Inorganic Chemistry (Well Organised)

Chapter one
THE PERIODIC TABLE
TABLE OF CONTENTS
Chapter 1: ................................................................................................................................... 1
1. ATOMIC STRUCTURE ................................................................................................ 1
1.1. Cathode rays ..................................................................................................................... 1
1.2. Neutrons ........................................................................................................................... 4
1.3. Electronic structure of atoms ............................................................................................ 4
1.4. Bohr’s theory of electronic energy levels ......................................................................... 5
1.5. Electronic configuration of atoms .................................................................................... 7
1.6. Rules/principles for writing electronic configurations of atoms ...................................... 9
Chapter 2 .................................................................................................................................... 1
2. THE MODERN PERIODIC TABLE OF ELEMENTS ................................................. 1
2.1. Variation of physical properties in the periodic table (periodic properties) ..................... 2
2.2. Atomic radius ................................................................................................................... 2
2.2.1. Factors that determine the magnitude of 1st ionization energy ...................................... 9
2.2.2. Variation in 1st ionisation energy across period 2 and period 3 in the periodic table .. 11
2.2.3. Determination of ionization energies of an atom of an element .................................. 13
2.2.4. Importance of ionization energies in understanding the chemistry of atoms of
elements: .................................................................................................................................. 15
Variation of electron affinity ................................................................................................... 23
2.3. Electronegativity: ........................................................................................................... 25
2.3.1. Variation in electronegativity across the period 2 and period 3 .................................. 27
2.3.2. b) Variation in electronegativity down a group. .......................................................... 27
2.4. Electropositivity ............................................................................................................. 28
2.5. Factors affecting electropositivity value of an atom of an element ................................ 29
2.6. Variation in electropositivity across the short periods in the periodic table .................. 30
2.7. Melting point .................................................................................................................. 31
i
The simplified version of inorganic chemistry by SSERWADDA Walter
Chapter one
2.7.1. Factors determining melting point ............................................................................... 31

2.7.2. Trends in melting point across period 2 elements ........................................................ 32
2.7.3. Trends in melting points across period 3 ..................................................................... 32
2.7.4. Trend in melting points of Group IIA elements ........................................................... 35
Chapter 3 .................................................................................................................................... 2
3. BONDING AND STRUCTURES. ................................................................................. 2
3.1. Electrovalent or ionic bonding ......................................................................................... 2
3.1.1. Properties of ionic compounds ....................................................................................... 4
3.1.2. Polarization of ionic compounds.................................................................................... 4
3.1.3. Diagonal relationship ..................................................................................................... 5
3.2. Covalent bonding.............................................................................................................. 8
3.2.1. Polarization of covalent bonds ..................................................................................... 10
3.3. Coordinate (dative) bonding ........................................................................................... 11
3.4. Metallic bonding ............................................................................................................. 12
3.5. Van der Waal’s forces of attraction ................................................................................ 13
3.5.1. Evidence for the existence of hydrogen bonding ......................................................... 16
3.6. Shapes of Molecules and Ions; ....................................................................................... 17
Chapter 4 .................................................................................................................................. 27
4. GROUP IA ELEMENTS (ALKALI METALS) and GROUP IIA
ELEMENTS(ALKALINE EARTH METALS) ...................................................................... 27
4.1. Physical properties of group IA elements ...................................................................... 27
4.2. Physical properties of Group II elements. ...................................................................... 27
4.3. Chemical properties of group II and II elements ............................................................ 28
4.4. Compounds of group I and II elements .......................................................................... 33
4.5. Complex formation by group II cations ......................................................................... 40
4.6. Analysis of Magnesium, Calcium and Barium ions in solutions ................................... 41
CHAPTER 5 ............................................................................................................................ 47
5. GROUP (IV) ELEMENTS ........................................................................................... 47
5.1. Physical properties of group (IV) elements .................................................................... 47
Chemical properties of group (IV) elements ........................................................................... 49
5.2. Compounds of group (IV) elements ............................................................................... 52
5.3. Test for Lead (II) ions in solution................................................................................... 62
ii
Chapter one
CHAPTER 6 ............................................................................................................................ 66
6. GROUP (VII) ELEMENTS (HALOGENS) ................................................................. 66
6.1. Physical properties of the halogens. ............................................................................... 66
6.2. General methods for preparing the halogens (excluding fluorine) in the laboratory ..... 72
6.3. Reactions of halogens ..................................................................................................... 72
6.4. Compounds of the halogens ........................................................................................... 74
6.5. Test for Chlorides, Bromides and Iodides ions .............................................................. 79
CHAPTER 7 ............................................................................................................................ 83
7. PERIOD 3 ELEMENTS IN THE PERIODIC TABLE ................................................ 83
7.1. Compounds of period 3 elements ................................................................................... 83
7.1.1. ALUMINIUM .............................................................................................................. 89
CHAPTER 8 ............................................................................................................................ 94
8. TRANSITION ELEMENTS ....................................................................................... 94
8.1. General characteristics of transition elements ................................................................ 96
8.2. CHROMIUM ................................................................................................................ 104
8.2.1. Similarities between the chemistry of Chromium & Aluminium .............................. 113
8.2.2. Test for Cr3+ ions in solution ...................................................................................... 114
8.3. MANGANESE ............................................................................................................. 116
8.3.1. Reactions of manganese ............................................................................................. 116
8.3.2. Compounds of manganese ......................................................................................... 117
8.4. IRON ............................................................................................................................ 125
8.4.1. Similarities in the chemistry of zinc and iron ............................................................ 131
8.4.2. Differences in the chemistry of Zinc & Iron .............................................................. 132
8.4.3. Test for Iron (ii) and Iron (iii) ions in aqueous solution ............................................ 132
8.5. COBALT ...................................................................................................................... 134
8.5.1. Reaction of Cobalt (II) ion, Co2+................................................................................ 134
8.6. NICKEL ....................................................................................................................... 135
8.6.1. Test for Nickel (II) Ion, Ni2+ in solution .................................................................... 135
8.7. COPPER ....................................................................................................................... 137
8.7.1. Extraction from malachite .......................................................................................... 138
8.7.2. Uses of copper ............................................................................................................ 139
8.7.3. Reactions of copper .................................................................................................. 140
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Chapter one
8.7.4. Compounds of copper ................................................................................................ 141

8.7.5. Determination of copper (ii) ions in copper (ii) salts ................................................. 142
8.7.6. Test for Copper(II) ions. ............................................................................................ 143
8.8. ZINC AND ITS COMPOUNDS .................................................................................. 145
8.8.1. Extraction process from sulphide ore ......................................................................... 145
8.8.2. Reactions of zinc ........................................................................................................ 146
8.8.3. Compounds of zinc .................................................................................................... 147
8.8.4. Test for zinc ions ..................................................................................................... 148
8.8.5. Uses of zinc ................................................................................................................ 149
iv
Chapter one
Figure 1: Cathode ray tube with anode directly opposite the cathode...................................... 1
Figure 2: Cathode ray tube with the anode at the side of the tube ............................................ 1
Figure 3: Cathode ray tube with an electromagnetic field ........................................................ 2
Figure 4: Rutherford’s α – Particles scattering Experiment ..................................................... 3
Figure 5: Hydrogen Spectral lines ............................................................................................ 5
Figure 6: Hydrogen spectral series ........................................................................................... 6
Figure 7 Aufbau pattern of filling energy levels ........................................................................ 9
Figure 8 Determination of ionization energies of an atom of an element ............................... 13
Figure 9: ionization energy of potassium against no of electron removed ............................. 16
Figure 10 ionization energy of sodium against no of electron removed ................................. 17
Figure 11: Variation of melting melting point down group II ................................................. 35
v
Chapter one
Table 1 Periodic table ................................................................................................................ 1

Table 2 Atomic radius of group 1 elements ............................. Error! Bookmark not defined.
Table 3 Variation in first ionization energy down group I in the periodic table .................... 10
Table 4: Variation of electronegativity in the periodic table .................................................. 28
Table 5: Structure of simple molecules ................................................................................... 18
Table 6: Physical properties of group 1 elements .................................................................... 27
Table 7: physical properties of group IV elements .................................................................. 47
Table 8: Physical properties of group VII elements ................................................................ 66
Table 9: Melting and boiling points of group VII elemts......................................................... 67
Table 10: Oxy acids of chlorine ............................................................................................... 77
Table 11: Physical properties of group III elements ............................................................... 83
Table 12: The Hydrides of period 3 elements .......................................................................... 88
vi
Chapter one
Chapter 1:
1. ATOMIC STRUCTURE
1.1. Cathode rays

When a gas is enclosed in a glass tube at very low pressure and high voltage is
applied across it, there is observed a stream of rays moving from the cathode
towards the anode.
Gas at low pressure
Cathode (-ve)
Anode (+ve)
Cathode rays
Figure 1: Cathode ray tube with anode directly opposite the cathode
The rays are not necessarily attracted by the anode because when the anode is
placed at a different position, these rays just pass the anode.
Gas at low pressure
Cathode (-ve)
Cathode rays
Anode (+ve)
Figure 2: Cathode ray tube with the anode at the side of the tube
However when an electric field is applied across the path of these rays, they get
deflected towards the positive plate.
1
Chapter one
Gas at low pressure
Cathode (-ve)
Cathode rays
Plate (+ve)
Figure 3: Cathode ray tube with an electromagnetic field
These shows that the rays consist of negatively charged particles
Properties of cathode rays
When they strike the opposite end of the cathode ray tube, the tube
fluorescences with a green light.
An object placed in the path of the rays gives a sharply defined shadow at the
extreme end of the tube. This shows the rays travel in straight lines.
When the blades of a paddlewheel mounted on an axle are placed in the path of
the rays the wheel is made to rotate in a direction away from the cathode. This
shows that the cathode rays are particles that have mass and momentum
They are deflected by a magnetic field and an electric field in a direction that
shows that the particles are negatively charged. JJ Thompson later determined
the charge and mass of these particles as follows; mass of one particle was
found to be equal to 1/1837 of the mass of a hydrogen atom and a charge of 1.6
× 10-19coulombs hence cathode rays are a stream of electrons.
Protons
When different gases are used in a cathode ray tube and perforated cathode is
used, positively charged particles are observed to move in a direction of the
electrons. Using hydrogen gas resulted in production of positively charged
protons which were identical in mass and charge to the positive particles in the
nucleus. These positive ions are produced by collision of electrons from the
cathode with the gaseous atoms of the molecules in the tube.
2
Chapter one
Rutherford’s α-particles scattering experiment
Geirgerand Marden working under Rutherford bombarded thin metallic foils of

gold with α-particles. Many of the particles were observed to pass through the
foil undeflected and a few were deflected at different angles
Diaphragm
Source of α - particles
Thin metal foil screen
α - Particles
Figure 4: Rutherford’s α – Particles scattering Experiment
The many undeflected α-particles were explained to have passed in a space

within an atom where there are no positive particles (protons) hence most of the
atom has space occupied by electrons. The small volume occupied by the
3
Chapter one
nucleus is where there is high concentration of positive charge and it’s located
at the centre of the atom.
1.2. Neutrons
These were later discovered when Beryllium was bombarded with α-particles
𝟗 𝟒 𝟏𝟐
𝟒𝑩𝒆 + 𝟐𝑯𝒆 𝟔𝑪 + 𝟏𝟎𝒏
These were found to have no charge but had mass of the magnitude of the
proton. These particles are also found within the nucleus.
1.3. Electronic structure of atoms

Atomic / Emission spectrum of hydrogen
When electricity passes through a discharge tube containing hydrogen gas at

low pressure, many of the hydrogen molecules break up into single atoms.
These atoms emit both visible and invisible radiations; some of the later being
in the infrared and some in the ultraviolet part of the spectrum. If the radiations
are analyzed by the spectrograph, they yield a line spectrum which can be
photographed. Each line in the spectrum represents a definite wave length of the
radiation. The series of lines in the visible part of the spectrum is called
Balmerseries.
The 3 most prominent lines in these series are labeled Hγ, Hβ and Hα.
Another series of lines in the spectrum is called the Lyman series and is found
in the ultraviolet part of the spectrum and a further series occurs in the infrared
part.
Measurement of wave length corresponding to lines in the various series shows

regularity and wave lengths are in accordance with the simple equation.
1  1 1 
= RH  2  2 
  n1 n2 
Where λ=wave length,
RH=Rydberg’sconstant,
n1 and n2are simple whole numbersindicating the quantum shell.

4
Chapter one
Where n1=1, and n2=2, 3, 4 etc
Figure 5: Hydrogen Spectral lines
Explanation
1.4. Bohr’s theory of electronic energy levels

Electrons in an atom could only rotate in certain passable orbits/ energy levels.
The energy levels are given principal quantum numbers 1, 2, 3, 4 etc with 1
being the one nearest to the nucleus. Under normal conditions, electrons occupy
quantum no. 1 first. This is a state of lowest energy called the ground state /
stationary state.
When atoms absorb energy, they become excited i.e. electrons are promoted to
energy levels with higher energy. But the excited state is an unstable state and
so electrons get rid of the energy as electromagnetic radiations and fall to lower
energy levels frequently in a series of discrete steps. The wave length of the
radiation emitted is determined by the energy difference of the electron in the 2
levels and is given by
𝐸1 − 𝐸2 = ℎ𝑣
Where E1 and E2 are energies of the electron in the higher and lower levels
respectively
H is the Planck’s constant, and

5
Chapter one
𝑐
V is the frequency of light in a vacuum. 𝑉 =
𝜆
Where c is the velocity of light in a vacuum

ℎ𝑐
Therefore 𝐸1 − 𝐸2 =
𝜆
Figure 6: Hydrogen spectral series
The spectral lines get closer and closer as the wave length of the emission
decreases. This is an indication that within an atom of hydrogen or any other
element, the energy levels become closer and closer and their energy of
separation decreases as the distance from the nucleus increases.
The very much close lines got from emission from higher levels to lower levels
are studied and are an indication that within the energy levels where electrons
are distributed, the energy difference between these sub energy levels is very
small since they are so close.
Significance of line spectrum of hydrogen
The line spectrum gives the fact that within atoms, electrons exist in different
energy levels (sub shells, shells, orbitals) in which electrons are distributed
around the nucleus. These energy levels become closer and closer as the
distribution from the nucleus increases.
Question
6
Chapter one
Hydrogen has only one electron but why does its line spectrum consists of
several lines?
Answer
Hydrogen has a single electron. When the electron absorbs sufficient energy, it
gets excited and moves from its ground state to higher energy levels. But the
excited state is an unstable state and so the electron gets rid of the excess energy
as electromagnetic radiations and fall to its lowest energy level in a series of
discrete steps.
Each emission has a different wave length corresponding to the different energy
levels and hence a single electron emits the absorbed energy in stages in definite
amounts giving rise to the different lines in the spectrum.
1.5. Electronic configuration of atoms

This refers to the arrangement of electrons in atoms.
Within an atom, electrons are arranged in shells/energy levels around the central
part of an atom called the nucleus.
The energy levels are numbered 1,2,3,4etc or k, l, m, n etc, with 1 or k being the
shell nearest to the nucleus.
With the exception of the first energy level, the other shells are divided into
sub- energy levels/sub shells.
Within the shells, electrons are found in what is referred to as orbitals.
An orbital is a volume or space around the nucleus within the shells where
electrons are found.
1) The 1st energy level has only one orbital called 1s and like any other s
orbital can take a maximum of two electrons.
7
Chapter one
1S2
Therefore when full the 1st shell has 2 electrons.

2) The 2nd energy level has 2 sub shells called the 2s and 2p.
The 2ssub shell takes in a maximum of two electrons because it has one
orbital.
The 2p sub shell like any other p sub shell has 3 orbitals and when full
has 6 electrons.
2s2 2p6
Therefore the maximum number of electrons that can be accommodated

within the 2nd shell of any atom is 8.
3) The 3rd energy level has 3 sub shells, namely 3s, 3p and 3d.
The 3s takes in up to 2 electrons.
The 3p takes in up to 6 electrons and the 3dshell like any other d sub shell
consists of 5 orbitals and takes up to 10 electrons when full. Therefore the
maximum number of electrons that can be accommodated within the 3rd
shell of an atom is 18.
3s2 3p6 3d10
4) The 4th energy level has 4 sub shell given denoted by 4s, 4p, 4d and 4f.
The 4s sub shell has one orbital and takes in up to 2 electrons.
The 4p sub shell has 3 orbitals and can have up to 6 electrons when full.
The 4d sub shell has 5 orbitals and has up to 10 electrons when full.
The 4f sub shell has 7 orbitals and when full accommodates 14 electrons.
4s2 4p6 4d10 4f14
Therefore the maximum number of electrons that can be accommodated

within the fourth shell of an atom is 32.
Note:
From the 4th shell onwards there are 4 sub shells, namely s, p, d and f.
For e.g. the 5th shell has 5s, 5p, 5d and 5f as its sub shells.
8
Chapter one
1.6. Rules/principles for writing electronic configurations of atoms

a) Pauli Exclusion Principle.
States that, an orbital can take in a maximum of two electrons on a
condition that the electrons have parallel and opposite spins.
b) Hund’s Rule of maximum multiplicity
States that when electrons are present in a number of degenerate orbitals
(orbitals with more or less equal energy) they occupy all the orbitals
singly first with parallel spins before pairing can occur in any one orbital.
c) Aufbau pattern of filling energy level
States that electrons fill in the sub shells in order of increasing energy
beginning with the sub shell of lowest energy, followed by that of second
lowest energy and so and so forth.
After filling the second energy level, the first in a series of several
overlap of energy occurs between the 3d and the 4s sub energy levels.
The 3d sub shell though nearer to the nucleus than the 4s sub shell
happens to be at a higher energy level therefore the electrons are filled in
the 4s sub energy level before filling the 3d sub shell. After filling the 4s
sub shell, electrons then fill the 3d energy level.
Figure 7Aufbau pattern of filling energy levels
9
Chapter one
1
Chapter Two
Chapter 2
2. THE MODERN PERIODIC TABLE OF ELEMENTS
Table 1 Periodic table
The Modern periodic table consists of two major divisions; namely
a) The horizontal arrangement of elements referred to as the periods
b) The downward arrangement of elements referred to as the groups.
Usually atoms within the same group in the periodic table have similar chemical
properties because they contain equal number of electrons in their outer most
shell.
In the periodic table the elements are arranged in accordance with their atomic
numbers. Earlier attempts to arrange the elements in accordance with their
atomic weights failed due to the existence of isotopy amongst some elements
like hydrogen, carbon, chlorine etc.
1
Chapter Two
By using atomic numbers as a basis for placing elements on the periodic table,
the different isotopes of the same element do take up a common place on the
periodic table.
2.1. Variation of physical properties in the periodic table (periodic

properties)
2.2. Atomic radius

Definition:
This is the distance of closest approach from the nucleus of an atom to another
identical atom in a bonding situation.
The bonding situation is metallic bonding if the two identical atoms are metal
atoms and if the identical atoms are non metals, the bonding situation is
covalent bonding.
At such a distance, the sum of the inter-electronic and the inter-nuclear

repulsions just balance the nuclear–electronic attraction.
Alternative definition
Atomic radius is half the inter-nuclear distance between two identical atoms in a
bonding situation.
Still the bonding situation is covalent if the two identical atoms are covalent
atoms and metallic bonding for two identical metallic atoms.
Factors determining the radius of an atom of an element
1) Nuclear charge.
This is determined by the number of protons present in the nucleus of an

atom e.g. sodium atom with 11 protons in its nucleus has a nuclear charge
of +11.
If an atom has a high nuclear charge, then the attraction the nucleus of
that atom has for the electrons present in its shells will also be high
leading to a decrease in atomic radius. If the nuclear charge is low, then
the attraction the nucleus of the atom has for the electrons present will be
low as well leading to an increase in atomic radius.
2
Chapter Two
2) Screening (shielding) effect.
This refers to the shielding (screening) of the electron(s) in the outer most shell
by the electrons present in the inner shell(s). Increase in screening effect on the
outer most electrons increases the atomic radius and the decrease in screening
effect on the outermost electrons decreases the atomic radius.
Variation in atomic radius down a group in the periodic table
Example;
Consider Group IA elements (Alkali metals)..
Table 2 Atomic radius of group 1 elements
Element Atomic Electronic configuration Atomic

Number radius /0A
Li 3 1s22s1 1.52
Na 11 1s22s22p63s1 1.54
K 19 1s22s22p63s23p64s1 2.27
Rb 37 1s22s22p63s23p64s23d104p65s1 2.48
Cs 55 -------- 2.65
Trend
Atomic radius increases from one atom to the next one down any group in the
periodic table.
Explanation.
Down any group in the periodic table, both the nuclear charge and screening
effect increase but the increase in screening effect outweighs that due to nuclear
charge because of an extra shell of electrons added from one element to the next
one.
3
Chapter Two
The effective nuclear charge therefore decreases i.e. the nuclear attraction for
the outer most electron(s) decreases leading to an increase in atomic radius.
Variation in Atomic radiuscross period 2 and period 3 in the periodic table
Trend
Atomic radius decreases across the periods.
Explanation
Across the period from one atom to the next, there is an increase in nuclear
charge but the shielding effect of the electrons in the inner shell(s) on the
electron(s) in the outermost shell remains more or less constant because
electrons are being added to the same shell.
As a result the effective nuclear charge increases and electron(s) in the outer
most shell is increasingly attracted more strongly to the nucleus leading to a
decrease in atomic radius.
Elements Li Be B C N O F Ne
At. 3 4 5 6 7 8 9 10
Number
Electronic 1s22s 1s22s 1s22s22p 1s22s22p 1s22s22p 1s22s22p 1s22s22p 1s22s22p

configuratio 1 2 1 2 3 4 5 6
Atomic 1.52 1.13 0.83 0.77 0.71 0.66 0.71 1.60

radius /0A
The diagrams below show how the atomic radius changes across Period 3.
4
Chapter Two
The figures used to construct this diagram are based on:
Metallic radii for Sodium, Magnesium and Aluminium;
Covalent radii for Silicon, Phosphorus, Sulphur and Chlorine;
Note
The atomic radius of argon is higher than expected because its atoms do no
form covalent bonds because they already have stable electronic configuration.
Instead the atoms have inter atomic van der Waal’s forces and the inter nuclear
distance between any neighboring argon atoms is therefore larger.
Variation in atomic radius among the d-block elements in periodic table
Element At. Elect. Conf

no
Scandium (Sc) 21 1s2

2s22p63s23p63d14s2
Titanium(Ti) 22 1s22s22p63s23p63d24s
2
Vanadium (V) 23 1s22s22p63s23p63d34s

2
Chromium 24 1s22s22p63s23p63d54s
(Cr) 1
Manganese 25 1s22s22p63s23p63d54s
(Mn) 2
Iron (Fe) 26 1s22s22p63s23p63d64s

2
5
Chapter Two
Cobalt (Co) 27 1s22s22p63s23p63d74s

2
Nickel (Ni) 28 1s22s22p63s23p63d84s

2
Copper (Cu) 29 1s22s22p63s23p63d104

s1
Zinc (Zn) 30 1s22s22p63s23p63d104

s2
Trend
Among the d-block elements, the atomic radius remains more or less the same.
Explanation
This is because the increase in nuclear charge from one atom to the next one is
roughly balanced by the increase in screening effect on the outer 4s electrons by
the electrons in the inner shells as the number of electrons on the inner 3d sub
shell (penultimate shell) increases. The effective nuclear charge therefore is
more or less the same throughout and the atomic radius as well.
Cations
A Cation (a positive ion) is formed by the removal of one or more electrons

from an atom. The radius of a cation is smaller than the radius of the atom from
which it’s formed.
Explanation
After the loss of one or more electrons by an atom, the number of protons in the
nucleus becomes greater than the number of remaining electrons, thus the
proton-electron ratio increases.
As a result, the nuclear attraction for the remaining electrons increases leading
to a decrease in cationic radius.
6
Chapter Two
Anions
An anion (a negatively charged ion) is formed by the gain of one or more

electrons by an atom. The radius of an anion is larger than that of the atom from
which it is formed.
Explanation
After the gain of one or more electrons by an atom, the number of electrons
present becomes more than the number of protons present in the nucleus thus
the proton-electron ratio decreases.
The nuclear attraction for the now electrons now decreases leading to increase
in anionic radius.
Isoelectronic ions
Isoelectronic ions are ions that have the same number of electrons and the same
electronic structure.
For example:
Isoelectronic ions Number of electrons

1. Al3+,Mg2+, Na+, F- 10 electrons
2. S2-, Cl-, K+, Ca2+ 18 electrons
Isoelectronic ions
Al3+,Mg2+,Na+, F- all have 10 electrons):
Al3+ has 13 protons; a nuclear charge of +13.
Mg2+ has 12 protons; a nuclear charge of +12.
Na+ has 11 protons; a nuclear charge of +11.
F- has 9 protons; a nuclear charge of +9.
Al3+, with the largest nuclear charge has the highest proton-electron ratio
and therefore the strongest nuclear attraction for the electrons.
Therefore Al3+ has the smallest ionic radius.

