complete modul sem 1 2021 2022

SCI 1054 CHEMISTRY SEMESTER 1

Ionization energy is the minimum energy required to remove one mole of electron from
one mole of gaseous atom or ion.

E (g)  E+ (g) + e-(g)

where E represents an element. All ionization energies are positive because energy always
required to remove an electron.

Ionization energy for hydrogen : ni = 1, nf = 

ΔE = 2.18 x 10-18 (1/12 – 1/) x NA
= 2.18 x 10-18 (1/1 – 0) x NA
= 2.18 x 10-18 J.e-1 x 6.023 x 1023 e/mole
= 1.313 x 106 J mole-1

2.1.5 The Weakness of Bohr’s Atomic Postulates

1. Bohr‟s approach did not account for the emission spectra of atoms and ions containing
more than one electron.

2. It can only explain the hydrogen spectrum or any spectrum of ions contain only one
electron such as He+ and Li2+.

2.2 Orbitals and Quantum Number

2.2.1 Orbitals

It is not possible to locate the position of an electron in an atom. However, quantum mechanics
does define where the electron might be given time. The concept of electron density gives
the probability that an electron will be found in a particular region of an atom. Region of high
electron density represent a high probability of locating the electron.

According to quantum mechanics, an atomic orbital is defined as the space in which there a
high possibility (99%) in finding electron. The wave functions describe the allowed shapes and
energies of the electron waves. Each of this different possible waves is called an orbital. Each
orbital in an atom has a characteristic energy and is viewed as describing a region around the
nucleus where the electron can be expected to be found.

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SCI 1054 CHEMISTRY SEMESTER 1

2.2.2 Quantum Numbers

Quanta are the discrete amounts of energy that an electron absorbs as it moves up an energy
level or releases when it moves down to lower energy level. Energy levels are certain regions
surrounding the nucleus of an atom where electrons are likely to be found.

The quantum mechanical model of the atom is a mathematical model which predicts the
probability of electron location and paths in electron clouds. The positions and orbits of
electrons are referred to as energy states and are described by FOUR quantum numbers :

I) The Principle Quantum Number (n)

The principle quantum number (n) indicates the energy level of the electron and can
have an integral value of 1,2,3, …

The energy levels in an atom are arranged into main levels, or shells, as determined by the
principle quantum number, n.

In a hydrogen atom, the value of n determines the energy of an orbital. The principle quantum
number also relates to the average distance of the electron from the nucleus in a particular
orbital. The larger the value of n, the greater the average energy of the levels and the greater
the average distance of an electron in the orbital from the nucleus and therefore the larger (and
less stable) the orbital.

Principle quantum number 1234
Letter designation K L MN

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SCI 1054 CHEMISTRY SEMESTER 1

II) The Angular Momentum or Azimuthal Quantum Number (l)

The angular momentum quantum number – indicates the shape of the orbital. The
values of l depend on the value of the principle quantum number, n.

For a given value of n, l has a possible integral values from 0 to (n – 1)

The value of l is generally designated by the letter s,p,d,f,…… as follow

l 012345
Name of orbital s pdf gh

The first four orbital numbers are indicated by the letter s (has a spherical shape), p (has a
dumb-bell shape), d and f (d and f have complex shapes).

A collection of orbitals with the same value of n is frequently called shell.

One or more orbitals with the same n and l values are referred to as sub-shell.

Shell, n Sub-shell, l Sub-shell are called
2 as
0 2s
1
2p

Note. The number of sub-shells in any given shell is simply equal to the values of n for that
shell.

III) The magnetic Quantum Number (m)

The magnetic quantum number (m) describes the orientation of the orbital in space.
For a certain values of l, there are (2l + 1) integral values of m.

For l = 1, m = -1, 0, 1
l = 2, m = -2, -1, 0, 1, 2

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SCI 1054 CHEMISTRY SEMESTER 1

IV) The Electron Spin Quantum Number (s)

Electron behaves as if it were spinning. Since the electron can only spin in either of two
direction, s can have only two values. These turn out to be + ½ and – ½ .

Notes :
Using these four quantum numbers, any electron in any element can be described.
Table below gives examples of all the orbital, within the first four principle quantum
numbers:

Principle Angular Subshel Megnetic Quantum Electron Number Num
Quantu Momentu l Number, m spin of ber
of
m m designa 0 quantum orbitals elect
Number, Quantum tion 0 number, in ron
n (shell) Number, l -1, 0, +1 2
(subshell) 1s 0 s subshell 8
1 2s -1, 0, +1
2 0 2p -2, -1, 0, +1, 2 ½ 1 18
0 3s 0 ½ 1
3 1 3p -1, 0, +1 ½ 3 32
0 3d -2, -1, 0, +1, 2 ½ 1
4 1 4s -3, -2, -1, 0, +1, 2, ½ 3 2n2
2 4p 3 ½ 5
n 0 4d ½ 1
1 ½ 3
2 4f ½ 5

3 ½ 7

Table 2.2 The four quantum numbers of electron.

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2.2.3 The Shapes of Atomic Orbitals

I) The shape of s orbitals

At any particular distance from the nucleus, the electron density is the same. The shape of s
orbital is a spherical.

Figure below compares the way electron density varies for 1s, 2s and 3s orbitals. The larger the
value of the principle quantum number of an orbital, the larger size is its size.

Figure 2.4 The shape of s orbital with different values of n.
II) The shape of p orbitals
The shape of 2p orbital are illustrated in Figure 3.5. The p orbitals with larger n values of their
principle quantum number are larger than a 2p orbital. The p orbital can be represent as a pair
of dumb-bell shaped.
A p sub-shell consists of three p orbitals, each with the same size, shape and energy; they
differ from one another only in orientation. Because the p orbitals can be drawn on set of x, y
and z axes, it is common practice to identify the orbital as px, py, and pz.

