6.0 Chemical Equilibrium 5
7.0 Ionic Equilibria 12
12.0 Hydrocarbons 8
15.0 Hydroxy compounds 3
18.0 Amines 5
20.0 Polymers 1
Chapter Topic Hour  
Paper 2 (Part A- structured) (Part B-long structured) -100%
 
CHEMISTRY ,9th Ed. – Raymond Chang, McGraw-Hill
CHEMISTRY –The Molecular Nature of Matter and Change, 3rd Ed.–  Martin Silberberg, McGraw Hill
CHEMISTRY – The Central Science, 9th Ed. Theodore L.Brown, H.Eugene LeMay,Jr, Bruce E Bursten, Pearson Education
 
08/16/1108/16/11 matter matter  77
GENERAL CHEMISTRY – Principle and Modern  Applications, 8th Ed. Ralph H. Petrucci, William S. Harwood, Prentice-Hall
ORGANIC CHEMISTRY, 7th Ed – T.W.Graham Solomon,Craig B.Fryhle, John Wiley and Sons
ORGANIC CHEMISTRY, 4th Ed – L.G. Wade, Jr , Prentice Hall
ORGANIC CHEMISTRY, 6th Ed – John McMurry, Thompson – Brooks/Cole
 
able to:
charge.
number, A and isotope.
(c) Write isotope notation.
mass.
ions.
 
 A substance is a form of matter that has a definite or constant composition and distinct properties.
Example: H 2 O, NH
3 , O
2
 A mixture is a combination of two or more substances in which the substances retain their identity.
Example : air, milk, cement
08/16/1108/16/11 matter matter  1313
 
 An element is a substance that cannot be separated into simpler substances by chemical means.
Example : Na, K, Al,Fe
 A compound is a substance composed of atoms of two or more elements chemically united in fixed proportion.
Example : CO 2 , H
element/compound.
- Proton (p)
- Neutron (n)
- Electron (e)
Electrons move around the region of the atom.
 
 All neutral atoms can be identified by the number of protons and neutrons they contain.
Proton number (Z) is the number of protons in the nucleus of the atom of an element (which is equal to the number of electrons). Protons number is also known as atomic number .
 
Neutron (n) 1.67 x 10-24 0 0
 
element that have the same proton number in
their nucleus but different nucleon number .
 
 A = Nucleon number of X
= p + n
notation ( atomic symbol )
electrons
number of protons is more than number of
electrons
electrons
and charge in each of the following species:
Hg200
80
Cu63
29
 217
 A 2 2 2
B 1 2 0
C 1 1 1
D 7 7 10
following nuclide :
atoms joined together by bonds.
b) Molecules
 A polyatomic molecule
 
 At the end of this topic, student should be able
to :
relative molecular mass, Mr based on
the C-12 scale.
element given the relative abundance of
isotopes or a mass spectrum.
 
 A mass of one atom of an element
compared to 1/12 mass of one atom of 12C
with the mass 12.000 amu
Cofatomoneof MassX 12
mass unit, amu (or u).
 Atomic mass unit, amu is defined to be one
twelfth of the mass of 12C atom
Mass of a 12C atom is given a value of exactly 12
amu
The relative isotopic mass is the mass of an
atom, scaled with 12C.
Determine the relative atomic mass of an
element Y if the ratio of the atomic mass of Y to
carbon-12 atom is 0.45
 A mass of one molecule of a
compound compared to 1/12 mass of one
atom of 12C with the mass 12.000amu
C of  atom one of  Mass  x 12
1
relative atomic masses of all atoms in
a molecular formula.
C5H5N,
 An atom is very light and its mass cannot be
measured directly
measure the precise masses and relative
quantity of atoms and molecules
 
Modern mass
spectrum converts
shows that Mg consists of 3
isotopes: 24Mg, 25Mg and 26Mg.
The height of each line is
proportional to the abundance
the 3 isotopes
   b   u   n
63
Learning Outcomes
 At the end of this topic, student should be able At the end of this topic, student should be able
to :
element given the relative abundances of
isotopes or a mass spectrum.
 
 
spectrum.
environment is the relative atomic mass, A r  of
the atom.
from the mass spectrum.
 
69.09% of 63Cu and 30.91% of 65Cu. The
isotopic masses of 63Cu and 65Cu are 62.93
u and 64.93 u respectively. Calculate the
relative atomic mass of copper.
 
 
of two isotopes, 191Ir and 193Ir in the ratio of
5:8. The relative isotopic mass of 191Ir and 193Ir are 191.021 u and 193.025 u
respectively. Calculate the relative atomic
mass of Iridium


 
in a mass spectrometer. How many peaks
would be expected in the mass spectrum of
chlorine?
spectrum of X2 which consists of 3 isotopes?
 
able to:
carbon-12 and Avogadro’s constant, NA
 
to count
to count chemical substances by weighing
them
exactly 12.00 g of carbon-12 isotope
The number of atoms in 12 g of 12C is called
Avogadro’s number, NA = 6.02 x 1023
 
O atoms
N atoms
H atoms
6.026.02 10102323
Cl- ions
6.02 1023
Mass of 1 mol C atoms = 12.01 g
Mass of 1 mol C atoms consists of 6.02 x 1023 C atoms
= 12.01 g
= 1.995 x 10-23 g
1 amu =
The mass of 1 N atom =
The mass of 1 mol of N atoms =
The molar mass of N atom =
The molar mass of nitrogen gas =
The nucleon number of N =
14.0114.01
1414
The mass of 1 mol of CH4 molecules =
The molar mass of CH4 molecule =
16.0516.05
 
able to:
(a) Interconvert between moles, mass, number of particles, molar volume of gas at STP and room temperature.
(b) Define the terms empirical & molecular formulae
 
3.011 x 1023 molecules of oxygen gas.
Solution:
= 0.5000 mol of O2 molecules
1 mol of O1 mol of O22 moleculesmolecules
molecules106.02
1.204 x 1023 molecules of nitrogen gas.
Solution:
2 mol of N atoms
1.204 x 1023 molecules of N2
= 0.4000 mol of N atoms
1 mol of N1 mol of N22 moleculesmolecules
molecules
Solution:
= 0.25 mol x (2 x 35.45 g mol-1)
= 18 g
methane, CH4
7.528 x 1023 CH4 molecules
= 20.06 g
at the same temperature and pressure contain
equal number of molecules
Molar volume is a volume occupied by 1 mol of gas
At standard temperature and pressure (STP), the
molar volume of an ideal gas is 22.4 L mol1
Standard Temperature and Pressure
0 C 101325 N m-2
101325 Pa
volume of a gas = 24 L mol-1
 
08/16/1108/16/11 MATTERMATTER 6969
Examp le 1: 
Calculate the volume occupied by 1.60 mol ofCalculate the volume occupied by 1.60 mol of
ClCl22 gas at STP.gas at STP.
Solution:
= 35.8 L= 35.8 L
08/16/1108/16/11 MATTERMATTER 7070
Exampl e 2: 
Calculate the volume occupied by 19.61 g ofCalculate the volume occupied by 19.61 g of
NN22 at STPat STP
Solution:
1 mol of N1 mol of N22 occupiesoccupies 22.4 L22.4 L
of Nof N22 occupiesoccupies
= 15.7 L= 15.7 L
08/16/1108/16/11 MATTERMATTER 7171
Examp le 3: 
0.50 mol methane, CH0.50 mol methane, CH44 gas is kept in a cylinder atgas is kept in a cylinder at
STP. Calculate:STP. Calculate:
(a)(a) The mass of the gasThe mass of the gas
(b)(b) The volume of the cylinderThe volume of the cylinder
(c)(c) The number of hydrogen atoms in the cylinderThe number of hydrogen atoms in the cylinder
Solution:
(a)(a) Mass of 1 mol CHMass of 1 mol CH44 ==
Mass of 0.50 mol CHMass of 0.50 mol CH44 ==
= 8.0 g= 8.0 g
(b)(b)At STP;At STP; 1 mol CH1 mol CH44 gasgas occupiesoccupies
0.50 mol CH0.50 mol CH44 gasgas occupiesoccupies
= 11 L= 11 L
(c)(c) 1 mol of CH1 mol of CH44 moleculesmolecules 4 mol of H atoms4 mol of H atoms
0.50 mol of CH0.50 mol of CH44 moleculesmolecules   2 mol of H atoms2 mol of H atoms
1 mol of H atoms1 mol of H atoms  
2 mol of H atoms2 mol of H atoms  
1.2 x 101.2 x 102424 atomsatoms
22.4 L
mol1
mol0.50L4.22  
Exercise 
A sample of CO2 has a volume of 56 cm3 at STP. Calculate:
a) The number of moles of gas molecules (0.0025 mol)
a) The number of CO2 molecules (1.506 x 1021 molecules)
a) The number of oxygen atoms in the sample (3.011x1021atoms)
Notes: 1 dm3 = 1000 cm3
1 dm3 = 1 L
elements in a molecule.
the actual number of atoms of each
element in a molecule.
and molecular formula is :
 
carbon and 14.3% hydrogen by mass. Its
molar mass is 56. Determine the empirical
formula and molecular formula of the
compound.
 