7
Chapter Two
Therefore the radius of the ions increase in the order Al3+<Mg2+<Na+<F-

Ion Al3+ Mg2+ Na+ F-
Ionic radius /nm 0.050 0.065 0.095
Ionization Energy
Definition:
Ionization energy is the energy required to remove one mole of electrons from
one mole of gaseous atoms of a given element to form one mole of gaseous
ions.
i.e.
M (g) →M+(g) +e ΔH= Ionization energy
First Ionisation energy
This is the enthalpy change when one of electrons is removed from one mole of
gaseous atoms to form one mole of unipositively charged gaseous ions of the
element.
M (g) → M+ (g)+ e ΔH= 1st ionization energy.
For example
For a sodium atom,
Na (g) ) → Na+(g) + e H=+494kjmol-l
Second ionization energy
Is the minimum energy required to remove one mole of electrons from one mole
of unipositively charged gaseous ions to form one mole of dipositively charged
gaseous ions of the element.
M+(g)→M2+(g) + e H= 2nd ionization

energy.
8
Chapter Two
Example; for a sodium atom;
Na+(g)→Na2+ (g) + e H= +4560KJmol-1
The higher the ionization energy value, the more difficult it is to remove the
electron.
For example, it is more difficult to remove the 2nd electron in sodium than the 1st
electron as can be seen from the 1st and 2ndionisation energies values above.
2.2.1. Factors that determine the magnitude of 1st ionization energy

1) Nuclear charge of the atom
2) Shielding/Screening effect of the electrons in the inner shells on the
electrons in the outer shell
3) Atomic radius
4) Electronic configuration of the atom
Explanation
Nuclear charge.
If the nuclear charge of an atom is high, then the electrons present in the outer
most shell of that atom experience a high nuclear attraction.
Therefore removing an electron from the atom requires more energy leading to
high 1st ionization energy.
For a low nuclear charge, the electrons present in the atom experiences a low
attraction from the nucleus of such an atom leading to low 1st ionization energy.
Shielding effect of the electrons in the inner shells
If the screening effect by the electrons in the inner shells on the electrons in the
outermost shell is high, then the outermost electron experiences a low nuclear
attraction leading to low 1st ionization energy.
If the screening effect on the outermost electron is low, then electron

experiences a high nuclear attraction leading to a high 1 st ionization energy
value.
Atomic radius.
9
Chapter Two
If the radius of an atom is small, then the outer most electrons are closer to the
nucleus and thus experience a high nuclear attraction leading to high 1st
ionization energy.
If the radius of an atom is large, the outer most electrons are further away from
the nucleus and hence experiences a low nuclear attraction leading to low 1 st
ionization energy value.
Electronic configuration of the outer most shell
When the first electron being removed is from an atom whose electronic
configuration is stable (e.g. N: 1s22s22p3 or Ne; 1s2 2s2 2p6), a lot more of
energy is needed to remove the 1st electron leading to a high 1st ionization
energy.
But if the 1st electron being removed is from an atom whose electronic
configuration is thermodynamically less stable (e.g. O; 1s22s22p4) then less
energy is required to remove the 1st electron from such an atom i.e. the 1st
ionization energy value is low.
Variation in first ionization energy down a group in the periodic table
For example consider group IA elements.
Table 3Variation in first ionization energy down groupI in the periodic

table
Element Atomic Electronic First ionisation energy/

number configuration KJmol-1
Li 3 1s22s1 513
Na 11 1s22s23s22p63s1 496
K 19 1s22s22p6s23p64s1 418
Rb 37 403
Cs 55 376
10
Chapter Two
Trend
Ionisation energy generally decreases down any group in the periodic table
Explanation
Down a group in the periodic table from one atom to the next one, both the
nuclear charge and screening effect increase but the increase in screening effect
on the electrons in the outer most shell by the electrons in the inner shells
outweighs the increase in nuclear charge.
This leads to a decrease in the effective nuclear charge and hence the nuclear
attraction for the outer most electrons decreases making the removal of the outer
most electron require less energy. i.e. 1stionisation energy generally decreases.
2.2.2. Variation in 1stionisation energy across period 2 and period 3 in

the periodic table
Consider across period 2 elements
Element Li Be B C N O F Ne
1stIonisation
energy /kjmol-1 513 899 800 1086 11402 1313 1681 2080
11
Chapter Two
Trend
Ionization energy generally increases across period 2 in the periodic table

except that the one for Beryllium is unexpectedly lower than for Boron and also
that for Nitrogen unexpectedly lower than for Oxygen.
Explanation
Across period 2 in the periodic table from one atom to the next one, an electron
is being added to shell number 2 and also the next atom has one more proton in
its nucleus compared with the previous atom. The nuclear charge therefore
increases while the screening effect of the electrons in the inner shell on the
outer most electrons remains approximately constant.
These leads to an increase in the effective nuclear charge leading to an increase

in the nuclear attraction for the outer most electron and hence increase in
1stionisation energy.
Beryllium, 1s22s2 has a higher 1st ionization energy than Boron because the
outer most 2s sub shell is full of electrons and energetically stable.
Boron, 1s22s2 2p1loses the outer most p-electron more easily than Beryllium
despite an increase in the nuclear charge because the shielding effect of the
interposing complete inner s-shells increases thus reducing the effective nuclear
charge considerably i.e. Boron has an unstable electronic configuration.
There after the nuclear charge increases from boron through carbon 1s22s22p2 to
nitrogen 1s22s22p3, in line with an increase in nuclear charge while electrons are
being added to the same shell.
At nitrogen, the 2p-subshell is half full, the three electrons being unpaired
experience minimum repulsion thus thermodynamically stable.
In the case of oxygen 1s22s22p4, electron is being paired in one of the 2p-
orbitals as such there is mutual repulsion between the two paired electrons thus
a decrease in 1st ionization energy is observed.
12
Chapter Two
Further increase in ionization energy is observed on traversing the period from

oxygen to neon. This is in line with increase in nuclear charge as the 2p sub-
shell is building up and reaches a maximum at neon, 1s22s22p6, which has a
complete stable configuration.
A similar trend is observed on traversing the 3 rd short period from sodium to

argon.
Note:
In every period, noble gases have the highest 1st ionization energy. This is
because noble gases have full outer most p sub shell and thus
thermodynamically stable.
Helium: 1s2 has the highest 1st ionization energy of all atoms.
This is because Helium atom has the smallest atomic radius therefore electrons
in its outer most shell is closer to the nucleus and is attracted more strongly.
Also its outermost electronic configuration is thermodynamically stable.
2.2.3. Determination of ionization energies of an atom of an element
G (+)
F
P (-)
Grid
Figure 8 Determination of ionization energies of an atom of an element

By means of the valve above, ionization energies of an atom can be determined.
Procedure
The valve is evacuated by means of a vacuum pump and the gaseous atoms of
the element whose ionistaion energies are required is introduced into the valve
and the inlet closed. The filament F is heated by passing an electric current
through it and it gives off electrons; a process referred to as thermo ionic
emission.
13
Chapter Two
The grid G is charged to varying positive potentials and the plate P is negatively
charged. When the potential on G is set at zero, the electrons emitted by F do
not move, but if the potential on G is gradually increased, the electrons
accelerate towards G. If the potential on G is high enough, as the electrons
accelerate towards G, they collide with the gaseous atoms in the valve and the
atoms ionize according to the equation below
A (g) A+ (g) + e
The emitted electron by A is attracted to the grid G while positive Ion A + move
to plate P. The circuit is then complete and current thus flows and the minimum
grid potential for the current to just flow is noted and it’s called the 1 st
ionization potential, measured in electron volts (eV). 1eV=1.6×10-19 J
The grid potential is then again increased further so that the 2 nd, 3rd, 4thetc
electrons are ejected. Each time an increase in the amount of current flowing
indicates that more electrons are being lost by A+. The 2nd, 3rd, 4thetc ionization
potentials of A are then measured.
Trend
Ionization energies of an atom of an element increases as more electrons is lost

by the atom or its ions i.e. the 2nd I.E is greater than the 1st , the 3rd I.E greater
than the 2nd I.E, the 4th I.E greater than the 3rd etc.
Example
Consider the ionization energies of Beryllium in KJmol-1

1st 2nd 3rd 4th
900 1768 14905
21060
Explanation
14
Chapter Two
The 2nd I.E is greater than the 1st I.E because after a gaseous atom of an element
has lost an electron the number of protons in the nucleus of the ion formed is
one more than the number of electrons remaining in the shells. Therefore the
nuclear attraction for the remaining fewer electrons increases making the lost of
a 2nd electron require more energy.
Note:
Still for Beryllium
Difference
2 I.E
nd
1 I.E
st
1,768 900 868

3rd I.E 2nd I.E
14,905 1,768 13,137
4th I.E 3rd I.E
21,060 14,905 6,155
The largest difference between successive ionization energies is 13,137KJmol-

1
and occurs between the 3rd and 2nd I.Es. This means the shells from which the
two electrons bringing about the largest difference in I.Es come from are
different i.e. the shell from which the 3rd electron is removed is different from
the shell from which the 2nd electron was removed. The first two electrons in
Beryllium therefore were removed from the same outer most shell before
removing a 3rd electron from another shell. Beryllium therefore belongs to
group II in the periodic table.
2.2.4. Importance of ionization energies in understanding the
chemistry of atoms of elements:
Ionization energy provides a basis for understanding the chemistry of elements.
The following information is provided:
1) Atomic number of the element.

This is given by the number of successive ionization energies an atom has
got e.g. sodium with 11 successive ionization energies has atomic number
11.
2) The arrangement of electrons in the shells and distribution of energy

level.
A plot of successive ionization energies of potassium shows distinct
breaks. The arrangement starts with two electrons with fairly similar
ionization energies which are both near the nucleus and difficult to
15
Chapter Two
remove. These are followed by 8 electrons with fairly similar variation in

ionization energies. These have less ionization energies than the first two
electrons. The next are eight electrons also with similar energies but are
easily removed than the previous.
Figure 9: ionization energy of potassium against no of electron removed
16
Chapter Two
In potassium atom there are 4 energy levels and the electrons are arranged
as:
Shell number: 1 2 3 4
No. of 2 8 8 1
electrons:
Since potassium atom has 4 energy levels and therefore belongs to period
4 in the periodic table.
Since it has one electron in the outer most shell, it belongs to group I in
the periodic table.
Note
For a sodium atom with 11 electrons, a similar graph appears as shown in
the figure below.
Figure 10 ionization energy of sodium against no of electron removed
Note
Ionization energies also provide information about the presence of sub
energy levels.
A careful plot of successive ionization energies in the 2nd energy level of
potassium shows that there are 2 electrons with fairly similar ionization
energies which are nearer to the nucleus and they are followed by 6
electrons with less energy but similar. This shows that the electrons in the
2ndenergy level are arranged as:
17
Chapter Two
Sub S P
energy:
No. of 2 6
electrons:
3) Determination of metallic or non metallic character of an atom of an

element.
The magnitude of ionization energy is used as a measure of the metallic
character of an element.
The first ionization energies of metals is all nearly below 800kj per mole
while those of non metals are all nearly above 800kj per mole
Example:
The first three ionization energies for elements A, B, C and D are given in the
table below.
Elements Ionization energies in kJ/mol
First Second Third
A 780 1500 7730
B 500 4560 6900
C 580 1815 4620
D 1310 3460 5300
18
Chapter Two
From the first ionization energies of the elements it can be deduced that
elements B and C have typical metallic properties since their 1st ionization
energies are well below 800kj/mol
Element A with first ionization energy near 800kj/mol also would show some
metallic properties.
Moving from second to third ionization energy for A there is an increase of

about 5 times. This means the third electron of A comes from a different shell
hence A has 2 electron on its outer most shell. A is therefore a group 2 element
with a valence of 2,
For element B, there is an increase of about 9 times from first to second

ionization energies and an increase of about 1 ½times from second to third.
Therefore element B has one electron on its outer most shell and thus is a group
1 element and has a valence of, therefore would react by giving away one
electron on its outermost shell.
In element C, from 1st to 2nd Ionization energy, there is an increase of about 3

times and from second to third by about Therefore there is similar rise in
ionization energy. This implies that the three electrons in C 2 ½ timesare from
the outer most shell and C therefore has 3 electrons in its outermost shell and
hence belongs to group III.
Element D with first ionization energy well over 800kjmol-1 is a non metal.
Exercise 1
The first 8 ionization energies in KJ per mole of an element Y are shown below
1st 2nd 3rd 4th 5th 6th 7th 8th
786 1580 3230 4360 16000 20000 23600 29100
a) Plot a graph of ionization energies against ionization number

b) Explain the shape of the graph
19
Chapter Two
c) State with reasons the group and the period in which the element belongs
in the periodic table
Exercise 2
Element A B C D E F
1st I.E 1013 1000 1255 1519 418 590
2nd I.E 1904 2255 2297 2665 3067 1146
3rd I.E 2916 3389 3853 3933 4393 4916
The table below shows the successive ionization energies of elements A-F.
a) Which element is a noble gas? Give reason for your answer.
b) Which element belongs to;
i) Group I
ii) Group II.
Give a reason for your answer in each case.
Penetrating power of electrons
In an atom with many electrons, up to the 4thshell and beyond may contain
electrons. Within the 4th shell, the sub shells are 4s, 4p, 4d and 4f with the 4s
electrons being nearest to the nucleus followed by the 4p, 4d and finally 4f
electrons being furthest to the nucleus of the atom.
The 4s electrons therefore experience the greatest nuclear attraction and hence
tend to move towards the nucleus. We say the 4s electrons have the greatest
penetrating power and the 4f electrons being furthest from the nucleus
experience the least nuclear attraction and have the least penetrating power of
all the electrons within the 4th shell of the atom.
20
Chapter Two
Electron affinity
Electron affinity is defined as the enthalpy change when one mole of gaseous
atoms gains one mole of electrons to form one mole of univalently charged
gaseous anions.
M (g) + e→ M-(g) H= 1st Electron Affinity.
Non-metallic electronegative elements are more likely to accept one or more

electrons to form anions with a noble gas like structure as such the concept of
electron affinity is more useful or important than ionization energy among non
metals.
The electron affinity is a measure of the attraction for the incoming electron by
the nucleus of the atom gaining the electron.
The greater the attraction, the higher the electron affinity value.
First electron affinity:
First electron affinity is defined as the enthalpy change when one mole of
gaseous atoms gains one mole of electrons to form one mole of univalently
charged gaseous anions.
X (g) + e → X-(g) H= 1st electron affinity
After the gain of an electron by a gaseous atom, the negatively charged gaseous
ion formed repels any further electron to be gained.
Second electron affinity
21
Chapter Two
Is defined as the enthalpy change when one mole of univalently charged

gaseous anions gains one mole of electrons to form one mole of divalently
charged gaseous anions.
X-(g) + e → X2-(g)H=2nd electron affinity
Note
1) First electron affinity is an exothermic process for most gaseous atoms.
Explanation:
This is because the incoming electron experiences a greater attraction
from the nucleus than repulsion from the electrons already present in the
atom.
2) Second electron affinity is always an endothermic process.
Explanation:
This is attributed to the greater repulsive force which the electron being
gained experiences from the electrons already present in the univalently
charged anion; therefore work must be done to overcome the effect of this
repulsion. This work involves input of heat energy.
Because of the same reason, the 3rd, 4th, 5th etc electron affinity will
have a positive H signs.
Factors determining the value of 1st electron affinity of an atom of an element

3) Atomic radius
4) Electronic configuration of the atom.
Explanation
Nuclear charge
22
Chapter Two
If the nuclear charge is high, the nuclear attraction for the incoming electron
will be high. As the atom gains the electron, a lot of energy is released.
If the nuclear charge is low, the attraction for the incoming electron will be low
leading to a low electron affinity.
Atomic radius
If the radius of an atom is small, the incoming electron experiences a high

attraction from the nucleus of the atom. As a result a lot of energy is given out
as the atom gains the electron giving rise to a high electron affinity.
If the radius of an atom is large, the incoming electron experiences a weak

attraction from the nucleus of the atom. As such a lot of energy is given out as
the atom gains the electron giving rise to a high electron affinity.
If the screening effect of the electrons in the inner shells is high, then electron
being gained experiences a low attraction from the nucleus leading to a low
value of electron affinity.
However, if the screening effect is low, the incoming electron experiences a

high attraction from the nucleus leading to a high electron affinity value.
Electronic configuration of the outer shell.
If an electron is being added to an atom with a stable outer electronic

configuration, the addition will not occur easily leading to a low electron
affinity value.
Variation of electron affinity

i) Across period 3 in the periodic table.
Elements Na Mg Al Si P S Cl
electron affinity in kj -20 67 -30 -135 -60 -200 -364

/mol
23
Chapter Two
Trend
First electron affinity generally increases from one atom to the next one across
period 2 and period 3 in the periodic table.
Explanation:
Across the period from one element to the next one, an electron is being added
to the same energy level as the nuclear charge increases by 1 unit. The nuclear
charge increases while the screening effect of the electrons in the inner shells on
the electron to be gained is roughly constant.
The nuclear attraction for incoming electron increases leading to an increase in

the1st electron affinity.
ii) Down a group in the periodic table
On descending any group in the periodic table, the electron affinity decreases.
Explanation.
Down a group from one element to the next one, both the nuclear charge and
shielding effect increase but the increase in shielding effect outweighs that due
to nuclear charge as a result of the more inner shell of electrons added.
The nuclear attraction for the electron being gained decreases leading to a
decrease in 1st electron affinity.
Example.
Consider the first electron affinity in Kjmol-1 for group VII elements.
Element 1st electron affinity
Fluorine -328
Chlorine -349
Bromine -325
Iodine -295
24
Chapter Two
Trend
The 1st electron affinity generally decreases from Chlorine to Iodine but the one
for fluorine is unexpectedly lower than expected.
Explanation
Down the group, the nuclear attraction for the incoming electron decreases.
As a result the electron being gained experiences a weaker attraction from the
nucleus leading to a decrease in 1st electron affinity.
Fluorine, however, has a low value for its electron affinity, numerically less
than that for chlorine.
This is due to very its small atomic radius. The 7 electrons in the outermost
shell are much closer to each other that they repel each other more strongly than
in the other halogens and hence the electron being added experiences a great
repulsive force from the electrons already present leading to a lower than
expected 1st electron affinity.
2.3. Electronegativity:
This is the tendency by an atom in a covalent bond to attract the bonding
electrons more towards itself.
If two atoms with same electronegativity value are covalently bonded to each
other, the resultant bond is 100% covalent and the resultant molecule is non
polar. This happens when the bonded atoms are atoms of the same element.
Example
Cl-Cl, Br-Br, H-H, O═O etc
If two atoms with different electronegativity values are covalently bonded, the
bond electrons are not shared equally. The more electronegative atom attracts
the bond electrons more towards itself and in the process acquires a partial
negative charge and the less electronegative atom acquires a partial positive
charge. The resultant bond molecule is said to be polar and the bond has some
ionic character.
25
Chapter Two
Example
Hδ+ _ Clδ-, Hδ+- Fδ-, Oδ-=Cδ+=Oδ-, H2O, NH3 etc.
Factors determining electronegativity values

3) Atomic radius
Explanation
Nuclear charge
If the nuclear charge of an atom covalently bonded to another atom is high, the
nuclear attraction for the bonding electrons will be high leading to a high
electronegativity value.
For a low nuclear charge, the attraction by the nucleus for the bonding electrons
will be low leading to a low electronegativity value.
Atomic radius
If the radius of an atom is small, the bonding electrons are nearer to the nucleus
and as such experience a high nuclear attraction leading to a high
If the radius of an atom is large, the bonding electrons are far away from the
nucleus and hence experience a low nuclear attraction leading to a low
If the screening effect of the electrons in the inner shells is high, the electrons in
the covalent bond experience a low attraction from the nucleus leading to a low
However, if the screening effect is low, the electrons in the covalent bond
experience a high attraction from the nucleus leading to a high electronegativity
value.
26
Chapter Two
2.3.1. Variation in electronegativity across the period 2 and period 3
Consider across period 3 elements.
Element Na Mg Al Si P S Cl
Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3
Trend
Across the period, electronegativity increases from one element to the next one.
Explanation:
From one element to the next one, an additional electron is added to shell
number 3.
As a result, the nuclear charge increases progressively while the screening effect
of the inner shells of electrons remains almost unchanged.
As a result, the effective nuclear charge increases leading to increase in the

nuclear attraction for the covalent bond electrons and hence an increase in the
2.3.2. b) Variation in electronegativity down a group.
27
Chapter Two
For group VII elements
Element Electronegativity
F 4.0
Cl 3.0
Br 2.8
I 2.5
On descending any group in the periodic table from one element to the next, the
screening effect of the inner shells of electrons outweighs the increase in the
nuclear charge due to an extra inner shell of electrons added.
The effective nuclear charge therefore decreases leading to a decrease in the

nuclear attraction for the covalent bond electrons and hence a decrease in
electronegativity.
Table 4: Variation of electronegativity in the periodic table
2.4. Electropositivity
This is the tendency of an atom to lose one or more electrons from its outer
most shell to form a positively charged ion.
28
Chapter Two
Electropositive elements are those which easily lose one or more electrons to
become positively charged ions e.g. Na, Mg, Ca, K, Rb, Ba etc.
2.5. Factors affecting electropositivity value of an atom of an

element
3) Atomic radius
4) Electronic configuration of the atom.
Explanation:
Atomic radius
If the radius of an atom is small, the outer most electrons are nearer to the
nucleus and experience a high nuclear attraction. It’s then not easy for the atom
to lose one or more electrons leading to a decrease in electropositivity.
If the radius of an atom is large, the outer most electrons are far from the
nucleus and hence experience a low nuclear attraction leading to a low
electropositivity value.
Nuclear charge
If the nuclear charge is high, the electrons in the outer most shell are attracted
more strongly to the nucleus as a result removing an electron from such an atom
is difficult leading to a decrease in electro positivity.
For a low nuclear charge, the electrons in the outer most shell are attracted less
strongly to the nucleus as a result removing an electron is relatively easy
leading to an increase in electropositivity.
Shielding effect of inner shells
If the shielding effect of the electrons in the inner shells on the outer most
electrons is high, the nuclear attraction for the outer most electrons is low
leading to an increase in electropositivity.
29
Chapter Two
For a low screening effect, the nuclear attraction for the outer most electrons
increases leading to a decrease in electropositivity.
Electronic configuration of the atom
If an atom has thermodynamically stable electronic configuration, the lost of an

electron will not occur easily leading to a decrease in electropositivity.
On the other hand if an atom has a less stable electronic configuration, then it
easily loses an electron leading to a high electropositivity.
Variation of electropositivity down a group in the periodic table
Trend
Electropositivity increases in moving down a group in the periodic table.
Explanation:
Down any group in the periodic table, the increase in the screening effect
outweighs that due to nuclear charge due to an extra shell of electrons added
from one element to the next one.
This decreases the effective nuclear charge and as a result the nuclear attraction
for the outer most electrons reduces leading to an increase in electropositivity.
2.6. Variation in electropositivity across the short periods in the

periodic table
Trend
Electropositivity decreases in moving across period 2 and 3 in the periodic

table.
Explanation
Across the period from one element to the next one, the nuclear charge
increases while the screening effect of the electrons in the inner shells remains
30
Chapter Two
more or less unchanged because the additional electron is added to a shell with
similar energy.
The effective nuclear charge therefore increases and as a result the electrons in
the outer most shell are attracted more strongly by the nucleus leading to an
increase in electro positivity.
2.7. Melting point

Melting point is the constant temperature at which the pure solid and liquid
phases of a substance co- exist in equilibrium at a given pressure.
2.7.1. Factors determining melting point
1) Among metals
The melting points of metals depend on the following factors:
a) The number of electrons available for metallic bonding (number of
delocalized electrons per metal atom).
The higher the number of electrons each metal atom contributes for
metallic bonding, the stronger the metallic bond formed and the
higher is the melting point.
b) The atomic radius.
If the radius of a metallic atom is small, the bonding electrons are
attracted more strongly by the nucleus making the inter-atomic
bond length to be shorter and stronger and thus a higher melting
point.
When the radius of a metallic atom is large, the metallic bonding
electrons are weakly attracted by the nucleus as a result the inter-
atomic bond becomes longer and weaker leading to a low melting
point.
c) The crystal structure of the element.
2) Among non-metals (Molecular substances)

Melting points of non-metals depend on:
a) Molecular mass
b) Shape of molecules
c) Type of intermolecular forces of attraction.
31
Chapter Two
2.7.2. Trends in melting point across period 2 elements

Trend
The Melting point of elements in period 2 increases from lithium to carbon and
then decreases abruptly for the non metals nitrogen to neon.
Explanation
The increase in melting point or boiling point from Lithium to Carbon is

attributed to:
Increase in number of electrons available for metallic bonding.
(1 for Lithium, 2 for Beryllium,)
The greater the electrons available, the stronger the bond and the higher the
melting point.
Decrease in atomic radius from Lithium to Beryllium.
The smaller the atomic radius, the closer are the bonding electrons to the
nucleus and thus the shorter and stronger are the metallic bonds.
Change in the crystal structure of the elements across the period.
Lithium has a body centred cubic (b.c.c) structure which has its atoms less
efficiently packed than Beryllium with hexagonal closed packed (h.c.p).
Boron and carbon have giant (macro-) molecular structure composed of large
number of covalently bonded atoms with carbon atoms more strongly bonded
than boron.
However, the non metallic elements form simple molecular structures held by
weak van der Waals forces of attractions as such have very low melting points.
2.7.3. Trends in melting points across period 3

The chart shows how the melting and boiling points of the elements change as
you go across the period. The figures are plotted in Kelvin rather than °C to
avoid having negative values.
32
Chapter Two
The melting points increase from Na to Si and drops abruptly for the non
metallic elements phosphorus to argon.
Melting and boiling points rise across the three metals because of the increasing
strength of the metallic bonds.
The number of electrons which each atom can contribute to the metallic
bonding increases.
The atoms also get smaller (atomic radius decreases) as you go from sodium
through magnesium to aluminium.
The nuclei of the atoms are getting more positively charged and the bonding
electrons are getting progressively nearer to the nuclei and so more strongly
attracted.
Silicon has the highest melting and boiling points because it has a giant covalent
structure. Here strong covalent bonds have to be broken before it melts or boils.
33
Chapter Two
Phosphorus, sulphur, chlorine and argon have simple molecular structures held
by weak van der Waals forces of attractions. Thus their melting points are much
lower.
The melting and boiling points of non metals are governed entirely by the sizes
of the molecules.
Phosphorus
Phosphorus contains smaller P4 molecules. To melt phosphorus you don't have

to break any covalent bonds - just the much weaker van der Waals forces
between the molecules.
Sulphur
Sulphur consists of larger S8 rings of atoms. The molecules are bigger than
phosphorus molecules, and so the van der Waals forces of attractions will be
stronger, leading to a higher melting and boiling point.
Chlorine
Chlorine, Cl2, is a much smaller molecule with comparatively weak van der
Waals forces of attractions, and so chlorine will have a lower melting and
boiling point than sulphur or phosphorus.
34
Chapter Two
2.7.4. Trend in melting points of Group IIA elements

The melting and boiling points of its group IIA metals are higher than those of
corresponding group I elements.
Graph of physical data
Figure 11: Variation of melting melting point down group II
Explanation
Melting points generally decrease going down Group II in the periodic table.
The Group 2 elements are all metals with metallic bonding. In metallic bonding,
metal cations in a metal lattice are attracted to delocalized electrons.
Therefore going down Group II:
The number of delocalized electrons remains the same.
The charge on each metal cation stays the same at +2, but.
35
Chapter Two
The atoms become larger so that the positive nucleus gets further away from the
delocalized electrons.
So the force of attraction between the delocalized electrons and the metal
cations decreases.
Although in general the melting point decreases going down the group, the
melting point for magnesium is anomalously low.
This is because magnesium has a different metallic structure from the other
elements in the group apart from beryllium.
Beryllium and magnesium have a hexagonal close-packed structure (h.c.p)
Calcium and strontium have a face-centred cubic structure (f.c.c) and
Barium has a body-centred cubic structure (b.c.c).
Variation in melting points among the group VII elements (halogens)
Element Molecular Melting point Boiling point

mass /0c /0c
Fluorine 38 -220 -188.
Chlorine 71 -101 -34.
Bromine 160 -7 58.
Iodine 254 114 183.
Trend
Both the melting and boiling points of the halogens increase from one element
to the next on e down the group.
Explanation
The halogens exist as diatomic molecules i.e. F2, Cl2, Br2and I2.
36
Chapter Two
Down the group the molecular mass increases. The increase in molecular mass
leads to an increase in the magnitude of the intermolecular van der Waal’s
forces and hence increases in both melting and boiling points.
Example
Consider the boiling points of the hydrides of group IV elements.
Hydride boiling point / 0c
CH4 -161
SiH4 -112
GeH4 -90
SnH4 -52
PbH4 -12
Trend
The boiling points of the hydrides increase down the group.
Explanation
37
Chapter Two
Down the group the magnitude of the intermolecular van der waal’s forces
increase with increase in molecular mass.
38
Chapter Three
1
Chapter Three
Chapter 3
3. BONDING AND STRUCTURES.
Atoms take part in chemical bonding to attain stable electronic configuration of

noble gases. They do so either through transfer of one or more electrons from
one atom to another; a process referred to as ionic or electrovalent bonding or
through sharing of electrons; a process referred to as covalent bonding.
In all bonding situations, the electrons involved are those in the outer most shell
of the atoms concerned.
The following types of bonding exist;
Ionic or electrovalent bonding
Covalent bonding
Dative or co-ordinate bonding
Metallic bonding
Hydrogen bonding
Van der waal’s forces of attraction.
3.1. Electrovalent or ionic bonding