Figure 2.5 The shape of p orbitals

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2.3 Electronic Configuration

2.3.1 The Electronic Configuration of the Elements

The four quantum numbers n, l, m and s enable us to label completely an electron in any orbital
in any atom. For example, the four quantum numbers of a 2s orbital electron are n = 2, 1 = 0,
m = 0 and s = + ½ or – ½ .

Simplified notation (n, l, m, s) of the quantum numbers are either (2, 0, 0, + ½) or (2, 0, 0, -
½ ). Electronic configuration of the atom is how the electrons are distributed among the various
atomic orbitals.

The electron in a ground-state hydrogen atom must be in the 1s orbital, so its electron
configuration is 1s1.

1 s1 The number of electron in the orbital or subshell.
The angular momentum quantum

Number 1 The principle quantum number n

The electronic configuration can also be represented by an orbital diagram that shows the spin
of the electron.

H↿ notes: please used ‘half-arrow’ for an electron

1s

‘ half-arrow ‘ ( ↿ ) is used to show an electron

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2.3.2 Aufbau Principle, Hund’s Rule and Pauli Exclusion Principle.

I) Aufbau Principle (The Building Up Principle)

The building up principle states that electrons are arranged in its atomic orbitals in order
of increasing energy.

Z=1 H ↿
1s

Z=2 H ↿⇂

1s

Z=3 Li ↿⇂ ↿

1s 2s

Z=4 Be ↿⇂ ↿⇂

1s 2s

~Old chemists never die, they just stop reacting ~

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SCI 1054 CHEMISTRY SEMESTER 1

The order in which atomic subshells are filled in a atom start with the 1s orbital and move
downward, following the arrows.

The sequence of filling electron will be: 1s 2s 2p 3s 3p 4s 3d 4p and so on

II) Pauli Exclusion Principle

This principle states that no two electrons in an atom can have the same four quantum
numbers.
Electrons in a given orbital must have the same values of n, l and m but they must have
different values of s. Only two values of s are possible, + ½ and – ½ . As a result it can be
concluded that an atomic orbital can accommodate only two electrons, and these electrons
must have opposite spin.

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SCI 1054 CHEMISTRY SEMESTER 1

Consider the helium atom which has two electrons. The quantum numbers for these two
electrons are (1, 0, 0, + ½) and (1, 0, 0, - ½). Thus, the helium atom has following
configuration:

He ↿⇂

1s

III) Hund’s rule

Hund‟s rule states that the most stable arrangement of electrons in electrons in
subshells is the one with greatest number of parallels spins.

Hund‟s rule is applied where orbitals of equal energy are available, for example 2px, 2py and
2pz. Each of these orbitals will first fill with one electron with parallel spins before a second
electron is added with one opposite spin.

Example 2.3.2 :

C ↿⇂ ↿⇂ ↿↿

1s 2s 2px 2py 2pz
1s2 2s2 2p2

N ↿⇂ ↿⇂ ↿↿↿

1s 2s 2px 2py 2pz
1s2 2s2 2p3

O ↿⇂ ↿⇂ ↿⇂ ↿ ↿

1s 2s 2px 2py 2pz
1s2 2s2 2p4

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SCI 1054 CHEMISTRY SEMESTER 1

F ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿

1s 2s 2px 2py 2pz
1s2 2s2 2p5

Ne ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂

1s 2s 2px 2py 2pz
1s2 2s2 2p6

2.3.3 General procedure for writing electronic configurations.

1. Determine the number of electrons to appear in the electron configuration. This is
simply the proton number of the element.

2. Add electrons to sub-shells in order of increasing sub-shell energy. Applied Aufbau
principle.

3. Observe the Pauli Exclusion Principle. There can be no more than two electrons in a
orbital and the two electrons must have opposite spins.

4. Observe Hund‟s rule. Orbital in a subshell are singly occupied whenever possible and all
unpaired electrons have the same spin direction (parallel spins).

Example 2.3.3:

State the electronic configurations for sulphur, using both the s, p, d, f notation and an orbital
diagram. Sulphur (Z=16)

Electronic configuration: S 1s2 2s2 2p6 3s2 3p4

Orbital diagram of S: ↿⇂ : ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ ↿

1s 2s 2p 3s 3p

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SCI 1054 CHEMISTRY SEMESTER 1

2.3.4 Exception to the Aufbau Principle

The electronic configuration of chromium (Cr) and copper (Cu) in the following table have been
established by experiment. They are not configurations that might have been expected based
on the order or filling of subshells and other aspects of the Aufbau process. The electronic
configurations of these two elements are shown below:

Element Expected Observed
Cr (Z = 24) [Ar] 3d4 4s2 [Ar] 3d5 4s1
Cu (Z = 29) [Ar] 3d9 4s2 [Ar] 3d10 4s1

The structure of chromium will be:

Cr [Ar] ↿ ↿ ↿ ↿ ↿ ↿

3d 4s

instead of: ↿⇂
↿⇂
Cr [Ar] ↿ ↿ ↿ ↿
4s
3d

This unexpected result because a half-filled or completely filled sub-shell possesses an extra,
added stability.
It is predicted copper to has the electronic configuration

Cu [Ar] ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿ ↿⇂

3d 4s

The actual electronic configuration of the ground state is given by:

Cu [Ar] ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿

3d 4s

Which is [Ar] : ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂ ↿⇂

1s 2s 2p 3s 3p

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SCI 1054 CHEMISTRY SEMESTER 1

ANSWER

Exercise 2.1.1:

1. Calculate the photon of energy emitted by an electron that produces a wavelength of

4.342x10-7 m. (Ans: 4.58 x 10-19 J)

E= = 4.58 x 10-19 J
=

2. A light emitted from a discharge tube has a frequency of 7.31 x 1014 Hz. What is the

wavelength of this light? (Ans : 410 nm)

= =
= 4.10 x 10 -7 m @ 410 nm

Exercise 2.1.4:
1. Calculate the energy liberated when an electron from the fifth energy level falls to the
second energy level in a hydrogen atom. (Ans: -4.58x10-19J)