Exercise:
A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the sample is 148, what is the molecular formula of the sample?
Answer : C6H12O4
08/16/1108/16/11 MATTERMATTER 8080
At the end of this topic, students should beAt the end of this topic, students should be
able to:
(a)(a) Define and perform calculation for each offor each of thethe following concentration measurements :following concentration measurements :
i) molarity (M)
Lear ning Out come 
A solution is a homogeneous mixture of
two or more substances:
in various ways :
a) Molar i t y
Molarity is the number of moles of solute in 1 litre of solution
Units of molarity: mol L-1
mol dm-3
 
containing 29.22 g of sodium chloride, NaCl
in a 2.00 L solution.
 
Example 2:
How many grams of calcium chloride, CaCl2 should be used to prepare 250.00 mL
solution with a concentration of 0.500 M
 

massmolarxn CaClofmass 2CaCl2 
= 13.9 g
b) Mola l i t y 
Molality is the number of moles of solute dissolved in 1 kg of solvent
Units of molality:mol kg-1
 
prepared by dissolving 32.0 g of CaCl2 in
271 g of water?
prepared by dissolving 24.52 g of sulphuric
acid in 200.00 mL of distilled water.
(Density of water = 1 g mL-1)
Ans = 1.250 mol kg-1
c) Mole Fr act i on (X) 
Mole fraction is the ratio of number of moles of one component to the total number of moles of all component present.
For a solution containing A, B and C:
T
A
CBA
(solution) is equal to one.
XA + XB + XC + X….. = 1
Mole fraction hasMole fraction has no unit (dimensionless)
since it is a ratio of two similar quantities.
 
5 OH contains
Calculate the mole fraction of 
(a) ethanol
(b) water
percentage of the mass of solute per mass
of solution.
100x solutionofmass
soluteofmass w
chloride, KCl is dissolved in 54.362 g of
water. What is the percent by mass of
KCl in the solution?
sodium chloride, NaCl in 100.00 mL of
water. Calculate the mass percent of
sodium chloride in the solution.
Answer = 4.0%
Percentage by volume is defined as the
percentage of volume of solute in milliliter per
volume of solution in milliliter.
Note:
solutionofvolume
of acetone. Calculate the volume percent
of benzene solution.
as 35.5% (v/v) ethanol. How many
milliliters of ethanol does the solution
contain?
solution has has a density of 1.33 g mL-1
at 20 ºC. Calculate the concentration
NaOH in:
(a) molarity
NaOH
V
n  
6.25 m of NaOH  there is 6.25 mol of NaOH in 1 kg of water
for a solution consists of 6.25 mol of NaOH and 1 kg of water;
Vsolution = solution
= 6.25 mol  (22.99 + 16.00 + 1.01) g mol 1
  = 250 g
= 1250 g
nwater   = water of massmolar 
NaOH
massmass
mass
  100
ammonia has a density of 0.9651 g mL-1.
Calculate the
(a) molality
(b) molarity
b) 4.53 mol L-1
Learning Outcome
At the end of the lesson, students should be able to:
a) Determine the oxidation number of an
element in a chemical formula.
b) Write and balance :
 
A chemical equation shows a chemical
reaction using symbols for the reactants and
products.
products are on the right.
 
Reactants Products
08/16/1108/16/11 MATTERMATTER 116116
A chemical equation must have an equal number of atoms of each element on each side of the arrow
The number x, y, z and w, showing the relative number of molecules reacting, are called the stoichiometric coefficients.
A balanced equation should contain the smallest possible whole-number coefficients
The methods to balance an equation:
a) Inspection method b) Ion-electron method
 
Inspect i on Met hod 
1. Write down the unbalanced equation. Write the correct formulae for the reactants and products.
1. Balance the metallic atom, followed by non- metallic atoms.
1. Balance the hydrogen and oxygen atoms.
 
inspection method.
 
inspection method.
 
The substance undergoes oxidation
increase in oxidation number
Half equation representing oxidation:
The substance undergoes reduction
decrease in oxidation number
Half equation representing reduction:
determined by applying the following rules:
1. For monoatomic ions,
e.g: ion oxidation number
08/16/1108/16/11 MATTERMATTER 124124
2. For free elements, e.g: Na, Fe, O2, Br2, P4, S8
oxidation number on each atom = 0
1. For most cases, oxidation number for
O = -2
H = +1
Halogens = -1
Exception:
1. H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1
1. Halogen bonded to oxygen (e.g: Cl2O7) oxidation number for halogen = +ve
1. In a neutral compound (e.g: H2O, KMnO4) the total of oxidation number of every atoms that made up the molecule = 0
1. In a polyatomic ion (e.g: MnO4 -, NO3
 
Exercise 
1. Assign the oxidation number of Mn in the following chemical compounds.
i. MnO2 ii. MnO4 -
1. Assign the oxidation number of Cl in the following chemical compounds.
i. KClO3 ii. Cl2O7 2-
1. Assign the oxidation number of following:
i. Cr in K2Cr2O7
 
Redox reaction may occur in acidic and basic solutions.
Follow the steps systematically so that
equations become easier to balance.
 
Solut i on 
reactions: reduction reaction and oxidation reaction
i. Fe2+ → Fe3+
ii. MnO4 - → Mn2+
1. Balance atoms other than O and H in each
half-reaction separately
Add H+ to balance the H atoms
i. Fe2+ → Fe3+
ii. MnO4 - + → Mn2+ +
charges
4H2O8H+
3.Multiply each half-reaction by an integer, so that
number of electron lost in one half-reaction equals the number gained in the other.
i. 5 x (Fe2+ → Fe3+ + 1e)
5Fe2+ → 5Fe3+ + 5e
 
sides of the equation.
i. 5Fe2+ → 5Fe3+ + 5e
 ___________________________________
and the same total charge on both sides.
5Fe2+ + MnO4 - + 8H+
→ 5Fe3+ + Mn2+ + 4H2O
Total charge reactant
= + 10 - 1 + 8
= + 10 - 1 + 8
Exer ci se: In Acid i c Solut i on
C2O4 2- + MnO4
08/16/1108/16/11 MATTERMATTER 135135
Bal ancing Redox React i on In Basic Solut i on 
1. Firstly balance the equation as in acidic solution.
1. Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O.
 
Cr(OH)3 + IO3 - + OH-
(basic medium)
(acidic medium)
 
St oichiomet r y 
Stoichiometry is the quantitative study of reactants and products in a chemical reaction.
 
1 molecule of C3H8 reacts with 5 molecules of O2
to produce 3 molecules of CO2 and 4 molecules of
H2O
6.02 x 1023 molecules of C3H8 reacts with 5(6.02 x
1023) molecules of O2 to produce 3(6.02 x 1023)
molecules of CO2 and 4(6.02 x 1023) molecules of
H2O
C3H8 + 5O2  3CO2 + 4H2O
1 mol of C3H8 reacts with 5 moles of O2 to
produce 3 moles of CO2 and 4 moles of H2O
44.09 g of C3H8 reacts with 160.00 g of O2 to
produce 132.03 g of CO2 and 72.06 g of H2O
5 moles of C3H8 reacts with 25 moles of O2 to
 
22.4 dm3 of C 3 H
8 reacts with 5(22.4 dm3) of O
2
 
oxidation of 0.125 mol of glucose?
C6H12O6(s) + O2(g)  CO2(g) + H2O(l)
0.125 mol C6H12O6 produce H2O
mass of H2O = (0.125 x 6) mol
x (2.02 + 16.00) g mol-1
= 13.5 g
carbon dioxide gas and water vapour.
(a) Write a balance equation of the
reaction
produced in the reaction at STP, how
many grams of ethene are used?
 
(b) 22.4 dm3 is the volume of 1 mol CO2
20.0 dm3 is the volume of CO2
2 mol CO2 produced by 1 mol C2H4
mol CO2 produced by
able to:
percentage yield
reactant and percentage yield.
the amount of product formed
Excess reactant is the reactant present in
quantity greater than necessary to react with
the quantity of limiting reactant
 
3H2 + N2  2NH3
If 6 moles of hydrogen is mixed with 6 moles of nitrogen,
how many moles of ammonia will be produced?
Solution:
6 mol H2 reacts with
mol3
mol1mol6  
limits the amount of products formed
3 mol H2 produce 2 mol NH3
6 mol H2produce
6 mol N2 react with mol NH3
mol1
mol3mol6  
limiting reactant
6 mol N2 produce mol NH3
= 4 mol NH3
2 Al(s) + 3Cl2(g)  2 AlCl3(s)
A mixture of 2.75 moles of Al and 5.00 moles of Cl2 are
allowed to react.
(b) How many moles of AlCl3 are formed?
(c) How many moles of the reactant remain at
the end of the reaction?
 
equation is the theoretical yield
The theoretical yield is never obtain because:
1. The reaction may undergo side reaction
2. Many reaction are reversible
3. There may be impurities in the reactants
 
form other product
5. It may be difficult to recover all of the
product from the reaction medium
The amount product actually obtained in a
reaction is the actual yield
 
08/16/1108/16/11 MATTERMATTER 157157
Percentage yield is the percent of the actual yield of a product to its theoretical yield
100 x yield 
C6H6 + Br2  C6H5Br + HBr
In an experiment, 15.0 g of benzene are mixed with excess bromine
(a) Calculate the mass of bromobenzene, C6H5Br that would be produced in the reaction.
 
 
At the end of this topic students should be able
to:-
levels in an atom.
(orbit) using.
Learning Outcomes
hydrogen atom.
during transition.
electron that produces a particular
wavelength during transition.
 J 
equation for Lyman, Balmer, Paschen,
Brackett and Pfund series:
atom from the Lyman series.
1
m −  1
n1n2
model.
Broglie’s postulate and Heisenberg’s
uncertainty principle.
Bohr’s Atomic Model In 1913, a young
Dutch physicist, Niels Böhr proposed a theory of atom that shook the scientific world.
 
