of light gases such as hydrogen and helium
exceeds the escape velocity from the Earth and as a result they have
left the atmosphere of the Earth.

Diffusion
-The spreading of a certain type of molecules throughout space
occupied by another type of molecules is called diffusion.
Example :- When a substance with a scent is kept inside a room
diffusion takes place until the scent is distributed uniformly
throughout the room.
-Rate of diffusion - solids < liquids < gases
-It has been experimentally found that different gases diffuse with
different rates.
-The production of ammonium chloride by the diffusion of ammonia
and hydrogen chloride molecules through air can be demonstrated by
the following apparatus.
-From this it is clear that the rate of diffusion of ammonia molecules
with a low molecular mass is higher than that of hydrogen chloride
molecules.
-Variation of the mean velocity of a gas with temperature is shown by
the following Maxwell - Boltzmann curves.
HCl Ammonia (Conc.)

Dalton’s law of partial pressure
-The contribution made by the constituent gases towards the total
pressure of a mixture of gases is called their partial pressures.
-The pressure that a constituent gas of a mixture of gases would exert
if it alone occupied the volume of the container of the mixture is
called the partial pressure of that gas.
-In a mixture of gases, the total pressure is equal to the sum of the
partial pressures of each of the constituent gases.
-If the partial pressures of individual gases in a mixture of gases A, B
and C are PA, PB and PC respectively, total pressure of the mixture
PT = PA+ PB + Pc
Derivation of Dalton’s law of partial pressure from the ideal gas
equation
In a mixture of gases A and B there are nA and nB moles of each gas
respectively.
Total pressure = PA + PB + PC +……..+ PN

Compressibility factor
-z is the compressibility factor. For ideal gases z = 1. But the fact that
this value is not a constant for real gases is revealed by experimental
data.
-The graph of the product PV against P for different gases at the
temperature 273 K
-According to this graph it is clear that real gases approach ideal
behaviour under conditions of low pressure.
-The graph of PV/RT against P for a mole of hydrogen at various
temperatures.
It is clear from the above graph that at high temperatures the real
gases approach ideal behaviour.

Van der Waals equation
P = Pressure
V = Volume
n = amount (moles) of substance
R = Universal gas constant
T = Absolute temperature
a and b are constants (Van der Waals constants) for real gases
Calculations using Van der Waals equation are not necessary
Calculation of molecular energy
-The transition kinetic energy of a particle is given as
E =½mNC-2
PV =⅟3mNC-2
E =½mNC-2
Therefore 2E = 3 PV & 2E = 3nRT

Unit 4 – Energetic
-The portion of the universe selected for the study is called the system
-All the rest other than the portion selected for the study is called the
environment
-The dividing line that separates the system and the environment is
called the boundary.
-Systems where there is an exchange of energy and matter across the
boundary are called open systems.
-Systems where only energy is exchanged across the boundary are
called closed systems
-Systems where there is no exchange of both energy and matter
across the boundary are called isolated systems.
Enthalpy
-The quantity of heat supplied to a system or given out from a system
under the condition of constant pressure is called the enthalpy change
(∆H).
-This is a thermodynamic property and a function of state.
-The enthalpy change (∆H ) associated with a reaction is given by the
difference in enthalpy of the products and reactants.
∆H= HProduct - HReactant

-Enthalpy change associated with a reaction;
If ∆H < 0 the reaction is exothermic
If ∆H > 0 the reaction is endothermic
-Standard conditions ∆Hø
Temperature – 298k / 25o
c
Pressure - 1 x 105
Nm-2
/ 1Pa / 1 atm
Standard enthalpy change.
Standard enthalpy of formation -∆Hø
F
-It is the enthalpy change that occurs when one mole of the
compound is formed in the standard state from the constituent
elements in the standard state.
Example: – Standard enthalpy of formation of H2O (l) is = 286 kJ
mol-1.
Standard enthalpy of combustion -∆Hø
C
-It is the enthalpy change that occurs when one mole of an element
or a compound in the standard state undergoes complete combustion
in an excess of oxygen.
Standard enthalpy of bond dissociation -∆Hø
D
- It is the enthalpy change that occurs when a gaseous species in the
standard state is converted into gaseous components by breaking a
mole of bonds (this is stated with respect to a specified bond in a
specified element or a compound).

Standard enthalpy of neutralization -∆Hø
NEU
-It is the enthalpy change that occurs when a mole of H+ ions in an
aqueous solution reacts with a mole of OH - ions in an aqueous
solution under the standard state to form a mole of H2O.
H+
(aq) + OH-
(aq) H2O(l) -∆Hø
= -57kJmol-1
Standard enthalpy of solution -∆Hø
SOL
- It is the enthalpy change that occurs when a mole of gaseous ions
under the standard conditions changes into the solution form in the
presence of an excess of the solvent.
Mn+
+ Excess solvent Mn+
(solvent)
Standard enthalpy of hydration -∆Hø
HDY
-It is the enthalpy change that occurs when a mole of gaseous ions
under the standard conditions changes into the solution form in the
presence of an excess of water.
Na+
(g) + H2O (l) Na+
(aq)
Standard enthalpy of transition -∆Hø
TRA
-It is the enthalpy change that occurs when a substance under
standard conditions gets changed from one phase to another phase of
the same substance (Here the phase means especially an allotropic
form.
-The enthalpy changes related to changes in physical state which are
considered as phase changes will be described separately.)
C (graphite) C (diamond)
Standard enthalpy of dissolution -∆Hø
DES
-It is the enthalpy change that occurs when a mole of a substance
under the standard conditions is dissolved in a solvent to form 1dm3
of solution.
NaCl(aq) + H2O(l) NaCl(aq) ∆Hø
DES= 1kJmol-1
Hess’s law
-The enthalpy change that takes place in a chemical reaction takes a
constant value through whatever the route the reaction was
performed.

A+B ∆H 1
C+D
∆H 2 ∆H 3
E
∆H 1 = ∆H 2 + ∆H 3
Enthalpy diagrams
-The below diagram represent a enthalpy diagram
Example – Dram enthalpy diagram for HBr

Enthalpy of formation of ionic compounds.
Standard enthalpy of sublimation-∆Hø
sub
-It is the change in enthalpy that occurs when a mole of a solid
element or a mole of a solid compound under the standard conditions
is converted completely into the gaseous state.
Ca(s) Ca(g) ∆Hø
sub= 193 kJmol-1
Standard enthalpy of evaporation -∆Hø
vap
-It is the change in enthalpy that takes place when a mole of a liquid
compound or element under the standard conditions is converted
into a mole of a gaseous compound/element.
Br2(l) Br2(g) ∆Hø
vap= 30.91kJmol-1
Standard enthalpy of fusion -∆Hø
fus
-It is the change in enthalpy that takes place when a mole of a solid
compound or element under the standard condition is converted into
a mole of liquid compound/ element.
Al(s) Al(l) ∆Hø
fus= 10.7kJmol-1
Standard enthalpy of atomization -∆Hø
atm
-It is the change in enthalpy that takes place when a mole of an
element under the standard conditions is converted into a mole of
atom in the gaseous state.
½Cl2(g) Cl(g) ∆Hø
atm= 121kJmol-1
Standard enthalpy of first ionization -∆Hø
Ie1
-It is the change in enthalpy that takes place when a mole of
unipositive ions are formed under standard conditions by removing
an electron each that is most weakly bonded to the nucleus from a
mole of gaseous atoms of an element.
Na(s) Na+
(s) ∆Hø
Ie1= 496kJmol-1
Standard enthalpy of electron affinity (or electron gain) -∆Hø
ea
- It is the change in enthalpy that takes place when a mole of
uninegative ions are formed in the gaseous state under the standard
conditions when electrons are given to a mole of atom in the gaseous
state. (All the time second EA is positive)
Cl(g) Cl-
(g) ∆Hø
ea= -352kJmol-1