Electrovalent or ionic bonding arises when there is transfer of one or more
electrons from a metallic atom to one or more non metallic atom.
Example
Consider the formation of sodium chloride.
Before bonding
2
Chapter Three
Na
Cl
After bonding:
Na+
Cl-
Sodium atom looses the single electron in its outermost shell to form
Na+(isoelectronic to Neon).
A chlorine atom on the other hand gains the electron lost by the sodium atom to
form a chloride ion (Cl-)( has the same electronic configuration as argon).
The sodium and chloride ions are then attracted to each other by strong
electrostatic force of attraction leading to the formation of crystalline compound
with a high melting point.
Examples of other compounds formed through ionic bonding include;
Calcium chloride; CaCl2
3
Chapter Three
Potassium chloride; KCl
Potassium iodide; KI
Magnesium oxide; MgO
Calcium carbonate; CaCO3etc
3.1.1. Properties of ionic compounds

a) They are usually crystalline solids of high melting and boiling points.
b) They conduct electricity in both aqueous and molten state.
c) They are usually soluble in water.
3.1.2. Polarization of ionic compounds

Consider an ionic compound AB with A+ and B- as its constituent ions. The
positive ion A+ attracts the electron clouds of the negatively charged ion B-
while at the same time repelling the positively charged nucleus of B-. This
causes distortion of the electron distribution in the B- ions. This distortion is
called polarization and leads to the covalent characteristics in the ionic
compound AB.
The factors leading to polarization and hence covalent character in ionic

compounds are summarized by the Fajan’s rule. The rule states that an ionic
compound will have appreciable covalent character if;
a) Either the cation or the anion is highly charged. A highly charged anion
is highly polarizable and a highly charged cation is highly polarizing.
b) The cation has a small radius making it have a high charge density and
hence a high polarizing power.
Example
Consider the melting points of sodium chloride, magnesium chloride and

aluminium chloride
Chloride Melting point /

0
C
4
Chapter Three
NaCl 800
MgCl2 714
AlCl3 180
Trend
Sodium chloride has the highest melting point followed by magnesium chloride
and lastly by aluminium chloride.
Explanation
The polarizing power of the cations within the chlorides increase in the order
Na+, Mg2+ and Al3+ with increase in cationic charge and decrease in cationic
radius. Therefore Al3+ is the most polarizing and Na+ the least polarizing. The
electron clouds of the chloride ion in sodium chloride are insignificantly
polarized while the electron clouds of the chloride ions in AlCl3 are greatly
polarized. This makes aluminium chloride predominantly covalent with a low
melting point.
The sodium chloride being the most ionic has the highest melting point
followed by magnesium chloride.
Note
Polarization is important in explaining the diagonal relationship of elements in

the periodic table.
3.1.3. Diagonal relationship
Group I II III IV V
Period 2 Li Be B C N
Period 3 Na Mg Al Si P
5
Chapter Three
This is relationship within the periodic table where by certain elements in the
second period have a close chemical similarity to their diagonal neighbours in
the next group of the third period.
This is particularly noticeable with the following pairs.
a. Lithium and Magnesium:
b. Beryllium and Aluminium:
c. Boron and Silicon
Causes of diagonal relationship:
The reasons why the chemistry of the above 3 pairs of elements resemble can be
summarized as follows:
a) The cations in each pair have similar charge densities & similar polarizing
powers.
b) Similar electropositivities of the elements.
c) Similar hydration energies of the cations,
d) Similar electrode potentials,
Resemblance between Lithium & Magnesium

The chemistry of lithium and magnesium resemble in the following ways:
Both react directly with nitrogen on heating to form nitrides.
6Li(s) + N2(g) 2Li3N(s)
3Mg(s) + N2(g) Mg3N2(s)
Both react with oxygen gas when heated to form normal oxides.
4Li(s) + O2(g) 2Li2O(s)
2Mg(s) + O2(g) 2MgO(s)
Both their nitrates decompose on heating to form normal oxide, nitrogen

dioxide and oxygen gas.
6
Chapter Three
4LiNO3 (g) 2Li2O(s) + 4NO2 (g) + O2(g)
2Mg (NO3)2(g) 2MgO(s) + 4NO2 (g) + O2(g)
Both form carbonates that decompose on heating.
Li2CO3(s) Li2O(s) + CO2 (g)
MgCO3(s) MgO(s) + CO2(g)
Both react with carbon when heated to form ionic carbide.
4Li(s) + C(s ) Li4C(s)
2Mg(s) + C(s) Mg2C(s)
Both form chlorides and bromides that hydrolyze slowly and are soluble in
ethanol;
MgCl2(s) + 2H2O (l ) Mg(OH)Cl(l) + H3O+(aq) + Cl -(aq)
The oxonium ions produced make the resultant solution acidic.
Resemblance between Beryllium and Aluminium:
The chemistry of beryllium and aluminium resemble in the following ways;
1) Both elements are passive to concentrated nitric acid.

2) Both elements react with concentrated alkalis to form complexes. They
too react with acids to form salts i.e. they are amphoteric.
Be(s) + 2OH -(aq) + 2H2O (l) Be(OH)42 -(aq) + H2(g)
2Al(s) + 2OH -(aq) + 6H2O (l) 2 Al(OH)4 -(aq) + 3H2(g)
Be(s) + 2H+(aq) Be2+(aq) + H2(g)

2Al (s) + 6H+(aq) 2Al3+(aq) + 3H2(g)
7
Chapter Three
3) Both elements form chlorides which are partly covalent and exist as
dimers in vapour phase.
Cl ClClCl
ClBe BeCl Al Al
Cl ClClCl
4) Their oxides are amphoteric as are their hydroxides.

Oxides
BeO(s) + 2OH -(aq) + H2O (l) Be (OH)42-(aq)
Al2O3(s) + 2OH\ -(aq) + 3H2O (l) 2 Al(OH)4 -(aq)
BeO(s) + 2H+(aq) Be2+(aq) + H2(g)

Al2O3(s) + 6H+(aq) 2Al3+(aq) + 3H2O(l)
Hydroxides
Be (OH) 2(s) + 2OH -(aq) Be (OH)42 -(aq)
Al (OH) 3(s) + OH -(aq) Al (OH) - 4(aq)
Be(OH)2(s) + 2H+(aq) Be2+(aq) + 2H2O(l)

Al(OH)3(s) + 3H+(aq) Al3+(aq) + 3H2O(l)
5) Both their carbides hydrolyze in water to form methane gas.
Be2C(s) + 4H2O (l) 2Be(OH)2(s) + CH4(g)
Al4C3(s) + 12H2O (l) 4Al (OH) 3(s) + 3CH4 (g)
3.2. Covalent bonding

The two Fluorine atoms pair up in order to achieve the same electronic
configuration as neon.
8
Chapter Three
Each atom in the molecules has its outer shell electrons. The atoms in the
molecules are held together by the attractive force of the two nuclei.
N.B;
It’s not only atoms of the same element which may form covalent bonds. Even
atoms of different elements form covalent bonds e.g.
Consider the formation of hydrogen chloride.
Some atoms achieve the noble gas electronic structure by forming double or
triple bonds. e.g;
Consider the formation of oxygen and nitrogen molecules.
OXYGEN
Note
9
Chapter Three
Although atoms share electrons in an attempt to attain noble gas electronic

configuration, they are not always successful e.g.
Consider the formation of Boron trifluoride.
B 3F F B F
F
Note
In BF3 each of the fluorine atoms has attained a stable configuration of Neon
but Boron has only six electrons in its outer most shell
Properties of covalent compounds

 They consists of molecules
 They have low melting and boiling points.
 They are usually insoluble in water but soluble in non polar solvents like
tetra chloromethane ( CCl4), carbon disulphide(CS2), and benzene (C6H6)
etc
 They are poor conductors of electricity in molten state.
3.2.1. Polarization of covalent bonds

Polarization of ions in some ionic compounds leads to a degree of covalent
character being developed and so it would also seem feasible that polarization
of covalent bonds should give them some ionic character.
Polarization of covalent bonds occurs due to atoms having different

electronegativities. If two atoms of the same element are covalently bonded, the
electrons in the bond will be equally shared between them and the resultant
bond is 100% covalent.
However, if the two bonded atoms are different, the bond electrons will be
displaced towards the atom with the higher electronegativity and the bond is
said to be polarized and will have some ionic character. It is apparent that bonds
10
Chapter Three
between atoms with a large difference in electronegativity will be highly

polarized i.e. have a considerable ionic character.
Note
There are few molecules that form pure covalent bonds. Even formation of pure
ionic compounds is extreme cases. Most compounds have intermediate type of
bond i.e. partially ionic and partially covalent but with one character
dominating.
3.3. Coordinate (dative) bonding

This is a special type of covalent bonding in which the two electrons being
shared in the bond come from only one of the participating atoms (ions). The 2 nd
atom (ion) only provides space to accommodate the 2 electrons. Although the
mode of formation is different, once formed, a coordinate bond is the same as
any covalent bond.
Note
Coordinate bond is sometimes indicated by an arrow to show that they have

been formed differently from covalent bonds. The arrow points away from the
donor towards the recipient.
Example
I. Consider the formation of ammonium ion from ammonia and

hydrogen ion
II. Consider the formation of hydroxonium ion from water and hydrogen
ion.
11
Chapter Three
III. Consider the reaction between BF3 and NH3
Question
The carbon-chlorine bonds in tetrachloromethane molecule are polar and yet

tetrachloromethane molecule as a whole is non polar. Explain.
Answer
Because chlorine is more electronegative than carbon, the electrons in the

carbon-chlorine bonds in CCl4 areattracted more towards the chlorine atoms.
The chlorine atoms develop partial negative charges and the carbon atom
partial positive charge i.e. the carbon-chlorine bonds are polar. But because the
4 bonds are in opposite directions and are all equal, the polar effect cancel each
other and overall CCl4 is non polar.
3.4. Metallic bonding

Practically metals are silverish white in colour and are bright and shiny because
they reflect all frequencies of light falling on them.
Note
Light absorption and reflection are related to energy levels and so the deduction
is that metals utilize similar energy levels.
Other common features metals have in common include low ionization energies
and vacant orbitals in their outer most shells. The presence of vacant orbitals in
their outer most shell is essential for metallic bonding. Metals may be regarded
as consisting of ions occupying fixed positions with the valency electrons
12
Chapter Three
moving in orbitals around the fixed ions. The valency electrons are said to be
delocalized and the conduction of electricity and heat by metals is due to these
delocalized electrons. The strength of the metallic bond and hence the melting
point of metals increases as the number of valency electrons increases.
3.5. Van der Waal’s forces of attraction

Van der Waal’s forces arise from induced fluctuating dipoles in atoms and
molecules. When one atom or molecule approaches another, the electrons of
one or both are temporarily displaced owing to their mutual repulsion and
polarization is said to occur. The slightly positive end of one molecule attracts
the slightly negative end of the neighbouring molecule. It is this attraction
which is the origin of van der Waal’s forces of attraction.
As the atoms/ molecules move on, their electron clouds return to normal until
their next encounter.
The magnitude of the intermolecular Van der Waal’s forces increase with
increase in the number of electrons in the molecules which depends on
molecular mass.
Note
Both the melting and boiling points of molecules depend on the magnitude of
the intermolecular Van der Waal’s forces.
13
Chapter Three
Example 1
Variation in melting points among the group VII elements (halogens)
Element Molecular Melting point Boiling point

mass /0c /0c
Fluorine 38 -220 -188.
Chlorine 71 -101 -34.
Bromine 160 -7 58.
Iodine 254 114 183.
Trend
Both the melting and boiling points of the halogens increase from one element
to the next one down the group.
Explanation
The halogens exist as diatomic molecules i.e. F2, Cl2, Br2and I2.
Down the group the molecular mass increases. The increase in molecular mass
leads to an increase in the magnitude of the intermolecular van der waal’s forces
and hence increases in both melting and boiling points.
Example 2
Consider the boiling points of the hydrides of group IV elements.
Hydride boiling point / 0c
CH4 -161
SiH4 -112
GeH4 -90
SnH4 -52
14
Chapter Three
PbH4 -12
Trend
The boiling points of the hydrides increase down the group.
Explanation
Down the group the magnitude of the intermolecular Van der Waal’s forces
increase with increase in molecular mass.
Hydrogen Bonding
This is a dipole –dipole attraction between a hydrogen atom bonded to an

electronegative atom and a second electronegative atom having a lone pair of
electrons. The atoms of the following elements are electronegative;
 Fluorine is the most electronegative element.

 Oxygen is the 2nd most electronegative after fluorine.
 Chlorine
 Bromine
 Iodine
 Nitrogen
For intermolecular hydrogen bonding to exist, at least one hydrogen atom

within the molecule being considered must be directly bonded to an
electronegative atom. The following molecules therefore exhibit intermolecular
hydrogen bonding.
H2O, NH3, CH3CH2OH, HCl, HBr, HI, CH3NH2, CH3COOH etc
Note
Hydrogen bonds are infact similar to Van der Waal’s forces except that they are
much stronger than Van der Waal’s forces in molecules of comparable
molecular masses.
15
Chapter Three
3.5.1. Evidence for the existence of hydrogen bonding

Example 1
Consider the hydrides of group VI elements
Hydride Molecular boiling point/

mass/g 0
c
H2O 18 100
H2S 34 -60
H2Se 81 -41
H2Te 130 -2
H2Po 211
Trend
Except for water, the boiling point increases down the group with increase in
molecular mass of the hydride.
Explanation
Save for water, the boiling point of the hydrides increase down the group with
increase in the magnitude of the intermolecular Van der Waal’s forces with
increasing molecular mass.
16
Chapter Three
Note
Water with the lowest molecular mass has the highest boiling point due to the
very strong intermolecular hydrogen bonds between its molecules.
Example 2
Consider the hydrides of group VII
Hydride Boiling point /0c
HF 20
HCl -85
HBr -67
HI -35
Trend
Except for HF, the boiling point increases down the group with increase in
molecular mass of the hydrides.
Explanation
From hydrogen chloride through hydrogen bromide to hydrogen iodide, the

boiling point increases down the group due to the increase in magnitude of the
intermolecular van der waal’s forces with increasing molecular mass of the
hydrides.
Note
In hydrogen fluoride, the intermolecular hydrogen bonds are quite strong

because fluorine is highly electronegative. This explains the very high boiling
point of hydrogen fluoride in comparison to the other hydrides in the group.
3.6. Shapes of Molecules and Ions;

The shape of a given molecule or ion is determined by number of electron pairs
present in the outermost shell of the central atom within the molecule or ion.
17
Chapter Three
These electron pairs repel one another one and so they try to get as apart as
possible.
 In the outer shell of the central atom there could be;

o Only bond pairs of electrons e.g. in methane; CCl4
o Bond and lone pairs of electrons e.g. in NH3, H2O, H2Setc
 A lone pair of electrons is closer to the nucleus of the central atom than
the bond pairs and therefore the repulsion of the lone pairs of electrons
have for the bond pairs is so strong i.e. the bond pairs are pushed closer to
each other by a lone pair.
 Multiple (double or triple) bonds occupy only one position in space and
so from the point of view of molecular shapes act as a single bond
Table 5: Structure of simple molecules
Numbe Structure Shape Bond Example

r of angle
bond s
pairs
on the
outer
most
shell of
the
central
atom
2 (All Linear 1800 CO2 , HCN
bond
Pairs)
3 (All Trigonal Planar 1200 BF3 , BCl3
bond
pairs)
18
Chapter Three
4 (All Tetrahedral 109.5 CCl4, CH4

bond 0
pairs)
5 (All Trigonalbipyramid 1200, PCl5

bond al 900
pairs)
6 (all Octahedral 900 [Al(H2O)6]3

bond +
pairs) [Fe(H2O)6]3
+
The above structures look like the figures shown below in three dimensions.
19
Chapter Three
Example 1
CH4
Central atom is C. Its atomic number is 6 (2:4)
The central carbon atom contributes a total of 4 electrons. (4×1)
The 4 hydrogen atoms contribute a total of 4 electrons.
Total number of electrons is 8 electrons (4 pairs, all bond pairs.)
The H-C-H bond angle is 109.50
Note.
Other molecules that adopt tetrahedral shape include CCl4, SiCl4 e.t.c
Example 2
PCl5
central atom is P
Its atomic number is 15 (2:8:5)
20
Chapter Three
P contributes 5 electrons.
The 5 chlorine atoms contribute a total of 5×1 =5 electrons.
The total of electrons is 10 electrons = (5 pairs, all bond pairs)
Structure: shape:
Trigonalbipyramidal
Molecules having both lone and bond pairs of electrons.
Example 1
NH3
The central atom is nitrogen
Its atomic number is 7 (2:5)
Nitrogen atom contributes 5 electrons.
The 3 hydrogen atoms contribute (3×1) 3 electrons.
Total = 8 electrons (4 pairs; 3 bond pairs and1 lone pair).
Exercise
Draw the structure and name the shape of the above molecule
Note.
In NH3there are 4 electron pairs in the outer shell of the central nitrogen atom.
Given 4 electron pairs, the angle between any two bonds in a molecule is
expected to be 109.50 but in ammonia, the H-N-H bond angle is only 1070.
Explanation.
21
Chapter Three
Of the 4 electron pairs present in the outermost shell of the central nitrogen
atom in ammonia, 1 pair is alone pair and the other 3 are bond pairs.
The lone pair being closer to the nitrogen nucleus repels the 3 bond pairs more
strongly than the bond pairs repel themselves and hence the H-N-H angle
reduces to 1070.
Example 2
H2O
Central atom is oxygen.
Its atomic number is 8 (2:6 or 1S22S22P4)
Oxygen contributes 6 electrons
The 2 Hydrogen atoms contribute a total of 2 electrons.
Total number of electrons = 8 electrons (4 pairs; 2 lone and 2 bond pairs).
Structure.Shape
V- Shaped or angular or bent
O
H H
Note
The H-N-H bond angle is 104.50 instead of the expected 109.50
Explanation.
The 2 lone pairs push the 2 bond pairs more strongly than the bond pairs push
themselves.
The repulsion between the bond pairs of electrons will therefore be minimal
when the H-N-H bond angle is 104.50
22
Chapter Three
Example 3
NH4+ (NH3 + H+ NH4+)
Central atom is N 7 (2:5)
N atom contributes 5 electrons
3 hydrogen atoms contributes 3 ×1= 3 electrons
H+ contributes 0 electrons
Total number of electrons = 8 electrons (4 pairs; all bond pairs).
Structure: shape:
Tetrahedral
Example 5
PCl3
Central atom is phosphorous. 15 (2:8:5)
Phosphorus contributes 5 electrons
3 Chlorine atoms contribute a total of 3 ×1 =3 electrons
Total number of electrons =8 electrons (4pairs; 3bond and1lone pair)

23
Chapter Three
Structureshape Trigonal
pyramidal
Example 6
CO2
Central atom is C, 6 (2:4) (1S22S22P2)
Carbon contributes 4 electrons.
The two 02 atoms contribute (2×2) = 4 electrons.
Total number of electrons =8 electrons (4 pairs.)
Structure shape: Linear
c) S2O42-( Tetrahedral)
O C O
Shapes of some common ions.
a) CO32 -(Trigonal Planar)
Structure:
.
d) NO3-(Trigonal Planar)
Structure:
24
Chapter Three
f) CrO42-
Structure
g) MnO4-
e) NO2-(V – shape)
Structure,
Structure:
Question
Both H2O and H2S molecules adopt v- shaped but the bond angle in H2O is
greater than that in H2S.
Answer
Oxygen atom being more electronegative than sulphur pulls the bond electrons
more towards its self and in the process drawing the bond pairs of electrons
nearer. This increases the repulsion between the bond pairs and to reduce the
repulsion to a minimum, the bond angle increases.In H2O therefore the two
bond electrons repel each other more strongly than those in H2S. The greater
the repulsion between the bonds pairs the greater the bond angles.
25
Chapter Four
26
Chapter Four
Chapter 4
4. GROUP IA ELEMENTS (ALKALI METALS) and

GROUP IIAELEMENTS(ALKALINE EARTH METALS)
4.1. Physical properties of group IA elements

Table 6: Physical properties of group 1 elements
Element Atomic Outer 1stI.E Atomic Melting pt

no elect conf /kjmol-1 radius / / 0c
0
A
Lithium 3 -2s1 520 0.15 181
Sodium 11 -3s1 500 0.19 98
Potassium 19 -4s1 420 0.23 63
Rubidium 37 -5s1 400 0.24 39
Caesium 55 -6s1 380 0.60 29
The alkali metals are solids with body centred cubic (b.c.c) arrangement.
They have one valence electron with outer most shell electronic configuration
of ns1 (where n= 2, 3, 4, 5,6).
They have low melting and boiling point (only one electron per atom is
available for metallic bond formation).
They have the largest atomic radii and lowest ionization energies than any
element within the period.
They are highly electropositive and the most reactive metals.
GROUP IIA ELEMENTS (ALKALINE EARTH METALS)
4.2. Physical properties of Group II elements.

Element Atomic no Outer elect Ionic 1st ionization
conf radius/0A energy/kjmol-1
Beryllium 4 -2s2 0.31 899
Magnesium 12 -3s2
0.65 738
Calcium 20 -4s2
0.97 589
27
Chapter Four
Strontium 38 -5s2 1.13 549

Barium 56 -6s2 1.35 502
The alkaline earth metals are electropositive metals with 2 valence electrons
(i.e. two electrons in their outer most shell).
The general electronic configuration of their outer shell is ns2 (n= 2, 3, 4, 5, 6)
They have smaller atomic radii than corresponding group I metals.
They have higher melting and boiling points than corresponding group I
metals.
They have higher 1st ionization energies than corresponding elements in group
I.
They are less electropositive, less reactive and form less ionic compounds than
group I.
4.3. Chemical properties of group II and II elements

1. Reaction with Hydrogen.
The more electropositive metals of group I & II react with hydrogen when
heated to form ionic hydride except Beryllium and Magnesium which form
covalent hydride.
2Na(s) + H2(g 2NaH(s)
Ca(s) + H2(g) CaH2(s)
The ionic hydrides are hydrolyzed by water to form hydroxide and hydrogen
gas.
NaH(s) + H2O(l) NaOH(aq) + H2(g)
2. Reaction with Water.
28
Chapter Four
All group I metals react with water more vigorously than their group II
counterparts to form corresponding hydroxide and hydrogen gas. The vigor of
reaction increases down the group.
Explanation
A group I element loses one electron while a group II element loses two
electrons. Therefore less energy is required to remove one electron than two
from the outer most shell and as such group I metals react faster with water
than their neighbouring metals of group II.
As the atomic radius increases down the group, ionization energy decreases
therefore increasing reactivity in each group. The lower the ionization energy,
the lower the activation energy for reaction and the faster the reaction.
2X(s) + H2O(l) 2XOH(aq) + H2(g). X = Li, Na, K, Rb, Cs.
Beryllium does not react with water.
Magnesium reacts with steam to form Magnesium oxide (white solid) and
hydrogen gas.
Mg(s) + H2O(g) MgO(s) + H2(g).
Calcium, Strontium, and Barium react with increasing vigour with cold water
forming corresponding hydroxide and hydrogen gas.
X(s) + 2H2O(l) X(OH)2(aq) + H2(g)
X= Ca, Sr, & Ba.
Note
Solubility of group II hydroxides in water
Hydroxide Solubility per 100g of

water at 250c
29
Chapter Four
Be(OH)2 0.00
Mg(OH)2 0.01
Ca(OH)2 0.15
Sr(OH)2 0.89
Ba(OH)2 3.32
Trend
The solubility of group II hydroxides increases down the group.
Explanation
Down the group, the radius of the cations increases while the charge on them
remains the same. This leads to a decrease in both the lattice and hydration
energies.
But the decrease in lattice energy of the hydroxides is more rapid than the
decrease in hydration energy.
As such the lattice energy is easily superseded by the hydration energy thus
facilitating solubility of the hydroxides.
3. Reaction with acids.
a) Dilute and concentrated hydrochloric acid
Group II metals react with hydrochloric acid forming corresponding salt

and hydrogen gas.
M(s) + 2H+(aq)M2+(aq) + H2(g) (M= Be, Mg, Ca, Sr, Ba)
b) Dilute sulphuric acid.
30
Chapter Four
Beryllium reacts with hot dilute sulphuric acid to form salt and hydrogen
gas.
Be(s) + 2H+(aq) Be2+(aq) + H2(g)
Magnesium reacts with cold dilute to form salt and hydrogen gas.
Mg(s) + 2H+(aq) Mg2+(aq) + H2(g)
Calcium, Strontium & Barium react to form a sparingly soluble sulphate

which renders the reaction passive.
c) Reaction with conc. sulphuric acid
Be reacts in the same way as with dilute sulphuric acid.
However other members of the group react with conc. sulphuric acid to
form salt, sulphur dioxide and water.
M(s) + 2H2SO4(l) MSO4(aq) + SO2(g) + 2H2O(l) (M = Mg,

Ca, Sr, Ba.)
d) Reaction with nitric acid.
Beryllium does not react with nitric acid at all conditions.
Magnesium reacts with dilute nitric acid to form salt and hydrogen gas.
Mg(s) + 2HNO3(aq) Mg(NO3)2(aq) + H2(g)
Magnesium reacts with conc. nitric acid to form salt, nitrogen dioxide and
water.
Mg(s) + 4HNO3(aq) Mg(NO3)2(aq) + 2NO2(g) + 2H2O(l)
Calcium, Strontium & Barium react with nitric acid to form salt, nitrogen
dioxide and water.
Ba(s) + 4HNO3(aq) Ba(NO3)2(aq) + 2NO2(g) + 2H2O(l)
4. Reaction with Oxygen.
The s-block elements react with oxygen to form 3 types of oxides namely
31
Chapter Four
a) Normal oxide (O2-)
b) Peroxide (O22-)
c) Superoxide (O2-)
All group I metals form normal oxide with oxygen.
4M(s) + O2(g) 2M2O(s) (M= Li, Na, K, Rb, Cs.)
Elements sodium to caesium in addition form peroxides.
2M(s) + O2(g) M2O2(s)(M= Na, K, Rb, Cs.)
Elements K to Cs in addition form superoxide.
M(s) + O2(g) MO2(s)(M= K, Rb ,Cs)
Note
Lithium is not capable of forming higher oxides due to the fact that Li + is very
small, with high charge density & high polarizing power. Thus the larger oxides
cannot be accommodated around it otherwise the compound would be highly
covalent & unstable.
All group II metals when heated in air form normal oxides.
2M(s) + O2(g) 2MO(s) (M= Be, Mg, Ca, Sr, Ba)
The more electropositive metals form peroxides e.g Strontium & Barium.
M(s) + O2(g) MO2(s)(M = Sr, Ba)
Note:
The tendency to form peroxide increases down the group. This is because the
radii of the cations increase down the group, as such their charge densities and
polarizing powers decrease. Therefore the larger peroxide can easily be
accommodated around the cation with least polarization forming stable lattice.
Beryllium oxide is amphoteric, shows both basic and acidic properties.
32
Chapter Four
BeO(s) + 2OH-(aq) + H2O(l) Be(OH)42 -(aq)
BeO(s) + 2H+(aq) Be2+(aq) + H2(g)
The other oxides are ionic and basic in nature.
5. Reaction with halogens.
All the metals of group I & II combine directly under heat to form chlorides.
2Li(s) + Cl2(g) 2LiCl(s)
Mg(s) + Cl2(g) MgCl2(s)
4.4. Compounds of group I and II elements