Bohr’s Atomic Model
 
Electron moves in circular orbits about the nucleus.
 
[orbit = energy level=shell]
Orbit is a pathway where the electron is move around the nucleus.
Orbit
Second Postulate
 
The energy of an electron in its level is given by:
R H
n (principal quantum number) = 1, 2, 3 …. ∞ (integer)
Note:
Energy is zero if electron is located infinitely far from
nucleus
Third postulate
At ordinary conditions, the electron is at the ground state (lowest level).
 
Electron at its excited states is unstable.
It will fall back to lower energy level and released a specific amount of energy in the form of light (photon).
 
Electron is excited from lower to higher
energy level.
E = h ν = E 3 -E
1
A photon is emitted.
3
Electron is excited from lower to higher
energy level.
E = h ν = E 3 -E
1
A photon is emitted.
3
 
Radiant energy emitted when the electron moves from higher-energy state to lower-energy state is given by:
Where: E=  E
 
The amount of energy released by the electron is called a photon of energy.
A photon of energy is emitted in the form of
radiation with appropriate frequency and
wavelength.
Thus :
λ 
v = frequency (s-1)
Rydberg Equation
Wavelength emitted by the transition of electron between two energy levels is calculated using Rydberg equation:
1
m
 
Exercises: 1. Calculate the energy of hydrogen electron in the:
(a) 1st orbit
(b) 3rd orbit
(c) 8th orbit
1. Calculate the energy change (J), that occurs when a
electron falls from n = 5 to n = 3 energy level in a
hydrogen atom.
radiation emitted in question 2.
 
Continuous Spectrum
A spectrum consists of radiation distributed over all wavelength without any blank spot.
Example : electromagnetic spectrum, rainbow
 
 
A spectrum consists of discontinuous & discrete lines with specific wavelength.
It is composed when the light from a gas
discharge tube containing a particular element
is passed through a prism.
 
Formation of Atomic / Line Spectrum
The emitted light (photons) is then separated into its components by a prism.
Each component is focused at a definite
position, according to its wavelength and forms
as an image on the photographic plate.
The images are called spectral lines.  
Formation of Atomic / Line Spectrum
 
Example : The line emission spectrum of hydrogen atom
Line spectrum are composed a few wavelengths giving a series of discrete line separated by blank areas
 
Formation of Line Spectrum When electron absorbed radiant energy, they
will move from lower energy level to higher energy level (excited state).
This excited electron is unstable and it will fall back to lower energy level.
 
Continuous Spectrum
A spectrum that contains all wavelength without any blank spots.
Example: Rainbow.
Line Spectrum
A spectrum of discrete lines with certain wavelengths.
Example: Emission spectrum an element.
 
n  = 1
n  = 2
n  = 3
n  = 1
n  = 2
n  = 3
Lyman Series
 
 
The following diagram is the line spectrum of hydrogen atom. Line A is the first line of the Lyman series.
Specify the increasing order of the radiant energy, frequency and wavelength of the emitted photon.
Which of the line that corresponds to
i) the shortest wavelength?
ii) the lowest frequency?
v
Example
Describe the transitions of electrons that lead to the lines
W, and Y, respectively.
W Y
 
corresponds to the longest wavelength of
photon?
Solution
 
Example
With refer to the second line in the Balmer series of the hydrogen spectrum, Calculate;
a)the wavelength in nm
Example
Refer to last line of hydrogen spectrum in Lyman series, Calculate:
a) Wavelength
b) Frequency

For Lyman series; n1 = 1 & n2 = ∞ Ans: i. 9.116 x10-8 m ii. 3.29 x1015 s-1
iii. 1.0970 X 107 m-1
 
Ionization Energy
Defination : Ionization energy is the minimum energy required to remove an electron in its ground state from an atom (or an ion) in gaseous state.
M (g) → M+ (g) + e H = +ve
 
removed from its ground state (n = 1) to n =
At n = ∞, the potential energy of electron is
zero, here the nucleus attractive force has no
effect on the electron (electron is free from
nucleus)
= 2.18 x 10-18 (1 – 0)
= 2.18 x 10-18 J
= 1.312 x 106 J mol-1
= 1312 kJ mol-1
(a) a hydrogen atom.
 
1 mol H atom
= 1.31 x 106 J
The energy to ionized 1 mol of hydrogen atom is
1.31 X 106 J
The Lyman series of the spectrum of hydrogen is
shown above. Calculate the ionisation energy of
hydrogen from the spectrum.
Solution
E = h X c/ λ  = h x c x wave no.
= (6.626x10-34 Js)(3x108 ms-1)(10.97x106 m-1)
= 218.06 x 10-20 J
= 1.312 x 106 J mol-1
= 1312 kJ mol-1
The weaknesses of Bohr’s Theory
It can only explain the hydrogen spectrum or any spectrum of ions contain one electron. example: He+, Li2+.Therefore, it did not account for the emission spectrum of atom containing more than 1 electron.
Electron are wavelike, we can’t define the
precise location of a wave because a wave
extends in space.
 
de Broglie’s Postulate In 1924 Louis de Broglie proposed that not only
light but all matter has a dual nature and possesses both wave and particle properties.
Electron is both particle and wave.
Tiny particle such as electron does have wave
properties.
properties are related by the expression:
 
Example
 
m = particle mass (kg)
=   h
mµ
momentum p (defined as mass times velocity) and the
position of a particle with certain.
Stated mathematically,
p = uncertainty in measuring the momentum
= mv
Learning Outcomes
At the end of this topic students should be able to:-
Define the term orbital.
 
dimensional region in
space around the
nucleus where there
finding an electron.
Quantum Numbers
Each of the electrons in an atom is described and characterised by a set of four quantum numbers, namely
principal quantum number, n 
magnetic quantum number, m 
 
Principal Quantum Number, n 
n determines the energy level (electron shell) and size of an orbital.
The principal quantum number n, may have +ve
value starting from n =1, 2, 3, …, ∞.
As n increase :
 
Orbital size 
Energy increases 
Alternative name:
atomic orbital.
 
Letters are assigned to different numerical values of
Value of l  Symbol 
value is depend on n. (i.e., 0 ≤ l < n).
If n = 1, l  = 0 (s -orbital)
If n = 2, l  = 0 (s -orbital)
= 1 (p -orbital)
= 1 (p -orbital)
= 2 (d -orbital)
One subshell  (s orbital )
Describe the orientation of orbitals in space.
Possible values of m depend on the value of l. For a given l, m can be : -l , …, 0, …, +l 
Example:
If l = 0, m = 0 » 1 orientation of s orbital
If l = 1, m = -1,0,+1 » 3 orientation of p orbital (p
x , p
y , p
z )
If l = 2, m = -2,-1, 0,+1,+2 » 5 orientation of d orbital (
d xy
,d xz
,d yz
Electron Spin Quantum Number,
The value of s represent the direction of an electron rotation on its own axis.
either clockwise or
 
When l = 0 , m = 0 , only 1 orientation of s
orbital.
The larger value of n, the size of s orbital gets
larger.
1s 2s 3s
When l = 1, m = -1, 0, +1
3 orientation of p-orbitals p x , p
y , and p
 
When l = 2 , m = -2, -1, 0, +1, +2.
There are five orientation of d orbitals.
 
xy
Set of Four Quantum Numbers
4 quantum number n ,l,m and s enable us to label completely an electron in any orbital of an atom. Example:
4 quantum numbers of 2s orbital electron are
n = 2 , l = 0 , m = 0 and s = +½ and -½
 
1 1 0 1s 0 1
2 2 0 2s 0 1
1 2p -1,0,+1 3
 
Exercise
Predict the following quantum numbers whether they are allowed or not
(a) (1,0,0,-½)
(b) (2,0,1,1)
(c) (0,1,1,+½)
(d) (4,1,0,-½)
 
Learning Outcomes At the end of this topic students should be able to:-
State and apply Aufbau principle, Hund's rule
Pauli exclusion principle in filling of electrons in
orbitals of an atom.
monoatomic ions.
 
Introduction
The electronic configuration of an atom show how electron are filled in the orbital.
Electronic configuration describes the
 
 
Electronic Configuration
To enable us to do electronic configuration, we have to obey the following rules:
a) The Aufbau Principle
c) The Hund rule
Aufbau Principle
State that electrons are filled in the orbitals in order of increasing energy.
Electrons should occupy the orbital with the
lowest energy first before enters the one with
higher energy.
Orbital energy levels in a many-electron atom 
2s  2p 
 
 
The order of filling orbitals is:
1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s
Start with the 1s orbital and move downward,
following the arrows. Example:
(a) 4 Be (b)
Electronic configuration 10
Pauli Exclusion Principle
No two electrons in an atom can have the same four quantum numbers (n, l , m, s).
Eg : Li (3 electrons)
Hund’s Rule
States that when electrons are added to the orbital of equivalent energy (degenerate orbitals), each orbital are filled singly with electron of the same spin first before it is paired.
The electron in half-filled orbitals have the same
spins, that is, parallel spins.
 