Standard lattice enthalpy of an ionic compound -∆Hø
L
-It is the change in enthalpy that takes place when one mole of an
ionic compound in the solid state under the standard conditions are
formed from gaseous positive ions and negative ions.
Na+
(g) Cl-
(g) ∆Hø
L= -780kJmol-1
Average Standard bond enthalpy
-Average bond enthalpies are approximate values.
-When you want to assign a value to a standard enthalpy of
dissolution of C-H bond in methane, the problem is different the
energy required to break first C-H is not same as that, it require to
remove H from CH3
-
radical
Example
CH4 CH3
.
+H -∆Hø
=426kJmol-1
CH3
.
CH2
..
+H -∆Hø
=436kJmol-1
CH2
..
CH ...
+H -∆Hø
=451kJmol-1
CH ...
C ....
+H -∆Hø
=347kJmol-1
Average bond enthalpy = (426+436+451+347)/4
= 415kJmol-1
Bond S.A.B.E (347kJmol-1
)
C-C +347
C=C +611
N-H +389
O-H +464
C-H +414
C-Br +292
C-O +360
For any reaction
∆Hø
R = [∆Hø
of breaking bonds] – [∆Hø
of bond formation]

Born Haber Cycle
-The born Haber cycle is a technique which occurs when an ionic
compound is formed.
-The Born–Haber cycle involves the formation of an ionic compound
from the reaction of a metal (often a Group I or Group II element)
with a non-metal. Born–Haber cycles are used primarily as a means
of calculating lattice energies (or more precisely enthalpies) which
cannot otherwise be measured directly.
- The lattice enthalpy is the enthalpy change involved in formation of
the ionic compound from gaseous ions.
-A Born–Haber cycle calculates the lattice enthalpy by comparing the
standard enthalpy change of formation of the ionic compound (from
the elements) to the enthalpy required to make gaseous ions from the
elements. This is an application of Hess's Law.
Example – Formation of lithium fluoride
The enthalpy of formation of lithium fluoride from its elements
lithium and fluorine in their stable forms is modeled in five steps in
the diagram ;
-Atomization enthalpy of lithium
-Ionization enthalpy of lithium
-Atomization enthalpy of fluorine
-Electron affinity of fluorine
-Lattice enthalpy
- The same calculation applies for any metal other than lithium or
any non-metal other than fluorine.

- The sum of the energies for each step of the process must equal the
enthalpy of formation of the metal and non-metal, ΔHf
-V is the enthalpy of sublimation for metal atoms (lithium)
-B is the bond energy (of F2). The coefficient 1/2 is used because the
formation reaction is Li + 1/2 F → LiF.
-IEM is the ionization energy of the metal atom
-EAX is the electron affinity of non-metal atom X (fluorine)
-UL is the lattice energy (defined as exothermic here)
The net enthalpy of formation and the first four of the five energies
can be determined experimentally, but the lattice energy cannot be
measured directly. Instead, the lattice energy is calculated by
subtracting the other four energies in the Born–Haber cycle from the
net enthalpy of formation.
- The word cycle refers to the fact that one can also equate to zero the
total enthalpy change for a cyclic process, starting and ending with
LiF(s) in the example. This leads to
Enthalpy changes when Ionic compounds dissolve
-When an ionic solid dissolve in a solvent , two types of enthalpy
terms are involved.
i) The ion must be separated - Ionic lattice energy is required.
ii) The separate ions interact with molecules of solvent
(If solvent is polar , ions can be attracted)
∆Hø
Dissolution = ∆Hø
Lattice dissolution + ∆Hø
solvation

Free energy And Entropy
-A reaction which happens of it’s own accord without any external
help is known as SPONTANEOUS REACTION.
-The difference of enthalpy between product and reactants cannot be
the only factor which decides whether the reaction is happened or
not.
-Entropy of a system is a measure of the randomness of the system.
-Entropy is a function of state and it depends only on the initial and
final state if the system and is independent of the path of the change.
-Spontaneous changes in an isolated system takes place with an
increase in entropy.
-As the entropy related to a certain system is a function of state, the
change in entropy can be calculated by subtracting the initial value of
entropy from the final value of entropy.
-The entropy of perfect crystal at 0K is 0
-When degree of order decreases then entropy also decreases
∆S = ∆SProduct - ∆SReactants
When Pressure Entropy
Volume Entropy
Temperature Entropy
-When disorder increases the entropy +ve
CaO(s) + H2O(l) Ca(OH)2(s) ∆S –ve
CaCO3(s) CaO2(s) + CO2(g)
-Both entropy and enthalpy change is important to decide weather a
chemical reaction has occur or not.
G = H-ST
Where H-Enthalpy
S-Entropy
T-Temperature in K
G-Free energy
-The sigh of G represents the nature of the reaction
If G > 0 ; Not spontaneous
G < 0 ; Spontaneous
G = 0 ; Reaction under equilibrium

Unit 5 – S, P, D Blocks of
Periodic table
PHYSICAL PROPERTIES OF S AND P BLOCKS
Atomic radius
-Atomic radius increases down the group
Covalent radius
- When two atoms of the same
element are covalently bonded, half
the internuclear distance of these two
atoms is called its Covalent radius.
-The covalent atomic radius increases
down a group and decreases from left to right
along a period.
Covalent radius

Van der Waals radius
-When two molecules are placed as close together as possible,
half the distance between the two nuclei which are close to
each other is called the Van der Waals radius.
Van der Walls radius = d/2
Metallic radius
-Half the distance between two adjacent cation nuclei in the metallic
lattice is the metallic radius.
Metallic radius = d/2
-Nuclear charge and the shielding effect affect the atomic radius.
Valency and oxidation numbers
-Formation of anions and cations depends on the number of electrons
in the valency shell and ionization energy.
-The elements of the group I, II and III, form cations while the
elements of the groups V, VI and VII form anions.
-The elements in the group IV do not form M4+ ions. The reason is
the high aggregate of first, second, third and fourth ionization
energies.
-When the atomic number increases along a period the highest
oxidation number increases.
-The highest oxidation number that an element can have in a
compound is equal to its number of valency electrons.

Ionization Energy
- the minimum
amount of energy
required to remove
the most loosely
bound electron
from an isolated
gaseous atom to
form an ion with a
1+ charge.
Electron Affinity
-The electron affinity (EA) of an element may be defined as
the amount of energy absorbed when an electron is added to an
isolated gaseous atom to form an ion with a -1 charge.

-Across a period from left to right the nuclear charge and atomic
radius decreases and the ionization energy increases. Therefore, the
tendency to form cations decreases and also the ability for reduction
decreases across a period.
-Similarly the ability to form anions increases and also the ability for
oxidation increases from left to right
across a period
Electronegativity
-The electronegativity (EN) of an element is a measure of the relative
tendency of an atom to attract electrons to itself when it is chemically
combined with another atom.
-Elements with high electronegativities (nonmetals) often gain
electrons to form anions.
-Elements with low electronegativities (metals) often lose electrons to
form cations.