Compounds of group II elements are less ionic than their corresponding group I
compounds.
Explanation
The cations of group II have smaller ionic radius than corresponding group I.
Also group II cations are doubly charged while those of group I are singly
charged.
Therefore cations of group II have higher charge density and more polarizing
power than corresponding group I cations. As such compounds of group II are
less ionic than corresponding group I compound. e.g. Magnesium chlorideisless
ionic than Sodium chloride.
Solubility of ionic salts in water
Solubility of an ionic salt in water is determined by two energy terms:
a) Lattice energy of the salt.
b) Hydration (solvation) energies of the constituent ions of the salt.
Lattice energy
This is the energy required to decompose 1 mole of an ionic salt into its
constituent gaseous ions.
NaCl(s) → Na+ (g) + Cl -(g)H = Lattice energy
Or
33
Chapter Four
It is the amount of energy given out when 1 mole of an ionic salt is formed from
its constituent gaseous ions.
Na+(g) + Cl -g) → NaCl(s)H = Lattice energy.
Hydration energy
This is the amount of energy released when 1 mole of gaseous ions is fully
dissolved in water at a given temperature.
Na+ (g) + aq → Na+ (aq)H = Hydration energy.
Hydration energy has a negative ΔH sign because it involves release of heat

energy.
Example
1) Each ion has its own hydration energy.
Mg2+(g) + aq → Mg2+(aq)HθHyd(298k) = -696kj/mol.
Na+(g) + aq → Na+(aq)HθHyd(298k) = -406kj/mol.
Cl-(g) + aq → Cl-(aq)HθHyd(298k) = -377kj/mol.
Therefore the HθHyd(298k) of NaCl = (-406 + -377) = -783kj/mol and that

of MgCl2 = (-696 + 2x-377)= -1450.
2) The smaller the ionic radius the higher the charge density and so the
greater (more negative) the hydration energy.
3) Water is a suitable solvent for dissolving ionic salts because:
a) it is a polar solvent,
b) has a high dielectric constant.
c) has a large dipole moment, so that ion-dipole interaction is high.
If the hydration energy of a substance is greater than its lattice energy, the
substance will dissolve exothermically in the solvent.
34
Chapter Four
If the hydration energy is less than the lattice energy then the substance
dissolves endothermically.
However, if lattice energy is much larger than hydration energy then the salt
does not dissolve in water.
Enthalpy of solution
This is defined as the heat change when 1 mole of an ionic salt is completely
hydrated in water.
NaCl(s) + aq Na+ (aq) + Cl -(aq)H = enthalpy of solution of
sodium chloride.
Na+(g) +Cl-(g)
H lattice H hydration
NaCl(s)
H solution
Na+(aq) + Cl-(aq)
Figure shows Born-Haber cycle for solubility of sodium chloride in water.
35
Chapter Four
H solution = H lattice + H hydration
Note:
Whenever this formula is being applied in calculation, the value of H lattice

must be positive. This is because the formula is only valid when the salt is being
decomposed as shown in the cycle above.
Exercise 1
The lattice and hydration energies of MgCl2 are -2644 kJmol-1 and -2653 kJmol-1
respectively.
a) Draw an energy diagram (Born-Haber cycle) for the solubility of

MgCl2inwater and indicate the energy changes that occur.
b) Calculate the enthalpy of solution of the salt.
Exercise 2
a) Using potassium iodide, draw energy diagram (Born Haber Cycle) to

show the energy changes during the solubility of the ionic salt in water.
b) The enthalpy of solution and lattice energy of potassium iodide are

+21kJmol-1 and -642kJmol-1 respectively. Calculate the hydration energy
for potassium iodide.
Factors determining the lattice energy of a given ionic salt
a) Ionic charge (charge on the ion)
b) Ionic radius
Explanation
Ionic charge
Lattice energy is directly proportional to the ionic charge.
36
Chapter Four
If ionic charge is high, then the electrostatic forces of attraction between the
oppositely charged ions are stronger leading to high lattice energy.
If ionic charge is small, the electrostatic forces of attraction between the

oppositely charged ions are weaker leading to low lattice energy.
Ionic radius
Lattice energy is inversely proportional to ionic radius.
If ionic radius is large, the electrostatic forces of attraction between the

oppositely charged ions are weaker leading to low lattice energy.
On the other hand if the ionic radius is small, the electrostatic force of attraction
between the oppositely charged ions is greater leading to high lattice energy.
Salts of group 2 elements are less soluble in water than the corresponding
group 1 salts.
Explanation
The cations of group II have smaller ionic radius than corresponding group I.
Group II cations are doubly charged while those of group I are singly charged.
Therefore the lattice energy of group II salts are much higher than those of
corresponding group I salts.
The higher lattice energy of group II salts makes them less soluble than their
corresponding group I salts.
Sulphates of group I& II elements
Sulphates of group I are white solids which are stable to heat and are readily
soluble in water.
Sulphates of group II are white solids and sparingly soluble in water.
Solubility of group II sulphates in water
37
Chapter Four
Sulphate Solubility per 100g of water at 250c
BeSO4 43
MgSO4 36
CaSO4 0.20
SrSO4 0.011
BaSO4 0.0024
Trend
The solubility of the sulphates decrease down the group
Explanation
Down the group, the radius of the cations increases while the charge on them
remains the same. This leads to a decrease in both the lattice and hydration
energies.
Therefore the decrease in solubility of the sulphates down the group is attributed
to the fact that the hydration energy of the suphates decreases more rapidly than
the lattice energy down the group.
Note
Solubility of the nitrates carbonates and chromates follow a trend similar to

those of the sulphates.
Carbonates of group I & II elements
The carbonates of group I metals are white solids; readily dissolve in water and
are unaffected by heat except lithium carbonate which decomposes on heating.
Li2CO3(s) Li2O(s) + CO2(g)
38
Chapter Four
Explanation
Li+ ion has a very small radius giving it a high charge density and a high
polarizing power. As such the Li+ ion is capable of forming a stable compound
with an oxide ion.
However, as the group is descended the radii of the cations increase and their
charged densities decrease. Thus they cannot easily be accommodated around
the smaller oxide ion due to its high polarizing effect.
Carbonates of group 2 decompose on heating to form the metal oxide and

carbon dioxide gas.
MCO3(s) MO(s) + CO2(g)( M= Be, Mg, Ca, Sr, Ba)
The thermal stability of the carbonates increases as the group is

descended.
Carbonate Decomposition
temp. 0C
BeCO3 100
MgCO3 350
CaCO3 900
SrCO3 1290
BaCO3 1350
Explanation
On descending the group, the charge densities of the metal cations progressively
decrease due to increase in cationic radius.
Since an oxide is smaller than a carbonate ion, the decomposition of metal

carbonates becomes less energetically favourable down the group.
39
Chapter Four
Nitrates of group I & II elements.
Group I nitrates (except lithium nitrate) decompose on heating

forming nitrites and oxygen.
2MNO3(s) 2MNO2(s) + O2(g)( M= Na, K, Rb, Cs)
However lithium nitrate decomposes to form metal oxide, nitrogen

dioxide and oxygen.
4LiNO3(s) 2Li2O(s) + 4NO2(g) + O2(g)
Explanation
The oxide (O2-) ion is much smaller than the NO3- ion. Li+ ion has a very small
radius giving it a high charge density and a high polarizing power.
As such the smaller oxide ion (O2-) approaches the Li+ ion more closely and thus
forming a stable oxide.
However, as the group is descended the radii of the cations increase and their
charge densities decrease. Their nitrates decompose to relatively smaller nitrites.
The decrease from nitrate to nitrite stabilizes the compound.
4.5. Complex formation by group II cations

Complex ion formation is favoured by:
a) Presence of suitable empty orbitals on the cations to accommodate the

lone pairs of electrons donated by ligands.
a) Small cationic radius.
b) High cationic charge.
The tendency to form complex ions decreases rapidly down the group with
increasing cationic radius & decrease in the charge density of cations. This is
because the attraction for the lone pairs of electrons on the ligands decreases
with decrease in cationic charge density.
40
Chapter Four
Beryllium cation forms complex easily due to its very small ionic radius & high
charge density e.g. Be(OH)42-, Be(H2O)42+
Reactions in which Beryllium resembles other group II elements

include:
a) Burns in air to form normal oxide.
b) Reacts with chlorine gas when heated to form chloride.
c) Reacts with dilute hydrochloric acid liberating hydrogen gas.
Reactions in which Beryllium differs from other group II metals include:
a) Beryllium reacts with conc. alkalis but other group II elements do not.
b) Beryllium oxide is amphoteric i.e. shows both basic and acidic properties
but other oxides of the group II elements are basic in nature.
c) Beryllium carbide reacts with water to form methane.
d) Beryllium chloride is partly covalent and exists as dimer in vapour state.

Other members form ionic chlorides.
4.6. Analysis of Magnesium, Calcium and Barium ions in solutions
1) Use of sodium hydroxide solution.
Procedure
To the solution containing the unknown cation is added sodium hydroxide

solution dropwise until in excess.
Observation
a) With Mg2+, a white precipitate of magnesium hydroxide, insoluble in

excess sodium hydroxide is observed.
41
Chapter Four
Mg2+ (aq) + 2OH- (aq) Mg(OH)2(s)
With Ca2+, a white precipitate of calcium hydroxide, insoluble in

excess sodium hydroxide is formed
Ca2+ (aq) + 2OH -(aq) Ca(OH)2(S)
b) With Ba2+ a white precipitate of barium hydroxide,insoluble in excess

sodium hydroxide is formed
Ba2+ (aq) + 2OH -(aq) Ba(OH)2(S)
2) Aqueous ammonia solution
Procedure
To the solution containing the unknown cation is added aqueous ammonia

Observation
a) With Mg2+, precipitate of magnesium hydroxide,insoluble in excess

ammonia is observed.
b) With Ca2+ or Ba2+ ions there is no observable change.
Explanation
Aqueous ammonia being a weak base does not have hydroxide ions in
sufficient concentration to exceed the solubility products of Ca(OH) 2
and Ba(OH)2 which are quite high.
3) Use of dilute suphuric acid or any solution containing sulphate ions.
Procedure
To a few cm3 of the solution containing the unknown cation is added

dilute sulphuric acid (sodium sulphate solution or potassium sulphate can
also be used).
Observation
a) There is no observable change with Mg2+
42
Chapter Four
b) With either Ca2+ or Ba2+ a white precipitate of either calcium sulphate

or barium sulphate is formed.
Equation
Ca2+(aq) + SO42 -(aq) CaSO4(s).
Ba2+(aq) + SO42 -(aq) BaSO4(s)
4) Use of potassium chromate solution
Procedure
To the solution containing the unknown cation is added potassium

chromate solution followed by ethanoic acid.
Observation
a) With Mg2+ there is no observable change.
b) With Ca2+ a yellow precipitate of calcium chromate soluble in ethanoic

acid is formed.
Ca2+(aq) + CrO42-(aq) CaCrO4(s)
c) With Ba2+ a yellow precipitate of barium chromate insoluble in

ethanoic acid is formed.
Ba2+(aq) + CrO42-(aq) BaCrO4(s)
5) Use of ammonium oxalate solution
Procedure
To the solution containing the unknown cation is added ammonium

oxalate solution followed by ethanoic acid.
Observation
43
Chapter Four
a) With Mg2+ there is no observable change.
b) With Ca2+ a white precipitate of calcium oxalate insoluble in ethanoic

acid is formed.
Ca2+(aq) + C2O42 -(aq) CaC2O4(s)
c) With Ba2+ a white precipitate of barium oxalate soluble in ethanoic

acid is formed.
Ba2+(aq) + C2O42 -(aq) BaC2O4(s)
6) Use of disodium hydrogen phosphate solution
Procedure
To the solution containing the unknown cation is added little aqueous

ammonia followed by ammonium chloride and then a few drops of
disodium hydrogen phosphate.
Observation
a) With Mg2+ a white precipitate is formed
b) There is no observable change with Ca2+ and Ba2+ ions.
Exercise I
A chloride of beryllium Z contains 11.25% beryllium and 88.75% of chlorine.
a) Calculate the empirical formula of Z.
b) Determine the molecular formula of Z (vapor density of Z=80).
c) Write the structural formula of Z.
Exercise II
A chloride of aluminium X, contains 20% aluminium and 80% chlorine.
a) Calculate the empirical formula of X.
b) Determine the molecular formula of X (vapor density of X= 133.5).

44
Chapter Four
c) Write the structural formula of X.
Exercise III.
When 0.13g of a chloride of iron was vaporized at 600k and 1 atmosphere

pressure, 20cm3 of vapor was formed.
a) Calculate the relative mass of iron chloride.
b) Determine the molecular formula of the iron chloride.
c) Write the structural formula of the iron chloride.
Exercise IV.
a) Compare the reaction of beryllium and barium with sulphuric acid under
various conditions.
b) Explain how the solubility and basicity of the hydroxides of the elements
of group II in the Periodic Table vary down the group.
Exercise V.
Describe the reactions, if any, between each of the following elements, Be, Mg
and Ca with:
a) Warm dilute sulphuric acid.
b) Warm conc. sodium hydroxide.
45
Chapter Five
c)
46
Chapter Five
CHAPTER 5
5. GROUP (IV) ELEMENTS
Table 7: physical properties of group IV elements
Element Atomic no Outer elect conf Atomic Electronegativity

radius /0A
Carbon 6 -2s22p2 0.077 2.50
Silicon 14 -3s23p2 0.177 1.75
Germanium 32 -4s24p2 0.122 2.00
Tin 50 -5s25p2 0.140 1.70
Lead 82 -6s26p2 0.154 1.55
5.1. Physical properties of group (IV) elements

All group (IV) elements have four electrons in their outermost shell with a
general outer configuration of ns2np2 (where n= 2, 3, 4, 5, 6).
Metallic character increases down the group as the radius of the atoms become
larger. Carbon and Silicon are non metals, Germanium is a metalloid i.e. it
exhibits properties characteristic of metals and non metals while Tin & Lead are
weak metals.eg they form amphoteric oxides
All elements show oxidation states of +2 and +4. The +2 oxidation state arise
due to the inability of the outer most paired ‘s’ electrons to take part in bond
formation. This is called the inert pair effect.
The stability of the +2 oxidation state increases down the group from carbon to
lead. This is due to increasing metallic character and increasing inert pair effect
as the group is descended.eg Carbon and Silicon form extremely unstable
compounds in +2 oxidation state.
Germanium and tin compounds in +2 oxidation state are predominantly

covalent and are readily oxidized ie they are powerful reducing agent.
Lead invariably forms more stable compounds in +2 oxidation state than +4

oxidation state. Also compounds of lead in +2 oxidation state are ionic in
47
Chapter Five
nature. Therefore lead (IV) compounds are unstable, and readily decompose to
(II) compouds.eg
2 PbO2(s) 2PbO(s) +O2 (g)
PbCl4 (l) PbCl2(s) + Cl2 (g)
The stability of +4 oxidation state decreases down the group. Carbon, silicon
and germanium form more stable compounds in +4 oxidation state than in +2.
This could explain why their compounds in +2 oxidation states are good
reducing agents.
Carbon is the only member of the group whose covalency is restricted to a

maximum of 4.This is due to absence of the vacant‘d’ orbitals. Other elements
have a maximum of 6 due to the availability of vacant d-orbitals which allows
them to expand their octet.
Hoverer, all the elements exhibit a covalency of 4 but compounds of the type
M4+ or M4- are rare. This is because the sum of the first four ionization energies
or first four electron affinities would be highly endothermic. An ion of the type
M4+ would be very small with high charged density as such would exert high
polarizing effect on the surrounding anion thus making the compound formed
highly covalent & unstable.
Note:
Carbon, however, being the most electronegative group (IV) element with the
smallest atomic radius and least number of electrons is most likely to form
compounds of the type M4+ or M4-
Carbon exhibits a unique property called catenation. This is the ability of an

element to form covalent bonds with its own atoms so as to produce long chains
and rings.
Carbon therefore forms limitless hydrides i.e. alkanes, alkenes, alkynes etc due
to its catenation ability. This unique property of carbon is attributed to the small
atomic radius of carbon. The electrons involved in bond formation between two
carbon atoms are closer to the nuclei and are strongly held by it thus making
48
Chapter Five
bonds very strong and stable. However in descending the group, atomic radius
increases. Therefore electrons between the atoms become more distant from the
nuclei making the bond weaker and unstable.
The melting points of group (IV) elements decrease as the group is descended.
Carbon, however, has a much higher melting point than others. This is because
carbon has a giant molecular (macro-molecular) structure composed of many
atoms due to its ability to catenate. It has abnormally high melting point because
large number of short and strong carbon –carbon bonds have to be broken
before it melts.
Melting point then decreases gradually from silicon to lead in line with increase
in atomic radius down the group. The inter-atomic bond length (element-
element bonds) becomes longer and weaker thus melting point decreases.
Chemical properties of group (IV) elements.
1) Reaction with water.
Carbon reacts with steam when heated to form carbon monoxide and
hydrogen gas.
C(s) + H2O(g) CO(g) + H2(g)
Silicon reacts with steam when heated to form silicon dioxide and
hydrogen gas.
Si(s) +2H2O(g) SiO2(s) + 2H2(g)
Germanium and Tin do not react with water at any condition.
Lead reacts slowly with soft water forming lead (ii) hydroxide.
2Pb(s) + 2H2O(l) + O2(g) 2Pb(OH)2(s)
Note
The above reaction leads to lead poisoning and explains why lead pipes
are not used for transporting water for drinking.
2) Reaction with Oxygen (air).
49
Chapter Five
a) Carbon.
When carbon is heated in limited air, carbon monoxide is formed and in

excess air, carbondioxide is formed.
2C(s) + O2(g) 2CO(g)
C(s) + O2(g) CO2(g)
When the rest of the elements are heated inoxygen , they form dioxides
except lead which forms lead(ii) oxide.
Si(s) +O2 (g) SiO2(s)
Ge(s) + O2(g) GeO2(s)
Sn(s) + O2(g) SnO2(s)
2Pb(s)+ O2(g) 2PbO(s)
3) Reaction with acids
a) Carbon is attacked by all oxidizing acids e.g hot concentrated

sulphuric acid and hot concentratednitric acids.
C(s) +4HNO3 (aq) CO2(g) + 4NO2(g) + 2H2O(l)
C(s)+ 2H2SO4(aq) CO2(g) + 2SO2(g) +2H2O(l)
Silicon is resistant to all acids except hydrofluoric acid (HF)
Si(s) + 6HF(aq) H2SiF6(aq) + 2H2(g)
Germanium, Tin and Lead are oxidized by hot conc. nitric acid
3Ge(s) + 4HNO3(aq) 3GeO2(s) + 2H2O(l) +4NO(g)
3Sn(s) + 4HNO3(aq) 3SnO2(s) + 2H2O(l) +4NO(g)
Pb(s) +4HNO3(aq) Pb(NO3)2(aq) + 2H2O(l) + 2NO2(g)
50
Chapter Five
Lead is rendered passive in both cold dilute hydrochloric and sulphuric

acids due to formation of insoluble salt of lead (II) chloride & lead (II)
sulphate respectively.
However, lead reacts with hot dilute hydrochloric acid to form lead (ii)
chloride and hydrogen gas.
Pb(s) + 2HCl(aq) PbCl2(aq) + H2(g)
Tin reacts with hot conc. Sulphuric acid to form tin(iv) sulphate.
Sn(s) + 4H2SO4 (aq) Sn(SO4)2(aq) + 2SO2(g) + 4H2O(l)
Lead reacts with hot conc. Sulphuric acid to form lead (II) sulphate,
sulphur dioxide and water.
Pb(s) + 2H2SO4 (aq) PbSO4(s) + SO2(g) + 2H2O(l)
4) Reaction with Alkalis
Carbon is resistant to alkalis but silicon dissolves in conc. Solution to

form silicate.
Si(s) + 2NaOH(aq) + H2O(l) Na2SiO3(aq) + 2H2(g)
Germanium dissolves in conc. Alkalis to form hexahydroxogermanate (iv)
Ge(s) +2KOH(aq) +4H2O(l) K2Ge(OH)6(aq) + 2H2(g)
Ionically
Ge(s) + 2 OH-(aq) + 4H2O(l) Ge(OH)62-(aq) + 2H2(g)
5) Tin and Lead dissolve slowly in hot conc. Solution of alkalis to form
hexahydroxystannate(II) and plumbate(II) respectively.
6) Sn(s) +4KOH(aq) +2H2O(l) K4Sn(OH)6(aq) + H2(g)
7) Pb(s) +4KOH(aq) +2H2O(l) K4Pb(OH)6(aq) + H2(g)
51
Chapter Five
5.2. Compounds of group (IV) elements
Group (IV) elements form compounds in +2 and +4 oxidation states. The +2

oxidation state becomes more stable while the stability of +4 decreases as the
group is descended.
The oxides of group (IV)
These elements form oxides of the type XO, wherein the element shows an
oxidation state of +2 and also of the type XO2, wherein the element shows an
oxidation state of +4.
The oxides of the elements at the top of Group 4 are acidic, but acidity of the
oxides falls as you go down the Group. Towards the bottom of the Group, the
oxides become more basic - although without ever losing their acidic character
completely.
The trend is therefore from acidic oxides at the top of the Group towards
amphoteric ones at the bottom.
An oxide which shows both acidic and basic properties is said to be amphoteric.
Monoxides
The monoxide of carbon and silicon are usually treated as neutral oxides, but
in fact carbon monoxide is very, very slightly acidic. It doesn't react with water,
but it will react with hot concentrated sodium hydroxide solution to give a
solution of sodium methanoate.
NaOH(aq) + CO(g) HCOONa(aq)
The fact that the carbon monoxide reacts with the basic hydroxide ion shows
that it must be acidic
Those of germanium, tin and lead are amphoteric. The amphoteric monoxides
(GeO, SnO and PbO) react with acids to give corresponding salts and alkalis to
give germinate(II), stannate(II) and plumbate(II) complex ions.
SnO(s) + 2H+(aq) Sn2+(aq) + H2O(l)
52
Chapter Five
SnO(s) + 2OH-(aq) + H2O(l) Sn(OH)42-(aq)
Similar reactions are undergone by Germanium monoxide & lead (ii) oxide
Alternatively:
XO(s) + 2OH-(aq) XO22-(aq) + H2O(l)
GeO(s) + 2OH-(aq) GeO22-(aq) + H2O(l)
SnO(s) + 2OH-(aq) SnO22-(aq) + H2O(l)
PbO(s) + 2OH-(aq) PbO22-(aq) + H2O(l)
They all react with concentrated hydrochloric acid.
MO(s) + HCl(aq) MCl2(aq) + H2O(l)
where M can be Ge or Sn.
Lead(II) chloride is fairly insoluble salt in water and, , it would form an

insoluble layer over the lead(II) oxide if dilute hydrochloric acid is used -
stopping the reaction from going on
PbO(s) + 2HCl(aq) PbCl2(s) + H2O(l).
However, using concentrated hydrochloric acid the large excessof chloride ions
in the concentrated acid react with the lead(II) chloride to produce a yellow
soluble complex PbCl42-.
PbCl2(s) + 2Cl -(aq) PbCl42 -(aq)
Dioxides
The first four dioxide are prepared by heating the elements in oxygen but a
similar process with lead gives lead(ii) oxide.
C(s) + O2(g) CO2(g)
Si(s) + O2(g) SiO2(s)
53
Chapter Five
Ge(s) + O2(g) GeO2(s)
Sn(s) + O2(g) SnO2(s)
2 Pb(s) + O2(g) 2PbO(s)
Note
Lead (IV) oxide is made by action of hot dilute nitric acid on red lead, Pb 3O4.
Pb3O4(s) + 4HNO3(aq) PbO2(s)+ 2Pb(NO3)2(aq)+ 2H2O(l)
Brown solid
With warm conc. hydrochloric acid, red lead (trilead tetra oxide) reacts
according to the equation:
Pb3O4(s) + 14HCl(aq) 3H2PbCl4(aq) + Cl2(g) + 4H2O(l)
Yellow solution
Red lead is used in paints to prevent the corrosion of iron
The structure of carbon dioxide
There is an enormous difference between the physical properties of carbon

dioxide and silicon dioxide (also known as silicon (IV) oxide or silica).
Carbon dioxide is a gas whereas silicon dioxide is a hard high-melting solid.
The other dioxides in Group 4 are also solids. The fact that carbon dioxide is a
gas means that it must consist of simple or discrete molecules held by weak Van
der Waals forces of attraction. These forces are not strong enough to bind the
carbon dioxide molecules into solid.
Carbon dioxide has a linear structure because the double bonds repel each other
strongly to give a structure with bond angle 180 degrees as there are no lone
pairs available
The structure of silicon dioxide
54
Chapter Five
This is based on a diamond structure with each of the silicon atoms being
bridged to its other four neighbors via an oxygen atom. This means that silicon
dioxide exists as a giant covalent structure in which large number of strong
silicon-oxygen bonds have to be broken hence it is a solid . The strong bonds in
three dimensions make it a hard, high melting point solid
CO2& SiO2 are weaklyacidic and covalent in nature
Reaction With water
Silicon dioxide doesn't react with water, because of the difficulty of breaking up
the giant covalent structure.
Carbon dioxide does react with water to a slight extent to produce hydrogen
ions.
H2O(l) + CO2(g) H+(aq) + HCO32-(aq)
The solution of carbon dioxide in water is sometimes known as carbonic acid.
Reaction With bases
Carbon dioxide reacts with sodium hydroxide solution in the cold to give either
sodium carbonate or sodium hydrogen carbonate solution - depending on the
reacting proportions.
2NaOH(aq) + CO2(g) Na2CO3(aq) + H2O(l)
Na2CO3(aq) + H2O(l) + CO2(g) 2NaHCO3(s)
White solid
55
Chapter Five
Silicon dioxide is acidic and reacts with sodium hydroxide solution,

but only if it is hot and concentrated. Sodium silicate solution is
formed.
SiO2(s) + 2NaOH (aq) Na2SiO3 (aq) + H2O (l)
. GeO2, SnO2, and PbO2 are amphoteric in nature.
The basic nature of the dioxides
The dioxides react with concentrated hydrochloric acid first to give compounds
of the type XCl4e.g
PbO2(s) + 4HCl(aq) PbCl4(aq) + 2 H2O(l)
These will react with excess chloride ions in the hydrochloric acid to give
complexes such as XCl62-.
PbCl4(aq) + 2Cl-(aq) PbCl62-(aq)
If the reaction is carried out at elevated temperature, lead (IV) oxide oxidizes hot
conc. hydrochloric acid to chlorine gas.
PbO2(s) + 4HCl(aq) → PbCl2(aq) + 2 H2O(l) + Cl2(g)
The acidic nature of the dioxides
The dioxides will react with hot concentrated sodium hydroxide solution to give
soluble complexes of the form [X(OH)6]2-.
GeO2(s) + 2OH -(aq) + 2H2O(l) → Ge(OH)62 -(aq)
SnO2(s) + 2OH -(aq) + 2H2O(l) → Sn(OH)62 -(aq)
PbO2(s) + 2OH -(aq) + 2H2O(l) → Pb(OH)62 -(aq)
OR
GeO2(s)+ 2OH -(aq) → GeO32 -(aq) + H2O(l)
SnO2(s)+ 2OH-(aq) → SnO32 -(aq) + H2O(l)
56
Chapter Five
PbO2(s)+ 2OH-(aq) → PbO32 -(aq) + H2O(l)
The halides of group (IV) elements
+2 halides
Carbon and Silicon do not form +2 halides. Those of Germanium, Tin and lead
can be prepared by heating the tetrachlorides with the free metal.
GeCl4(l) + Ge(s) 2GeCl2(s)
SnCl4(l) + Sn(s) 2SnCl2(s)
PbCl4(l) + Pb(s) 2PbCl2(s)
Tin (II) chloride may be prepared in anhydrous form by passing dry hydrogen
chloride gas over heated tin.
Sn(s) + 2HCl(g) SnCl2(s) + H2(g)
Lead (ii) chloride can also be prepared either by action of hot concentrated
hydrochloric acid over lead(iv) oxide or by passing dry hydrogen chloride gas or
chlorine over heated metal.
PbO2(s) + 4HCl(aq) PbCl2(s) + 2H2O(l)+ Cl2(g)
Note:
Lead (ii) chloride is more soluble in concentrated hydrochloric acid than in

water. Explain this observation.
Explanation
Lead (II) chloride has a high lattice energy which is not over-come on addition
of cold water. However on heating, the high lattice energy is overcome and
therefore lead (II) chloride dissolves in hot water.
Lead (II) chloride dissolves in conc. Hydrochloric acid due to formation of

soluble complex, tetrachloroplumbate (II) ions.
PbCl2(s) + 2Cl -(aq)→ PbCl42-(aq)
57
Chapter Five
Yellow solution.
+4 halides
All the elements form tetra chlorides of the type XCl4, which are covalently
bonded and are volatile liquids.
All the tetra chlorides have a tetrahedral shape in which the element is
surrounded by 4 chlorine atoms.
Preparation of the tetra chlorides
CCl4
By reacting chlorine with carbon disulphide .
CS2(l) + 3Cl2(g) AlCl3/heat CCl4(l) + S2Cl2(l)
The two liquids are separated by fractional distillation.
SiCl4 , GeCl4& SnCl4
Are prepared by passing a stream of dry chlorine over the heated metal.
Si(s)+ 2Cl2(g) SiCl4(l)
Ge(s) + 2Cl2(g) GeCl4(l)
Sn(s) + 2Cl2(g) SnCl4(l)
If a stream of chlorine is passed over heated lead, only lead(ii) chloride is

formed.
Pb(s) + Cl2(g) PbCl2(s)
To prepare lead(iv) chloride, use ice cold concentrated hydrochloric acid on

lead(iv) oxide.
PbO2(s) + 4HCl(aq) PbCl4(l) + 2 H2O(l)
Note
58
Chapter Five
If the reaction is carried out at elevated temperature, lead (IV) oxide oxidizes hot
concentrated
hydrochloric acid to chlorine gas.
PbO2(s) + 4HCl(aq) → PbCl2(aq) + 2 H2O(l) + Cl2(g
Hydrolysis of the tetrachlorides
Carbon tetrachloride (CCl4) does not undergo hydrolysis in water. This is due to
the absence of a d sub shell in carbon and as such it cannot expand its octet.
However, other tetrachlorides undergo hydrolysis in water. This is attributed to

availability of the empty d-orbitals which allow the elements to expand their
octet and show a covalency of more than +4. These empty d-orbitals accept the
lone pairs of electrons from the oxygen atom of water molecule thus facilitating
the hydrolysis of the tetra chlorides.
SiCl4(l) + 2H2O(l) SiO2(s) + 4HCl(g)
White solid.
GeCl4(l) + 2H2O(l) → GeO2(s) +
SnCl4(l) + 2H2O(g) SnO2(s)+ 4HCl(g)
PbCl4(l) + 2H2O(l) PbO2(s) + 4HCl(g)
Brown solid
NB: In each case white fumes of hydrogen chloride gas is formed.
Stability of the tetrachloride