E = RH ( 1/n1 2 – 1/n2 2) = -4.58x10-19J
= (2.18 x 10 -18 J) ( 1/52 – 1/22)

2. Calculate the wavelength of the fourth line in Balmer series of hydrogen atom.
(Ans: 410 nm)

4th line Balmer series : nf = 2, ni = 6
1/ = RH (1/n12 – 1/n22)
; n1 < n2
= (1.097 x 107 m-1) (1/22 – 1/62)
 = 4.10 x 10-7 m @ 410 nm

3. Calculate the frequency of a line in the hydrogen spectrum corresponding to a transition
from n = 5 to n = 2. (Ans: 6.91 x 1014 Hz)

E = RH ( 1/n1 2 – 1/n2 2) = 6.90 x 1014 Hz
= (2.18 x 10 -18 J) ( 1/52 – 1/22)
= 4.58x10-19 J

E = hv
v = (4.58x10-19 J) / (6.63 x 10-34 Js)

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CHAPTER 3 PERIODIC
TABLE

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SCI 1054 CHEMISTRY SEMESTER 1

3.0 PERIODIC TABLE

3.1 INTRODUCTION

3.1.1 Group and Period

The periodic table is a table that arranges all the known elements in order of increasing proton
number. This order generally coincides with increasing atomic mass. In the Periodic Table, a
vertical column of elements is called a group and a horizontal row is known as a period.

3.1.2 Metal, nonmetal and metalloids

All the elements on the left side and in the middle of the periodic table (except for hydrogen)
are metallic elements, or metals. Metals share many characteristic properties such as luster,
high electrical and heat conductivity.
The metals are separated from the nonmetallic elements by a diagonal step like line that runs
from boron (B) to polonium (Po) - refer Periodic Table.
Many of the elements that lie along the line that separates metals from nonmetals, such as
antimony (Sb) have properties that fall between those of metals and non-metals. These
elements are often referred to as metalloids.

3.1.3 Deduce the position of elements

I) Groups
The groups in the Periodic Table are numbered from 1 to 18. Elements in the same group have
the same number of valence electrons. For example, oxygen (O) and Sulphur (S) are both
found in group 16 which means that they both have 6 valence electrons.

Group number = number of valence electrons

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SCI 1054 CHEMISTRY SEMESTER 1

Example:

(i) Main Group (Representative Group) in Periodic Table
• Group 1 : alkali metals (except H)
• Group 2 : alkaline earth metals
• Group 3-12 : transition metals
• Group 16 : chalcogens
• Group 17 : halogens
• Group 18 : inert/ noble gases
The periods in the Periodic Table are numbered from 1 to 7. For example, hydrogen and helium
are in Row 1 or Period 1 because their principal quantum number of the main electron shell is

1.

(H: 1s1; He: 1s2)

Period number = Principle quantum number, n, of the electrons in the valence shell

II) Blocks
All the elements in the Periodic Table can be classified into 4 main blocks according to
their valence electron configuration. These main blocks are block s, p, d and f.

Block s:
* Group 1 and 2

* Filling of valence electron only involve orbital s
* Configuration of valence electron: ns1 to ns2
* Eg. 11Na: 1s2 2s2 2p6 3s1 → 3s1
20Ca: 1s2 2s2 2p6 3s2 3p6 4s2 → 4s2

Block p:
* Groups 13 to 18
* Configuration of valence electrons : ns2 np1 to ns2 np6
* Eg. 13Al: 1s2 2s2 2p6 3s2 3p1 → 3s2 3p1

52Te: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p4 → 5s2 5p4

Block d:
* Also known as transition element
* Groups 3 to 12
* Configuration of valence electron : (n-1)d1 ns2 to (n-1)d10 ns2
* Eg. 23V 1s2 2s2 2p6 3s2 3p6 4s2 3d3

=[Ar] 4s2 3d3  4s2 3d3 where [Ar] = 18 electrons
Block f:
* Involve elements in the series of lanthanides (Ce to Lu) and actinides (Th to Lr) in

which the filling of valence electron happens in the subshell of 4f and 5f.

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SCI 1054 CHEMISTRY SEMESTER 1

Example:

Classify the following elements into its appropriate group, period and block.

P ………… 1s2 2s2 2p6 3s2 3p6
Q …………. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5
R …………. 1s2 2s2 2p6 3s2 3p6 4s2
S …………. 1s2 2s2 2p6 3s2 3p6 3d3 4s2
T ………….. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

Answer

Element Group Period Class/block
P 18 3 Inert gas/block p
Q 17 4 Block p
R 2 4 Block s
S 5 4 Transition element/ block d
T 18 4 Inert gas/block p

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Figure 4.1: The

SEMESTER 1

e Periodic Table

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SCI 1054 CHEMISTRY SEMESTER 1

3.2 Periodicity

3.2.1 Periodic trends in the size of atom (atomic radii)
The size /radius of atom is difficult to be defined exactly because the electron cloud
has no clear boundary. To solve this, we measure the distance between the 2 nuclei
in a molecule.

Eg. Radius, r = a/2
a Size  volume
V = 4/3 r3  V  r

Notes:
*Down the group, atomic radii increases
*Across the period, atomic radii decreases

Two factors that contributing to the changes of atomic radii in the Periodic Table are

i) Effective nuclear charge (Zeff) ii) shielding effect/screening effect
felt by the valence electrons

i) Effective nuclear charge (Zeff)
Electrons moving across the nucleus do not experience the same nuclear attraction; those
electrons closer to the nucleus experience a greater force than those that are farther away. The
nuclear charge actually "felt" by an electron is called the effective nuclear charge, Zeff.
Zeff for a given electron is given by the true nuclear charge, Z, less the amount by which
electrons closer to the nucleus screen it, S.

Zeff = Z – S , S = no. of electrons filled at the inner orbital
Where Z = no. of proton

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SCI 1054 CHEMISTRY SEMESTER 1

Example:

Consider a sodium atom. If the 3s valence electron was (1s2 2s2 2p6 3s1) at all times completely
outside the region in which the ten inner /core electrons (1s2 2s2 2p6) are found, the 3s electron
would be perfectly screened from the positively charged nucleus. Thus, the valence electron
would experience an attraction to a net positive charge of only +11 – 10 = +1.