S-Block Elements
-The elements which having electrons in last shell S
orbital, is known as S-BLOCK element.
-Their general electronic configuration is ns1
,ns2
n=1 to 7
-Fr , Ra are radio active elements.
-IA - alkaline metals, they react with water to form
alkali.
-IIA -alkaline earth metals, they oxides and react with
water to form alkali and found in soil or water
-The total number of S block elements – 14
-Cs and Fr are liquid elements.
Physical properties
FLAME COLORS
Li Red Be Colorless
Na Yellow Mg Bri: white
K Lilac Ca Brick red
Rb Red Sr Crimson
Cs Blue Ba Apple green

Chemical properties
Reactions with water
-All the elements of the first group react with water liberating
hydrogen and become hydroxides(Except Mg and Be).
Example: Na reacts rapidly with water liberating hydrogen.
-Group I metals are kept under non reactive medium to prevent
them from being react with water vapour in the air.
-When the Mg is reacted with warmed water, it react slowly.
- As the reactivity shown by Mg with water is lower compared to Na,
it can be said that the metals of group II compared to metals of group
I show a lower reactivity.
-Ca, Sr, and Ba react with water liberating hydrogen and forming the
hydroxides.
-Be and Mg react with steam to form the oxides.
Reactions with air
-There are several reactions with air of group I metals.
4Na(s)+O2(s) 2Na2O(s)
2Na(s)+O2(s) 2Na2O2(s)
-With CO2 in the air , Na2CO3 can form
Na2O(s) + CO2(g) Na2CO3(s)
-K, Rb and Cs react forming superoxides.
K(s) + O2(g) KO2(s)
Rb(s) + O2(g) RbO2(s)
Cs(s) + O2(g) CsO2(s)
Oxygen shows oxidation state O-2
in oxides
Oxygen shows oxidation state O-1
in peroxides
Oxygen shows oxidation state O2
-1
in superoxides
- When heated in air only Li of Group I reacts with nitrogen.
6Li(s) + N2(g) 2Li3N(s)

-When a clean piece of Mg ribbon and a small cut piece of Na are
exposed to air Na tarnishes faster than Mg. Hence it is clear that the
reactivity of Mg is lower than Na.
-Accordingly it can be said that relative to metals of group I, the
reactivity of group II metals with air is low
- Metals of the Group II when heated in air burn forming oxides and
nitrides.
2Mg(s) + O2(g) 2MgO(s)
3Mg(s) + N2(g) Mg3N2(s)
-For Be to react it should be heated to a very high temperature.
Reactions with acids
-The metals of s- block elements can act as a reducing agent when
reacting with acids.
-When group I elements react with acids to form particular salt, they
emits a large amount of heat and H2
-Reactions of group I elements with acids
-Reactions of group II elements with acids
P Block elements
-The elements which having
electrons in last shell P orbital,
is known as P-BLOCK
element.
-Their general electronic
configuration is np2
p1-6
Where n= 2 to 6
-P orbital can accommodate a maximum of six electrons.

-The zero group contain Nobel gases
-The total no of p – block elements in the periodic table is 30.
-There are 9 gaseous elements (Ne,Ar,Kr,Xe,Rn,F2,Cl2,O2 and N2)
-Gallium (Ga) and Bromine (Br) are liquids.
-B to At are metalloids.
Chemical properties
Reactions with water
- When the elements of p - block are considered it can be observed
that only the halogens react with water while reactivity decreases
down the group.
-Halogens dissolve in water and the reactivity with water decreases
down the group.
-Fluorine displaces oxygen in water.
2F2(g) + 2H2O(l) 4HF(aq) + O2(g)
-Chlorine reacts with water slowly because the activation energy is
high.
Cl2(g) + 2H2O(l) HCl(aq) + HOCl(aq)
Oxidation no (O) (-1) (+1)
- The halogens except F show a disproportionate reaction with water
as shown above.
Reactions with air
- When the elements of the p - block are considered, Al when heated
in air reacts forming the oxide and releasing a large quantity of heat.
4Al(s) + 3O2(g) 2Al2O3(s)
-Al does not react with air at the room temperature because it is
covered by an oxide layer.
- Carbon burns at high temperature forming CO2
C(s) +O2(g) CO2(g)
-Si forms oxides when heated to a very high temperature.
-N2 reacts with O2 only at very high temperatures to give NO.
N2(g) + O2(g) 2NO(g)

-White phosphorus reacts with air (O2).
In a limited supply of oxygen P4(s) + 3O2(g) 2P2O3(s)
In excess of oxygen P4(s) + 5O2(g) 2P2O5(s)
-S burns in air to form SO2
S(s) + O2(g) SO2(g)
Reactions with acids
-Elements of p block react with acids showing a greater diversity.
-C, S and P react with hot concentrated H2SO4.
- C, S and P react with hot concentrated HNO3
The properties of compounds and their trends associated with s
and p block elements.
-The solubility of a given pair of compounds can be compared in the
following way

-Their solubility can be compared by comparing the change in Gibbs
energy relevant to the above two occasions. Here, the change in
Gibbs energy for MgSO4 is a negative value whereas for BaSO4 it is a
positive value. Therefore, the solubility of MgSO4 which has a more
negative change in Gibbs energy is relatively higher than
that of BaSO4.
-When both values of Gibbs energy are negative values in a pair that
is compared, the solubility of the substance with a larger numerical
value is higher.
-When both values of Gibbs energy are positive in a pair that in
compared, the solubility of the substance with the smaller positive
value in higher
-When comparing the thermal stability of a given pair of compounds
also, the above method is suitable.

-By comparing ∆Gø
1 and ∆Gø
2 , it can be predicted that MgCO3 with
the smaller positive value would be subjected to thermal dissociation
easily.
- The dissociation temperature of MgCO3 is 540 0
C whereas that of
CaCO3 is about 900 0
C.
-Almost all the salts belonging to group I of s block are soluble in
water.
-Information about the solubility of salts formed by group II metals
belonging to the s- block is given in the following table.

- Acidic/basic/amphoteric nature of oxides, hydroxide and hydrides
of elements of the 3rd period
-The acidic nature increases from left to right across the table
Elements and compounds of s and p blocks
Properties of D block elements and compounds
PHOSPHOUROUS
- Allotropic forms of phosphorus
CO3
-
HCO3 NO2
-
NO3
-
S-2
SO3
-2
SO4
-2

.+.
- P4 is stored in water because it reacts with O2 in the air. But
nitrogen exists in the atmosphere as a free gas because of the strength
of N= N bond in it.
- Oxy- acids of phosphorus
Hypo phosphorous acid -H3PO2
Ortho phosphorous acid -H3PO3
Phosphoric acid -H3PO4
OXYGEN
- Allotropic forms of oxygen
:O = O: O
-1/2
:O: :O:-1/2
Oxygen Ozone
SULPHUR
-Allotropic forms of sulphur exist in two forms namely crystalline and
amorphous.
-Crystalline Sulphur
All crystalline forms of sulphur consist of S molecules.

-Amorphous sulphur
Examples are plastic sulphur and colloidal sulphur.
-Oxy - acids of sulphur
HYDROGEN PEROXIDE (H2O2)
-H2O2 as a reducing agent
-H2O2 as an oxidizing agent
SULPHUR DIOXIDE (SO2)
-It is a colorless gas with a higher density than air and with a pungent smell.
-It is very soluble in water.
-SO2 dissolves in water to form sulphuric(IV) acid. It is a weak acid.