Write the electronic configuration of the following atom or ion:
(a)C
(b)Ne
(c)Al
(d)Al3+
(e)Cl
Element Expected (Aufbau
3d4 1s22s22p63s23p6 4s 1
4s 2 1s22s22p63s23p6 3
4s 1
 
24Cr : 18[Ar]
24Cr : 18[Ar]
3d orbital with a half filled orbital arrangement are more stable.
Actual
3d 4s
3d 4s
 
Cu : 18
3d orbital with fully filled orbital arrangement is more stable.
Copper actual orbital notation
 
LEARNING OUTCOMES
• At the end of the lesson the students should be able to :
• (i) Indicate period, group and block (s, p, d, f ).
• (ii) Specify the position of metals, metalloids
and non-metals in the periodic table.
• (iii) Deduce the position of elements in the
periodic table from its electronic
configuration.
3.1 Classification of elements
• The periodic table is a table that arranges all the known elements in order of increasing proton number .
• This order generally coincides with increasing atomic mass.
 
4
• A vertical column of elements is called a group and a horizontal row is known as a period.
• Elements in the same group have the same number of valence electrons.
Group number = number of valence electrons (if the element is in block s and d)
 
5
For example, oxygen and sulphur are both found in group 16 which means that they both have 6 valence electrons.
Transition metals
 – Group 1 : alkali metals (except H)
 – Group 2 : alkaline earth metals
 – Group 3-11 : transition metals
 
7
• The periods in the Periodic Table are numbered from 1 to 7
• For example, hydrogen and helium are in Row 1 or Period 1 because their principal quantum number, n, of the main electron shell is 1. (H:1s1 ;He: 1s2)
Period number = Principle quantum number 
 
8
Blocks
• All the elements in the Periodic Table can be classified into 4 main blocks according to their valence electrons configuration.
 
• Configuration of the valence electrons :
• Eg:
20Ca : 1s2 2s2 2p6 3s2 3p6 4s2
ns1 to ns2
• The filling of valence electrons involve s and p orbital.
• The configuration of valence electrons:
• Eg.
13Al : 1s2 2s2 2p6 3s2 3p1
52Te : 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2
5p4
 
• The filling of valence electrons involve s and d orbitals.
• Group 3 to 11 known as Transition metal.
• Configuration of valence electron :
[Ar] 3d3 4s2
 
12
f-block 
• Involve the elements in the series of lanthanides (Ce to Lu) and actinides (Th to Lr).
 
velenc e
electro n
 
 
15
Example
• Classify the following elements into its appropriate group, period and block.
 A ……1s2 2s2 2p6 3s2 3p6
B …….1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
C …….1s2 2s2 2p6 3s2 3p6 4s2
D …….1s2 2s2 2p6 3s2 3p6 3d3 4s2
E …….1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6
 
(atomic radii)
 
V = 4/3 πr 3 , V r 

 
19
Two factors that influence the changes of atomic radii in the Periodic Table are:
i. Effective nuclear charge
experienced by the valence
the valence electrons
)
• Electrons around the nucleus experience different nucleus attraction.
• Those electrons closer to the nucleus experience a greater attraction than those that are farther away.
• The actual nuclear charge experienced by an electron is called the effective nuclear charge, Z
eff 
• Effective nuclear charge increase, nucleus attraction stronger, atomic radii decrease
•  Across the period, the effective nuclear charge increases as proton number increase.
 
valence electrons
• As we move down a group, the number of shells increases, more inner electrons are present to shield the valence electrons from the nucleus.
• The valence electrons are farther from the nucleus.
 
23
 
The graph shows that : •Atomic radius decreases when :
* Across a period (from left to right) * Moving up a group in the periodic table.
•Atomic radius increases when * Going down the group
• The greater the nucleus attraction, the
smaller the atomic radius.
26
Example
Solution • N and P are in the same group and N is above
P. • Atomic radius increases as we go down the
group.
• Therefore, the radius of N is smaller than that of P
• Both Si and P are in the third period and Si is to the left of P.
• Atomic radius decreases as we move from left to right.
• Therefore, the radius of P is smaller than Si. • Thus the order of increasing radius :
N<P<Si
 
3.2.2 Trends in the ionic radii
• When electrons are added to an atom, the mutual repulsions between them increase.
• This enlarge the domain of electron cloud.
• Therefore, negative ions (anions) are
larger than the atoms from which they are
formed.
• When electrons are removed from the valence shell, the electron-electron repulsions decrease but the nuclear charge remains the same.
• So the remaining electrons are to be pulled closer together around the nucleus.
 
30
• Isoelectronic species are groups of atoms and ions which have the same electronic configuration.
Within isoelectronic species: a) the more positive the charge, the smaller the species
E.g : • Na+, Mg2+, Al3+and Si4+ ions are isoelectronic
(10 e) with the electron configurations as 1s2 2s2 2p6.
Isoelectronic species
Isoelectronic species with electronic configuration 1s2 2s2 2p6 (10 electrons)
•When proton number increase, effective nuclear charge increase.
•The attraction between nucleus and remaining electron increase.
•Therefore, the ionic radii decrease.
•The ionic radii of Na+ > Mg2+ > Al3+ > Si4+
species Number of proton
 
Isoelectronic species with electronic 1s2 2s2 2p6 3s2 3p6 (18 electrons)
•When proton number increase, effective nuclear charge increase.
•The attraction between nucleus and remaining electron increase.
•Therefore, the ionic radii decrease.
•The ionic radii of Cl- < S2- < P3-
species Number of proton
33
Na+,Si4+ ,Mg2+, N3- ,O2- ,Al3+ and F- are isoelectronic with the electronic configuration as 1s2 2s2 2p6. Arrange in an descending order the size of those isoelectronic species.
Exercise
Answer :
 
energies
• The ionization energy (IE) is the minimum energy required to remove an electron from a gaseous atom in its ground state.
The first ionization energy (IE1) is the minimum energy required to remove the first electron from the atom in its ground state.
E.g: energy + X(g) → X+(g) + e-  Δ H = IE1
 
The effective nuclear charge increases, the atomic size decreases.
Electrons are held tightly to the nucleus thus it is difficult to remove the first electron.
Therefore the first ionisation energy is high.
It can be said that the first ionization energy
increases from left to right .
However, there are some irregularities in the trend.
 
a) Between group 2 and 13
• 5B : 1s2 2s2 2p1 in group 13 has a lower IE1 than 4Be: 1s2 2s2 in group 2.
• Be loses a 2s electron while B loses a 2p electron.
 
• O (group16) has lower IE1 than N (group 15)
• 7N :1s2 2s2 2p3 (the half-filled 2p orbital )
8O :1s2 2s2 2p4( the partially-filled 2p orbital)
• When N loses an electron it must come from the half-filled 2p orbital which is more stable than that of electron of the partially-filled orbital in O.
 
ii) Ionization energy going down the group
• Going down the group, the atomic size increases as the energy level, n increases.
• Therefore the outer electrons are farther from the nucleus and are held less tightly (weaker attraction) by the nucleus.
• Thus, it is easy to remove the first electron.
• Hence the Ionization Energy decreases down the group.
 
41
• Second ionization energy (IE2) is the minimum energy required to remove an electron from a positive gaseous ion.
X+(g) → X2+(g) + e-
• When an electron is removed from a neutral atom, the mutual repulsion among the remaining electrons decrease.
• Since the nuclear charge remain constant, the electron are held tightly to the nucleus.
 
IE1< IE2< IE3< IE4<…..
• Although the removal of a subsequent electron from an atom requires an increment amount of energy but it may not be consistence.
 
IE1 IE2  IE3  IE4 
899 1757 14850 21005
95.1 899
45
• A sharp increase in ionization energy occurs when an inner-orbital electron is removed.
 
Example 2
• Five successive ionization energies (kJmol-1) of atom M is shown below:
• Determine
electron.
IE1  IE2  IE3  IE4  IE5 
800 1580 3230 4360 16000
 
47
• The sharp increase is in IE5, this means the 5th electron occupies the inner shell.
• Therefore, there are 4 valence electrons.
 
Trends in the electronegativity
 
• Therefore, the electronegativity
• The atomic size increase
• Hence, weaker nuclear attraction
Trends in the melting or boiling point
 
elements
Element Na Mg Al Si P S C l Ar
Melting
point
(oC)
Boiling
point
(oC)
 
elements in the 3rd period can be discussed
as:
(b) Gigantic covalent structure (Si)
(c) Simple molecular structure (P to Ar)
 
a)Metallic structure (Na to Al)
Metal has positive metal ions attracted to the electrons sea which form the metallic bonding.
Strength of metallic bonding is proportional to the
number of valence electrons.
e e The more valence
electrons, the stronger the metallic bond and the higher the melting
/ boiling point
• Silicon has a gigantic covalent structure.
• Melting and boiling point of Si is very high
 
c) Simple molecular structure ( P to Ar)
• The non-metal that exist as molecules of P4, S8, Cl2 and Ar (monoatom).
• The covalent bond between the atoms is very strong but the intermolecular force (Van der
Waals), is very weak.
• The strength of Van der Waals force is proportional to molecular size (relative molecular weight)
 – Molecular size: Ar < Cl2 < P4 < S8
 – therefore melting / boiling point : Ar < Cl < P < S
 
Group 1
• The size increase, the attraction between nucleus and electron sea become weaker .
• Therefore, less energy is needed to overcome the attraction.
• Thus, melting and boiling point decrease.
 