59
Chapter Five
The tetrachlorides of carbon, silicon, and germanium are stable to

heat, but those of tin and lead decompose on heating to form
dichlorides.
SnCl4(l) SnCl2(s) + Cl2(g)
PbCl4(l) PbCl2(s) + Cl2(g)
Hydrides of group IV
CH4 Methane
SiH4 Silane
GeH4 Germane
SnH4 Stannane
PbH4 Plumbane
The boiling points of the hydrides increase from methane down to

stannane.
This is due to increase in relative molecular mass and thus the Van der Waals
forces of attraction holding the molecules together become progressively
stronger as the molecular mass increases.
60
Chapter Five
Methane, germane and stannane do not react with sodium hydroxide, but silane
reacts with dilute sodium hydroxide to form sodium silicate and hydrogen gas.
SiH4(l) + 2NaOH(aq) + H2O(l) → Na2SiO3(aq) + 4H2(g)
Ionically
SiH4(l) + 2OH -
(aq) + H2O(l) → SiO32 -(aq) + 4H2(g)
Hydrolytic behavior of hydrides of group IIV elements
Methane does not undergo hydrolysis due to absence of vacant d-orbitals.
Other hydrides are able to hydrolyze in water due to the availability of vacant d-
orbitals which can accept electrons from the water molecules.
SiH4(l) + 4 H2O(l) → Si(OH)4(s) + 4H2(g)
GeH4(l) + 4 H2O(l) → Ge(OH)4(s) + 4H2(g)
SnH4(l) + 4 H2O(l) → Sn(OH)4(s) + 4H2(g)
Carbon differs from other group IV elements in the following ways:
a) Its covalence is limited to 4 but others can have a maximum of 6. This is

due to absence of vacant d-orbitals in carbon but others have.
b) Its unique ability to catenate.
Due to its small atomic radius, the bonding electrons between two carbon
atoms are closer to the nuclei and are attracted more strongly.
Its unique ability to form multiple bonds such as: -C=C-, -
C≡C-, -C≡N,etc.
c) Its tetrachloride does not hydrolyze in water but tetra chlorides of other
elements hydrolyzedue to absence of vacant d-orbitals.
d) Carbon is the only element in group IV which forms only
gaseous oxides.
Why carbon differs from other elements in group IV
a) It has the smallest radius in the group

b) Its electronegativity is the highest in the group
61
Chapter Five
c) Carbon lacks a d-sub shell unlike other group members

d) Carbon is unable to show the inert pair effect and always
exhibit a valency of 4 in almost all its compounds
5.3. Test for Lead (II) ions in solution

1) Use of sodium hydroxide solution
Procedure
To the solution containing the unknown cation is added sodium hydroxide
Observation
A white precipitate of Lead (II) hydroxide is formed. The precipitate
dissolves in excess sodium hydroxide solution to form a cololourless due
to the formation of plumbate (II) ions.
Equation
Pb2+(aq) + 2OH -(aq) Pb(OH)2(S).
Pb(OH)2(S) + 2OH - (aq) Pb(OH)42-(aq)
2) Use of aqueous ammonia solution.

Procedure
To the solution containing the unknown cation is added aqueous ammonia
dropwise until in excess.
Observation
In the presence of Lead (II) ions, a white precipitate of Lead (II)
hydroxide is formed. The precipitate is insoluble in excess aqueous
ammonia.
Equation
Pb2+(aq) + 2OH -(aq) Pb(OH)2(S).
3) Use of dilute hydrochloric acid.

Procedure
To the solution containing the unknown cation is added a few drops of
dilute hydrochloric acid.
Observation
62
Chapter Five
A white needle like precipitate of lead (II) chloride is formed. The

precipitate dissolves on heating but reappears on cooling.
Pb2+(aq) + 2Cl -(aq) PbCl2(s).
4) Use of potassium chromate solution

Procedure
chromate solution.
Observation
A yellow precipitate of lead (II) chromate is formed in the presence of
Lead (II) ions.
Equation
Pb2+(aq) + CrO4 2-(aq) PbCrO4(s).
5) Use of Potassium Iodide solution.

Procedure
To the solution containing the unknown cation is added a few drops of
potassium iodide solution.
Observation
A yellow precipitate of lead (II) iodide is formed in the presence of lead
(II) ions.
Equation
Pb2+(aq) + 2I -(aq) PbI2(s).
6) Use of dilute sulphuric acid.

Procedure
To the solution suspected to contain lead (II) ions is added dilute
sulphuric acid.
Observation
A white precipitate of lead (II) sulphate is formed.
Equation
Pb2+(aq) + SO4 2-(aq) PbSO4(s)
7) Use of sodium carbonate solution
63
Chapter Five
In the presence of Pb2+ ions a white precipitate of lead (II) carbonate is

formed on addition of sodium carbonate solution.
Pb2+(aq) + CO3 2-(aq)PbCO3(s)
Question
When dilute hydrochloric acid is added to a solution of lead (ii) nitrate, a white
precipitate is formed, but if the concentrated acid is used, there is no precipitate.
Explain this observation and write an ionic equation for the reaction in each
case.
64
The Halogens
65
The Halogens
CHAPTER 6
6. GROUP (VII) ELEMENTS (HALOGENS)
Table 8: Physical properties of group VII elements
Element Atomic no Outer elect Melting Boiling Bond energy

conf pt/0c pt/0c /kcalmol-1
Fluorine 9 -2s22p5 -220 -188 37
Chlorine 17 -3s 3p
2 5
-101 -34 57.2
Bromine 25 -4s24p5 -7 58 45.4
Iodine 33 -5s25p5 114 183 35.6
6.1. Physical properties of the halogens.

The halogens exist as diatomic molecules i.e. F2, Cl2, Br2 andI2.
Fluorine is a pale green gas, chlorine is a yellowish green gas, bromine is a

reddish brown gas while iodine is a black (dark purple) solid at room
temperature which sublimes to give a purple gas.
Halogens are highly reactive non-metals. They exist as diatomic molecules X2

containing a covalent bond.
Due to their reactivity the elements do not occur in the free state but are always
combined either among themselves or with other elements.
All the elements have 7 electrons in their outer most shells with general outer
most electronic configuration of ns2np5 (n= 2, 3, 4, 5)
They can complete their octet by either gaining an electron to form an ionic
bond or sharing electrons with other elements or themselves in a covalent bond.
The melting and boiling points of the halogens are low but increase down the
group.
Explanation
The diatomic molecules of halogens are held by weak intermolecular van der
Waals forces of attraction. These forces become stronger with increasing
molecular weight/ masses down the group.
66
The Halogens
In bromine and iodine, the forces are strong enough to bind their molecules to
exist as liquid and solid respectively at room temperature. However, in fluorine
and chlorine, the forces are not strong enough and as a result they exist as gases
at room temperature.
Table 9: Melting and boiling points of group VII elemts
Bond dissociation energy
Bond energy is the energy required to break one mole of a covalent bond into its
constituent gaseous atoms.
Or it’s the energy released when one mole of a covalent bond is formed from its
constituent gaseous atoms.
Bond dissociation energy of the diatomic molecules decreases down the group
from chlorine to iodine. However, Fluorine has abnormally low bond
dissociation energy.
67
The Halogens
This is due to small atomic radius of fluorine atom; the non bonding p- electrons
present in Fluorine molecule are closer to one another as such repel each other
strongly thus weakening the F-F bond.
The decrease in bond dissociation energy from Chlorine to Iodine is attributed to

increase in atomic radius as the group is descended; the bonding electrons in the
halogen-halogen bond are far from the nuclei of the halogens& are attracted less
strongly due to the screening effect of the inner shells of electrons thus
weakening the bond progressively
68
The Halogens
NB: The high reactivity of fluorine may be attributed to its low bond
dissociation energy.
Electron affinity of halogens
E.A is the energy change when one mole of electrons is added to one mole of
gaseous atoms to form one mole of negatively charged gaseous ions.
X (g) + e X(g) -
The electron affinity is a more useful concept for the halogens than first
ionisation energy because the halogens need only one electron to complete their
octet to attain a noble gas like structure.
69
The Halogens
The electron affinity decreases numerically from Chlorine to Iodine. As the

atomic radius increases down the group, the electron being added experiences a
low attraction from the nucleus leading to a decrease in electron affinity.
Fluorine, however, has a low value for its electron affinity, numerically less than
that for chlorine.
Explanation
This is due to small atomic radius of fluorine atom; the 7 electrons in the outer
most shell are too close to one another therefore the electron being added
experiences greater repulsive forces from the electrons already present.
The electron affinity is a measure of the attraction between the incoming

electron and the nucleus. The higher the attraction, the higher the electron
affinity.
Oxidizing properties of the halogens
Halogens are very good oxidizing agents with fluorine being the most oxidizing.
This is because fluorine is the most electronegative and has the highest positive
value for its standard electrode potential.
It oxidises water to oxygen.
2F2(g) + 2H2O(l) O2(g) + 4HF(aq)
70
The Halogens
The standard electrode potential (Eθ) becomes less positive as the group is
descended and so oxidising power decreases down the group.
Fluorine> Chlorine> Bromine> Iodine.
Oxidation states of the halogens
They all exhibit oxidation state of -1 by gaining an extra electron to complete

their octet Fluorine is always o univalent due to absence of d-orbitals.
Other elements can show higher oxidation states due to availability of vacant d-
orbitals which allow them to expand their octet.
Differences in the chemistry of Fluorine from other members of its group are:
 Fluorine is has a fixed oxidation state of -1 while the other halogens have
more than one oxidation states. This is because fluorine lacks a d sub
shell while the other halogens have.
 Hydrogen fluoride is a liquid while other hydrogen halides are gases at
room temperature. This is due to the strong hydrogen bonds between
hydrogen fluoride molecules as a result of high electronegativity of
fluorine atom.
 Fluorine forms the least acidic hydride whereas the hydrides of other
members are fairly more acidic in aqueous solution.
This is due to the high strength of the H-F bond.
 Fluorine is the most reactive halogen due to its low bond dissociation
energy and high electronegativity.
 Fluorine forms compounds with more ionic character due to its high
electronegativity.
 Fluorine forms hydride with the highest boiling point due to extensive
hydrogen bonds.
Why does Fluorine show a difference?
 Fluorine is the most electronegative halogen.

 The bond dissociation energy of fluorine molecule is lower than
expected.
 Fluorine has a high positive value of standard electrode potential.
71
The Halogens
 Fluoride ion having the smallest radius compared to other halides has the
highest charge density and enthalpy of hydration.
 Fluorine lacks a d sub shell.
6.2. General methods for preparing the halogens (excluding

fluorine) in the laboratory
a) By the reaction between manganese (IV) oxide and concentrated halogen
acid e.g.
4HCl(aq) + MnO2(s) MnCl2(aq) + 2H2O(l) + Cl2(g)
4HBr(aq) + MnO2(s) MnBr2(aq) + 2H2O(l) + Br2(g)
4HI(aq) + MnO2(s) MnI2(aq) + 2H2O(l) + l2(g)
b) By the reaction between potassium manganate (VII) solution

and hot concentrated halogen acid.
e.g.
2KMnO4(s) + 16HCl(aq) 2KCl(aq) + 2MnCl2(aq) +8H2O(l)

+5Cl2(g).
2KMnO4(s) + 16HBr(aq) 2KBr(aq) + 2MnBr2(aq) +8H2O(l)

+5Br2(g).
6.3. Reactions of halogens

In all their reactions the halogens act as oxidising agents. Hence, their
reactivity decreases down the group.
1) Reaction with water

a. Fluorine oxidises H2O to O2 gas in a very exothermic
reaction.
2F2(g) + 2H2O(l) O2(g) + 4HF(aq)
b. Chlorine dissolves in H2O to form ‘chlorine water'

which is a mixture of two acids. No oxygen gas is
evolved.
72
The Halogens
Cl2(g) + H2O(l) HCl(aq) + HOCl(aq)
c. Bromine is only slightly soluble in water.

Br2(l) + H2O(l) HBr(aq) + HOBr(aq)
d. Iodine is virtually insoluble in water. It is however

soluble in potassium iodide solution due to the formation
of a soluble complex the tri-iodide anion.
I2(s) + I -(aq) I3 -(aq)
Iodine is sparingly soluble in water because the energy
involved in disrupting the weak van der walls forces in
molecular iodine crystal lattice and the hydrogen bonds
in water exceeds the energy changes involving the weak
attraction between iodine and water molecules.
Note: All halogens are more soluble in non-polar
solvents such as CCl4. Cl2 gives a colourlesssolution.
Br2 a red solution and I2 a violet one
2) Reaction with alkalis (sodium hydroxideor potassium

hydroxide solution)
This reaction depends on the conditions:
a) With cold and dilute alkalis the reactions are as below:

2KOH(aq) + F2(g) 2KF(aq) + F2O(g) + H2O(l)
2NaOH(aq) + Cl2(g) NaCl(aq) + NaClO(aq) + H2O(l)
2NaOH(aq) + Br2(l) NaBr(aq) + NaOBr(aq) + H2O(l)
2KOH(aq) + I2(s) KI(aq) + KOIO(aq) + H2O(l)
In hot solution the ClO - ions undergo disproportionation

reaction.
Disproportionation reaction being a reaction in which is a
species under go self oxidation& reduction reactions.
73
The Halogens
e.g.
3ClO -(aq) 2Cl- (aq) + ClO3 -(aq)
6OH-(aq) + 3I2(s) 5I -(aq) + IO3 - + 3H2O(l)
Iodine is reduced to iodide ions as its oxidation state decreases

from zero to -1 and at the same time its oxidised to iodate ions
as its oxidation state increases from 0 to +5.
b) With hot and concentrated alkali, reactivity decreases down

the group.
4NaOH(aq) + 2F2(g) 4NaF(aq) + O2(g) + 2H2O(l)
6KOH(aq) + 3Cl2(g) 5KCl(aq) + KClO3(aq) + 3H2O(l)
Note
Other halogens react in a similar manner.
6.4. Compounds of the halogens
Hydrides of group VII elements
a) Boiling points of hydrides increase from Hydrogen chloride through

Hydrogen bromide to Hydrogen bromide but that of Hydrogen fluoride is
abnormally high.
Hydrogen halide Boiling point/ 0C
HF 19
HCl -85
HBr -66
HI -36
74
The Halogens
The increase in boiling point from Hydrogen chloride to Hydrogen iodide is

attributed to increasing molecular mass of the halogens.
This in turn increases the strength of van der Waals forces of attraction between
their molecules hence increasing the boiling point of Hydrogen chloride to
Hydrogen iodide.
Hydrogen fluoride however, has abnormally high boiling point because Fluorine
atom has a very small atomic radius and high electronegativity, as a result it
forms a more polar Hydrogen fluoride bond and its molecules associate through
strong intermolecular hydrogen bonds which require greater amount of energy to
break.
The acidity of the hydrides
The acidity of the hydrides is in the order H-F <H-Cl< H-Br< H-I.
H-F is the weakest acid because the H-F bond is very strong as a result of small
atomic radius and high electronegativity of Fluorine atom.
It is weakly dissociated in solution releasing fewer H+ ions.
In addition, the presence of hydrogen bonding between HF molecules and water

molecules inhibits the ionization of HF in aqueous solution.
However, as the group is descended, electronegativity decreases whereas atomic

radius increases. As such the H- halide bonds become progressively longer and
weaker. They dissociate in solution releasing more hydrogen ions.
Bond energy/enthalpy of the hydrides generally decreases in moving from

Hydrogen fluoride to Hydrogen iodide i.e.
75
The Halogens
H-F > H-Cl > H-Br > H-I
As the halogen atom gets bigger, the bonding pair gets more and more distant
from the nucleus. The attraction is less, and the bond gets weaker.
Fluorine is the most electronegative atom with the smallest atomic radius thus
the H-F bond is the most polar. Iodine is the least electronegative with the
largest atomic radius as a result the H-I bond is the least polar. The greater the
extent of polarity, the greater the ionic character in the bond and therefore the
stronger the bond.
Reactions of the halogen acids
All the halogens acids undergo reactions typical of all acids
e.g.
They react with carbonates and hydrogen carbonates giving off carbon dioxide
gas
They react with metals producing corresponding salts and hydrogen gas
They neutralise bases forming salts and water
Note
For hydrofluoric acid, as well as undergoing reactions typical of all acids, it

reacts with silicon dioxide and silicon forming fluorosilic acid
76
The Halogens
6HF(aq) + SiO2(s) H2SiF6(aq) + 2H2O(l)
6HF(aq) + Si(s) H2SiF6(aq) + 2H2(g)
Reaction of halogen acids with concentrated sulphuric acid
Hydrofluoric acid and hydrochloric acid not react with concentrated sulphuric
acid
Hydrobromic acid and hydroiodic acid react with hot concentrated sulphuric
acid
HBr(aq) + H2SO4(aq) Br2(g) + SO2(g) + 2H2O(l)
Hl(aq) + H2SO4(aq) l2(g) + SO2(g) + 2H2O(l)
Oxy acids of chlorine
Table 10: Oxy acids of chlorine
Formula structure Name

HOCl H-O-Cl Hypochlorous acid
(chlorous (I) acid)
HClO2 HOCl=O Chlorous acid

(chlorous (III) acid)
HClO3 Chloric acid
HOClO2 (chlorous (V) acid)
HClO4 HOClO3 Perchloricacid

(chlorous (VII) acid
Note
Salts of hypochlorous acid (HOCl) are called hypochlorites e.g.
a) Sodium hypochlorite (NaOCl) present in liquid bleaches

e.g. JIK
77
The Halogens
b) Calcium hypochlorite (Ca(OCl)2 ) present in bleaching

powder
The percentage of available chlorine in either JIK or bleaching

powder can be determined in the laboratory.
Procedure
The JIK or bleaching powder is added to ethanoic acid or

hydrochloric acid. Chlorine gas is liberated and its passed through a
solution of potassium iodide solution and iodine is liberated.
The liberated iodine is then titrated against a standard solution of

sodium thiosulphate using starch as indicator.
The volume of the thiosulphate solution used is noted.
Equation of reactions taking place
JIK ( NaOCl and NaCl)
NaOCl(aq) Na+(aq) + ClO -(aq)
NaCl(aq) Na + (aq) + Cl -(aq)
On addition of an acid to JIK the reaction below takes place,
OCl-(aq) + Cl -(aq) + H +(aq) H2O(l) + Cl2(g)
Available chlorine
Then,
Cl2(g) + 2I -(aq) 2Cl -(aq) + I2(aq)
Cl2(g) + 2I -(aq) 2Cl -(aq) + I2(aq)
S2O32-(aq) + I2(aq) S4O62 -(aq) + 2I -(aq)
Acid strength of the oxyacids
Trend
78
The Halogens
The acidity of the oxyacids in aqueous solution increase with increase in the
number of oxygen atoms present i.e. hypochlorous acid (HOCl) is the least
acidic oxyacid and chloric (vii) acid (HClO4)themost acidic in aqueous solution.
Explanation
Order of acidity is HOCl<HOClO< HOClO2< HOClO3
As the number of the oxygen atoms increase, the H-O bond becomes weaker and
in aqueous solution the bond easily ionise producing more H+ ions in solution.
6.5. Test for Chlorides, Bromides and Iodides ions

1) Use of silver nitrate solution
Procedure
To the solution containing the unknown anion is added silver nitrate
solution.
Observation
a) In the of chloride ions a white silver chloride is formed. The

precipitate dissolves in excess aqueous ammonia due to the formation
of a soluble complex called diammine silver (I) ions.
Equation
Ag +(aq)+ Cl -(aq) AgCl(s) white precipitate.
AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl -(aq)
Soluble complex.
b) In the presence of bromide ions a pale yellow precipitate of silver

bromide is formed. The precipitate dissolves in excess aqueous
ammonia due to the formation of a soluble complex.
Equation
Ag +(aq) + Br -(aq) AgBr(s)pale yellow precipitate.
79
The Halogens
AgBr(s) + 2NH3(aq) Ag(NH3)2+(aq) + Br -(aq)
Soluble complex.
c) In the presence of iodide ions, a yellow precipitate of silver iodide is

formed. The precipitate is insoluble in excess aqueous ammonia.
Equation
Ag +(aq)+ I -(aq) AgI(s) yellow precipitate.
2) Use of lead (II) nitrate or lead (II)ethanoate solution.