SO2(g) + H2O(l) H2SO3(aq)
-SO2 as an oxidizing agent
-SO2 as a reducing agent
-As a bleaching agent
SO2(g) +2H2O(l) H2SO4(aq)+2H+
(aq) + 2e
X + 2H+
+ 2e XH2
(Colored dye) ( Colorless compound)
HYDROGEN SULPHIDE (H2S)
- It is a colorless gas slightly soluble in water.
-It has the smell of rotten eggs.
-An aqueous solution of H2S is somewhat acidic.
- Evidence for the acidic nature
(i) Reactions with sodium
Excess
2H2S(g) + 2Na(s) 2NaHS(s) + H2(g)
2Na(s) + H2S(g) Na2S(s) + H2(g)
(ii) Reaction with sodium hydroxide
When the base is in excess
2NaOH(aq) + H2S(g) Na2S(s) + 2H2O(l)
When H2S is in excess

H2S(g) + NaOH(aq) NaHS(s) + H2O(l)
-H2S reacts with many metal ions to give sulphides. This is used as a test for
identification of metal ions.
Pb2+
(aq) + H2S(g) PbS(s) + 2H+(aq)
Cu2+
(aq) + H2S(g) CuS(s) + 2H+(aq)
-H2S as a reducing agent
2KMnO4(aq)+3H2SO4(aq)+5H2S(g)
K2SO4(aq)+5S(s)+2MnSO4(aq)+8H2O(l)
K2Cr2O7(aq)+4H2SO4(aq)+3H2S(g)
K2SO4(aq)+Cr2(SO4)3(aq)+3S(s)+7H2O(l)
-As3+ reacts with aqueous S2-
ions to form a precipitate.
As3+
(aq) + 3S2-
(aq) As2S3(s)
-H2S as an oxidizing agent
Cu(s) + H2S(g) CuS(s) + H2(g)
Distinguishing between SO2 and H2S
-When the gases are passed through aqueous H+
/K2Cr2O7 , SO2 and
H2S turn the orange colour to green but due to the formation of S
with H2S the solution will not be clear.
-A filter paper moistened with aqueous Pb(CH3COO)2 turns
glistening black with H2S.
-When passed through aqueous H+
/KMnO4
both gases turn the
purple color to colorless but with H2S the solution will not be clear
due to S formed.
-Petals of flowers are bleached by SO2 but not by H2S.

Halogens
F2 - Pale yellow poisonous gas
Cl2 - Yellowish light green poisonous gas
Br2 - Reddish brown liguid
I2 - Shining black solid. Sublimes.
At - A radioactive element.
- Oxy acids of chlorine
-Hydrogen halides (HX)

Noble gases and their compounds
-Boiling points are very low.
-The boiling points increase with the increase in atomic number.
-Polarizability appears in large atoms.
-It was found by 1962 that Xe forms compounds with two
electronegative elements.
Examples : XeF2, XeF4, XeF6, XeO3
Acidity of hydrogen halides in aqueous solution
-Under dry conditions hydrogen halides are not acidic. However,
their aqueous solutions are acidic.
HCl(g) + H2O(l) H3O+
(aq) + Cl-
(aq)
- Hydrofluoric acid is a weak acid and all the other hydrogen halides
are strong acids. The reason for this is the specially strong H - F
bond.
Disproportionation in chlorine
Cl2 (g) + dil. 2NaOH(aq) NaCl(aq) + NaOCl(aq) + H2O(l)
(0) (-I) (+I)
Cl2 (g) + conc. 6NaOH(aq) 5NaCl(aq) + NaClO3(aq) + 3H2O(l)

D-block Elements
-D block elements contain partially filled or fully d sub orbital.
Ionization energy
Oxidation number
- Oxidation numbers of elements from Sc to Zn (Common oxidation
numbers are shown in bold letters)

Instances where d block elements and their compounds are used in
industries as catalysts.

Boiling and Melting points
Properties of compounds of d-block

-Using the solubility in acids and bases of the types of oxides stated
above, they can be classified as acidic, basic or amphoteric.
- The reaction of MnO4 in an acidic medium as an oxidizing agent
with SO2
- The reaction of MnO4
-
as and oxidizing agent in a basic medium
with SO2
-Chromium

Properties of complex compounds of d – block
- Complex ions formed by the elements Cr, Mn, Fe, Co and Cu with
the ligands H2O, NH and Cl –
-The color of the complex varies depending on the central metal
atom.
Examples : [Cr(H2O)6]3+
- Blue - violet
[Fe(H2O)6]3+
- Yellow brown
-The color of the complex varies depending on the oxidation state of
the central metal atom.
Examples : [Fe(H2O)6]2+
- Light green
[Fe(H2O)6]3+
- Yellow brown
[Mn(H2O)6]2+
- Light pink
[Mn(H2O)6]3+
- Violet
-The colour changes also when the ligands change.
Examples : [Co(H2O)6]2+ Pink
[Co(NH3)6]2+ Yellow brown
-The reaction of hydrochloric acid with aqueous Cu2+
ions.
Cu2+
(aq) + 4HCl(aq) [Cu(Cl)4]2-
(aq) + 4H+(aq)
Blue Yellow
-The reaction of hydrochloric acid with Co2+
(aqueous)
Co2+
(aq) + 4HCl(aq) [Co(Cl)4]2-
(aq) + 4H+
(aq)
Pink Blue
-Oxy- anions of chromium and their conversion

CrO-2
4(aq) CrO-2
7(aq)
Yellow Orange
IUPAC nomenclature of complex compounds
(D block)
-The complex compounds are considered simply under two
categories.
-Cations are simple while the anions are complex.
-Cations are complex while the anions are simple.
-Whatever the complex compound considered, common set of rules
has to be followed in nomenclature.
As in the case of a simple inorganic compound, frist the cation
is named and then the anion afterwards. A space is kcpt between the
name of the cation and the name of the anion.
The most important step is the identification of the complex
ion part of the compound. This can be positively or negatively
charged.
In the identified complex ion, the names of the groups or the ligands
surrounding the central metal atom have to be found out. The name
used depends on the charge of these ligands.
1) When there are neutral ligands specific names should be used for
them.
Examples : H2O - aqua
NH3 - ammine
CO - carbonyl
NO - nitrosyl
2) When there are negative ligands the suffix - 'o' is added to their
English name.
Example : Cl-
- chloro
CN -
- cyano
NO2
-
- nitro
OH -
-hydroxo
SCN-
- thiocyanato
H+
- hydrido
O 2
-
- oxo

3) When there are positive ligands the suffix" ium" is added to their
English name.
Example : +
NH3 - NH2- Hydrazinium
When there are more than one ligand of the same type. in
order to indicate the number of such ligands the name of the relevant
number is used as a prefix before the name of the ligand. When there
are 2, 3, 4, 5 and 6 ligands of the same type, the prefixes di-, tri,- terta-
penta- and hexa- are used respectively.
When the complex ion is positively charged or neutral, name
of the metal is used and without leaving a space, the oxidation
number should be shown by a capital Roman numeral within
parantheses after the name.
Examples :
Co3+
- cobalt(III)
Fe2+
- iron(II)
Cr6+
- chromium(VI)
Now the complex ion can be named. Here, the ligands are
named first and the metal next. There should be no spaces between
the words in writing the name.
Examples :
[Co(NH3)6]3+
- hexaamminecobalt(III) ion
[Fe(H2O)6]2+
- hexaaquairon(II) ion
[Cu(NH3)4]2+
-tetraamminecopper(II) ion
When several ligand types are joined to a complex ion in
naming the ligands they should be listed in the English alphabetical
order.
Examples :
[Fe(CN)2 (NH3)4]+
- tetraamminedicyanoiron(III) ion
The complex ion part may be positively charged, negatively
charged or neutral. Depending on this the name also changes.
-When the complex ion part is positively charged or neutral, it is
named using the name of the metal. Here also, it is necessary to
indicate the oxidation number of the metal by a capital Roman
numeral in parantheses.