Atoms Electronic Zeff
12Mg configuration
13Al 1s2 2s2 2p6 3s2 12-10 = +2
14Si 1s2 2s2 2p6 3s2 3p1 13-10 = +3
15P 1s2 2s2 2p6 3s2 3p2 14-10 = +4
1s2 2s2 2p6 3s2 3p3 15-10 = +5

Across the period, the nuclear charge increases and as a result, the outer electrons feel an
increase in positive charge which causes them to be drawn inwards, and thereby causes the
sizes of atom to decrease.

ii) Shielding effect/ screening effect

Going from top to bottom within a group, the principal quantum number of the valence shell
increases.

Example:

Elements of Group1:
lithium, valence shell configuration is 2s1
sodium, valence shell configuration is 3s1
potassium, valence shell configuration is 4s1
the value of n for the valence shell increases

The larger the value of n, the larger the orbital. This is because the electrons in inner shells
shield the electrons in outer shells from the nuclear charge known as the screening effect.
Therefore, the atoms become larger as we go down a group.

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SCI 1054 CHEMISTRY SEMESTER 1

300

250 Cs ([VALUE][

Atomic Radius/ pm K ([VALUE][ Rb ([VALUE][

200

Na ([VALUE][

150 Li ([VALUE][ I ([VALUE][ At ([VALUE][

100 Br ([VALUE][
Cl ([VALUE][

50 F ([VALUE][

0
Atomic Number (Z)

Figure 4.2
This graph shows that when down the group, sizes of atoms increase. It can be said that
atomic radius decreases when across a period and up a group in the periodic table. The
greater the force of attraction, the smaller the radius.

3.2.2 Periodic trends in the ionization energies

The ionization energy, IE, is the energy required to remove the outermost electron from a
gaseous atom or ion. The first ionization energy, IE1, is the energy for the removal of one
mole electron from one mole of neutral, gaseous atom:

X (g) X+ (g) + e-

Metallic atoms tend to lose enough electrons to gain the electronic configuration of the
proceeding noble gas. There are periodic trends in the ionization energies, also related to the
Zeff. As the Zeff increases, it requires more energy to remove the outermost electron from an
atom. Consequently, ionization energy is also related to the atomic radius, with ionization
energy increasing as atomic radius decreases. Therefore, the first ionization energy
increases across a period and up the group.

Na < Al < Mg < Si

Second ionization energy (IE2) is the energy required to remove one mole electron from
one mole positive ion in the gaseous state.

X+(g) X2+(g) + e-

The trends in IE are just opposite that of the trends in atomic size within the periodic table;
when size increases, IE decreases. This is because the same factors that affect atomic size also
affect ionization energy.

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Going down the group, the atomic size increases, the outer electrons are farther from the
nucleus and held less tightly by the nucleus, therefore IE decreases.
There is a gradual overall increase in IE as we move across a period, although the variation is
irregular. The reason for the overall trend is the increase in effective nuclear charge felt by the
electrons. This draws the electrons closer to the nucleus and causes the valence electrons to be
held more tightly, which makes it more difficult to remove them.

Variation of first IE (kJ/mol) across period 2:

Group 1 2 13 14 15 16 17 18
Be B C N O F Ne
Elements Li 900 801 1086 1402 1314 1681 2081

IE (kJ/mol) 520

In general, IE increases from left to right.

ANOMALOUS CASES
The irregularity between groups 2 and 13 can be explained. A 2p electron is
more easily removed from the (1s2 2s2 2p1) electronic configuration of a boron
atom than the 2s electron from the (1s2 2s2) configuration of a beryllium atom.
The 2p electron is at a higher energy than the 2s electron. As a result, IE is
smaller for B than for Be.
The irregularity between groups 15 and 16 can be explained by the repulsions between
electrons. This repulsive tendency makes it easier to remove one of the paired electrons in a
filled 2p orbital of an oxygen atom ([He] 2s2 2p2x 2p1y 2p1z) than an unpaired electron from a
half-filled 2p orbital of a nitrogen atom ([He] 2s2 2p1x 2p1y 2p1z).

3.2.3 Periodic trends in the electronegativity.
Electronegativity is the relative tendency of an atom to attract electrons to itself when
chemically combined with another atom. Example: Na < Al < S < F
In HCl, for example, chlorine is more electronegative than hydrogen. The electron pair of the
covalent bond spends more of its time around the more electronegative atom.
Electronegativity increases up a group, and across a period. This follows the trends for
ionization energy also.

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3.2.4 Periodic trends in the melting / boiling point
The melting / boiling point of a substance depends greatly on the type of intermolecular forces
or bondings involved.

Element Na Mg Al Si P S Cl Ar

Melting point (oC) 97.8 651 660 1410 44 119 -101 -189

Boiling point (oC) 892 1107 2467 2680 280 445 -34.5 -186
Table 4.1 Melting and boiling point of elements in period 3

Variation of melting and boiling point of elements in period 3 can be discussed in 3 parts:

a) Metallic structure (Na to Al)

Metallic crystals have positive metal ions attracted by electrons in a cloud -metallic bonding.
Strength of metallic bonding is proportional to the number of valence / delocalized
electrons.