Group 17
 
- malleable and ductile
• Generally, metallic character :
60
• The easier to the electrons to be removed from an atom, the more metallic the element.
• Therefore metallic character increases down a group and decreases across a period.
metal Metalloids
(semimetal) nonmetal
For Period 3:
• When react with oxygen : (a) Na & Mg form basic oxide
(b)  Al form amphoteric (both acidic and basic) oxide.
(c) Si, P, S & Cl form acidic oxide
Na Mg Al Si P S Cl
 
4Na (s) + O2 (g)→2Na2O (s)
• The oxide will produce base solution when react with water.
Na2O (s) + H2O (l) → 2NaOH (aq)
• Mg burns in oxygen to form a basic oxide, MgO. 2Mg (s) + O2 (g) →2MgO (s)
MgO (s) + 2HCl (aq)→MgCl2 (aq) + H2O (l)
base acid
63
•  Al forms amphoteric oxide, can react either with an acid or a base.
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3 H2O (l)
base acid
 
64
Si, P, S & Cl burn in oxygen to form acidic oxide.
Si :
SiO2 (s) + NaOH (aq) → Na2SiO3(aq)+H2O (l)
acid base
phosphorus acid
sulfurous acid
 
molecule or compound
chemical bonds
4.4 Intermolecular Forces
4.5 Metallic Bond
dots or cross is used to represent the
valence electrons in an atom of the
element.
Group 1 2 13 14 15 16 17 18
Valence
electron
Lewis dot
same valence electronic configurations
shell
 
through:
 
 Anions
electrons to form cations with noble gas
configurations
Example:
Ca : 1s22s22p63s23p64s2
 
configurations
Example:
pseudonoble gas configuration.
d block element can also donate electrons to
achieve the stability of half-filled orbitals
Example:
Fe : 1s22s22p63s23p64s23d6
Fe3+: 1s22s22p63s23p63d5
 
 
electrons areelectrons are transferredtransferred betweenbetween
atomsatoms ((metal to nonmetalmetal to nonmetal)) to giveto give
electrically charged particleselectrically charged particles thatthat attract each other attract each other ..
 
Example 1: NaCl
 
Cl- produce ionic bond
+
 
Cl: 1s2 2s2 2p6 3s2 3p5
(Has seven outer electrons)
Calcium Chloride
If Ca atom transfer 2 electrons, one to each chlorine atom, it become a Ca2+ ion with the stable configuration of noble gas.
 At the same time each chlorine atom to achieve noble gas configuration gained one electron becomes a Cl- ion to achieve noble gas configuration.
The electrostatic attraction formed ionic bond between the ions.
 
Calcium Chloride
+ +
compounds:
2. Hard and brittle
3. Can conduct electricity when they
 
below.
electrons are shared by two atoms.
ii.The electrostatic force between the electrons
being shared the nuclei of the atoms.
Why should two atoms share electrons?
To gain stability by having noble gas configuration (octet)
 
bonds:
two pairs of electrons
8e- 8e- 8e-
double bondsdouble bonds
three pairs of electrons
Dative bond is a bond in which the pair of
shared electrons is supplied by one of the two bonded atoms
Involve overlapping of a full orbital and
an empty orbital
one lone pair electrons
electrons should have empty
1. Count total number of valence e- of atoms involved.
2. Add 1 for each negative charge. Subtract 1 for each
positive charge.
electronegative element in the center.
4. Complete an octet for all atoms except hydrogen
5. If structure contains too many electrons, form double
and triple bonds on central atom as needed.
 
following compounds:
i. HF
ii. CH4
Total: 26 e
Center atom: N
Compare the bond length between single, double and triple bond
Bond length : The distance between nuclei of the atoms involves in the
bond
C C C C C C 1.54 Å 1.34 Å 1.20 Å
 As the number of bonds between the carbon increase, the bond length decreases because C are held more closely and tightly together 
 
 The sumof formal charge on
each atom should equal: i.zero for a molecule ii.the chargeon the ion for a polyatomic ion
 
2 .
EXAMPLE
Formal charge is determined before
completing a Lewis structure to predict the most stable structure because
formal charge closest to zero.
1) 2)
2 .
2 .
2 SO
4 . Explain
your answer.
Elements that can form incomplete octet are: Boron,B , Beryllium, Be &
 Aluminium, Al This is due to elements being
relatively small in size but having high nuclear charge.
 
Formed by non-metals that have d orbitals
OR Non-metals of the 3rd, 4th, 5th….rows
in the periodic table
electrons.
 
The use of two or more Lewis structures to represent a particular molecule.
 
CCl 4
CO 3
 
be able to;
2. Draw the basic molecular shapes : linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.
3. Predict and explain the shapes of molecule and bond angles in a given species.
4. Explain bond polarity and dipole moment.
 
i. VSEPR theory
 
The Valence-Shell Electron Repulsion theory states that:
 
pairs
Bonding pair-bonding pair repulsion
Lone pair-bonding
pair repulsion > >
Note:
 
4.2.2 Shape of a molecule
Basic shapes are based on the repulsion between the bonding pairs.
Tips to determine the molecular shape : Step 1 Draw Lewis structure of the molecule
Step 2 Consider the number of bonding pairs
 
Example: BeCl2
Lewis structure
Example: BCl3 Lewis structure
120°
Example: CH4
Lewis structure
109.5°
Tetrahedral
Example: PCl5 Lewis structure
Example: SF6
Lewis structure
 
 
4 electron pairs in the valence shell of central atom:
109.5o
 
 
4.2.3 Effect of lone pairs on molecular shape
The geometries of moleculesmolecules and polyatomic ionspolyatomic ions, with oneone or more lonemore lone pairspairs around the central atom can be predicted using VSEPR.
 
Bonding pair- bonding pair repulsion
Lone pair- bonding
> >
82
Electrons in a bond are held by the attractive forces exerted by the nuclei of the two bonded atoms therefore, they take less space of repulsion.
LoneLone-- pair electronspair electrons in a molecule occupy more space; therefore they experienceexperience greater repulsiongreater repulsion from neighboring lone pairs and bonding pairs
 
Example : SO2
 
Example : NH3
 
Example : H2O
Example : SF4
seesaw
Example : ClF3
Example : I3 -
Example : BrF5
Example : XeF4
Bond angle : < 120o
 
 
 
 
 
 
180o
 
 
109.5o 107.3o 104.5o
The repulsionrepulsion between the bonding pairs electrons are equalequal.
The bond anglesbond angles are all 109.5109.5oo
 
102
has 3 bonding pairs3 bonding pairs electron and 11 lone pairlone pair electron.
according to VSEPR, lone pairlone pair -- bonding pair >bonding pair > bonding pairbonding pair -- bonding pair repulsionbonding pair repulsion.
Lone- pair repels the bonding-pair more stronglymore strongly, the three NH bonding-pair are pushed closerpushed closer together, thus HNHHNH angle in ammonia become smaller, 107.3107.3oo.
b) NH3
103
Has 2 bonding pairs2 bonding pairs electrons and 2 lone pair2 lone pair electrons.
According to VSEPR, lone pairone pair –– lone pair >lone pair > lone pairlone pair –– bonding pair > bonding pairbonding pair > bonding pair –– bonding pair repulsionbonding pair repulsion.
Lone-pair tend to be as farfar from each other as possible.
Therefore, the two OHOH bonding-pairs are pushed toward each other.
Thus, the HOHHOH angle is 104.5104.5oo.
c) H2O
4.2.4 POLAR AND NONPOLAR MOLECULES
A quantitative measure of the polarity of a bond is its dipole moment ( µ ).
 µ µ  = Qr
Where : µ = dipole moment Q = the product of the charge from
electronegativity r = distance between the charges.
Dipole moments are usually expressed in debye units(D)
µ 
Hydrogen fluoride is a covalent molecule with a polar bond.
F atom is more electronegative than H atom, so the electron density will shift from H to F.
The symbol of the shifted electron can be represented by a crossed arrow to indicate the direction of the shift.
H F
δ + : partial positive charge
δ - : partial negative charge
107
Diatomic molecules containing atoms of different elements (e.g. : HCl, NO and CO) have dipole moments and are called polar molecules.
 
108
For polyatomic molecules, the polarity of the bond and the molecular geometry determine whether there is a dipole moment.
 
 –  Carbon dioxide, CO2
 –  Ammonia, NH3
(a) Carbon dioxide, CO2
 
(b) Carbon tetrachloride, CCl4
 
112
- molecular geometry : tetrahedral - Cl is more electronegative than C, C is more
electronegative than H - Dipole moment cannot cancell each other - has a net dipole moment (µ   ≠ 0)
- therefore CH3Cl is a polar molecule.
( c) Chloromethane, CH3Cl
- molecular geometry : tetrahedral - N is more electronegative than H,
- Dipole moment cannot cancell each other - has a net dipole moment (µ   ≠ 0)
- therefore NH3 is a polar molecule.
 
Factors that affected theFactors that affected the polarity of moleculespolarity of molecules
molecular geometry
 
Symetrical moleculesSymetrical molecules - basic molecular shape with the same terminal atom - molecules with lone pairs linear
(from trigonal bipyramidal) and square planar with the same terminal atom
NonNon--symetrical moleculessymetrical molecules - basic molecules with different terminal atom - molecules with lone pairs except linear and square planar
NON-POLAR MOLECULES
116
SO2 ; HBr ; SO3 ; CH2Cl2 ; ClF3 ; CF4 ; H2O ; XeF4 ; NF3 ; Cis-C2H2Cl2 ; trans-C2H2Cl2
Exercises :
 
1.1. Formation Covalent BondFormation Covalent Bond
2.2. Formation Hybrid orbitalsFormation Hybrid orbitals
3.3. Orbital OverlappingOrbital Overlapping
118
Objectives
 At the end of this subtopic, students should be able to:
1. Draw and describe the formation of sigma( and pi() bonds from overlapping of orbitals.
2. Draw and explain the formation of hybrid orbitals of a central atom: sp, sp2, sp3, sp3d, sp3d2
using appropriate examples.
 