Example :
[Fe(CN)3 (NH3)4]
This is neutral. Hence its name is tetraamminetricyanoiron.
[Cu(H2O)6 ]2+
This is positively charged. Its name is hexaaquacopper(II) ion
-When the complex ion part is negatively charged, the suffix 'ate' is
added to the end of the name of the metal. Here also the oxidation
number of the metal shoud be indicated by a capital Roman numeral
inside parantheses.
Examples :
[CoCl4]2-
- tetrachlorocobaltate(II) ion
[Co(CN)6]3-
- hexacyanocobaltate(III) ion
[CuCl4]2-
- tetrachlorocuprate(II) ion
[Fe(CN)6]4-
- hexacayanoferrate(II) ion
[Fe(CN)6]3-
- hexacayanoferrate(III) ion
[Ag(CN)2]-
- dicyanoargentate(I) ion
[Cr(Br)6]3-
- hexabromochromate(III) ion
The IUPAC name of any compound can be developed by following
the rules studied so far systematically. Try to do this under two main
groups.
-It should be emphasized how a space is placed between the
positively charged part and the negatively charged part and the name
of the complex part is written as a single word without leaving spaces.
Examples :
K3[Fe(CN)5 NO] -Potassium
pentacyanonitrosylferrate(II)
Na2[ZnCl4] -Sodium tetrachlorozincate(II)
-When there is a complex cation and a simple anion there should be a
space between the positively charged part and the negatively charged
part.
Examples :
[Ag(NH3)2]Cl2 - Diamminesilver(I) chloride
[Fe(OH)2(H2O)4]Br -Tetraaquahydroxoiron(III) bromide
[CoCl(NH3)5](NO3)2 -Pentaamminechlorocobalt(III)nitrate(V)

Writing formula of a complex compound when its name is given.
-As in the general method show the positively charged ion first and
the negatively charged ion afterwards.
-Always the complex ion part of the compound should be written
within square brackets.
-If there is a charge, it should be indicated outside the square bracket
at the upper end of the right side. The sign of the charge and its
numerical value should be given.
Example : [Fe(CN)5NO]
-When the formula of the complex ion part is written first, the metal
should be indicated and then the ligands.
- In writing the ligands, the order of the English alphabet used during
the nomenclature is not used here. Instead ligands are written
according to their charges. That is in the order of negatively charged
ligands first, neutral ligands next and the positively charged ligands
last.
However, the various ligands present in a similar group should be
arranged alphabetically.
This is explained below.

Unit 6 – Organic Chemistry
-Carbon has the ability to form a large number of compounds.
Because the Carbon
-form single, double and triple bonds
-having higher bonding energies. (Higher than Si–Same
group)
Bond Bonding energy- kJmol-1
C-C 346
C=C 610
C=C 835
C-H 413
Si-Si 226
Si-H 318
-Carbon
-Can form 4 covalent bonds.
-Stable in room temperature.
-Reaction of organic compounds with oxygen is highly
Exothermic.
-Carbon compounds are thermodynamically unstable. But
the values of activation energies associated with the reactions
of organic compounds with oxygen are very high. Hence,
according to chemical kinetics, organic compounds are stable
and a large number of them occur naturally.
Functional Groups
-Compound which contain only C and H are known as
hydrocarbons.
-On the basis of the structure, hydrocarbons are divided into two
main groups

-The set of hydrocarbons consisting of open carbon chains only are
named as aliphatic (acyclic) hydrocarbons.
-The aliphatic organic compounds are classified as alkanes, alkenes,
and alkynes.
-The cyclic organic compounds which are stabilized by forming a
delocalized cloud of electrons are called aromatic compounds.
-Benzene which is indicated by the molecular formula C6H6 is the
simplest of aromatic hydrocarbon compounds.
-In many organic compounds, when hetero atoms such as nitrogen
and oxygen combine with the carbon chain, due to the difference in
electro negativity between the carbon and the combined atoms, this
group of atoms will impart the compound a characteristic reactivity.
Such a group of atoms is called a functional group.
-The compounds are classified according to the functional group
present in a molecule.

* In the IUPAC nomenclature, the halogen is not considered as a
functional group
Naming aliphatic organic compounds
-The International Union of Pure and Applied Chemistry -IUPAC,
presented systematic method to name the organic compounds in
1949.
-The name given to a compound according to the IUPAC
nomenclature consists of several parts.

1. The suffix that is used to indicate the main functional
group of the structure
2. Name of the chain that is used to identify the main carbon
chain of the compound
3. The prefixes that are used to indicate the substituent
groups
4. The numbers that are used to indicate the places at which
the substituent groups, additional groups and the main
functional groups are attached to the chain
-In the nomenclature of aliphatic compounds, by following the steps
stated in the given order, the IUPAC name of a compound can easily
be developed.
 Identifying the principal functional group
 Selecting the main chain
 Selecting the root name for the principal chain
 Addition of the suffix for the remaining groups (additional
groups) to the name of the chain
 Addition of the suffix used to indicate the principal functional
group to the name of the chain
 Naming the substituent groups
 Adding the names of the substituent groups to the name of
the chain
 Numbering the carbon chain
 Showing the numbers that are used to indicate the positions
of the main functional group and the substituent groups in
front of these groups.
The root names used for the compounds according to the number
of carbon atoms and the name of corresponding hydrocarbon.

The series of functional groups arranged in the decreasing order of
their priority
Summary of IUPAC Rules for Naming Alkanes
1. Find the longest chain of C atoms. Choose the base name that
describes the number of C atoms in this chain, with the ending -ane .

The longest chain may not be obvious at first if branches of different
sizes are present.
2. Number the C atoms in this longest chain beginning at the end
nearest the first branching. If necessary, go to the second branch
closest to an end, and so on, until a difference is located. If there is
branching at equal distances from
both ends of the longest chain, begin
numbering at the end nearest the
branch that is first in alphabetical
order.
3. Assign the name and position
number to each substituent.
Arrange the substituent’s in
alphabetical order. Hyphenated
prefixes, such as tert- and sec-, are not
used in alphabetization of the
substituents.
4. Use the appropriate prefix to group
like substituent’s: di- = 2,
tri- = 3, tetra- = 4, penta- = 5, and so
on. Don’t consider these prefixes
when alphabetizing attached groups.
5. Write the name as a single word.
Use hyphens to separate numbers and
letters (plus some hyphenated
prefixes) and commas to separate
numbers. Don’t leave any spaces.
Summary of IUPAC Rules for Naming Alkenes and Cycloalkenes
1. Locate the C atoms in the longest C chain that contains the double
bond. Use the base name prefix with the ending -ene.
2. Number the C atoms of this chain sequentially beginning at the
end nearer the double bond. Insert the number describing the
position of the double bond (indicated by its first carbon location)
before the base name. (This is necessary only for chains of four or

more C atoms, because only one position is possible for a double bond
in a chain of two or three carbon atoms.)
3. In naming alkenes, the double bond takes positional precedence
over substituents on the carbon chain. The double bond is assigned
the lowest possible number.
4. To name compounds with possible geometric isomers, consider the
two largest groups within the carbon chain that contains the double
bond—these are indicated as part of the base name. The isomer in
which the largest groups at each end of the CUC are located on
opposite sides is called trans. If the largest groups are on the same
side, the molecule is referred to as cis. Insert the prefix cis- or trans-
just before the number of the double bond to indicate whether the
largest groups are on the same or opposite sides, respectively, of the
double bond.
5. For cycloalkenes, the double bond is assumed to be between C
atoms 1 and 2, so no position number is needed to describe it.