• Na : 1 valence electron The more valence electrons, the stronger
• Mg : 2 valence electrons the metallic bonding and the melting /
• Al : 3 valence electrons boiling point

b) Gigantic Covalent Structure (Si)

Silicon is made up of atoms that are covalently bonded to other atoms at neighboring lattice
sites. The result is a crystal that is essentially one gigantic / network molecule. Melting and
boiling point of Si is very high because of the strong attractions between covalently bonded
atoms. High energy is needed to break the strong covalent bonding in melting or boiling.

c) Simple molecular structure (P to Ar)

These are mainly the non-metal that exist as molecules of P4, S8, Cl2 and Ar. The covalent bond
between the atoms is very strong but the intermolecular forces, that is, London dispersion force
(Van der Waals), is very weak. Van der Waals force is proportional to molecular size (relative
molecular mass)
Molecular size: Ar < Cl2 < P4 < S8
Therefore, melting / boiling point : Ar < Cl < P < S

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3.2.5 Periodic trends in the metallic character

Metallic character is closely related to atomic radius and ionization energy. The easier it is to
remove electrons from an atom, the more metallic the element. Easy removal of electrons
corresponds to large atomic radius and low ionization energy. Metallic character increases down
a group and decreases across a period of the periodic table.

metal metalloids nonmetal

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SCI 1054 CHEMISTRY SEMESTER 1

CHAPTER 4 CHEMICAL
BONDING

The strong bond of friendship is not always a balanced equation; friendship is not always
about giving and taking in equal shares. Instead, friendship is grounded in a feeling that

you know exactly who will be there for you when you need something, no matter what or
when.

Simon Sinek

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SCI 1054 CHEMISTRY SEMESTER 1

4.0 CHEMICAL BONDING

4.1 Valence Electron and Lewis Structure

4.1.1 Lewis dot symbol
Elements in the same group have similar outer electron configurations and hence similar Lewis
dot symbols.
A Lewis dot symbol consists of the symbol of an element and one dot for each valence electron
in an atom of the element.
The Lewis dot symbol shows the valence electron in an atom.
Example:
The Lewis dot symbol of atom.

Group 1 2 13 14 15 16 17 18

Valence 1 2 3 4 5 6 7 8
electron

Lewis C N O F Ne
dot Li Be B

symbol

4.1.2 Octet Rule
- Atoms combine in order to achieve a more stable electronic configuration.
- Maximum stability results when an atom is isoelectronic with a noble gas.
- Atom can achieve noble gas configuration through:

a) transfer of electron
b) sharing electron

4.1.3 Formation of ionic bond.
An ionic (electrovalent) bond is formed as a result of the transfer of one or more
electron(s) from a metal atom to a non-metal atom until both the ions have achieved
octet structure.
As a result of formation of oppositely charged cation and anion, these two ions are held
together by strong electrostatic force called electrovalent or ionic bond.

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SCI 1054 CHEMISTRY SEMESTER 1

Example: Na  Na+ + e-

A sodium atom, Na forms the cation, Na+ by donating or releasing its electron, which has the
same electron configuration as the noble gas argon.
A chlorine atom, Cl by gaining an electron, forms the anion Cl- which has the same electron
configuration as the noble gas argon.

Cl + e-  Cl-

As these processes occur together, we can use an equation to represent the net result.

transfer of e-

x Cl Na + -
xCl
Na +

As another example, consider the reaction of lithium with oxygen.

Li x

+ O 2 Li + 2-
Li x xOx

The formation of electrovalent bonds:
- An ionic bond is formed by the electrostatic forces that hold ions together in an ionic
compound.
- Ionic bond form between two ions with different charges ( + ) and ( - ) and through
electron transfer.
- In the formation of the ionic compound, metal elements will donate electron while
non metal elements received electrons to achieve stability.
- This happen because metals are more electropositive while non-metals more
electronegative.

4.1.4 Lewis structures
Lewis structure for ionic compound consists of:
a) Cation  valence electron in neutral atom is removed to achieve noble gas
configuration.
b) Anion  The valence shell is fully filled with valence electrons (after gaining
electron from other elements)

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Example:
a) KF

Kx + F K + xF -

The electron configuration of potassium is 1s2 2s1 ancdattihonat of falunioornine is 1s2 2s2 2p5. When
potassium and fluorine atoms come in contact with each other, the outer 2s1 valence electron of
potassium is transferred to the fluorine atom. Both of the atoms then becoming cation and
anion and achieve noble gas configuration.

b) BaO

Ba xx + O Ba 2+ xOx 2-

The electron configuration of barium is 1s2 2s2 and that of oxygen is 1s2 2s2 2p4. When barium
and oxygen atoms come in contact with each other, the outer 2s2 valence electron of barium is
transferred to the oxygen atom.

c) Na2O

Na x O 2 Na + xOx 2-

+
Na x

The electron configuration of sodium is 1s2 2s2 2p6 3s1 and that of oxygen 1s2 2s2 2p4. When
sodium and oxygen atoms come in contact with each other, the outer 3s1 valence electron of
sodium is transferred to the oxygen atom.

4.1.5 The properties of ionic compound

1) Solid at room temperature.
2) High melting / boiling point.
3) Soluble in water (polar solvent), insoluble in organic solvent (non-polar)
4) Molten ionic compounds conduct electricity because they contain mobile ions

(cations and anions).

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4.1.6 The formation of covalent and dative bonds

i) Covalent bond

It is formed by sharing one or more pairs of valence electrons between the bonded atoms,
involving only nonmetal atoms.

H +H H H or H H

This type of electron pairing is an example of a covalent bond. For simplicity, the shared pair of
electron is often represented by a single line. Covalent bonding between many electron atoms
involves only the valence electrons. For example, there is only one unpaired electron on
Fluorine, so the formation of the F2 molecule can be represented as follows:

F +F F F or F F

Note that only two valence electrons participate in the formation of F2
The other non-bonding electrons are called lone pairs (pairs of valence electrons that are not
involve in covalent bond formation).
The formation of these molecules illustrates the octet rule: An atom other than H2 tends to
form bonds until it is surrounded by eight valence electrons. In other words, a covalent bond
forms when there are not enough electrons for each individual atom to have a complete octet.
Atoms can form different types of covalent bonds. In a single bond, two atoms are held
together by one electron pair. If two atoms share two pairs of electrons, the covalent bond is
called a double bond.