Procedure
To the solution containing the unknown anion is added lead (II) nitrate or
lead (II) ethanoate solution.
Observation
a) In the presence of chloride ions a white precipitate of lead (II) is
formed. The precipitate dissolves on cooling but reappears on cooling.
Pb2+(aq) + 2Cl -(aq) PbCl2(s)
b) In the presence of bromide ions a white precipitate of lead (II) is

formed. The precipitate dissolves on warming but reappears on cooling
Pb2+(aq) + 2Br -(aq) PbBr2(s)

c) In the presence of iodide ions a pale yellow precipitate of lead (II)
iodide is formed. The precipitate dissolves on warming but reappears
on cooling
Pb2+(aq) + 2I -(aq) PbI2(s)
3) Use of chlorine water and tetrachloromethane

Procedure
To the solution containing the unknown anion is added Chlorine water
followed by 2-3 drops of tetrachloromethane
Observation
a) In the presence of chloride ions, there is no observable change.
80
The Halogens
b) In the presence of bromide ions, a brown colour appears in the

organic liquid.
c) In the presence of iodide ions, a purple/violet colour appears in the

organic liquid.
4) Use of concentrated sulphuric acid

Procedure
To the a few grams of the solid containing the unknown anion is added
concentrated sulphuric acid
Observation
a) In the presence of chloride ions, hydrogen chloride gas is evolved
(white fumes).
b) In the presence of bromide ions, bromine gas (brown) and hydrogen
bromide gas is evolved.
c) In the presence of iodide ions, iodine vapour (purple) and hydrogen
iodide gas is evolved.
5) Use of copper ( II ) sulphate solution

Procedure
When the reagent is added to a solution containing iodide ions, a white
precipitate of copper (I) iodide and a brown solution of iodine is formed.
Equation
Cu 2+
(aq) + 2I -
(aq)Cu2I2(s) + I2(aq)
81
Chapter Seven
82
Chapter Seven
CHAPTER 7
7. PERIOD 3 ELEMENTS IN THE PERIODIC TABLE
Table 11: Physical properties of group III elements
Element Na Mg Al Si P S Cl Ar
Atomic no 11 12 13 14 15 16 17 18
Outer electronic -3s1 -3s2 -3s2 - - - - -3s2

conf 3p1 3s23p2 3s23p3 3s23p4 3s23p5 3p6
Melting 97.8 650 660 1410 44 119 -101 -189

point/0C
Atomic 1.57 1.36 1.26 1.17 1.10 1.04 0.99

radius/0A
1st I.E/kjmol-1 496 737 577 786 1012 999 1255 1521
1st E.A/kjmol-1 -20 67 -30 -135 -60 -200 -364
Revision question
State and explain the trend in the variation of the following across the
3rd period in the periodic table
a) Atomic radius
b) 1st ionization energy
c) 1st electron affinity
d) Electronegativity
e) Electropositivity
f) Melting point
7.1. Compounds of period 3 elements
1) Chlorides
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Chapter Seven
Element Na Mg Al Si P S Cl
Formula NaCl MgCl2 AlCl3 SiCl4 PCl3,PCl5 S2Cl2 Cl2

of chloride
Type of Ionic Ionic Covalent Covalent Covalent Covalent Covalent

bonding in
the
chloride
Sodium chloride and magnesium chloride are ionic chlorides with giant
structures. They have high melting points.
The melting point of sodium chloride is higher than for magnesium chloride
which in turn is higher than for aluminium chloride.
Explanation
Ionic compounds tend to have higher melting points than covalent compounds.
From NaCl, through MgCl2 to AlCl3, the cationic radius decreases from Na+,
through Mg2+ to Al3+ as the charge increases from +1 to +3.
This leads to an increase in cationic charge density from Na+ to Al3+.
Aluminiumion with the highest charge density greatly polarizes the electron
clouds of the chloride ions to which it is bonded making aluminium chloride
predominantly covalent.
Magnesium ion also polarizes the electron clouds of the chloride ions to which
it is bonded but the extent of polarization is less than in aluminium chloride.
Magnesium chloride is therefore predominantly ionic.
Sodium ion with the least polarizing power makes sodium chloride the most
ionic chloride in the period and with the highest melting point, followed by
magnesium chloride and then aluminium chloride.
Note
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Chapter Seven
The other chlorides in the period; namely SiCl4, PCl3, PCl5, S2Cl2 and Cl2are
covalent molecules with weak intermolecular van der waals forces of attraction
in them. The magnitude of these forces increasing with increase in molecular
mass. They all therefore have low melting and boiling points.
Reactions of the chlorides with water
a) Sodium chloride dissociates in water to give a neutral solution

i.e the PH of the resultant solution is 7.
NaCl (aq) + aq Na +(aq) + Cl -(aq)
b) Magnesium chloride hydrolyses in water to form an acidic

solution
MgCl2(s) + H2O (l) Mg(OH)Cl(s) + HCl(aq)
c) Aluminium chloride in water hydrolyses to form a solution

whose PH is less than 7
AlCl3(s) + 6H2O (l) Al(H2O)63+(aq) + 3Cl -(aq)
[Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5 OH]2+(aq) + H3O+(aq)
[Al(H2O)5OH]2+(aq) + H2O(l) [Al(H2O)4 2OH]+(aq) + H3O+(aq)
[Al(H2O)4 2OH] +(aq)+ H2O(l) [Al(H2O)3 3OH] (s) + H3O+(aq)
Over all equation:
[Al(H2O)6]3+(aq)+3H2O(l) [Al(H2O)3 3OH] (s) + 3H3O+(aq)
White ppt
Silicon (IV) chloride reacts with water producing silicondioxide and

hydrochloric acid. The PH of the resultant solution is therefore less
than 7.
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Chapter Seven
White solid
a) The rest of the chlorides react with water to form acidic solutions.
PCl3(l) + H2O(l) H3PO3(aq) + 3HCl(aq)
PCl5(l) +4H2O(l) H3PO4(aq) + 5HCl(aq)
S2Cl2(l) + 2H2O(l) 3S(s) + 4HCl(aq) + SO2(g)
Cl2(g) + H2O (l) HCl(aq) + HOCl(aq)
2) The oxides of period 3 elements.
Eement Na Mg Al Si P S Cl
Oxide Na2O,Na2O Mg Al2O SiO2 P2O3, SO2, SO3 Cl2O
2 O 3 P2O5 Cl2O7
Type of Ionic ionic Ionic covalen covalen Covalen covalen
bondin t t t t
g
The oxide of sodium, magnesium and aluminium are ionic compounds and
conduct electricity in molten state due the presence of free ions. They have giant
ionic structures with high melting points.
Silicon dioxide has a giant molecular structure and therefore has a very high
melting point.
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Chapter Seven
The rest of the oxides have simple molecular structures with weak
intermolecular van der waals forces and hence have low melting and boiling
points.
Reactions of the oxides with water

a) The oxides of sodium and magnesium react with water forming alkaline
solutions. Aluminium oxide is however insoluble in water.
Na2O(s) + H2O (l) 2NaOH(aq)
Na2O2(s) + H2O (l) 2NaOH (aq) + H2O2(aq)
MgO(s) + H2O(l) Mg2+(aq) + 2OH-(aq)
b) Silicon dioxide does not react with water. Being acidic, it however reacts with
alkalis forming silicates
SiO2(s) + 2NaOH (aq) Na2SiO3 (aq) + H2O (l)
c) The rest of the oxides react with water to form acidic solutions
SO2(g) + H2O (l) H2SO3(aq)
SO3(g) + H2O(l) H2SO4(aq)
P2O3(g) + 3H2O(l) H3PO3(aq)
P2O5(g) + 3H2O(l) H3PO4(aq)
Cl2O(g) + H2O (l)2HOCl(aq)
Cl2O7(g) + H2O (l)2HClO4(aq)
87
Chapter Seven
The Hydrides of period 3
Table 12: The Hydrides of period 3 elements
Hydride Formula
Sodium hydride NaH
Magnesium hydride MgH2
Aluminium hydride AlH3
Silane SiH4
Phosphine PH3
Hydrogen sulphide H2S
Hydrogen chloride HCl
Hydrolysis of the hydrides in water
NaH(s) + H2O (l) NaOH (aq) + H2(g
MgH2(s) + H2O (l) Mg (OH)2(s) +H2(g)
AlH3(s) +3 H2O(l) Al(OH)3(s) +3H2(g)
SiH4 (g)+ 2H2O(l) SiO2(g) + 4H2(g)
H2S (g) + 2H2O (l) S 2-(aq)+ 2 H3O+(aq)
HCl (g) + H2O (l) Cl -aq) + H3O+ (aq)
Note
PH3is not affected by water because it is non polar. This is because

phosphorous and hydrogen have similar electronegativity values.
88
Chapter Seven
7.1.1. ALUMINIUM
The ore of aluminium
a) Bauxite; Al2O3.2H2O
b) Cryolite; Na3AlF6
Extraction of aluminium
Aluminium is commonly extracted from its ore called bauxite.
Impurities found in bauxite are:
 Silica, SiO2
 iron(iii) oxide,Fe2O3
The extraction process is divided into two stages.
1) Purification of the impure bauxite
This involves purifying the impure bauxite to form pure aluminium oxide.
The ore is first roasted at a low temperature to convert the iron (ii) oxide
to iron (iii) oxide. The roasted product is then crushed to powder form and
dissolved in concentrated sodium hydroxide solution.
The dissolution in concentrated sodium hydroxide is to separate iron (iii)

oxide from the ore.
Being an amphoteric oxide, Al2O3 and the acidic oxide SiO2dissolve in

the sodiumhydroxide solution as shown by the equations below
Al2O3(s) + 2OH-(aq) + 3H2O(l) 2 Al(OH)4-(aq)
SiO2(s) + 2NaOH(aq) Na2SiO3(aq) + H2O(l)
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Chapter Seven
The iron (iii) oxide being basic does not dissolve and is eliminated by
filtering off.
Aluminium hydroxide is then obtained from the sodium aluminate

solution by either:
i. Adding a little prepared aluminium hydroxide to aluminate solution

to precipitate more aluminium oxide; a process referred to as
seeding.
Al(OH)4 -(aq) Al(OH)3(s) + OH -(aq)
ii. bubbling CO2 gas through the filtrate ( aluminate solution)
2Al(OH)4 -(aq) + CO2(g) 2Al(OH)3(s) + CO32 -aq)
The aluminium hydroxide from either method is filtered off, washed and
dried. The dry aluminium hydroxide is then heated to form aluminium
(III) oxide.
2Al(OH)3 Al2O3(s) + 3H2O(g)
2) Electrolysis of the pure aluminium oxide
The pure oxide is 1st dissolved in molten cryolite (sodium hexafluoride

aluminate - Na3AlF6) to lower its melting point.
The molten aluminium oxide is then electrolyzed at a temperature

between 800-900 0c at a high current density between graphite electrodes.
Molten aluminium sinks to the bottom at the cathode where it is tapped

off while oxygen is liberated at the carbon anode.
Cathode: Al3+(l) + 3e Al(l)
Anode: 2O2–(l) O2(g) + 4e
The electrolysis cell
90
Chapter Seven
Note
The oxygen gas produced at the anode oxidizes the graphite (carbon) anode and
therefore it is replaced (renewed) from time to time.
Reactions of aluminium
a) Reaction with water.
Aluminium reacts slowly with steam to form the oxide and hydrogen gas.
2Al(s)+ 3H2O(g) Al2O3(s) + 3H2(g)
b) Reaction with chlorine.
Aluminium when heated reacts with dry chlorine to form a chloride which
sublimes. The chloride is predominantly covalent.
2Al(s) + 3Cl2(g) 2AlCl3(s)
Explanation
Aluminium forms a smaller cation with high charge density & high
polarizing power. It distorts the electron cloud of surrounding anions. The
larger the anion, the greater the extent of polarization. Therefore the large
chloride ion is polarized greatly and as such a covalent character is
induced.
The chloridedimerises,Al2Cl6, at high temperature with a chloride bridge

structure. Aluminium bromide & iodide have similar structures.
91
Chapter Seven
c) Reaction with alkalis.
Aluminium is an amphoteric metal. It reacts with concentrated alkalis to

form aluminates complex and hydrogen gas.
2Al(s) + 2OH -(aq) + 6H2O(l) 2 Al(OH)4 -(aq) + 3H2(g)
d) Reaction with acids.
With dilute mineral acids except nitric acid, aluminium forms salt and
hydrogen gas.
2Al(s) + 6H+(aq) 2Al3+(aq) + 3H2(g)
With hot conc. H2SO4, aluminium forms salt, sulphur dioxide gas and
water.
2Al(s) + 6H2SO4(aq) Al2(SO4)3(aq) + 3SO2(g) + 6H2O(l).
Aluminium does not react with dilute nitric acid. The metal is rendered
passive due to the formation of an impervious layer of oxide.
The acidity of the hexaaqua ions
In aqueous solution, soluble salts of aluminium i.e. AlCl3,Al2(SO4)3etc exist as

[Al(H2O)6]3+.
The resultant solution is acidic due to hydrolysis reactions:
[Al(H2O)6]3+(aq) + H2O(l) [Al(H2O)5 OH]2+(aq) + H3O+(aq)
[Al(H2O)5OH]2+(aq) + H2O(l) [Al(H2O)4 (OH)2]+(aq) + H3O+(aq)
[Al(H2O)4 (OH)2]+(aq)+ H2O(l) [Al(H2O)3 (OH)3](s) + H3O+(aq)
Over all equation:
[Al(H2O)6]3+(aq)+3H2O(l) [Al(H2O)3 (OH)3](s) + 3H3O+(aq)
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Chapter Seven
The hydroxonium ions produced make the resultant solution acidic with P H less
than 7 and a white solid is formed.
If sodium carbonate solution is bubbled through aqueous solution of aluminium

salts, bubbles of a colorless gas that turns lime water milky is formed.
The oxide and hydroxide of aluminium are amphoteric. They react with both
conc. alkalis and dilute acids.
Al2O3(s) + 2OH-(aq) + 3H2O(l) 2 Al(OH)4 -(aq)
Al2O3(s)+ 6H+(aq) 2Al3+(aq) + 3H2O(l)
Al(OH)3(s) + OH-(aq) Al(OH)4 -(aq)
Some uses of aluminium include:
Aluminium is used for: Useful properties

1. Aircraft light, strong, resists
corrosion
2. other transport such as ships' light, strong, resists
superstructures, container corrosion
vehicle bodies, tube trains
3. Overhead power cables . light, resists corrosion, good
conductor of electricity
4. Saucepans light, resists corrosion, good
appearance, good conductor
of heat
93
Chapter Eight
CHAPTER 8
8. TRANSITION ELEMENTS
(Scandium, Titanium, Vanadium, Chromium, Manganese, Iron, Cobalt, Nickel,

Copper, Zinc)
d-block elements:
A d-block element is one in which the inner 3d sub-shell is in the

process of filling up with electrons after the outermost4s sub-
shell(which has a lower energy) has already been filled.
In the periodic table, they are found between the very reactive s-block
elements and the less reactive p-block elements.
94
Chapter Eight
Transition Elements.
A transition element is one that forms at least one ion with a partially
filled 3d sub shell. All transition elements are d-block elements.
The sub shells are filled up in order of increasing energy and so the
4s- orbitals are filled 1st before the 3d-orbitals.
NB:
For chromium, the electronic configuration of [Ar]3d54s1 is a more

stable arrangement than [Ar] 3d44s2
This is because the former configuration has 3d sub shell half filled
with electronsand therefore energetically more stable than the latter
configuration where the 3d has only 4 electrons.
Unpaired electrons experience minimum repulsion between them.
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Chapter Eight
For copper, the configuration 3d104s1 is a more stable arrangement

than 3d94s2. The explanation is because the 3d-subshell is full
therefore energetically stable
Configuration of selected ions
Sc3+ 1s22s22p63s23p6
Cr3+ 1s22s22p63s23p63d3
Mn2+ 1s22s22p63s23p63d5
Mn3+ 1s22s22p63s23p63d4
Fe2+ 1s22s22p63s23p63d6
Fe3+ 1s22s22p63s23p63d5
Cu+ 1s22s22p63s23p63d10
Cu2+ 1s22s22p63s23p63d9
Zn2+ 1s22s22p63s23p63d10
When a transition metal atom loses electrons to form ions, the 4s-electrons are
lost first before 3d-electrons.
This is because the 3d sub -shell is situated closer to the nucleus than the 4s sub
shell.
Thus when the 3d sub shell is occupied, the electrons in it repel the 4s-electrons
further from the nucleus and in the process the energy of the 4s electrons
increase more than for the 3d electrons. Therefore the 4s electrons are lost
before the 3d electrons.
8.1. General characteristics of transition elements

1) High melting and boiling points.
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Chapter Eight
Melting points of metals depend on the number of electrons available

for metallic bond formation.
Transition elements use both the 4s and 3d-electrons in metallic bond

formation. The greater the number of electrons contributed to the
metallic bond, the higher the melting point. Transition elements
therefore have high melting and boiling points.
A dip in the melting point is observed at manganese [Ar]4S 23d5. This is

attributed to the stability of half-filled 3d-orbitals, electrons therefore being less
available for metallic bond formation.
Zinc [Ar]4S23d10 with a full 3d-orbitals uses only the outer most 4s-electrons in
metallic bond formation as such has the lowest melting point.
The melting points of these metals rise to a maximum value and then decrease
with increase in atomic number
Exercise
97
Chapter Eight
The atomic numbers and melting points of some d-block elements are as
follows:
Element Sc Ti V Cr Mn Fe Co Ni Cu
Atomic.No 21 22 23 24 25 26 27 28 29
M.Point(0 154 168 1900 1890 1240 154 1500 1450 1080
C) 0 0 0
i) Plot a graph of the melting points against atomic numbers. (3mks)
ii) Explain the shape of your graph.(4marks)
2) Variable oxidation states.
The variable oxidation states exhibited by transition elements are attributed to

the fact that both the 4s and 3d-electrons are energetically similar and they can
both be used in bond formation.
All elements with exception of scandium and zinc show variable oxidation
states. For an oxidation state of +2, only the outer most 4s-electrons are used in
bond formation. For an oxidation state greater than +2, both the 4s and 3d-
electrons are involved in bond formation.
The maximum oxidation state is achieved when all the 4s and 3d electrons are
involved in bonding.
98
Chapter Eight
EXERCISE
Workout the oxidation state of the transition metal in each of the following
compounds or ions:
i) MnO4-
ii) MnO42-
iii) MnO2
iv) Cr2O72-
v) K2CrO4
vi) Fe(CN)63-
3. Catalytic activity.
Transition metals and their compounds behave as catalysts due to the presence
of partially filled d-orbitals .
In reactions involving gases, the reacting gases are adsorbed onto the surface of
the catalyst where they form weak bonds by either
99
Chapter Eight
donating or accepting electrons.
This weakens the internal bonding of the reactant molecules and in the process
reducing the activation energy for the reaction.
Also there is a higher concentration of the reactants at the surface of the catalyst
as a result the rate of reaction increases.
Some common examples include:
I.Haber process in the manufacture of ammonia where finely divided iron is used.
II.Contact process in the manufacture of sulphuric acid, vanadium pent oxide is

used.
III.Decomposition of hydrogen peroxide, manganese (IV) oxide is used.
IV.Hydrogenation of alkenes and alkynes, nickel metal is used.
4. Complex ion formation.
A complex ion basically consists of a central metal atom or ion surrounded by

electron donating groups called ligands capable of forming
dative covalent bonds with the metal atom or ion.
Complex ion formation in transition metals is affected by the following factors:
 Presence of vacant d-orbitals.
 The high charge on the metal cation.)
 The small radius of the cation.
Example of a complex ion is [Ni(H2O)6]2+, and is said to have a coordination

number of 6.
Coordination number is the number of ligands directly bonded to the central

metal ion. A complex species can be cationic, anionic or neutral.
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Chapter Eight
Cationic complexes are those that carry an overall positive charge
Formula Colour Name of complex ion Coordination Ox.

number St. of
metal
ion
Cu(H2O)4 2+
Blue Tetra-aqua copper(ii)ion 4
+2
[Cu(NH3)4]2+ Tetra-ammine 4
copper(ii)ion +2
[Co(H2O)6]2+ Pink Hex-aqua cobalt(ii) ion 6 +2
[Ni(H2O)6]2+ Green Hex-aqua nickel(ii) ion 6 +2
[Cr(H2O)6]3+ Violet Hex-aqua 6 +3

chromium(iii)ion
[Cr(NH3)6]3+ Yellow Hex-ammine 6 +3

chromium(iii)ion
[Cr(NH3)4Cl2]+ Tetra-ammine 6 +3
dichlorochromium(iii)ion
Anionic complexes are those that carry a negative charge.
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Chapter Eight
Formula Color Name of complex Coordination Oxidation

number state of
cation
[CuCl4]2- Yellow Tetra chlorocuprate(ii)ion 4 +2
[CoCl4]2- Blue Tetra 4 +2

chlorocobaltate(ii)ion
[Zn(OH)4]2- Colorless Tetra 4 +2

hydroxozincate(ii)ion
[Fe(CN)6] 3- Hexacyanoferrate(iii) ion 6 +3
[Fe(CN)6] 4- Hexacyanoferrate(ii) ion 6 +2
[Ni(CN)4] 2- Tetracyanonickelate(ii)ion 4 +2
Neutral complexes are those that carry no charge.
Ni(CO)4, Tetracarbonylnickel(O)
Fe(CO)5, Pentacarbonyliron(O)
102
Chapter Eight
Exercise
Give the name, coordination number, and the oxidation state of the
metal ions in the following complexes:
i) Cr(H2O)6Br3 ii) [ Cr(H2O)6]Cl3 iii)[Cr(H2O)5Cl]2+
iv) [Co(NH3)5Br]2+SO42- v) [Cr(H2O)4Cl2]+.
5)Form colored compounds.
Transition metal ions are colored because they have partially filled
3d’sub shell.
A transition metal ion exhibits color when it absorbs light from the
visible region.
Light in the visible region has sufficient energy to promote d-

electrons from a lower energy level to a higher energy level.
6)Magnetic Properties
Molecules with one or more unpaired electrons are attracted into a

magnetic field. The more unpaired electrons in the molecule the
stronger the attraction.
This type of behavior is called paramagnetism.
Substances with no unpaired electrons are weakly repelled by a magnetic field.

This property is called diamagnetism.
Many transition metal complexes exhibit simple paramagnetism. In such

compounds the individual metal ions possess some number of unpaired
electrons.
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Chapter Eight
8.2. CHROMIUM
QUESTION
a) Write the electronic configuration of the following species

(atomic number Cr is 24):
i) Cr
ii) Cr2+
iii) Cr3+
iv) Cr6+
b) State the common oxidation states of chromium.
c) Aqueous sodium hydroxide was added drop wise to a solution containing

Chromium (iii) ions, state what was observed and write equation for the
reaction that took place.
d) Discuss using equation where applicable the reactions of

chromium with:
i) Air
ii) Water
iii) Nitric acid, hydrochloric acid and sulphuric acid.
ANSWERS
a)
i) Cr: 1S22S22P63S23P63d54S1
ii) Cr2+:1S22S22P63S23P63d4
iii) Cr3+: 1S22S22P63S23P63d3
iv) Cr6+:1S22S22P63S23P6
Note
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Chapter Eight
Chromium adopts the above electronic configuration rather than the

expected 1s22s22p63s23p63d44s2 because the 3d is half filled unlike in the
expected configuration therefore it’s more energetically stable.
b) +2, +3, +6
c) Observation
Grey green precipitate, soluble in excess forming a green solution.
Equations
Cr3+(aq) + 3 OH -(aq) Cr(OH)3 (s)
Cr (OH)3 (s) + 3 OH -(aq) Cr(OH)63 -(aq)
d)
With Air (oxygen)
Chromium does not react with oxygen at room temperature, when heated
it forms green chromium (iii) oxide.
4Cr(s) + 3O2(g) 2Cr2O3(s)green solid.
With water.
Chromium reacts with steam to form green chromium (iii) oxide and
hydrogen gas.
2Cr(s) + 3 H2O(g) Cr2O3(s) + 3H2(g)
With acids.
Chromium reacts with dilute hydrochloric acid forminggreen

chromium(iii) chloride solution andliberating hydrogen gas.
2Cr(s) +6HCl(aq) 2CrCl3(aq)+ 3H2(g)
Chromium reacts with dilute sulphuric acid forming blue solution of

chromium (ii) sulphate and liberating hydrogen gas.
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Chapter Eight
The blue solution turns to green when exposed to air. This is due to aerial
oxidation of Cr2+ to Cr3+ ions.
Cr(s) + H2SO4(aq) CrSO4(aq) + H2(g)
4CrSO4(aq) + O2(g) + 2H2SO4(aq) 2Cr2(SO4)3(aq) + H2O(l)
Ionically:
Cr(s) + 2H+(aq) Cr2+(aq) + H2(g)
4Cr2+(aq) + 4H+(aq)+ O2(g) 4Cr3+(aq) +2H2O(l)
Chromium reacts with dilute sulphuric acid on heating to form chromium

(iii) sulphate and hydrogen gas
2Cr(s) + 3H2SO4 (aq) Cr2(SO4)3(aq)+3H2(g)
Chromium reacts with hot conc. sulphuric acid to form chromium (iii)
sulphate, sulphur dioxide and water.
2Cr(s) + 6H2SO4 (aq) Cr2(SO4)3(aq)+ 6H2 O (aq) + 3SO2(g)
Note:
Chromium does not react with nitric acid due to the formation of a thin
layer of chromium (vi) oxide which renders it passive towards nitric acid.
With chlorine
If chlorine gas is passed over heated chromium, green solid of chromium (iii)
chloride is formed.
2Cr(s) + 3Cl2(g) 2CrCl3(s)
NB: If hydrogen chloride is used instead, chromium (ii) chloride is formed.
Cr(s) + 2HCl (g) CrCl2(s)+ H2(g)
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Chapter Eight
With sodium hydroxide solution
Chromium dissolves in sodium hydroxide solution giving:
 Effervescence of a colorless gas that burns with a pop sound.
 a green solution.
2Cr(s) + 2OH -(aq) + 6H2O(l) 2Cr(OH)4 -(aq) + 3H2(g)
Compounds of Chromium
Chromium shows principal oxidation states of +2, +3, & +6.
The most stable oxidation state of chromium is +3.
The acidity of the hexaqua ions
In aqueous solution, soluble salts of chromium in +3 oxidation state e.g. CrCl3,

Cr (SO4)3etc exist as [Cr(H2O)6]3+.
The resultant solution is acidic due to hydrolysis reaction
.i.e.
[Cr(H2O)6]3+(aq)+H2O(l) [Cr(H2O)5OH]2+(aq) +H3O+(aq)
[Cr(H2O)5OH]2+(aq)+H2O(l) [Cr(H2O)4 (OH)2]+(aq) + H3O+(aq)
[Cr(H2O)4 2OH]+(aq)+ H2O(l) [Cr(H2O)3(OH)3](s) + H3O+(aq)
Over all equation:
[Cr(H2O)6]3+(aq)+3H2O(l) [Cr(H2O)3(OH)3](s) + 3H3O+(aq)
Green ppt.
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Chapter Eight
The hydroxonium ions produced make the resultant solution acidic with P H less
than 7 and a green solid is observed due to the formation of hydrated chromium
(iii) hydroxide.
If sodium carbonate solution is added to an aqueous solution of chromium (iii)

salts, bubbles of a colorless gas that turns lime water milky is observed due to
the carbon dioxide gas produced by the reaction between the hydroxonium ions
and the carbonate ions.
CO32 -(aq) + 2H+(aq) CO2 (g) + H2O (l)
If magnesium ribbon is added to an aqueous chromium (iii) salt, a colorless gas

that burns with a pop sound is liberated due to the reaction between the acidic
solution and magnesium metal.
Mg(s) + 2H3O+(aq) → Mg2+(aq) + H2(g) + 2H2O (l)
or
Mg(s) + 2H+(aq) → Mg2+(aq) + H2(g)
Note
Like chromium (iii) salts in aqueous solution, aqueous solution of aluminium

and iron (iii) salts also undergo hydrolysis reactions to produce acidic solutions.
Chromium (iii) oxide, Cr2O3
It’s a green solid which can be obtained by heating chromium (iii) hydroxide.
2Cr (OH) 3(s) Cr2O3(s)+ 3H2O (l)
It is an amphoteric oxide.
It reacts with dilute mineral acids to form corresponding Cr3+ salt and with
sodium hydroxide to forming the soluble complex Cr (OH)63-(aq)
Cr2O3(s) + 6H+(aq) 2Cr3+(aq) + 3H2O(l)
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Chapter Eight
Cr2O3(s) + 6OH-(aq) + 3H2O(l) 2Cr(OH)63-(aq)
Green solution
Chromium (iii) hydroxide
It’s a green solid formed when little sodium hydroxide is added to a solution
containing Cr3+ions. It’s an amphoteric hydroxide; reacting with dilute
acidsforming corresponding Cr3+ salts and with aqueous alkalis forming the
complex Cr(OH)63-(aq).
Cr(OH)3(s) + 3H+(aq) Cr3+(aq) + 3H2O(l)
Cr(OH)3(s) + 3OH -(aq) Cr(OH)63 -(aq)chromite ion
Hydrated Chromium (iii) Chloride.CrCl3.6H2O
This salt of chromium exhibits hydration isomerism i.e. existence of compounds

with the same molecular formula but different structural formulae due to the
difference in the number of water molecules directly bonded to the central
chromium (iii) ion.
The following are the isomers of hydrated chromium (iii) chloride:
i. [Cr(H2O)6]3+ 3Cl-
(Hexaquachromium (iii) chloride)
ii. [Cr(H2O)5Cl]2+.2Cl-.H2O
(Chloropentaqua chromium (iii) chloride monohydrate)
iii. [Cr(H2O)4Cl2]+.Cl-.2H2O
(Dichlorotetraaqua chromium (iii) chloride dihydrate)
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Chapter Eight
Identification/test for the isomers
1) Using silver nitrate solution:
 The first isomer precipitates 3moles of silver chloride because all

the 3Cl- ions can be precipitated because they are not directly
bonded to the central metal cation,
 The 2nd isomer precipitates 2moles of silver chloride per mole of

the isomer.
 The 3rd isomer precipitates 1mole of silver chloride per mole of the
isomer.
2) Using conductivity measurement:
 This test provides the number of moles of conducting ions in

solution for every mole of isomer.
 The 1st isomer contains 4moles of conducting ions i.e. [Cr(H2O)6]3+

and 3Cl-
 The 2nd isomer contains 3moles of conducting ions i.e.