Summary of IUPAC Rules for Naming Alkynes
Alkynes are named like the alkenes except for the following two
points.
 The suffix -yne is added to the characteristic root.
 Because the linear arrangement about the triple bond does
not lead to geometric isomerism, the prefixes cis- and trans-
are not used.
Drawing compound according to IUPAC
-Drawing the structural formula of a compound according to the
IUPAC nomenclature can be done by following the steps given
below.
 Identifying the chain and drawing the chain according to that
name
 Numbering the chain
 Identifying the principal functional group and the remaining
groups according to the IUPAC name given and joining these
groups to the correct places of the chain according to the
number in front of these groups.
 Placing hydrogen atoms in the chain structure so that each
carbon atom has a valency of four.
Isomerism
-Isomers are different compounds that have the same molecular
formula; they have the same number and kinds of atoms arranged
differently.
-There are two major classes of isomers: structural (constitutional)
isomers and stereoisomerism. For coordination compounds, each can
be further subdivided as follows.

STRUCTURAL ISOMERISM
Chain isomerism
-Chain isomerism occurs when the nature of the carbon chain
changes for the same molecular formula in the same homologous
series.
Examples:
Position isomerism
-Though there is the same molecular formula, the same functional
group and the same carbon chain when there is a change in the
carbon atom to which the functional group is attached or a change in
the location of the active position, then there occurs position
isomerism.
Examples:
Functional group isomerism
-Functional group isomerism is the existence of structures with
different functional groups for the same molecular formula.
Examples:

STEREOISOMERS
-Compounds that contain the same atoms and the same atom-to-
atom bonding sequences, but that differ only in the spatial
arrangements of the atoms relative to the central atom, are
stereoisomers. Complexes with only simple ligands can exist as
stereoisomers only if they have coordination number 4 or greater. The
most common coordination numbers among coordination complexes
are 4 and 6, and so they will be used to illustrate stereoisomerism.
Geometric (cis–trans) Isomers
- Geometric isomerism is one occasion where diastereo isomerism is
seen.
- In a C = C double bond due to the bond which exists in addition to
the σ bond, these carbon atoms cannot freely rotate about the σ bond.
Therefore, although the carbon atoms are bonded in the same way,
due to their three dimensional position, different configurations can
exist. This is referred to as geometrical isomerism.
-The different structures obtained here are called geometrical
isomers.
- Geometrical isomers can be identified by the structural formula of
the alkenes.
- If a ≠ b and p ≠ q, that compound exhibits geometrical isomerism. -
-In any of the above occasions if two identical groups are on the same
side with reference to the axis of the double bond, that structure is

called the cis- isomer and if the groups are on opposite sides it is
called the trans - isomer.
-When all the four groups a, b, p and q are different also it shows
geometrical isomerism. But these isomers cannot be named as cis or
trans. The nomenclature of these compounds is not relevant to the
syllabus.
Enantiomer isomerism
-The isomers of which one is the mirror image of the other are known
as enantiomers
-A compound having a carbon atom which is joined to four different
groups shows enatiomerism.
-When plane - polarized light is passed through a solution containing
only one enantiomer, the plane of polarization rotates. One
enantiomer rotates the plane of polarization in one direction and the
other enantiomer in the opposite direction. As the enantiomers rotate
the plane of polarization, they are also known as optically active
isomers.

Unit 7 – Hydrocarbons
Aliphatic Hydrocarbons.
Alkanes
-This type contain C-C single bond.
- If two consecutive members differ only by a CH2 unit, such a series
of compounds is called a homologous series.
-An alkane molecule is non polar or very weakly polar. The attractive
forces between two non -polar molecules are the very weak Van der
Waals forces.
-The first few members of the series are gases at room temperature.
-Liquid and solid members are met on going down the series.
-With the branching of the carbon chain Van der Waals forces
become weak
When alkane members with 5 carbon atoms are considered:
Alkenes
- The general formula CnH2n+2
- The simplest alkenes contain one carbon–carbon double bond,
C=C, per molecule. The general formula for noncyclic alkenes is
CnH2n. The simplest alkene is C2H4, ethene, which is usually called
by its common name, ethylene.
- In alkenes also the physical properties are very much similar to
those of alkanes.
-The hybridization (sp2
) and bonding at other double-bonded carbon
atoms are similar. Both carbon atoms in C2H4 are located at the
centers of trigonal planes. Rotation about C=C double bonds does not
occur significantly at room temperature.

Alkynes
- As the polarity of alkynes is also low, their physical properties, are
very similar to those of the corresponding alkenes and alkynes.
- The alkynes, or acetylenic hydrocarbons, contain one or more
carbon–carbon triple bonds, -C=C-. The noncyclic alkynes with
one triple bond per molecule have the general formula CnH2n-2.
The bonding in all alkynes is similar to that in acetylene.
Aromatic hydrocarbons
Benzene
- Benzene is the simplest aromatic hydrocarbon.
-Under the normal conditions benzene does not answer the tests for
unsaturation.
-Therefore, benzene cannot have a structure similar as that of a
simple alkene or an alkyne.
-Although the structure proposed for benzene by Kekule showed
three double bonds for the molecule, benzene is not similar to the
Kekule structure.
- Elemental analysis and determination of its molecular weight
showed that the molecular formula for benzene is C6H6. The formula
suggests that it is highly unsaturated. But its properties are quite
different from those of alkenes and alkynes.
- The bond length between two carbon atoms in benzene is the same
and its value is 1.39 x 10-10
m. Further, the benzene molecule is
completely planar.
-But the length of a C=C double bond is 1.34x 10-10
m and the length
of a C – C single bond is 1.54 x 10-10
m.
-Therefore it is clear that the structure of benzene should be a hybrid
of the resonance structures given below.

-The double – headed arrow, does not indicate that benzene changes
between these two structures or that it is in equilibrium. It shows
resonance.
-The structure of benzene is very clearly explained by the molecular
orbital theory
-All its C atoms have undergone sp2 hybridization.
Here, as the electrons in the unhybridized p- orbitals exist as
conjugate double bonds, they can overlap with the
unhybridized p orbitals present on both sides.
-From this, there is the ability to form a circular electron
cloud .There fore, due to the circular delocalization of the six
'p' electrons, the resonance structure of benzene is relatively
more stable than the structure with three real triple bonds.
-The data for the standard enthalpy of hydrogenation also help to
explain the stability of a benzene molecule.
+H2 ∆Hø
= -120 kJmol-1
C6H6 + H2 ∆Hø
= -207 kJmol-1
-But if benzene possesses three double bonds, its standard enthalpy of
hydrogenation should be 3 x (-120 kJ mol-1
), that is -360 kJ mol-1
.
Hence, it is clear that benzene is more stable than its Kekule
structures by an amount equal to (360-208) = 152 kJmol-1

-Therefore, benzene doesn’t contain C=C 3 bonds. the structure is
drawn as below,
Chemical reactions of Alkanes, Alkenes and Alkynes
Reactions of Alkanes
-The most important reaction in alkanes is combustion.
-Inn alkanes there’s no special functional group.
-All alkanes are covalent bonds
-Alkanes doesn’t react with common lab reagents. (CN-
,OH-
)
-Polarity is very low
-As by-products they form CO2 and H2O
2C8H18 + 25O2 8CO2 + 9H2O ∆Hø
= -11020kJmol-1
-The hydrocarbon gasoline (petrol) boils at 1500
C
The best fuel for best power is 2,2,4,Trimethylpentane ,This is known
as iso octane and has assigned octane no 100.
Heptane
-This fuel has low power and octane no is 0.
-Octane number is calculated by the percentage of
2,2,4,Trimethylpentane contain in fuel.
-Alkanes react with Cl2 , Br2 etc which undergo hemolytic fission.

-Homolysis of C-H bond is the first task of free radical generation.
During this process free radical containing a carbon is generated.
-The probability for a Cl free radical to collide with CH4 molecule is
high at the beginning and the before CH3Cl is formed.
-Because the termination step of the chain can also take place in
between the course of the reaction it is necessary to supply sunlight
for the reaction to proceed continuously.
Halogenations of alkanes
-The density of π electrons which lie above and below the plane of
the ethylene molecule is capable of attracting electrophilic reagents.