Molecule Lewis Structure

CO2 OCO or O C O

C2H4 HH HH
C2H2 CC
HH or C C

HH

CC or H C C H
HH

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ii) Coordinate covalent or dative bond.
Coordinate covalent or dative bond is formed when one atom can provide both electrons of
the shared pairs while the other atoms have unfilled orbitals. This type of bond has the same
properties as the normal covalent bond.

Species Lewis Structure

H3O+

H+ + OH HOH
H H
H
NH4+ ion

H+ + H N H HNH
H H

BH3 and NH3 HH
HBNH
H + H
HB NH HH
H
H

4.1.7 The Lewis structure of covalent compounds.

A Lewis structure indicates the properties in which atoms combine. A Lewis structure shows the
bonded atoms acquire the electron configuration of a noble gas, that is, the atoms obey the
octet rule. The basic steps are as follows.

1. Write the skeletal structure of the compound. In general, the least electronegative atom
occupies the central position ( H2 and F2 usually occupy by terminal positions).

2. Count the total number of valence electrons present ( for the anion, we added the
number of negative charges and for cation we subtract the number of positive charge
from this total).

3. Draw a single covalent bond between the central atom and each of the surrounding
atoms. Complete the octets of the atoms bonded to the central atom.

4. If the octet rule is not satisfied for the central atom, try adding double or triple bonds
between the surrounding atoms and the central atom using the lone pairs from the
surrounding atoms.

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4.1.7.1 Formal Charges

Formal charges are apparent charges on certain atoms in a Lewis structure that arise when
atoms have not contributed equal numbers of electrons to the covalent bonds joining them.

Formal charge = number of valence electrons – [number of non-bonding
in free atom electrons +
½ number of bonding
electrons]

An atom‟s formal charge is the difference between the valence electrons in an isolated atom
and the number of electrons assigned to that atom in a Lewis structure.

The sum of the formal charges for a neutral molecule is zero. For a polyatomic ion, the
sum of the formal charges must equal to the charge carried by the ion.

Formal charge is utilized to find the most plausible Lewis structure. In general, a Lewis
structure with formal charges of zero is preferable. Small formal charges are preferable to
large formal charges.

Which is Lewis structure for CO2? Based on the calculation of formal
charge, the structure of CO2 is
A)
B) O=C=o

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Step by step examples of drawing Lewis structure for covalent compounds

Species Step
NF3 1 N atom less electronegative then F. N will be at the centre.

The skeletal structure is,

FNF

F

2 The total valence electron of N and F are
3 N, 1 x 5 = 5
F, 3 x 7 = 21
HNO3 1 Total = 5 + 21 = 26
Draw a single bond connecting each N and F.
1 single bond = 2 electrons. Put the remaining electrons on
the electronegative atom (terminal atom). Complete the
octets of the atoms. If have any remaining electron, put at
the centre atom.

FNF

F

ONOH
O

2 The total valence electron of O, N and H are
N, 1 x 5 = 5
O, 3 x 6 = 18
H, 1 x 1 = 1
Total = 5 + 18 + 1 = 24

3 ONOH

O

All total valence electrons are distributed at the single
bonds and at terminal atoms. This structure satisfies the
octet rule for all the oxygen atoms but not for the nitrogen
atom.

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4 Move a lone pair from one of the end oxygen atoms to
form another bond with nitrogen.

ONOH forms ONOH
O O

Other example: 1 HCCH
C2H2 2 Total valence electrons,
(4 x 2 ) + 2 = 10 electrons
3 HCCH
C did not satisfied octet rule.
CO3 2- 4 HCCH
1
O
2 OCO
3 Total valence electrons,
4 + ( 3 x 6 ) + 2 = 24
4
2

O

OCO
C did not satisfied octet rule. Move a lone pair
from one of the end oxygen atoms to form
another bond with carbon.

2

O

OCO

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4.1.8 The properties of covalent compound
 Low melting / boiling point.
 Insoluble in water (polar solvent), but soluble in organic solvent (non-polar).
 Cannot conduct electricity.

4.1.9 Exception to the octet rule

A. Incomplete octets

Incomplete octets occur when central atom has less than 8 electrons in its valence shell. The
beryllium (Be), boron (B) and aluminium (Al) cannot achieve octet configuration even after
sharing electron with other atoms. E.g:

FF
B

F

Electron configuration for B :     _
1s 2s 2p
In this example the central atom B only has 6 electrons.

B. Expanded octets

In expanded octets the central atoms having 10 or even 12 valence electrons around them.
Molecule with expanded octets typically involves non-metal atom of the third period and beyond
that are bonded to highly electronegative atom.
The expansion has been rationalized by assuming that after the 3s and 3p sub-shells are filled
completely, extra electrons go into the empty 3d sub-shell.

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Examples:
(a) PCl5 Cl
Cl Cl

P
Cl Cl

Electron configuration for P

P         
3s 3p
1s 2s 2p 3d

Excited electronic configuration due to electron pairing for P

P        _ ___ _ ___ ___
3s 3p 3d
1s 2s 2p

the bonding electrons of the P atom appears to have up to 10 electrons (exceed the octet
rule).

(b) SF6

F
FF

S
FF

F

S :         
1s 2s 2p 3s 3p

Excited electronic configuration due to electron pairing for

S:            __ ___ ___

1s 2s 2p 3s 3p 3d

- the bonding electron of the S atom appears to have up to 12 electrons.

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C. Odd-electron molecules

Contain an odd number of valence electrons.

Most odd-electron molecules have a central atom from an odd-numbered group, such as
nitrogen (Group 15) and chlorine (Group 17).

Examples: (a) NO

NO

(b) ClO2

O Cl O

Draw the Lewis structure for SO42-?
Hint: calculate formal charge to determine the structure

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4.1.10 Basic Molecular Shapes - ideal bond angles = 180
1) Linear (AX2) - example: BeCl2
- atoms arranged in a straight line
180°

2) Trigonal planar (AX3) - ideal bond angles = 120
120°
- example : BF3
- central atom has 3 terminal atoms

3) Tetrahedral (AX4) - ideal bond angles = 109.5
109.5° - example : CH4
- central atom have 4 terminal atoms

4) Trigonal bipyramid (AX5) - ideal bond angles = 90,120

90° - example : PCl5
120° - central atom has 5 terminal atoms

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5) Octahedral (AX6)
90° - ideal bond angles = 90
- example : SF6
- central atom have 6 terminal atoms.