[Cr(H2O)5Cl]2+and 2Cl-
 The 3rd isomer contains 2moles of conducting ions

i.e.[Cr(H2O)4Cl2]+and Cl-
Compounds of chromium in +6 oxidation states
Because of the high charge density of Cr6+ ion, it can only exists as oxo- ions.eg
, CrO3
, CrO42- , Cr2O72-
Chromates, CrO42-
Chromates are salts of chromic acid; they have a characteristic yellow color. e.g.
K2CrO4 and Na2CrO4.
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Chapter Eight
A Chromate ion has a tetrahedral structure
Chromates are obtained when an alkaline solution is added to a solution

containing dichromate ions.
Cr2O72 -(aq)+ 2OH -(aq) 2 CrO42 -(aq) + H2O(l)
Orange yellow
Chromates are stable in alkaline medium; however Chromates react with acids
to form dichromate.
2CrO42 -(aq) + 2H+(aq) Cr2O72 -(aq) + H2O(l)
ALUMS
Alums are double sulphate salts with general formula M2SO4.M2(SO4)3.24H2O
The 1st M is a group (I) metal e.g. Li+, Na+, K+, NH4+, Rb+ , Cs+etc
The 2nd M is a group (III) metal or a metal bearing a +3 charge e.g. Al3+, Cr3+,
Fe3+, Mn3+, Co3+etc
Examples of common alums are:
Chrome alum;K2SO4.Cr2 (SO4)3.24H2O
Its solution in water is acidic because the chromium (iii) ions undergo hydrolysis
making the solution acidic.
The complex ion is acting as an acid by donating a hydrogen ion to water

molecules in the solution.
The water is, of course, acting as a base by accepting the hydrogen ion.
[Cr(H2O)6]3+(aq)+H2O(l) [Cr(H2O)5OH]2+(aq) + H3O+(aq)
[Cr(H2O)5OH]2+(aq)+H2O(l) [Cr(H2O)4 2OH]+(aq) + H3O+(aq)
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Chapter Eight
[Cr(H2O)4 2OH]+(aq)+ H2O(l) [Cr(H2O)3 3OH] (s) + H3O+(aq)
Over all equation:
[Cr(H2O)6]3+(aq)+3H2O(l) [Cr(H2O)3 3OH](s) + 3H3O+(aq)
The accumulation of hydroxonium ions renders the resultant solution acidic.
If aqueous sodium carbonate is added, an efferverscence of a colorless gas

which turns lime water milky is formed and a greeen solid.
2) Potash alum K2SO4.Al2(SO4)3.24H2O
3) Ammonium alum (NH4)2SO4.Al2(SO4)3.24H2O
Preparation of Chrome Alum
By bubbling sulphur dioxide gas through an acidified potassium dichromate

solution, chrome alum is formed.
SO2(g) + H2O (l) H2SO3 (aq)
H2SO3 (aq)2H+(aq) + SO32-(aq)
SO32-(aq) + H2O (l) SO42-(aq) +2H+(aq) 2e
Cr2O72-(aq) + 14H+ (aq)+ 6e 2Cr3+(aq) + 7H2O(l)
The ions present after the above reactions are in the correct proportion for the
formation of chrome alum on crystallizing.
Similarities between the chemistry of chromium and lead
Cr3+and Pb2+ form complex with sodium hydroxide solution.
Cr3+(aq) + 3 OH -(aq) Cr(OH)3(s)
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Chapter Eight
Cr(OH)3(s) + 3 OH -(aq) [Cr(OH)6] 3-(aq)
Pb2+(aq) + 2 OH -(aq) Pb(OH)2(s)
Pb(OH)2(s) + 2 OH -(aq) [ Pb(OH)4]2-(aq)
Both their oxides in higher oxidation states are acidic.
CrO3(s) + 2NaOH (aq)Na2CrO4(aq) + H2O(l)
PbO2(s) + 2NaOH (aq) Na2PbO3(aq) + H2O(l)
Their oxides in higher oxidation states are strong oxidizing agents.
CrO3(s) + Al(s) Cr(s )+ Al2O3(s)
PbO2(s) + 4HCl(aq) PbCl2(s) + 2H2O(l) + Cl2(g)
They both form compounds with variable oxidation statesie
Lead: +2 & +4 and Chromium: +2, +3 & +6
Chromium and Lead in higher oxidation states Cr6+and Pb4+ form covalent
compounds e.g. PbCl4 and CrCl6
8.2.1. Similarities between the chemistry of Chromium& Aluminium

1.
Cr3+ and Al3+ form complexes with Alkalis.
Cr3+(aq) + 3 OH-(aq) Cr(OH)3(s)
Cr(OH)3(s) + 3 OH-(aq) [Cr(OH)6] 3-(aq)
Al3+(aq) + 3 OH-(aq) Al(OH)3(s)
Al(OH)3(s) + OH-(aq) [ Al(OH)4]-(aq)
2. Oxides and hydroxides of Chromium and Aluminium are

amphoterici.e react with both acids and bases.
Al2O3(s) + 2OH-(aq) + 3H2O(l) → 2 Al(OH)4-(aq)
Cr(OH)3(s) + 3 OH -(aq) [Cr(OH)6] 3-(aq)
Al(OH)3(s) + OH -(aq) [ Al(OH)4] -(aq)
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Chapter Eight
Al2O3(s)+ 6H+(aq) 2Al3+(aq) + 3H2O(l)
8.2.2. Test for Cr3+ ions in solution

Use of sodium hydroxide solution:
Observation
Grey green precipitate, soluble in excess forming deep green solution.
Equations
Cr3+(aq) + 3 OH -(aq) → Cr(OH)3(s)
Cr(OH)3 (s) + 3 OH -(aq) → Cr(OH)63 -(aq)
Note:
Chromium (iii) hydroxide is oxidized by hydrogen peroxide solution in an

alkaline medium to chromate ions (CrO42-) which has a characteristic yellow
colour.
Procedure
Sodium hydroxide solutionis added drop-wise till excess to a solution containing

Cr3+ ions followed by a little H2O2 and the resultant solution warmed.
Observation:
A grey green precipitate soluble in excess alkali forming a green solution. The
solution turns yellow on addition of hydrogen peroxide.
Cr3+(aq) + 3 OH -(aq) Cr(OH)3(s)
Cr(OH)3(s) + 3OH -(aq) Cr(OH)63-(aq)
2 Cr(OH)63 -(aq)+ 3H2O2 (aq) 2CrO42 -(aq) + 2 OH -(aq) + 8H2O(l)
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Chapter Eight
The presence of the chromate ions produced can be confirmed by:
a) Addition of Lead(ii) ethanoate or Lead(ii) nitrate solution.
Observation
A yellow precipitate of lead (ii) chromate is formed.
Equation
Pb2+(aq) + CrO42-(aq) PbCrO4(s)
b) Addition of Barium nitrate or Barium chloride solution.
Observation
A yellow precipitate of barium chromate is formed
Equation
Ba2+(aq) + CrO42-(aq) BaCrO4(s)))
c) Addition of silver nitrate solution.
d) Observation
A brick red precipitate of silver (i) chromate is formed.
Equation
2Ag+(aq) + CrO42-(aq) Ag2CrO4(s)
e) Addition of a little amyl alcohol (eg Butan-1-ol) followed by dilute

sulphuric acid.
Observation
A blue color is observed in the organic layer.
Use of aqueous ammonia solution:

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Chapter Eight
Observation
Green precipitate slowly dissolves in excess concentrated ammonia solution to

form a purple solution.
Equations
Cr3+(aq) + 3 OH -(aq) → Cr(OH)3(s)
Cr(OH)3(s) + 6NH3(aq) → [Cr(NH3)6]3+(aq) + 3 OH -(aq)
Use of sodium carbonate / potassium nitrate mixture
Procudure
Fuse the solid suspected to contain Chromium (III) ions with (Na2CO3/KNO3)
mixture
Observation
A yellow mass of Chromate (CrO42-) is formed
8.3. MANGANESE
Atomic number is 25
Its electronic configuration is 1s22s22p63s23p64s23d5
8.3.1. Reactions of manganese

a) Reactions with acids
Manganese being a metal reacts with both dilute hydrochloric and

sulphuric acid producing corresponding manganese (ii) salts and hydrogen
gas.
Mn(s) + 2HCl(aq) MnCl2(aq) + H2(g)
Mn(s) + H2SO4(aq) MnSO4(aq) + H2(g)
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Chapter Eight
Note
The products when manganese react with dilute nitric acid are not well
defined.
b) Reaction with steam
Like magnesium, manganese reacts with hot water to produce manganese

(ii) hydroxide and hydrogen gas.
Mn(s) +H2O (l) Mn(OH)2(s) + H2(g)
c) Reaction with oxygen
Manganese burns in air when heated forming trimanganese tetra oxide
Mn(s) + O2(g) Mn3O4(s)
8.3.2. Compounds of manganese

Manganese exhibits the following oxidation states; +2, +3, +4, +5, +6 and +7.
The +2 oxidation state being the most stable state because the 3d sub shell is
half filled with electrons and therefore energetically stable.
Compound of manganese in the +2 oxidation state
i) Manganese (ii) oxide
On heating either manganese (ii) carbonate or manganese (ii) oxalate,

manganese (ii) oxide is produced as a green solid.
MnCO3(s) MnO(s) + CO2(g)
MnC2O4(s) MnO(s) +CO(g) + CO2(g)
ii) Manganese (ii) hydroxide
On addition of dilute sodium hydroxide solution or aqueous ammonia

solution to a solution containing manganese (ii) ion, manganese (ii)
hydroxide is formed as a white precipitate. The precipitate rapidly turns
brown due to the aerial oxidation of manganese (ii) oxide formed to
manganese hydrated manganese (iv) oxide
Mn2+(aq) + 2 OH -(aq) Mn(OH)2(s) white ppt

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Chapter Eight
2Mn (OH)2(s) + O2(g) 2MnO2.H2O(s) brown ppt
Note
Manganese (ii) compounds are pink in solid state but in aqueous solution,
the pink colour is sometimes too faint to be noticed and so manganese (ii)
compounds in aqueous solution may appear colourless.
Compounds of manganese in +4 oxidation state
The most important compound of manganese in this oxidation state is

manganese (iv) oxide (manganese dioxide).
It is a dark solid produced on heating manganese (ii) nitrate.
Mn(NO3)2(s) MnO2(s) + NO2(g)
It is insoluble in water and not attacked by dilute acids.
It reacts with cold concentrated hydrochloric acid to form the complex

hexachloro manganate (iv) ion.
MnO2(s) + 4H+ (aq) +6Cl -(aq) MnCl62 -(aq) +2H2O (l)
With hot concentrated hydrochloric acid, manganese (iv) oxide reacts

forming chlorine gas as one of the products.
MnO2(s) + 4HCl(aq) MnCl2(aq) + Cl2(g) +2H2O(l)
Warm concentrated sulphuric acid reduces manganese (iv) oxide to

manganese (ii) sulphate.
MnO2(s) + H2SO4 (aq) MnSO4(aq) + O2(g) + H2O(l)
Note
Manganese (iv) oxide is a powerful oxidizing agent. Some of the

oxidizing reactions of manganese (iv) include the following
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Chapter Eight
i) MnO2(s)+ HCl(aq) MnCl2(aq) + Cl2(g) + 2H2O(l)
ii) MnO2(s) + C2O42 -(aq) + H+(aq) Mn2+(aq) +2H2O(l) +2CO2(g)
Compounds of manganese in the +6 oxidation state
Potassium manganate (vi); K2MnO4 isthe only stable compound of

manganese in the +6 oxidation state.
It is a dark green solid prepared on fusing together manganese (iv) oxide

with potassium hydroxide in the presence of excess oxygen
MnO2(s) + KOH(s) + O2 (g) K2MnO4(s) +H2O(s)
It can also be prepared by heating manganese (iv) oxide, potassium

hydroxide and potassium chlorate.
MnO2(s) + 6KOH(aq) + KClO3(aq) K2MnO4(s) +KCl(aq)

+3H2O(l)
Potassium manganate (vi) is also a powerful oxidizing agent.
Behavior of potassium manganate (VI) in
a) Acidic medium
MnO42 -(aq) + H+(aq) 2MnO4 -(aq)+MnO2(s) +2H2O(l)
b) Neutral medium
3K2MnO4(s) + 2H2O(l) KMnO4(s) +MnO2(s) + 4KOH(aq)
Compounds of manganese in the +7 oxidation state
Potassium manganate (vii) (potassium permanganate) is the most important

compound of manganese in the +7 oxidation state.
Potassium permanganate is a powerful oxidizing agent and although not a

primary standard, it’s commonly used in volumetric analysis because of the
following;
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Chapter Eight
1) End points involving manganate (vii) titrations do not need indicators.

The persistence of the intense purple colour of the permanganate solution
when added from a burette indicates the end point of the titration has been
reached.
2) Permanganate titrations have sharp end points
3) Potassium permanganate is fairly soluble in water.
4) Potassium manganate (vii) has a fairly high formula mass and this reduces
errors due to weighing inefficiencies.
Note
1) Potassium manganate is not a primary standard because it aqueous

solution on standing is reduces to manganese (iv) oxide which appears as
brown solid.
2) In acidic medium in the presence of an oxidizing agent,

manganate (vii) ions react as shown by the equation below
MnO4 -(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l)
3) The purple colour of the manganate (vii) solution turns colourless due to
the formation of manganese (ii) ions.
4) Only dilute sulphuric acid is used to acidify the solution.
5) Dilute hydrochloric acid is not used because it is oxidizes by the

manganate (vii) solution forming chlorine gas which is an interference to
the desired reaction
2KMnO4 (aq) + 16HCl (aq) 2MnCl2 (aq) + 2KCl(aq) + 5Cl2(g) + 8H2O(l)
6) Dilute nitric acid is not used because itself like potassium manganate (vii)
is an oxidizing agent and therefore would interfere with the desired
reaction.
7) Some of the oxidizing properties of manganate (vii) ions ( MnO4-) in

acidic medium include:
a) 2MnO4-(aq) + 5C2O42 -(aq) + 16 H+(aq) 2Mn2+(aq) + 10CO2(g) +8H2O(l)

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Chapter Eight
b) MnO4 -(aq) + 5Fe2+(aq) + 8H+(aq) 5Fe3+(aq)+ Mn2+(aq) +4H2O(l)
c) 2MnO4 -(aq) + 5H2O2(aq) + 6H+(aq) 2Mn2+(aq) +8H2O(l)+5O2(g)
d) 2MnO4 -(aq) + 5Sn2+(aq) + 16H+(aq) 5Sn4+(aq)+2 Mn2+(aq) +8H2O(l)
Revision questions
a) Write the electronic configuration of the following species (atomic

number= 25):
i. Mn
ii. Mn2+
iii. Mn3+
iv. Mn7+
b) State the most stable oxidation state of Manganese. Give a reason for your
answer.
c) State the possible oxidation states of manganese.
d) Explain why manganese shows variable oxidation states.
e) Aqueous sodium hydroxide was added drop wise to a solution containing

Mn2+,
State what was observed and write equation for the reaction that took
place.
f) Give a formula of a compound or ion containing Manganese in an

oxidation state of +7 oxidation state?
g) Write equation to show the reduction of MnO4- in :
i. Alkaline medium.
ii. Acidic medium.
h) Discuss the chemistry of manganese and magnesium showing similarities.
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Chapter Eight
Answer
a)
i) 1s22s22p63s23p64s23d5
i) 1s22s22p63s23p63d5
ii) 1s22s22p63s23p63d4
iii) 1s22s22p63s23p6
b) The most stable oxidation state is +2. This is because after using the 4s-
electrons, a half filled 3d -orbital that is stable is left.
c) +2, +3, +4, +5, +6, +7.
Acidity of the oxide of manganese increases with the increasing oxidation

state.
d) The variable oxidation state of manganese is due to the fact that both the
4s and the 3d- electrons take part in bond formation.
This is because electrons are being removed from energetically similar 4s

and 3d- orbitals.
For +2 oxidation states, only the 4s-electrons are utilized for bond
formation, forming stable Mn2+. For higher oxidation states,both the 4s
and 3d- electrons are utilized.
e) Observation:
White precipitate insoluble in excessturns brown on standing due to aerial

oxidation to Manganese iv oxide.
Equation:
Mn2+(aq) + 2 OH-(aq) Mn(OH)2(s)white ppt
2Mn(OH)2(s) + O2(g) 2MnO2.H2O(s)brown ppt

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Chapter Eight
Note
A similar observation & reactions occur on addition of aqueous

ammonia solution to a solution containing Manganese (ii) ions.
f) MnO4 -
Manganese 1s22s22p63s23p64s23d5 shows an oxidation state of

+7 by using all the 4s and 3d- electrons in the bond
formationbecause they are energetically similar.
g) In alkaline condition, the change in oxidation state is from +7 to

+
4 since potassium manganate (vii) is a mild oxidizing agent in
alkaline medium
MnO4 -(aq) + 2H2O(l) + 3e MnO2(s) + 4OH -(aq)
Or
MnO4 -(aq) + 4H+(aq) + 3e MnO2(s) + 2H2O(l)
In acidic medium, the reaction is as shown below.
MnO4 -(aq) + 8H+(aq) + 5e Mn2+(aq) + 4H2O(l)
Note
KMnO4 has wide applications in the laboratory including the following:
 Being used as indicator in volumetric analysis titrations involving its

solution.
 To test for reducing agents e.g. sulphur dioxide.
 Estimating iron (ii) salts and oxalates.
 Preparation of gases e.g. chlorine gas is liberated when Conc.

hydrochloric acid is reacted with it.
Hydrochloric acid is not used to acidify KMnO4 during titration

because potassium permanganate solution is a very strong oxidizing
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Chapter Eight
agent, oxidizes chloride ions fromhydrochloric acid to chlorine which

interferes with the desired reaction.
2KMnO4( + 16HCl(aq) 5Cl2(g) +2KCl(aq) +2MnCl2(aq)

+8H2O(l)
h) They both react with chlorine to form chloride when heated.
Mn(s) + Cl2(g) →MnCl2(s)
Mg(s) + Cl2(g) →MgCl2(s)
They both react directly with nitrogen to form similar nitrides
3Mn(s) +N2(g) →Mn3N2(s).
3Mg(s) +N2(g) →Mg3N2(s)
They both react with dilute hydrochloric acid to liberate

hydrogen gas.
Mn(s) +2HCl(aq) →MnCl2(aq)+ H2(g).
Mg(s) +2HCl(aq) →MgCl2(aq)+ H2(g)
Manganese reacts with hot water to form oxide. Magnesium

reacts with steam in a similar way.
Mn(s) +H2O(l) → MnO(s)+ H2(g)
Mg(s) +H2O(g) → MgO(s)+ H2(g)
Note:
Confirmatory test of Manganese, Mn2+ ions in solution
Procedure
To solid manganese (II) salt is added concentrated nitric acid and

little solid lead (IV) oxide or sodium bismuthate solution and the
mixture
boiled.
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Chapter Eight
Observation
A purple solution forms due to the formation of manganate (vii) ions
Equations
2Mn2+(aq) + 5 BiO3 -(aq) + 14H+(aq) → 2MnO4 -(aq) + 5Bi3+(aq) + 7H2O(I)
2Mn2+(aq) + 5PbO2(s) + 4H+(aq) → 2MnO4 -(aq) + 5Pb2+(aq) + 2H20(1)
8.4. IRON
Ores of iron:
 Haematite , Fe2O3
 Magnetite , Fe3O4
 Iron Pyrite , FeS2
 Siderite , FeCO3
The 1st two ores are fed directly into a blast furnace but the last two must first be
roasted in air to convert them to stable iron (iii) oxide before feeding them into
the blast furnace. i.e.
4FeS2(s) + 11 O2(g) → 2Fe2O3(s) + 8SO2(g)
4FeCO3(s) + O2(g) → 2Fe2O3(s) + 4CO2(g)
Extraction of iron
Iron is commonly extracted from its ore called haematite (Fe2O3)using a blast
furnace
The raw materials used in the extraction are:
1) The ore, Haematite
2) Coke
3) Limestone( calcium carbonate)
4) Hot air
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Chapter Eight
The hot blast of air enters the blast furnace through narrow pipes called tuyeres
located at the lower part of the furnace.
The other materials are fed into the furnace from the top.
Reactions taking place in the blast furnace
In the presence of hot air coke burns to produce carbon dioxide gas
C(s) +O2(g) →CO2(g)
As the carbon dioxide ascends, it reacts with unburnt coke to form carbon
monoxide
CO2(g) + C(s) →2CO(g)
The carbon monoxide produced reduces the ore to iron.
Fe2O3(s) + 3CO (g) 2Fe(l) + 3CO2(g)
The role of the limestone is to remove impurities like silica (SiO2)
CaCO3(s)→ CaO(s) + CO2(g)
The quick lime then combines with silica and alumina present as impurities to
form slag.
CaO(s) + SiO2(s) → CaSiO3(l)
CaO(s) + Al2O3(s)→ CaAl2O4(l)
Molten iron and slag both sink to the bottom of the furnace and they can be
tapped off separately. The molten iron at the bottom is coveredby the slag to
prevent oxidation.
The molten iron solidifies on cooling to form PIG iron.
Properties of pig iron:
 Low melting point.
 Impure i.e. contain impurities such as carbon, silicon, sulphur, etc.
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Chapter Eight
Pig iron can be used in making gas stove, drainage pipes, bunsen burner bases,
boiler plates etc
Purification of pig iron
The Bessemer process:
This involves pouring the molten iron into a large tab called the converter and
air is blown into it to oxidize the impurities such as carbon, sulphur and
phosphorus to their gaseous oxides.
Reactions of iron
a) With air
Iron combines with oxygen when heated to form black solid of tri-iron tetra
oxide.
3Fe(s) + 2 O2 (g) → Fe3O4(s)
Ax
Iron reacts with moist cold air to form brown hydrated iron (iii) oxide or rust.
2Fe(s) + 6H2O (l) + 3 O2 (g) → 2Fe2O3.3H2O(s)
How rusting occurs:
Rusting is an electrochemical process which takes place when different parts of

iron act as anode and cathode.
At the cathodic region:
Fe(s) →Fe2+(aq)+2e
At the anodic region:
2 H2O (l) + O2 (g)+ 4e 4OH -(aq)
If the cathodic and anodic areas are close enough, iron (ii) hydroxide is formed.
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Chapter Eight
Fe2+(aq) + 2OH -(aq) → Fe(OH)2(s)
Moist air then oxidizes the iron (ii) hydroxide to rust.
4Fe (OH)2(s) + 2H2O(l) + O2(g) → 2Fe2O3.3H2O(s)
Prevention of rusting
a. Painting
b. Greasing
c. Galvanizing (coating with zinc)
b) With water
Heated iron reacts with steam to form black tri-iron tetra oxide.
3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)
c) With acid.
Iron reacts with dilute acids (HCl& H2SO4) to form hydrogen gas and iron (ii)
salts.
Fe(s) + 2HCl (aq)→ FeCl2(aq) + H2(g)
Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
Hot conc. sulphuric acid oxidizes iron to iron (iii) sulphate, sulphur dioxide and
water.
2Fe(s) + 6H2SO4(aq) → Fe2(SO4)3(aq) + 3SO2(g) + 6 H2O(aq)
d) With halogens.
Heated iron reacts with dry chlorine gas to form iron (iii) chloride.
2Fe(s) + 3Cl2(g) → FeCl3(s)
Compounds of iron
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Chapter Eight
+2 oxidation state
In aqueous solution exists as [Fe(H2O)6]2+ which is a green solution. It

undergoes hydrolysis making the resultant solution acidic.
Iron (ii) hydroxide
It’s a green ppt formed when sodium hydroxide solution is added drop wise to a
solution of Fe2+ solution. The ppt is insoluble in excess and turns brown due to
aerial oxidation on exposure to air.
Fe2+ (aq) + 2OH-(aq) Fe (OH)2(s)
4Fe (OH)2(s) + 2H2O(l)+ O2(g) 2Fe2O3.3H2O(s)
NB:
Similar reaction occurs on addition of ammonia solution to aqueous Fe 2+

solution
Hydrated iron (ii) sulphate.FeSO4.7H2O
It’s prepared by heating iron fillings with dilute sulphuric acid and crystallizing
the salt from solution.
During the crystallization process, hydrated iron (ii) sulphate forms as green
crystals.
Fe(s) + H2SO4(aq) FeSO4(aq) + H2(g)
Action of heat on hydrated iron (ii) sulphate
On gentle heating
FeSO4.7H2O (s) FeSO4(s) + 7H2O (g)
On further heating.
2FeSO4(s) Fe2O3(s) + SO3(g) + SO2(g)
White acidic fumes
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Chapter Eight
If conc. nitric acid is added to iron (ii) sulphate, the green solution turns to
yellow/ brown due to formation of iron (iii) ions.
Ammonium Ferrous Sulphate (Ammonium Iron (ii) sulphate)

(NH4)2SO4.FeSO4.6H2O
By mixingequimolar concentrated solutions of iron (ii) sulphate and ammonium

sulphate, ammonium ferrous sulphate is formed.
It is a green double salt preferred to iron (ii) sulphate in volumetric analysis

because it is not efflorescent and also not easily oxidized.
+3 oxidation state
This is the most stable oxidation state of iron.
The soluble salts in this oxidation state exist as [Fe(H2O)6]3+ and undergo
hydrolysis in water making resultant solution acidic.
e.g.a solution of iron (iii) chloride turns blue litmus red.
The Fe3+ ion is small and highly charged. It undergoes hydrolysis in aqueous
solution to release hydrogen ions or hydroxonium ions which make the resultant
solution acidic.
FeCl3(aq) → Fe3+ (aq) + 3Cl -(aq)
Fe3+(aq) + 6H2O(l) →[Fe(H2O)6]3+(aq)
[Fe(H2O)6]3+(aq) + 3H2O(l) [Fe(H2O)3(OH)3](s) +3H3O+(aq)
Brown ppt
Or
[Fe(H2O)6]3+(aq) [Fe(H2O)3(OH)3](s) + 3H+(aq)
On addition of zinc powder to a brown/ yellow solution of iron (iii), the color of
the solution changes to green.
2Fe3+(aq) + Zn(s) → 2Fe2+(aq) + Zn2+(aq)

130
Chapter Eight
The resultant solution can now be titrated with acidified solution of standard
potassium permanganate.
MnO4 -
(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq)+4H2O(l) +5Fe3+(aq)
On bubbling hydrogen sulphide gas to a solution containing Fe3+ ion, the color
of the solution changes from yellow to green with formationof a yellow solid.
H2S(g) + 2Fe3+(aq) → 2H+(aq) + 2Fe2+(aq) + S(s)
8.4.1. Similarities in the chemistry of zinc and iron
1) Both metals when heated (red hot) react with steam to form oxide and
hydrogen gas.
3Fe(s) + 4H2O (g) Fe3O4(s)+ 4H2(g)
Zn(s) +H2O (g)→ ZnO(s) + H2(g)
2) Both react with dilute acid (HCl& H2SO4) to liberate hydrogen gas.
Fe(s) + H2SO4(aq) → FeSO4(aq) + H2(g)
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
3) Both metal when heated react with oxygen to form oxide.