-When an electrophile gets attracted to the π electron cloud and
bonds with a carbon, the other carbon will be attached only to three
groups and hence another bond will be formed to complete the
valency.
Reactions of Alkenes
-Alkenes react with hydrogen halides. Here, the hydrogen ions act
like electrophiles and attack the double bond. During these
electrophilic addition reactions, intermediate carbocations are formed.
- Stability of carbocations is tertiary > secondary > primary.
R R H
R C+
R C+
R C+
R H H
- When alkyl groups are attached to the positively charged C atom of
the carbocation, the stability of the cation increases. The reason for
this is the repulsion of electrons by the alkyl groups through the
bonds towards the positively charged carbon atom to which they are
attached. Here what happens is the stabilization as a result of the
positive charge spreading completely throughout the ion (the alkyl
groups are represented by R groups).
-The are two types of alkyles
i) A symmetrical
ii) Symmetrical
Symmetrical Asymmetrical(Non alkyl)
- In the electrophilic addition reactions of a symmetric alkenes with
hydrogen halides, two asymmetric carbocations can be formed after
the bonding of the electrophile.
-Out of these the more stable carbocation forms easily.
-According to the stability of carbocations, the more stable
carbocation is obtained when the electrophile gets attached to the
carbon atom to which the highest numbers of hydrogen atoms are
attached.
-After studying the reactions of a large number of alkenes, this
observation was generalized as Markownikoff’s rule.

-Of the hydrogen halides, only hydrogen bromide adds in the
opposite way to this rule when there are peroxides in the reaction
medium. The reason for this is that in the presence of peroxides the
reaction of hydrogen bromide and the alkenes takes place via a free
radical mechanism.
Mechanism of addition of bromine to alkenes
-Due to the electron density present on the two sides of the plane of
the alkene, an induced dipole forms in the bromine molecule. It is
this induced dipole that acts as the electrophile.
Addition of sulphuric acid to alkenes and the hydrolysis of the
product obtained.
-In the presence of Hg 2+ and dilute sulphuric acid , one molecule of
water gets added on to alkynes.
-The immediate reorganization of the oxygen into the alkehyde is due
to the high stability of C = O

Reaction of alkenes with cold alkaline KMnO4
Catalytic hydrogenation of alkenes
-Under the normal conditions, hydrogen and alkenes do not react.
But in the presence of finely divided Pt/Pd/Ni metal catalysts
alkenes react with hydrogen to form the relevant alkanes.
Acidic nature of alkynes
- In the alkynes, H-C=C-H and R-C=C-H the H attached to the C
that forms the triple bond (terminal hydrogen) shows acidic
properties.
-The acidic H in these alkynes can be displaced by metals.
H-C=C-CH3
Na
Na+-
C =C-CH3 + H2
H-C=C-CH3
NaNH
3 Na+-
C =C-CH3 + NH3
H-C=C-CH3
NH
3
/CuCl
2 Cu-C=C-CH3
H-C=C-CH3
NH
3
/AgNO
3 Ag-C=C-CH3
Reaction types that Organic compounds can show
-Substitution reactions
An atom or a group replaces another.
-Addition Reactions
The molecule reacts to form another.
-Elimination Reactions
One molecule reacts to form another one or more.
-Rearrangement Reaction
Rearranging , whenever the structure permits.
Types of reagents
-In a covalent bond between A and B , If A is more electronegative ,
the distribution of electrons can be shown as A-β
– B- β
-Due to that one side of the molecule become slightly polarized.
-There are two types of reagents

1) Neucleophillic reagents
-This type of reagents like positive charge
-They attack the centre of the positive charge
Example- OH-
,CN-
2)Electrophilic reagents
-They attack the place where the electron density is high.
Example- NO2
+
, SO3
+
Benzene
-In benzene there’s a high electron density in both sides of the ring.
-Due to that electrophilic reagents get attached on to it and form a
carbo cation.
-A electron deficiency occurs and H+
is eliminated to obtain cyclic
delocalized electron clue.
NITRATION
-Replacement of –H by –NO2 using NO2
+
(Nitronium) as an
electrophile is known as nitration.
-Nitronium ion is formed by using nitrating mixture of conc H2SO4
and conc HNO3 .
- Conc H2SO4 being a strong acid than HNO3 pretending to
protonation of hydroxyl oxygen on HNO3
-Mechanism
H2SO4 + H-O-NO2 HSO4
-
+ H O+
NO2
H
NO2
+
NO2 + H+
H+
+ HSO4
-
H2SO4
+ A+
A
H
A
+ H+

CH3
FREDEL-CRAFTS ALKYLATION
-During the reaction of benzene with alkyl halides, reaction requires a
catalyst in form of Lewis acid.
-Anhydrous AlCl3 is a strongest Lewis acid it pulls the halide in
maximum strength.
+ R-X
-Mechanism
R-Cl + AlCl3 R-Cl+
-AlCl3
-
R-Cl+
+ AlCl3 R+
+AlCl4
Consider R as CH3
CH3
+
-Cl- ++gg
[AlCl4]-
HCl + AlCl3 +
HALOGENATION
-When benzene react with a halogen carrier, a substitution of halogen
in benzene ring take place.
-Halogen molecule is polarized by the halogen carrier, the molecule
function as a electrophile.
-Mechanism is similar to above diagram
OXIDATION
-Benzene doesn’t get oxidized by normal oxidizing agents like
KMnO4/H+
.
-But after a substitution it reacts.
Anhydrous AlCl3
R
CH3
H
+
+
CH3
H
Dry FeBr3/Febr2
Br
COOH
KMnO4/H+

Ni/1500
c
-Tertiary alkyl groups doesn’t oxidize under normal condition in
which primary and secondary alkyl groups get oxidized.
-Only in higher reactions they get oxidize.
ELECTROPHILIC ADDITION RERACTIONS
-Although, alkenes undergo electrophilic reactions easily Benzene
doesn’t show such reactions.
-But in presence of a catalyst at about 150o
C it react woth H2 gas to
form Cyclohexane.
+ H2
DIRECTING ABILITY OF A SUBSTITUTE GROUP TO A MONO
SUBSTITUTENTED BENZENE
Orth,Para directing groups
-If second substitution group get attached to Ortho or Para position
relative to first group, it’s a orthopara directing group.
-Ortho , Para directing groups are –OH , -R , -NH2 , -OCH3 ,
Halogens
Meta directing groups
-If substitutional group is a electron attracting group , it deactivate
the ring , due to that second substitutional group take place in meta
position
- Meta directing groups are –NO2, -CHO , -COR, -COOH, -COOR

Unit 8 – Alkyl Halides
Alkyl Halides
-Alkyl halides are named as primary, secondary or tertiary depending
on the nature of the carbon atom which carries the halogen atom.
- Alkyl halides are polar compounds.
-Solubility of alkyl halides is water is very low. One reason for this is
that they do not form hydrogen bonds with water.
-Due to the high eletronegativity of the halogen atom relative to the
carbon atom, the C – X bond gets polarized. As a result, there is a
deficiency of electrons in that carbon atom. Therefore, the
nucleophiles attack this position.
Neucleophillic substitution reactions
-Nuclephilic substitution reactions are characteristic of alkyl halides.
Here, the carbon atom forms a new bond with the nuclephile and the
halogen atom leaves as a halide ion.
- Any nucleophile can also act as a base. Therefore, there is an ability
to subject alkyl halides to an elimination reaction by removing an
acidic hydrogen atom.
-Although the hydrogen atoms attached to the carbon atom to which
the halogen atom is attached are more acidic in nature, a stable
compound is not formed by their elimination. Therefore, the
hydrogen atoms with a slight acidity attached to the adjoining carbon
atoms participate in elimination reactions.
H H H H
OH + H- C – C – X C = C + H2O +X
H H H H