90°

4.2 Covalent bond

4.2.1 Valence-Shell Electron-Pair Repulsion (VSEPR) Model

The major features of molecular geometry can be predicted on the basis of quite simple
principle electron-pair repulsion.
According to the VSEPR model, the valence electron pairs surrounding a central atom repel
one another. Consequently, the orbital containing those electron pairs are oriented to be as far
apart as possible. This minimizes the energy of repulsion and represents the lowest energy
configuration of the molecule or polyatomic ion.

The number of valence electron pairs (which may composed of bonding and nonbonding pairs)
gives the basic geometry of a molecule or polyatomic ion. The terminal atoms will then adopt
specific orientations about the central atom to which they are bonded.

Figure below shows the geometries predicted by the VSEPR model for molecules of the types
AX2 to AX6.

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Figure 4.1 Basic molecular shapes
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4.2.2 Predicting Molecular Geometries

In drawing Lewis structures, there are two types of valence-shell electron pairs: bonding
pairs, which are shared by atom in bonds and nonbonding pairs (or lone pairs).
We begin by considering BeF2, a molecule which has two bonding pairs. These electron pairs
will arrange themselves as far apart as possible to form a linear structure.

F Be F

180o

The molecule BF3 represents the AX3 type of molecule, which has three bonding pairs of
electrons. To have minimum repulsions these groups of electrons are situated at the corner of a
planar equilateral triangle.

FF

B

120o

F

In species that follow the octet rule, the central atom is surrounded by four electron pairs. If
each of these pairs forms a single bond with a terminal atom, a molecule of the type AX4
results. The four bonds are directed toward the corner of a regular tetrahedron. All the bond
angles are 109.5o, the tetrahedral angle. This geometry is found in many polyatomic ions such
as NH4+ and SO42 and in a wide variety of organic molecules, the simplest of which is methane,
CH4.
Molecule of the type AX5 and AX6 require that the central atom have an expanded octet. In PF5
, the five bonding pairs are directed toward the corner of a trigonal bipyramidal, a figure
formed when two trigonal pyramidal are fused together, base to base. Three of the fluorine
atoms are located at the corners of an equilateral triangle with the phosphorus atom at the
centre; the other two fluorine atoms are directly above an below the phosphorus atom.
In SF6 , the six bonds are directed toward the corners of a regular octahedral, a figure with
eight sides but six vertices. Four of the fluorine atoms in SF6 are located at the corners of a
square with the S atom at the centre; one fluorine atom is directly above the S atom, another
directly below it.

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4.2.3 Effect of lone pairs on molecular geometry

The VSEPR model can be extended to predict the geometries of molecules and polyatomic ions,
which have one or more lone pairs of electron around a central atom. The geometric
arrangement of atoms around the central atom is determined by the repulsions of electron pairs
in the valence shell of the central atoms.
Repulsion between electron pairs decreases in the order:

(Lone pair-lone pair > Lone pair-bonding pair > Bonding pair-bonding
pair)

The Lewis structure of ammonia reveals three bonding pairs and one nonbonding pair around
the nitrogen atom:

Non-bonding electron pair @ lone pair

HNH

H Bonding pairs

The tetrahedral arrangement of the electron pairs in ammonia is shown in table below.

Molecular Lewis structure Electron pair Molecular
formula geometry geometry

NH3 N H
H
HNH
H H

Basic shape is similarly as tetrahedral. But then we substitute one X (bonding pairs) with one E
(non-bonding/lone pairs)

Basic: AX4  presence of one cenral atom, four bonding pairs
Actual Basic shape: tetrahedral

AX3E  presence of one cenral atom, 3 bonding pairs, 1 lone pairs
Geometrical shape : trigonal pyramidal

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The following steps are used to predict molecular geometries with the VSEPR model:

Application of the VSEPR model to molecules that contain
multiple bonds reveals that a double or triple bond has
essentially the same effect on bond angles as does a single bond.
This observation leads to one further rule:

1. A double or triple bond is counted as one bonding pair when predicting
geometry.

OCO

XA X

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Figure 4.3 Summarizes the molecular geometries of AXnEm molecule. (The symbol of bonding
pairs X can also be replaced with B)
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4.3 POLARITY OF MOLECULE

Covalent compound can exist as polar and non-polar molecules
When two atoms of different electronegativities form a covalent bond, the electrons are not
shared equally between them. The atom with greater electronegativity draws the electron pair
closer to it, and a polar covalent bond results.

Figure 4.4 Unequal electronsharing/cloud. Bonding electrons are more attracted to the most
electronegative atom.

An example of such a polar covalent bond is the one in hydrogen chloride (HCl). The chlorine
atom, with its greater electronegativity, pulls the bonding electrons closer to it. This makes the
hydrogen atom somewhat electron deficient and gives it a partial positive charge (+). The
chlorine atom becomes somewhat electron rich and bears a partial negative charge ().

+ 

H Cl

Because the hydrogen chloride molecule has a partially positive end and a partially negative
end, it has a dipole moment ( ).
The dipole moment ( ) is a physical property that can be measured experimentally. It is
defined as the product of the magnitude of the charge ( ) and the distance (d ) that separates
the centre of positive and negative charge.

Dipole moment ( ) = charge ( ) x distance ( d )

if   0 , it is polar molecule
if  = 0 , it is non-polar molecule

The direction of polarity of a polar bond can be symbolised by a vector quantity
( ).
The crossed end of the arrow is the positive end and the arrow head is the negative end.