4.3.1 Valence Bond theory
• explains the formation of covalent bonds and the molecular geometry outlined by the VSEPR.
• States that a covalent bond is formed when the neighboring atomic orbitals overlap.
• Overlapping may occur between:
a) orbitals with unpaired electrons
 
120
Change in electron density as two hydrogen atoms approach each other.
10.3
(covalent bond formed)
FORMATION OF COVALENT BOND
• Valence bond theory - Covalent bond is formed when two neighbouring atomic half-filled orbitals overlap.
• Two types of covalent bonds are
a) sigma bond (σ)
a)   bond
• formed when orbitals overlap along its internuclear axis (end to end overlapping)
Example:
H +
 x x +  x
orientation overlap sideways  y  y
+
• Covalent bonds may form by:
a) overlapping of pure orbitals
b) overlapping of hybrid orbitals
 
O : 1s2 2s2 2p4
σ 
4.3.2 Formation Hybrid orbitals4.3.2 Formation Hybrid orbitals • Overlapping of hybrid orbitals and the pure orbitals
occur when different type of atoms are involved in the bonding.
•• HybridizationHybridization of orbitals:
mixingmixing of two or moretwo or more atomic orbitalsatomic orbitals to form a newform a new set of hybrid orbitalsset of hybrid orbitals
• The purpose of hybridisation is to produce newproduce new orbitalsorbitals which have equivalent energyequivalent energy
 
 
sp3 hybridization
• one s orbital and three p orbitals are mixed to form four sp3 hybrid orbitals
 
sp3
sp3
sp3
sp3
CH H H
H Lewis Structure
1s 2s 2p
Excited state :
sp3
sp3
sp3
sp2 hybridization
• one s orbital and two p orbitals are mixed to form three sp2 hybrid orbitals
 
 
 
sp hybridization
• one s orbital and one p orbital are mixed to form two sp hybrid orbitals
 
10.4
 
sp3d hybridization
• one s orbital, three p orbitals and one d orbital are mixed to form five sp3d hybrid orbitals.
 
 
sp3d2 hybridization
• one s orbital, three p orbitals and two d orbitals are mixed to form six sp3d2 hybrid orbitals
 
 
2
3
4
5
6
sp
sp2
sp3
sp3d
sp3d2
BeCl2
BF3
How do I predict the hybridization of the central atom?
Count the number of lone pairs AND the number  of atoms bonded to the central atom
10.4
 
168
Exercise:
• For each of the following, draw the orbital overlap to show the formation of covalent bond
a) XeF2
b) O3
c) ICl4
d) OF2
- dipole-dipole interactions or permanent dipole
- London forces or dispersion forces ii. Hydrogen bonding
2. Explain factors that influence the strength of van der Waals forces
3. Explain the effects of hydrogen bonding on i. boiling point ii. Solubili ty iii . Density of water compared to ice
1. Explain the relationship between : i. intermolecular forces and vapour pressure ii. Vapour pressure and boiling point
 
on the physical properties.
 
Have effects on these physical properties:
 
molecules
 
Forces that act between covalent molecules
Three types of interaction: i. Dipole-dipole attractive
forces - act between polar molecules
ii. London Dispersion forces - act between non-polar
molecules
Dipole-dipole forces (permanent dipole forces)
Exist in polar covalent compounds
Polar molecules have permanent dipole due to the uneven electron distributions
Example:
Chlorine is more electronegative, thus i t has higher electron density
 
result from the temporary (instantaneous) polarization of molecules
 
The formation of London forces
 At any instant, electron distributions in one molecule may be unsymmetrical.
The end having higher electron density is partially negative and the other is partially positive.
 
Br Br  
Br Br  
 -  +
Br Br  
The temporary dipole molecule induce the neighboring atom to be partially polar 
Temporary dipole molecule
180
Factors that influence the strength of the van der Waals forces.
The molecular size/molecular mass Molecules with higher molar mass have stronger van der Waals forces as they tend to have more electrons involved in the London forces. Example: CH4 has lower boiling point than C2H6
Note: However if two molecules have similar molecular mass, the dipole-dipole interaction will be more dominant. Example: H2S has higher boil ing point than
CH3CH3
4.4.1.3 Hydrogen intermolecular bond
Dipole-dipole interaction that acts between a Hydrogen atom that is covalently bonded to a highly electronegative atom ; F, O ,N in one molecule and F,O or N of another molecule.
Example:
 
Properties of compounds with Hydrogen intermolecular forces
Have relatively high boiling point than compounds having dipole-dipole forces or London forces
- the Hydrogen bond is the strongest attraction force compared to the dipole- dipole or the London forces.
Boiling point
Hydrogen bonds are highly polar. They may form interaction with
any polar molecules that act as solvent.
 
188
Example
NH3 dissolves in water because it can form Hydrogen intermolecular bond with water.
N H
189
Problem:
Explain the trend of boiling point given by the graph below:
HF
HCl
190
 Answer 
 
The effect of Hydrogen bond on water molecules
The density of water is relatively high compared to other molecules with similar molar mass.
Reason:
 
Ice (solid H2O) has lower density compared to its liquid. Refer to the structure of ice
Density
193
Hydrogen bond takes one of the tetrahedral orientation and occupy some space
Ice form tetrahedral arrangement
H2O(l) is denser than H2O(s) because
the hydrogen bond in ice arrange the H2O molecules in open hexagonal crystal
H2O molecules in water have higher kinetic energy and can overcome the hydrogen bond
V-shaped water molecules slide between each other.
 
a) the number of hydrogen bonds per molecule
b) the strength of H intermolecular forces which directly depends on the polarity of the hydrogen bond
Example: Explain the trend of boiling points given below:
The order of the increase in boil ing point is: H2O > HF > NH3 > CH4
Boiling points of substance with Hydrogen intermolecular bonds
 
199
by looking at the polarity of the bond, we have (Order of polarity: HF > H2O > NH3) but H2O has the highest boiling point. For H2O, the number of hydrogen bonds per molecule affects the boiling point.
Each water molecule can form 4 hydrogen bonds with other water molecules. More energy is required to break the 4 Hydrogen bonds.
HF has higher boil ing point than NH3 because F is more electronegative than Nitrogen.
CH4 is the lowest - it is a non polar compound and has weak van der Waals forces acting between molecules.
 Answer:
1)Boiling point
For molecules with similar size, the order of intermolecular strength:
Hydrogen bond > dipole-dipole forces > London dispersion forces
Strength of intermolecular forces ↑
Fluorine is more electronegative than oxygen, therefore stronger hydrogen bonding is expected to exist in HF liquid than in H2O.
 
On the other hand, H-F has only 2 hydrogen bonds.
 
Boiling point HF > NH3 Fluorine is more electronegative than
 
 Vapour Pressure
Molecules can escape from the surface of liquid at any temperature by evaporation
in a closed system :
vapour molecules which leaves the surface cannot escape from the system
 
 
The pressure exerted by those molecules is called vapour pressure (or maximum vapour pressure)
Vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid
phase.
Vapour molecules are trapped in the close container 
Some of the vapour molecules may collide and lose their energy. They
re-enter the liquid surface
Volume of liquid remains constant
Molecules have enough energy to overcome intermolecular forces
Rate of vaporisation is faster than the rate of condensation
Rate of vapourisation is equal to the rate of condensation
 
Rate of evaporation = rate of condensation
The vapour pressure at this stage is constant and known as the equilibrium vapour pressure.
Number of liquid molecules leaving the surface is the same as the number of vapour molecules entering the liquid surface.
Note:
= vapour pressure
i. Intermolecular forces
Molecules with weak intermolecular forces can easily vapourise. More vapour molecules will be present and exert higher pressure. the weaker intermolecular forces the higher is
the vapour pressure. ii. Temperature
Heating causes more molecules to have high kinetic energies that are higher than their intermolecular forces. More liquid molecules will form vapour.
vapour pressure increases with temperature.
 
Increasing the temperature will increase in the vapour pressure.
 As heat is applied, the vapour pressure of a system will increase unti l it reaches a point whereby the vapour pressure of the liquid system is equal to the atmospheric pressure.
Boil ing occurs and the temperature taken at this point is known as the boiling point.
 At this point, the change of state from liquid to gas occurs not only at the surface of the liquid but also in the inner part of the liquid.
Bubbles form within the liquid.
 
Boiling Point: the temperature at which the vapour pressure of a liquid is equal to the external atmospheric pressure.
 
1. Intermolecular forces
 A substance with weak intermolecular forces can easily vapourise and the system requires less heat to achieve atmospheric pressure, thus it boils at a lower temperature. 2. Atmospheric pressure
 
electron sea model.
i. malleability
ii. Ductility
3. Explain the factors that affect the strength of metallic
bond
4. Relate the strength of metallic bond to boiling point
LEARNING OUTCOMES
able to;
Metallic bond
 An electrostatic force between positive charge metallic ions and the sea of electrons.
Bonding electrons are delocalized over the entire crystal which can be imagined as an array of the ions immersed in a sea of delocalized valence electron.
217
Free moving electrons
Metallic Bond (Electron-sea Model) Metals form giant metallic structure Each positive ion is attracted to the
‘sea of electrons’. These atoms are closely held by the
strong electrostatic forces acting between the positive ions and the ‘sea of electrons’.
 