Chlorobenzene and vinyl chloride do not undergo nucleophilic
substitution under the conditions which alkyl halides undergo these
reactions. As reasons for this, mention the double bond nature of the
C- X bond and the decrease in bond length due to the existence of
the carbon atom on the sp2
hybridization in these compounds.
GRIGNARD REAGENT
-Alkyl halides react with Mg in the medium of dry ether to form the
Grignard reagent.
- Due to the polarization of bonds in RMgX, the carbon attached to
magnesium acts as a strong nucleophile and a very strong base.
-Therefore, to prevent the formation of alkanes by the donation of
protons by water a dry ether medium is used.
The strong nucleophilic features of the Grignard reagent can be
shown by the following reactions.
PROPERTIES OF ALKYL HALIDES
- To explain the nucleophilic substitution reaction of alkyl halides,
the time interval between bond breaking and bond making steps can
be considered.
-When the breaking of the bond and the formation of the new bond
take place simultaneously, the nucleophilic substitution reaction of
the alkyl halide is considered as a one step reaction.
-Accordingly, the one step reaction can be presented as follows:

- When the formation of the new bond takes place after the breaking
of the bonds the nucleophlic substitution reaction of the alkyl halide
is considered as a reaction that takes place by two steps.
- Accordingly, the reaction that takes place by two steps can be
presented as follows:
- The reaction that takes place by two steps goes through an
intermediate carbocation.
-On considering the stability of the carbocation formed, the tertiary
alkyl halides which are able to from a more stable tertiary carbocation
undergo nuclephilic substitution in two steps.
-The primary alkyl halides undergo nucleophilic substitution
reactions in one step as they are unable to form a stable intermediate
carbocation.
-The procedure adopted by secondary alkyl halides will be
determined by the nature of the reaction medium and the reaction
conditions supplied.

Unit 9 – Oxygen containing
organic compounds
- Monohydric alcohols can be classified into three types as primary,
secondary and tertiary
Physical properties
-In alcohols the –OH bond polarizes as R – 0 β-
- Hβ+
. Hence, due to
the inter molecular hydrogen bonds formed between alcohol
molecules, their boiling points have higher values compared to the
alkenes and ether with comparable relative molecular masses.
-The boiling point increases in going down the alcohol series.
-The above diagram shows how the intermolecular hydrogen bonds
exist in ethanol.
-Alcohols are soluble in water. The solubility of alcohols in water
which is a polar solvent is facilitated by the – OH group. The non
polar alkyl group in the alcohol molecule is a hindrances to the
solubility in water.
- In going down the homologous series of alcohols the size of the non-
polar alkyl group gradually increases compared to the –OH groups.
Accordingly the solubility in water gradually decreases.

REACTIONS INVOLVING CLEAVAGE OF THE O-H BOND
Reaction with sodium
- Alcohols behave as weak acids and react with sodium liberating
hydrogen and forming sodium alkoxides. The alkoxide ion is a strong
nucleophile and also a strong base.
Reaction with carboxylic acids (Acylation of alcohols)
Where G represent reversibility
- Alcohols react with carboxylic acids to form esters. For this
esterification reaction, concentrated H2SO4 acid acts as a catalyst.
- Nucleophilic substitution reactions that take place by the cleavage of
C-O bond in alcohols
-Alcohols react with PCl3 or PCl5 to give alkyl chlorides.
- Reaction with alkyl halides

Alcohols react with HBr to give corresponding alkyl bromides.
Here the Br- ion from HBr is the nucleophilic reagent.
Reaction with anhydrous ZnCl2 and conc. HCl (Lucas test)
-Here, ZnCl2 acts as a catalyst. Due to the alkyl halide formed as the
product, a turbidity appears in the reaction medium. In relation to
the time taken for turbidity to appear, the rates of reaction of primary,
secondary and tertiary alcohols with the Lucas reagent can be
compared.
-This test is restricted to primary, secondary and tertiary alcohols
which are soluble in water.
-The reaction rate of alcohols with the Lucas reagent is a follows:
-Under the supplied conditions the above reaction takes place in two
steps. Here, the tertiary alcohol forms a more stable intermediate and
therefore the tertiary alcohol in the presence of the Lucas reagent
forms a turbidity in a very short time.
Alcohols undergo an elimination reaction with conc. H2SO4
(Dehydration of alcohols)
-The reaction in which a molecule of water is eliminated from an
alcohol is the dehydration of alcohols. Here, an alkene is formed as
the product of the reaction.
Oxidation of alcohols

The product of oxidation depends on primary, secondary or tertiary
nature of the alcohol.
Primary alcohols
Secondary alcohol
-Normally the tertiary alcohols do not undergo oxidation in the
presence of condition that oxidize primary and secondary alcohols.
- It is because in tertiary alcohols due to the lack of hydrogen on the
carbon attached to the oxygen, it is the carbon - carbon bond that has
to be broken.
-Oxidation of alcohols is carried out with H+
/KMnO4 or
H+
/K2Cr2O7 or H+
/CrO3.
-In the presence of the above reagents the aldehydes undergo further
oxidation very easily. Therefore, it is not easy to obtain aldehydes
using these reagents. When there is a need to obtain the aldehyde,
PYRIDENIUM CHLOROCHROMATE [C5H5NH.CrO3Cl] can
be used.
-When primary alcohols are oxidized in the presence of pyridenium
chlorochromate the corresponding aldehyde is obtained as the
product. Here it does not get further oxidized to carboxylic acid.
Phenol
-In aromatic compounds, the compounds in which an -OH group is
joined to a carbon atom of the benzene ring instead of a hydrogen
atom are called phenols.
-The phenols are more acidic relative to the alcohols. It means that
the equilibrium point of alcohols shown above is more shifted
towards the right.
-The reason for this is that the stability of phenoxide ion relative to
phenol is greater than the stability of the alkoxide ion relative to the
alcohol.

-The phenoxide ion is more stable because its negative charge gets
delocalized by resonance. In alkoxide ion there is no such charge
dispersion.
-The higher acidity of phenols is confirmed by the following
examples too.
- Although an alcohol reacts with sodium it does not react with
NaOH. But phenol reacts with sodium as well as with NaOH
2C2H5OH + 2Na 2C2H5O-+
Na + H2
C2H5OH + 2NaOH No reaction
2C6H5OH + 2Na 2C6H5O-+
Na + H2
C6H5OH + 2NaOH 2C6H5O-+
Na + H2O
-Because the carbon atom in phenol exists in the sp2
hybridization
state and due to the double bond nature developed as a result of
resonance in the bond that oxygen forms with the π electron cloud
of the benzene ring, the carbon- oxygen bond in phenol is shorter
than the carbon- oxygen bond in an alcohol.
-Therefore, unlike alcohols phenols do not undergo nucleophilic
substitution reactions.
Effect of the -OH group on the reactivity of the benzene ring in
phenol
-Due to the delocalization of the lone pairs of electrons which were on
the oxygen atom of phenol in the benzene ring, the ring has become
very reactive towards electrophilic reagents.
-The O-H group of phenol is ortho, para directing.
-When the electrophilic substitution reactions of phenol are
compared with the corresponding reactions of benzene along with the
relevant conditions, it is clear that the benzene ring of phenol, had
become more reactive towards electrophiles.
-Consider the following example
(i) Reacts immediately with bromine water to give a white
precipitate of 2,4,6 –tribromophenol.

Chemistry Book By Supun Ayeshmantha

Chemistry Book By Supun Ayeshmantha

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