3Fe(s) + 2 O2(g) → Fe3O4(s)
2Zn(s) + O2(g) → 2ZnO(s)
4) Both metals when heated react with chlorine to form chloride.

2Fe(s) + 3Cl2(g) → 2FeCl3(s)
Zn(s) + Cl2(g) → ZnCl2(s)
5) 5)Both elements react with sulphur to form sulphide.

Zn(s) + S(s)→ ZnS(s)
Fe(s) + S(s)→FeS(s)
131
Chapter Eight
8.4.2. Differences in the chemistry of Zinc & Iron

1) Zn dissolves in conc. alkalis forming hydrogen gas but iron does not react
with alkalis.
Zn(s) + 2OH -(aq) + 2H2O(l) → Zn(OH)42 -(aq) + H2(g)
2) Zn shows a single valency of +2 while iron exhibits variable valencies of
+2,+3 and +6
3) Zn forms compounds that are white in color where as Fe forms
compounds which are colored.
Fe2+ -green
Fe3+ -brown
4) Iron is paramagnetic(due to unpaired electrons in the 3d-subshell) while
Zn is diamagnetic( due to paired electrons in its 3d-subshell)
5) Fe forms dimeric chloride, Fe2Cl6.
Cl ClCl
Fe Fe
Cl ClCl
6) Zinc does not react with cold moist air. Iron reacts with moist air to form
hydrated iron(iii) oxide (rust).
8.4.3. Test for Iron (ii) and Iron (iii) ions in aqueous solution
a) Test for Ferrous ions ( Fe2+)
Use of sodium hydroxide solution / aqueous ammonia solution
Procedure
To the solution containing the unknown cation is added dilute sodium
hydroxide solution or aqueous ammonia dropwise until in excess.
Observation
In the presence of Iron (ii) ions, a green precipitate of Iron (ii) hydroxide is
formed. The precipitate turns brown due to aerial oxidation to iron (iii)
hydroxide.
132
Chapter Eight
Equation
Fe2+(aq) + 2OH -(aq) Fe(OH)2(s)
Use of Potassium hexacyanoferrate (iii) solution
Procedure
hexacyanoferrate (iii) solution.
Observation
A dark blue precipitate (turnbull’s blue) is formed.
Equation
K+(aq) + [Fe (CN)6]3 -(aq) + Fe2+(aq)K[Fe(Fe(CN)6](s)
Test for Iron (iii) ions in solution

Use of sodium hydroxide solution/ aqueous ammonia solution
Procedure
To the solution containing the unknown cation is added dilute sodium
hydroxide or aqueous ammonia solution dropwise until in excess.
Observation
A reddish brown precipitate of iron (iii) hydroxide is produced. The
precipitate is insoluble in excess sodium hydroxide/ aqueous ammonia.
Equation
Fe2+(aq) + 3OH -(aq) Fe(OH)3(s)
Use of potassium hexacyanoferrate (ii) solution

Addition of a few drops of potassium hexacyanoferrate (ii) solution to
solution containing iron (iii) ions produces a dark blue precipitate (Prussian
blue).Equation
K+(aq) + [Fe (CN)4]4-(aq) + Fe3+(aq) K[Fe(Fe(CN)6](s)
Use of potassium thiocyanate solution

Procedure
Addtidion of potassium thiocyanate solution to a solution containing Fe3+ion
s produces a blood red coloration of thiocyanatopenta aqua iron(III)complex
ion.
[Fe(H2O)6]3+(aq) + SCN -(aq) [Fe(H2O)5SCN] 2+(aq) + H2O(l)
Question.
133
Chapter Eight
State the similarities in the chemistry of zinc & magnesium.
8.5. COBALT
8.5.1. Reaction of Cobalt (II) ion, Co2+

The +2 oxidation state is the most stable for simple cobalt salts.
Cobalt (ii) compounds are colored pink. The simplest ion that cobalt forms in
solution is the hexa aqua cobalt (II) ion - [Co(H2O)6]2+.which is pink in colour
A ligand exchange reaction involving chloride ions
If you add concentrated hydrochloric acid to a solution containing hexa aqua

cobalt(II) ions, the solution turns from its original pink colour to arich blue. The
six water molecules are replaced by four chloride ions.
The reaction taking place is reversible.
Co(H2O)62+(aq) + 4Cl-(aq) CoCl42-(aq)+ 6H2O(l)
If you add water to the blue solution, it turns back to the pink colour due to the
formation of hydrated cobalt (ii) ions.
a) Reaction of cobalt (ii) ions with sodium hydroxide solution

A blue precipitate formed insoluble in excess and turns pink on standing
Co2+(aq) + 2OH -(aq)→Co(OH)2(s)
With sodium hydroxide solution and hydrogen peroxide

Rather than relying on oxidation by the air, you can add an oxidizing
agent such as hydrogen peroxide
You get the cobalt (II) hydroxide complex when you add the sodium
hydroxide solution.
Addition of hydrogen peroxide produces lots of bubbles of oxygen and a
dark chocolate brown precipitate. The final precipitate contains cobalt in
the +3 oxidation state.
4Co (OH)2(s) + 2 H2O2(aq) 4 Co(OH)3(s)
134
Chapter Eight
b) Reaction of cobalt (ii) ions withammonia solution

A blue precipitate formed which dissolves in excess to form a brown
solution and turns to a deep red -brown solution on standing in air.
Co2+(aq) + 2OH-(aq) → Co(OH)2(s)
Co(OH)2(s) + 6NH3(aq) → Co(NH3)62+(aq) + 2 OH-(aq)
Co(NH3)62+(aq)→ Co(NH3)63+(aq) + e
c) With ammonia solution and hydrogen peroxide

Rather than relying on oxidation by the air, you can add an oxidizing
agent such as hydrogen peroxide
The reaction with ammonia solution followed by hydrogen peroxide
produces the same dark reddish-brown solution as before only that the
reaction is faster this time.
The equation for the oxidation of the ammine complex is:
2Co (NH3)62++H2O2(aq) 2Co (NH3)63+(aq)+ 2OH-(aq)
8.6. NICKEL
8.6.1. Test for Nickel (II) Ion, Ni2+ in solution
a) Use of of dilute sodium hydroxide solution

A green precipitate insoluble in excess formed.
Ni2+(aq) + 2OH -(aq)→ Ni(OH)2(s)
b) Use of aqueous ammonia Solution

A green precipitate formed which dissolves in excess to form a blue
solution
Ni2+(aq) + 2OH -(aq) → Ni(OH)2(s)
Ni(OH)2(s) + 6NH3(aq) → (Ni(NH3)62+(aq) + 2OH-(aq)
Note
135
Chapter Eight
The initial pale green precipitate of Ni(OH)2 is frequently not seen if Ni2+
solution is dilute.
c) Use of dimethylglyoxime
Procedure
To the solution containing Ni2+ is 1stadded aqueous ammonia solution
drop-wise until in excess followed by a drop of dimethylglyoxime
solution.
Observation
A green pptsoluble in excess ammoniaforming a blue solution. On
addition of dimethylglyoxime, a red precipitate of nickel
dimethylglyoxime complex is formed .
136
Chapter Eight
8.7. COPPER
Atomic number is 29
Electronic configuration:1s22s22p63s23p63d104s1
Ores of copper
1. copper pyrite CuFeS2
2. malachite CuCO3.Cu(OH)2
Extraction of copper
Copper is commonly extracted from its ore called copper pyrite.
The ore is 1st crushed to powder form. The finely divided ore is then mixed with
water and frothing agent such as pine oil is added. The earthly materials sink to
the bottom because of their high density whereas the ore particles float on top
from where they are removed, washed and dried.
The ore is then roasted to convert the copper pyrite to copper (i) sulphide,
iron(ii) oxide and sulphur dioxide.
2CuFeS2(s) +4O2(g)→Cu2S(s) +2FeO(s) +3SO2(g)
The solid product is transferred into a blast furnace and silica added.
On strong heating, the iron(ii)oxide reacts with silica to formslag.
FeO(s) +SiO2(s) →FeSiO3(l)
The copper (i) sulphide reacts with air forming copper (i) oxide.
2Cu2S(s) +3O2(g) → 2Cu2O(s)+2SO2(g)
The copper(i)oxide produced and the unroasted copper (i) sulphide react in the
heat of the blast furnace with limited air supply to form copper.
2Cu2O(s) +Cu2S(s) →6Cu(l) + SO2(g)
The copper formed solidifies on cooling and is called blister copper and
contains impurities. The last process in the extraction is purification of
blistercopper.
137
Chapter Eight
8.7.1. Extraction from malachite

Concentration process
The ore is ground to powder. The finely divided ore particles is mixed with
water and frothing agent such as pine oil and shaken for a while.
The earthly materials because of their high density sink to the bottom whereas
the ore particles because of their low density float on the surface, where they are
removed,washed and dried.
The ore is roasted to form copper(ii) oxide.
CuCO3.Cu (OH)2(s) → 2CuO(s) + CO2(g) + H2(g)
The copper(ii) oxide is reduced with carbon on heating to copper.
CuO(s) + C(s) → Cu(s) + CO2(g)
The copper formed is then purified by electrolysis.
Purification of Copper
–Electrolysis
 The anode is a block of made of the impure copper to be purified.

138
Chapter Eight
 The cathode is a thin piece of pure copper
 The electrolyte is copper (ii) sulphate solution.
 When electricity is passed through the cell,
At the Anode
Copper is dissolved by oxidationi.e Cu2+ionsgo into solution.
 Cu(s) → Cu2+(aq)+2e
At the cathode,
 copper is deposited by reduction.
 Cu2+(aq) + 2e → Cu(s)
 As copper ions move from the anode to the cathode

the anode gets smaller as the cathode gets bigger
8.7.2. Uses of copper

Amongst other things copper is used for:
 Electrical wiring. It is a very good conductor of electricity and is

easily drawn out into wires.
 Domestic plumbing. It doesn't react with water, and is easily bent into
shape.
 Boilers and heat exchangers. It is a good conductor of heat and doesn't

react with water.
 Making brass. Brass is a copper-zinc alloy. Alloying produces a metal

harder than either copper or zinc individually.
 Bronze is another
 Copper alloy - this time with tin.
 Coinage. Copper-colored coins, "silver" coins are also copper alloys -

this time with nickel. These are known as cupronickel alloys.
139
Chapter Eight
 UK pound coins and the gold-colored bits of euro coins are copper-
zinc-nickel alloys.
8.7.3. Reactions of copper

1) With air
Copper reacts with moist air to form a green outer layer of copper (ii)
carbonate (protective layer).
Heated copper reacts with oxygen enriched air (below 800 oC) to form
black copper (ii) oxide.
2Cu(s) + O2(g) → 2CuO(s)
At very high temperature> 800 oC, copper reacts with oxygen to form
copper (i) oxide.
4Cu(s) + O2(g) →2Cu2O(s)
2) With halogens.
Heated copper reacts with halogen to form copper (ii) halide.
Cu(s) + Cl2(g) → CuCl2(s)
However copper reacts with iodine to form copper (i) iodide.
3) With sulphur.
When heated, copper forms a sulphide .
Cu(s) + S(s) →CuS(s)
4) With acids
a. Dilute nitric acid reacts with copper to form copper (ii) nitrate,
nitrogen monoxide and water.
3Cu(s) +8HNO3 (aq) →3Cu (NO3)2(aq) +2NO(g) +4H2O(l)
b. Copper reacts with concentrated nitric acid to form

copper(ii)nitrate, nitrogen dioxide gas and water.
140
Chapter Eight
Cu(s) +4HNO3 (aq) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
c. Copper reacts with concentrated sulphuric acid to form copper (ii)

sulphate, sulphur dioxide and water.
Cu(s) + 2H2SO4 (aq) → CuSO4(aq) + SO2(g) + 2H2O(l)
8.7.4. Compounds of copper

Copper shows 2 oxidation states of +1 and +2, however compounds in +1
oxidation state are unstable.
Compounds in +2 oxidation states
This is the most stable oxidation state of copper and in aqueous solution exists
as [Cu(H2O)4]2+ which is a blue solution.
It slowly undergoes hydrolysis forming acidic solution.
[Cu (H2O)4]2+(aq) [Cu(H2O)3OH]+ (aq) + H+(aq)
1) Copper(ii) hydroxide
It’s a blue ppt formed when aqueous sodium hydroxide is added to copper
(ii) ions.
Cu2+(aq) + 2OH-(aq) Cu(OH)2(s)
Copper (ii) hydroxide does not dissolve in excess sodium hydroxide.
However dissolves in excess ammonia solution to form a deep
bluesolution.
Cu(OH)2(s) + 4NH3(aq) →[Cu(NH3)4]2+(aq) +2OH-(aq)
Tetraamminecopper (ii) ions
2) Copper (ii) chloride.
It’s prepared by adding a stream of dry chlorine gas over heated copper
metal.
Cu(s) +Cl2(g) → CuCl2(s)
In aqueous solution, it exists as [Cu (H2O)4]2+ .
141
Chapter Eight
It slowly undergoes hydrolysis forming acidic solution.
[Cu(H2O)4]2+(aq) [Cu(H2O)3OH]+ (aq) + H+(aq)
Copper(ii) chloride dissolves in conc. hydrochloride acid to form a

yellow solution of tetrachlorocuprate(ii) ions [CuCl4]2-
Note:
A complex ion may undergo ligand exchange whereby all its ligands are
replaced by a different ligand which is more willing to donate the pair of
electrons more readily. Altering a ligand around a central metal ion brings about
a marked color change.
E.g.[Cu(H2O)6]2+
[CuCl4]2- excess conc. HCl

[Cu(H2O)4]2+ excess ammonia
[Cu(NH3)4]2+
Yellow Blue Deep blue
8.7.5. Determination of copper (ii) ions in copper (ii) salts

Method1
Applicable to any soluble salt of copper.
(starting from copper ore, dissolve the ore in dilute sulphuric acid)
Procedure:
To a solution containing copper(ii) ions is added excess potassium iodide, a

white precipitate of copper(i) iodide is formed which is stainedbrown by the
liberated iodine.
2Cu2+(aq) + 4I -(aq) → Cu2I2(s) + I2(aq)
White ppt brown soln
The amount of iodine liberated is then determined by titrating the resultant

solution with a standard solution of sodium thiosulphate using starch as
indicator.
142
Chapter Eight
2S2O32 -(aq) +I2(aq) S4O62-(aq) + 2I-(aq)
The volume of thiosulphate solution used is noted.
Treatment of results
The overall equation of reaction is obtained by adding the two equations above:
2Cu2+(aq) +2S2O32 -(aq) 2I -(aq) Cu2I2(s) + S4O62 -(aq)
2moles of thiosulphate ions react with 2moles of Cu2+ ions.
Thus knowing the moles of thiosulphate ions that reacted, that of copper ions
present in the solution can be determined.
Method2
This method applies to copper (ii) iodate only Cu (IO3)2
i) Cu(IO3)2(s) + aq Cu2+(aq) + 2IO3 -(aq)
ii) IO3-(aq)+ 5I -(aq) + 6H+(aq) 3I2(aq) + 3H2O(l)
iii) I2(aq) + 2S2O32 -(aq) S4O62-(aq) + 2I -(aq)
To a solution containing copper (ii) iodate is added excess potassium iodide

solution followed by dilute sulphuric acid.
Iodine is liberated according to the equations above. The liberated iodine is then
titrated with a standard solution of sodium thiosulphate usingstarch as indicator.
The volume of thiosulphate used is noted.
8.7.6. Test for Copper(II) ions.
Use of Sodium Hydroxide solution
Procedure:
To the solution containing the cation is added sodium hydroxide dropwise till in
excess.
Observation
A blue precipitate of Cu(OH)2 insoluble in excess sodium hydroxide is formed.
143
Chapter Eight
Cu2+(aq) + 2OH -(aq) Cu(OH)2(s)
The precipitate turns black on heating due to the formation of copper(II)oxide.
Use of aqueous ammonia solution
Observation
A blue precipitate of copper(II)hydroxide soluble giving a deep blue solution

Tetraammine copper(II)ions complex.
Cu2+(aq)+4NH3(aq) [Cu(NH3)4]2+(aq)
Use of potassium Iodide Solution
Procedure
To the solution containing copper(II)ions is added Potassium Iodide Solution.
Observation:
A white precipitate of copper (I) iodide and a brown solution of Iodine is

produced.
2Cu2+(aq) + 4I - Cu2I2(s) + I2(aq)
Use of Potassium hexacyanoferrate(II) solution
Procedure
To the solution containing copper (II) ions is added Potassium

hexacyanoferrate(II) Solution.
A brown precipitate of copper hexacyanoferrate(II)
Cu2+(aq)+ [Fe(CN)6]4 -(aq) Cu2[Fe(CN)6](s)
144
Chapter Eight
8.8. ZINC AND ITS COMPOUNDS

Atomic number is 30.
Its electronic configuration is 1s22s22p63s23p63d104s2
Ores of Zinc
 Zinc blende :ZnS
 Zinc calamine:ZnCO3
8.8.1. Extraction process from sulphide ore

The ore contains traces of lead sulphide and silica as impurities which should be
removed.
Concentration of the ore
The ore is first crushed and mixed with water and oil (a frothing agent) so as to
remove the unwanted earthly materials.
The oil wets the sulphide particles and the water wets the impurities.
Air is then blown through the mixture producing froth. The oiled sulphide
particles float on the surface while the impurities sink to the bottom.The ore is
collected, washed and dried.
Roasting
The concentrated ore is then heated strongly in air to form zinc oxide.
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
Reduction
Zinc oxide, limestone (CaCO3) and excess coke are then heated in a blast
furnace.
Limestone decomposes into calcium oxide and carbon dioxide.
145
Chapter Eight
CaCO3(s) → CaO(s) + CO2(g)
The calcium oxide reacts with silica, one of the impurities to form calcium
silicate (slag).
CaO(s) + SiO2(s) → CaSiO3(l)
Coke burns to carbon dioxide, which is reduced by unburnt coke to carbon

monoxide. The carbon monoxide reduces zinc oxide to zinc.
C(s) + O2(g) → CO2(g)
CO2(g) + C(s) → 2CO(g)
ZnO(s) +CO(g) → Zn(l) + CO2(g)
The slag sinks to the bottom and can be removed whereas the zinc produced
leaves as vapour at the top of the blastfurnace where its cooled and allowed to
solidify.
8.8.2. Reactions of zinc

a) With air
On exposure to air, zinc develops a thin layer of zinc oxide. This layer
prevents further reaction with the oxygen present in air.
Zinc burns in air (oxygen) to form zinc oxide.
2Zn(s) + O2(g) → 2ZnO(s)
(Yellow when hot and white on cooling)
b) With water
Zinc (red hot) reacts with steam to form zinc oxide and hydrogen gas.
Zn(s) + H2O (g) →ZnO(s) +H2(g)
c) With acid
Zinc reacts with both dilute and conc. hydrochloric acid forming salt and
hydrogen gas.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Zinc reacts with dilute sulphuric acid to form salt and hydrogen gas.
Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)
146
Chapter Eight
d) Reaction of zinc with conc. H2SO4

Produces zinc sulphate, sulphur dioxide and water as the products.
Zn(s) +2H2SO4 (aq) → ZnSO4 (aq) +2 H2O (aq) +SO2(g)
NB:
The products for the reaction with nitric acid (both dilute and conc.) are
not easily defined.
e) With alkalis
Zinc reacts with aqueous alkalis to form zincates’ complex and hydrogen
gas.
Zn(s) + 2OH -(aq) + 2H2O(l) → Zn(OH)42-(aq) + H2(g)
8.8.3. Compounds of zinc

1) Zinc hydroxide
It is prepared as a white precipitate on addition of a little sodium
hydroxide or aqueous ammonia to a solution containing zinc ions.
Zn2+(aq) + 2OH -(aq) → Zn(OH)2(s)
White ppt
The hydroxide is amphoteric. It dissolves in excess sodium hydroxide to

form sodium zincates’ complex.
Zn(OH)2(s) + 2OH -(aq) →Zn(OH)42 -(aq)
In aqueous ammonia, zinc hydroxide dissolves forming a colorless
solution of tetra amine zinc ion.
Zn(OH)2(s) + 4NH3(aq) → Zn(NH3)42+(aq) + 2OH -(aq)
2) Zinc Sulphide
This is prepared as a white ppt on passing hydrogen sulphide through a
solution of zinc ions.
When H2S gas is passed through a solution of a metallic salt, a metallic
sulphide is formed but if the H2S is passed through an acidic solution of
a metallic salt, the sulphide may not be formed.
Explanation:
H2S(g) + aq 2H+(aq) + S2 -(aq)
147
Chapter Eight
Zn2+(aq) + S2 -(aq) → ZnS(s)
If before bubbling hydrogen sulphide the solution is acidified, the excess

H+ ions present suppresses the formation of sulphide ions in1st equation
above and therefore the concentration of sulphide ions will be too low to
precipitate zinc sulphide.
Note
Aqueous solution of zinc chloride is acidic due to hydrolysis of [Zn
(H2O)6]2+
ZnCl2(s) + 6H2O (l) → [Zn (H2O)6]2+(aq) + 2Cl -(aq)
[Zn(H2O)6]2+(aq) + H2O(l) [Zn(H2O)5OH] +(aq) +

H3O+(aq),
8.8.4. Test for zinc ions

1) Addition of aqueous sodium hydroxide drop wise to a solution of zinc
ions produces a white ppt which dissolves in excess alkali to form
acolourless solution.
Zn2+(aq) + 2OH -(aq) → Zn(OH)2(s) White ppt
Zn(OH)2(s) + 2OH -(aq) →Zn(OH)42-(aq) colorless solution
2) Addition of ammonia solution drop wise to a solution of zinc ions

produces a white ppt soluble in excess to form a colorless solution.
3)
Zn2+(aq) + 2OH -(aq) → Zn(OH)2(s)
Zn(OH)2(s) + 4NH3(aq) → Zn(NH3)42+(aq) + 2OH-(aq)
4) Addition of potassium ferrocyanide solution to a solution of Zinc ions

produces a white ppt soluble in sodium hydroxide.
148
Chapter Eight
5) Addition of ammonium sulphide solution produces a white precipitate

soluble in dilute hydrochloric acid.
6) Using disodium hydrogen phosphate forms white ppt with zinc ions.
8.8.5. Uses of zinc

 Making alloys e.g. Brass is alloy of zinc and copper.
 Galvanizing iron.
Zinc is higher than iron in electrochemical (reactivity) series.
If a galvanized piece of iron developed a small scratch and the iron is exposed,
iron does not rust.
This is because the zinc in the neighborhood of the exposed iron undergoes
oxidation in preference to iron forming a thin layer of Zinc oxide formed which
prevents further attack
149
Chapter Eight
H
A
halogens · 28, 38, 50, 66, 92, 94, 95,
Alkalis · 80, 130
96, 97, 98, 99, 100, 143, 153
alloys · 152, 160
hydrides · 39, 50, 51, 52, 53, 62, 78,
Aufbau · 8
88, 96, 99, 100, 109
Hydrolysis · 87, 109
B
Beryllium · 3, 18, 20, 21, 34, 38, 43, I

45, 61, 62, 64, 65, 72, 73
inert pair effect · 77, 89,
boiling point · 34, 36, 39, 50, 52, 53,
Ionic charge · 69
61, 96, 99
ionic compounds · 42, 47, 48, 62
BONDING · 40
Ionic radius · 15, 61, 69
C
L
Carbonates · 70, 71
Lattice energy · 66, 69
catenation · 78
ligand · 147, 154
Cathode ray · 1, 2
ligands · 72, 119, 154
Chromium · 114, 123, 124, 125, 126,
Linear · 53, 59, 83
127, 130, 131
Lyman series · 4
Coordinate (dative) bonding · 48
M
D
Metallic bonding · 40, 49
dative · 48, 119
dimethylglyoxime · 149
O
E Octaherdral · 54
oxidation state · 77, 78, 81, 117,
Electropositivity · 31, 33
118, 121, 122, 125, 127, 130, 143
150
Chapter Eight
P T
polar · 29, 47, 48, 49, 67, 98, 99, Tetrahedral · 54, 58, 59
100, 109 Transition elements · 114, 116
transition elemnts · 114
R Trigonal bipyramidal · 54, 56
Rutherford · 3
V
S Van der Waal’s forces · 49, 51, 52
Scandium · 114
Z
scattering experiment · 3
Shapes of Molecules · 53 zinc · 118, 142, 157, 158, 159, 160
Sodium hydroxide solution · 131
151

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Chapter one

THE PERIODIC TABLE

2.7.1. Factors determining melting point ............................................................................... 31

8.7.4. Compounds of copper ................................................................................................ 141

Table 1 Periodic table ................................................................................................................ 1

1.1. Cathode rays

Gas at low pressure

Gas at low pressure

Gas at low pressure

Figure 3: Cathode ray tube with an electromagnetic field

These shows that the rays consist of negatively charged particles

Properties of cathode rays

Rutherford’s α-particles scattering experiment

Geirgerand Marden working under Rutherford bombarded thin metallic foils of

Thin metal foil screen

Figure 4: Rutherford’s α – Particles scattering Experiment

The many undeflected α-particles were explained to have passed in a space

1.3. Electronic structure of atoms

When electricity passes through a discharge tube containing hydrogen gas at

Measurement of wave length corresponding to lines in the various series shows

Where λ=wave length,

n1 and n2are simple whole numbersindicating the quantum shell.

Where n1=1, and n2=2, 3, 4 etc

Figure 5: Hydrogen Spectral lines

1.4. Bohr’s theory of electronic energy levels

H is the Planck’s constant, and

Where c is the velocity of light in a vacuum

Figure 6: Hydrogen spectral series

Significance of line spectrum of hydrogen

1.5. Electronic configuration of atoms

Within the shells, electrons are found in what is referred to as orbitals.

Therefore when full the 1st shell has 2 electrons.

Therefore the maximum number of electrons that can be accommodated

Therefore the maximum number of electrons that can be accommodated

1.6. Rules/principles for writing electronic configurations of atoms

Figure 7Aufbau pattern of filling energy levels

2. THE MODERN PERIODIC TABLE OF ELEMENTS

Table 1 Periodic table

The Modern periodic table consists of two major divisions; namely

a) The horizontal arrangement of elements referred to as the periods

b) The downward arrangement of elements referred to as the groups.

2.1. Variation of physical properties in the periodic table (periodic

2.2. Atomic radius

At such a distance, the sum of the inter-electronic and the inter-nuclear

Factors determining the radius of an atom of an element

This is determined by the number of protons present in the nucleus of an

2) Screening (shielding) effect.

Variation in atomic radius down a group in the periodic table

Consider Group IA elements (Alkali metals)..

Table 2 Atomic radius of group 1 elements

Element Atomic Electronic configuration Atomic

Variation in Atomic radiuscross period 2 and period 3 in the periodic table

Atomic radius decreases across the periods.

Electronic 1s22s 1s22s 1s22s22p 1s22s22p 1s22s22p 1s22s22p 1s22s22p 1s22s22p

Atomic 1.52 1.13 0.83 0.77 0.71 0.66 0.71 1.60

The figures used to construct this diagram are based on:

Metallic radii for Sodium, Magnesium and Aluminium;

Covalent radii for Silicon, Phosphorus, Sulphur and Chlorine;

Variation in atomic radius among the d-block elements in periodic table

Element At. Elect. Conf