220
metals have high melting point high energy is required to overcome these
strong electrostatic forces between the positive ions and the electron sea in the metallic bond
Physical properties of metals
e e e e e e
e e e e e e
e e e e e e
 
221
The strength of the metallic bond increases with the number of valence electrons and the size of ions.
The smaller the size of positive ions the greater is the attractive force acting between the ions and the valence electrons
The strength of the metallic bonds
 
e ee e e e
e e e e e e
+2 +2 +2 +2 +2 +2 +2 +2
+2 +2 +2 +2 +2 +2 +2 +2
+2 +2 +2 +2 +2 +2 +2 +2
ee ee ee ee ee ee ee ee ee
ee ee ee ee ee ee ee ee
ee ee ee ee ee ee ee ee
Na Mg
Has one valence electron
the electrostatic force acting between posi tive ions and free moving electrons form metallic bonds
Has 2 valence electrons
Stronger metallic bond due to the size of Mg being smaller than Na and the strong electrostatic force between +2 ions and the two valence electrons,
Mg has higher boiling point than Na
 
223
Example:
 
224
The cationic size of Al is smaller compared to magnesium and its charge is higher (+3).
Mg has two valence electrons  Al has three valence electrons
involved in the metallic bonding. The strength of metallic bond in
 Aluminium is greater than that of Magnesium
 Al has higher boiling point
 Answer 
 
The strength of metallic bond is directly proportional to the boiling point.
The stronger metallic bond,the higher the boiling point.
225
 
 At the end of the lesson, student should be able to :
(a) Explain the general properties of gas in terms of arrangement of 
particle, density and compressibili ty.
(b) Explain qualitatively the basic assumptions of the kinetic
molecular theory of gases for an ideal gas.
(c) Define gas laws
(i) Boyle’s Law
(ii) Charles’s Law
(iii) Avogadro’s Law
(d) Sketch and interpret the graphs of Boyle’s and Charles’s laws
(e) Perform calculations involving gas laws.
 
(g) perform calculations using the ideal gas equation
(h) determine the molar mass and density of a gas using ideal gas
equation
(j) perform calculation using Dalton's law
(k) compare the ideal and non-ideal behaviours of gases in terms
of intermolecular forces and molecular volume
(l) explain the condit ions at which real gases approach
the ideal behaviour.
(m) explain qualitatively van der Waals equation and relate the
values of a and b to intermolecular forces and molecular volume
of a gas
General
Properties
of Gas Particles of gas are far apart and fill the available space.
Gases assume the volume and shape of their containers.
can be compressed –due to the particles being
so small and are relatively far apart from one another.
Gases have relatively low densities.
 
equationequation(PV = nRT)
 
The theory is based on the following assumptions:
1) Gas molecules are very tiny that their size are negligible
compared to the volume of the container.
(having mass but no volume)
2) Gas molecules move in straight lines and are at
a constant motion unless they collide.
3) Molecular collisions are elastic – no energy is lost during
collisions.
negligible.
5) The average kinetic energy of the particles is proportional
to the absolute temperature.
temperature is inversely proportional to the gas
pressure
PV = k
Where:
1
P
Where
Graph of P versus V Graph of P versus 1
V
proportional to 1
 
 A sample of chlorine gas occupies a volume of 2 L at a pressure
of 1 atm. Calculate the pressure of the gas if the volume is
increased to 5 L at constant temperature.
0.4 atm
Example 1
 
The pressure of a sample of hydrogen gas in a 50.0 mL container is
765 mmHg. The sample is then transferred into another container and
the measured pressure is 825 mmHg. What is the volume of the second
container?
 
a) Charles’s Law : The volume of a fixed amount of gas atvolume of a fixed amount of gas at
constant pressure is directly proportionalconstant pressure is directly proportional to the
absolute temperature of the gas (in Kelvin).absolute temperature of the gas (in Kelvin).
V  T (no of mole and pressure constant)
Where :
V = volume
Where V1 = initial volume
V
T(K)0
V
 At 125 °C. The sample is then cooled at constant
pressure until it contracts to 1.54 L. Calculate the
final temperature in degree Celsius.
-81.54 °C
Example 1
 
A sample of gas trapped in a capillary tube by a
plug of mercury at 22 oC has a volume of 4.5 mL.
Calculate the volume of the gas when the capillary
tube is heated to 60 oC.
5.08 mL
Example 2
Boyle’s law :
1
11
T
VP =
2
22
T
VP
298.15 K and 153.3 kPa. Find its volume at STP.
 Ans : 35.32 L
2 moles of chlorine gas kept in a cylinder with
piston occupies a volume of 49 L. When another 3
moles of chlorine gas is pumped into the cylinder
at constant temperature and pressure the piston
moves upwards to accommodate the gas.
Calculate the final volume of the gas.
Ans : 73.5 L
C)  Avogadro’s Law
 At constant pressure and temperature At constant pressure and temperature, the volumevolume of a
gas is directly proportionaldirectly proportional to the number of molesnumber of molesof the
gas present
V = k n
Combination of Boyle's law, Charles's law and Avogadro's law : 1
Boyle's Law: V P
Charles' Law: V T
Avogadro's Law: V n
unit of pressure
unit of volume
value of R
unit of R
atm L or dm3 0.08206 L atm mol 1 K 1
Nm 2
J mol 1 K 1
Value of R depend on the unit of pressure and
volume used in the equation.
 
Example 1
 A steel gas tank has a volume of 275 L and is filled
with 0.485 kg of O2. Calculate the pressure of O2 if
the temperature is 29 oC.
 Ans : 1.36 atm
Example 2
 A sample of chlorine gas is kept in a 5.0 L container
at 228 torr and 27 °C. How many moles of gas are
present in the sample?
Molar mass and density of a gas can be calculated
by rearranging the Ideal Gas
Equation:
r 
 
per litre (g/L) at 752 mmHg and 55 C.
 Ans : 0.625 gL-1
 A chemist has synthesized a greenish-yellow compound of
chlorine and oxygen and finds that its density is 7.71 g L-1
at 36 °C and 2.88 atm. Calculate the molar mass of the
compound.
2.40 L O2 gas that measured at a pressure of
1 atm and a temperature of 26 oC. The
reaction equation is
2KClO3(s) 2KCl(s) + 3O2(g)
 Ans : 8.0 g
 
Dalton’s Law of Partial Pressure  The total pressure of mixture of non reacting gasesmixture of non reacting gases is the
sum of the partial pressures exertedsum of the partial pressures exerted by each of the gas in
the mixture
((Partial pressurePartial pressure is the pressure of individual gas component in a
mixture).
PT = P A + PB + PC
 
pressure exerted by gas A = P A =
pressure exerted by gas B = PB =
V
 
V
 

and V
Total
P
P
 A
Total
n
n
 
 A gaseous mixture of 7.00 g N2and 3.21 g CH4 is placed in
a 12.0 L cylinder at 25 oC.
a) What is the partial pressure of each gas?
b) What is the total pressure in the cylinder?
 Ans :a) 0.501 atm , 0.41 atm
b) 0.92atm
Example 1
 A mixture of gases contains 4.53 moles of neon, 0.82
moles of argon and 2.25 moles of xenon. Calculate the
partial pressure of the gases if the total pressure is 2.15 atm
at a certain temperature.
 Ans : PNe = 1.28 atm,,P Ar = 0.079 atm, P Xe = 0.63 atm,
Example 2
 
 A sample of gas at 5.88 atm contains 1.2 g CH4, 0.4 g H2 and
0.1 g He. Calculate :
a) The partial pressure of CH4, H2 and He in the mixture.
b) What is the partial pressure of CH4 and H2 if He is removed?
 Ans : a) PCH4= 1.74 atm,PH2 = 3.92 atm, P He= 0.49 atm
b) PCH4= 1.6 atm,PH2 = 4.28 atm
Example 3
Vapour pressure of water, Pwater  = 23.8 torr 
One of the applications of Dalton’s Law is to calculate the pressure
of a gas collected over water ( for gases that not soluble in water) .
The gas collected is actuallyThe gas collected is actually a mixture of the gasa mixture of the gas andand water vapour.water vapour.
gas + water vapour gas
2KClO3 2KCl + 3O2
 A sample of 5.45 L of oxygen is collected over water at a total
pressure of 735.5 torr at 25 °C.How many grams of oxygen have
been collected?
 Ans : 7.011 g
 
Excess amount of hydrochloric acid is added to 2.5 g of pure
zinc. The gas produced is collected over water in a gas cylinder 
at 28 oC and 100.0 kNm-2. Calculate :
a) the number of mole of gas produced in the reaction.
b) the volume of gas collected in the cylinder.
 Ans : a) 0.038 mol
equation and has the properties as
outlined by the Kinetic Molecular
Theory
 
real gas (nonreal gas (non--ideal gas)ideal gas) : gases which do not obeydo not obey
ideal gas properties
Real gases do not behave ideally because: i) gas molecules do have its own volume and they occupy
some space.
between them
 
• The free volume where the molecules move about
is smaller than the volume of the container.
 
are less frequent
• The actual pressure exerted by the gas is lower than
calculated by ideal gas equation
 
low pressure and high temperature.
 