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In HCl, for example, we would indicate the direction of the dipole moment in the following way:
+ 
H  Cl

4.3.1 Polar and non-polar molecules

When more than one polar bond is present in the same molecule, the polarity of one bond may
cancel that of another. Thus the presence of polar bonds in a polyatomic molecule does not
guarantee that the molecule as a whole will have a dipole moment. So, for molecules that
contain more than two atoms, the combined effects of all the polar bonds must be considered.
To determine whether a molecule is polar or nonpolar, we can generally use the three-step
approach outlined below :

Examples: + 
I) Carbon dioxide (CO2)
CO


O

Nett dipole moment,  = 0
Non-polar molecule

i) Carbon dioxide is a linear molecule (as predicted by VSEPR theory).
ii) Oxygen is more electronegative than carbon so the two bonds should have a

dipole in which the carbon end is positive (+ ) and the oxygen end is negative (
). There is a bond dipole in each carbon-oxygen bond.
iii) Bond dipoles are equal in magnitude and they point in opposite direction, leading
to a cancellation of their effects.
iv) There is no molecular dipole ( = 0 ) and the CO2 molecule is nonpolar.

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II) Carbon tetrachloride (CCl4)

Cl

C Cl

Cl
Cl

Nett dipole moment,  = 0

Non-polar molecule

i) Carbon tetrachloride is tetrahedral

ii) Electronegativity of chlorine is greater than carbon, each of the bond dipole

/polar bond ( ) carbon-chlorine is polar.

iii) Has 4 polar bonds and in same orientation (pointing out to Cl atom). They are

said to exerts an equal attractive force in four opposing direction.

iv) The attractions exerted by the 4 polar bonds cancel each other because of the

symmetrical structures of these molecules.

v) The  = 0 (their vector sum is zero) and the molecule is considered nonpolar.

III) Chloromethane (CH3Cl)

Cl

C

HH
H

Nett dipole moment,  ≠ 0

Polar molecule

i) Chloromethane is tetrahedral

ii) Since carbon and hydrogen have electronegativities that are nearly the same, the
contribution of three CH bonds to the net dipole is negligible.

iii) The electronegativity difference between carbon and chlorine is large, and this highly
polar CCl bond accounts for most of the dipole moment of CH3Cl.

iv) Thus the molecule has a net dipole and is polar.

The dipole moment of chloromethane arises mainly from the highly polar carbon-
chlorine bond.

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Electron density in covalent bonds shifts toward the more electronegative
atom, producing partial charges on each atom and hence a dipole. In a
polyatomic molecule, bond dipoles must be added as vectors to obtain a
resultant which indicates molecular polarity. In the case of symmetric
molecules, the effects of individual bond dipoles cancel and a nonpolar
molecule results.

IV) Ammonia (NH3 )

N resultant
H

H

H

Nett dipole moment,  ≠ 0

i) Geometrical of Ammonia is a trigonal pyramidal molecule
ii) Large dipole moments because nitrogen is more electronegative than hydrogen

and because it has lone pair electrons.
iii) The lone pair electrons make large contributions to the dipole moments because

lone pair electrons have no other atom attached to it to partially neutralise its
negative charge.
iv) Thus in molecules with lone pairs, polarity usually occurs because there is no
opposite vertex to cancel.

Molecule Geometry Dipole moment

HF Linear 1.92
HCl Linear 1.08
HBr Linear 0.78
HI Linear 0.38
H2S Bent 1.10
NH3 Pyramidal 1.46
SO2 Bent 1.60

Table 5.1: Dipole moments of some polar molecules.

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Step 1: Draw the Lewis structure
Step 2: Draw the 3D molecular structure VSEPR rules
Step 3: Use symmetry to determine if the molecule is polar or non-polar.

4.4 Orbital Overlap and Hybridization

Valence Bond theory describes covalent bond formation as well as the electronic structure of
molecules. The theory assumes that electrons occupy atomic orbitals of individual atoms within
a molecule, and that the electrons of one atom are attracted to the nucleus of another atom.
This attraction increases as the atoms approach one another until the atoms reach a minimum
distance where the electron density begins to cause repulsion between the two atoms. This
electron density at the minimum distance between the two atoms is where the lowest potential
energy is acquired, and it can be considered to be what holds the two atoms together in a
chemical bond. Valence bond theory is based on the following four postulates:

Postulate 1:
A bond is formed when atomic valence orbitals overlap with each other.

Postulate 2:
Overlapping orbitals contain a pair of electrons.

Postulate 3:
Electron density concentrates between bonded atoms.

Postulate 4:
The strength of the bond depends on the degree of overlapping.

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Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently
bond with other atoms in order to gain more stability, which is gained by forming a full electron
shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell
and gain stability. Nonmetals will readily form covalent bonds with other nonmetals in order to
obtain stability, and can form anywhere between one to three covalent bonds with other
nonmetals depending on how many valence electrons they possess. Although it is said that
atoms share electrons when they form covalent bonds, they do not usually share the electrons
equally.

Only 2 electrons with their spins paired may be shared in one set of overlapping orbitals.
2 types of bond:

• sigma bond (σ)
• pi bond (π)

The simplest molecule we can discuss in terms of valence bond theory is a hydrogen gas, H2

a) Sigma bond

In chemistry, sigma bonds are the strongest type of covalent chemical bond. They are
formed by head-on overlapping between atomic orbitals on neighbouring atoms. Sigma
bonding is most simply defined for diatomic molecules using the language and tools of
symmetry groups.

(i) Result from the overlapping of TWO s orbitals.
Example: H2

H: 
1s

Figure 4.5 : Overlapping of two s orbitals hydrogen atoms.
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(ii) Result from the overlapping of s and p orbitals.
Example: HF

H:  The H electron and F electron
1s will pair up

F:    _
2s 2p

valence electrons

One of the 2p orbital of F atom is occupied by a single electron.
H electron and the F electron pair up and be shared between the two nuclei.

Figure 4.6 : Overlapping of sp orbitals.
iii) Result from overlapping of two p orbitals.

Example: Cl2
Cl :         
1s 2s 2p 3s 3p
Valence electron
One of the 3p orbitals is occupied by a single e

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