a container is increased.
another, hence the intermolecular forces
can be neglected.
• At a low pressure the size of a container is
extremely large compared to the size of
molecules, thus the size of molecules can
be neglected.
is very large
from one another. Hence the
intermolecular forces can be neglected.
• At low pressure, the volume of the
container is extremely large compared to
the size of the molecules, thus the volume
can be neglected
The molecules are able to free themselves
from the intermolecular forces that act
between them.
ideally.
at di fferent temperatures
 
Since real gas does not exhibitdoes not exhibit ideal gas behavior
at high pressure and low temperaturehigh pressure and low temperature:
the ideal gas equation (PV=nRT) needs to be
adjusted
be reconsidered :
√ volumevolumeof the gas molecules
make the molecules move slower 
give less impact to the wall
pressure exerted by the real gas is less compared to the ideal
gas
since Preal< Pideal
the term pressure need to be corrected by adding coefficient 2
2
V
an
V2
V = volu me of con tainer 
 
b = size factor 
• Since the gas occupy a sizeable portion
of a container, the space in which the
molecules are able to move are less
than the volume of the container.
Vreal < V container 
volume is: vnb
PV = nRT
P ideal = P real + n2a
V2
V2
 
• a is a posi tive constant which depends on the
strength of the attractive forces between molecules
• Molecules wi th a higher value of a have stronger
attracti ve forces.
Value of b :
by the molecules.
 
 
a) Explain the properties of liquids : shape, volume, surface tension, viscosity, compressibility and diffusion.
b)Explain vaporization process and condensation process based on kinetic molecular theory and intermolecular forces.
c) Define vapour pressure and boiling point
 
1. Volume and Shape
has a definite volumedefinite volume but not a definite shapenot a definite shape
√ the particles are arranged closely but not rigidly
√ held together by a strong intermolecular forces
but not strong enough to hold the particles
firmly in place
√ particles are able to move freely
 
2.2. Surface tensionSurface tension
Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area
• Surface tension is caused by the attractive
forces and the direction of the force acting
between
molecules
√ Molecules within a liquid experienced intermolecular attractive forces in allall directionsdirections by their neighbouring molecules
 
√ these intermolecular attractive forces will
pull the molecules into the liquidpull the molecules into the liquid
√ Thus, cause the surface to stretch and
tighten
forces, the higher the surface tensionhigher the surface tension
 
3.3. ViscosityViscosity
is a measure of the resistance of a liquidresistance of a liquid
to flowto flow
the greater the viscositygreater the viscosity, the more slowlymore slowly
the liquid flows.the liquid flows.
 
i) The strengthstrength of intermolecular forcesforces
Liquids that have strong intermolecular forces have higher viscosity
ii)ii) T T he size of the moleculeshe size of the molecules
Liquids with larger size (higher molar mass) is more viscous because it has stronger
intermolecular forces
 
iii) The temperature of the liquid At higher temperature, molecules have higher kinetic energy,
the molecules move faster molecules can overcome the intermolecular
 
together
between the molecules
compresscompress than gas
5.5. DiffusionDiffusion
Diffusion is the movement of liquid from one fluid through another.
Diffusion can occur in a liquid because molecules
are not tightly packed and can move randomly
around one another.
Vaporisation ProcessVaporisation Process
a process in which molecules escape fromescape from liquid into gaseousliquid into gaseous state through a surfacethrough a surface √ molecules in a liquid moves freely, collide
with each other and posses different magnitudes of kinetic energy
 
√√ i) Surface areai) Surface area
~ the surface area , the chances for the
molecules to escape from the surface
surface areasurface area , evaporation rate, evaporation rate
 
overcome the attractive intermolecular forces
and escape from the surface of the liquid
~ thus evaporation rateevaporation rate
Condensation ProcessCondensation Process a process whereby gaseous molecules turn to liquid
√√ some of the vapour molecules may losesome of the vapour molecules may lose their kinetickinetic energyenergy during the collision
√ they do not have enough energy to remain as vapour molecules
√ they reached the surface of the liquid and trapped by the attractive forces
√ if they cannot overcome the attractive forces, these vapour molecules return as liquid molecules
√ the process is known as condensationcondensation
 
: referred as vapour molecules
vapour molecules will diffused away until
the liquid dries up
 
in a closed system :
√ vapour molecules which leaves the surface are trapped in the close container
√ these vapour molecules are in constant random motion
√ the molecules strike the wall of container and exert some pressure
√ As the quantity of molecules in the vapour phase increase, some molecules may lose energy and condense
Eventually,
(The system achieved dynamic equilibrium)
 
• Temperature
– Applying the heat causes more molecules have high kinetic energy
– More molecules will be able to overcome the intermolecular forces and escape from the liquid to form vapour
– Temperature , vapour pressure
– More vapour molecules will exert pressure on the liquid surface
– Intermolecular forces , vapour pressure
temperature and at an atmospheric
pressure.
any temperature and pressure.
 
Boil ing Point: the temperature at which the vapour pressure of a liquid is equal to the external atmospheric pressure.
 
• Increasing the temperature will increase in the vapour pressure.
• As heat is applied, the vapour pressure of a system will increase until it reaches a point whereby the vapour pressure of the liquid system is equal to the atmospheric pressure.
• Boiling occurs and the temperature taken at this point is known as the boiling point.
• At this point, the change of state from liquid to gas occurs not only at the surface of the liquid but also in the inner part of the liquid.
• Bubbles form within the liquid.
 
1. Atmospheric pressure
When the external atmospheric pressure is low, liquid will boil at a lower temperature
2. Intermolecular forces
 
(b) Explain the process of:
- freezing - sublimation
- melting - deposition
(d) Describe the types of bonding and the interparticle/intermolecular forces involved in the following crystalline solids using appropriate examples.
i. metallic iii. molecular covalent
ii. ionic iv. giant covalent
OBJECTIVE…
2. Rigid arrangement- particles can vibrate, rotate
about fixed position and cannot move freely without
disrupting the whole structure.
5. Has high densities.
 
interconvertible
- its particles gain energy
- therefore able to vibrate more rapidly
- at certain temperature, the kinetic energy is higher enough to overcome the intermolecular forces of
attraction between solid particles.
 
when the temperature of a liquid is lowered, the
kinetic energy of the liquid particles decreases
the liquid particles vibrate at a slower rate
when the intermolecular forces are strong
enough to hold the particles together in a fixed
and orderly arrangement, the liquid freezes.
F r eezin g (S o lid ific atio n ) P r o c es s
 
• The process by which a substance changes
directly from solid to the gaseous state without
passing through the liquid state.
•Occurson solid with weak intermolecular forces
of attraction
• The process where molecules from vapourstate
change to the solid state.
• The opposite process of sublimation
 
amo r phous so l id
Solid that does not have a
regular three dimensional
structure where atoms, ions or
molecules show a regular
repetition in three dimensional
occupyspecific position
 
togetherby metallic bond
 The propertiesof metal:
Examples: all metallic elements : Na, Mg, Fe
 
bonds
- High melting point
- Hard but brittle
does conduct electricity in molten state or in
aqueous state.
intermolecular forces (van der Waals
and/orhydrogen bonds)
covalentbond
Learning Outcomes :
At the end of this lesson students should be able to:
(a) Define phase .
(c) Define triple point and critical point.
(d) Explain the anomalous behaviour of H2O.
(e) Describe the changes in phase with respect to
i. temperature (at constant pressure)
ii. pressure (at constant temperature).
 
Phase is a homogeneous part of a system in contact with other parts of the system but separated from them by a well-defined boundary
Phase Diagram is a diagram that describes the stable phases and phase changes of a substance as a function of temperature and pressure
 
 
Solid, liquid and gas
- A particular phase is stable for any combination of pressure
and temperature.
- Any point along a line shows the pressure and temperature
at which the two phases exist in equilibrium.
 
- Known as the triple point
- Triple point is the point at which the vapour, liquid and
solid states of a substance are in equilibrium.
- Triple point for water is 0.01°C and 0.06 atm.
2. Point C :
- Known as the critical point
- Critical point is the point on a phase diagram at which
the vapour cannot be condensed to a liquid.
- The liquid – gas line ends at the critical point.
- Above the critical point, the liquid cannot be
distinguished from its vapour form.
 
T
A
8/16/11
 At point T, the triple point; solid, liquid and vapour are in equilibrium
CO2(s)  CO2(l)  CO2(g)
Triple point for CO2 is at 5.2 atm pressure and temperature 57 OC.
for most compounds, the TB curve slant to
the right because solid is denser than liquid
TRIPLE POINT 
 The phase diagram for water is not typical.
 The melting temperature line , TB , slopes slant to the left(negative slope)
i.e. the melting temperature decreases with
pressure.
•  This is because ice is less dense than water, the
solid occupies more space than the liquid. An
inc re a se in p re ssure , fa vo urs the fo rm a tio n o f
liq uid H 2 O .( the p ha se tha t oc c up y le ss sp a c e ) 
Thu s, the hig he r the p re ssure , the low e r the
 
Exercise:
 
Exercises:
 
a) a non-reversible reaction
- occurs in one direction
b) a reversible reaction
- occurs in both direction
i.e (forward & reverse reaction)
consumed and amount of products formed
depends on the limiting reactant.
•  A single arrow (→) is used to represent
reaction.
and product reacts to form reactants.
•  Amount of products formed does not depend
on the amount of reactant used.
• Reaction will reach equilibrium when the
concentration of products & reactants
no observable change occurs.
a. Physical Equilib