Bonding - ionic covalent & metallic

STRUCTURE AND BONDING
Prepared by Janadi Gonzalez-Lord

TABLE OF CONTENTS

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 Syllabus requirements  Covalent bonding
 Review: atoms and the  Covalent nomenclature
periodic table  Properties of covalent
 Elements and bonding compounds
 Representing ions and  Metallic bonding
molecules  Properties of metallic
 Ion formation compounds
 CATION formation  Allotropes
 ANION formation  Atomic radius
 Ionic Bond formation  Electronegativity
 Representing ionic bonding  Thermodynamics of ion
 Ionic nomenclature formation
 Properties of ionic
compounds
2

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SYLLABUS REQUIREMENTS
3 BONDING

SYLLABUS REQUIREMENTS – IONIC BONDING

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The students should be able to :

 state that atoms like to achieve a stable state by electron gain or

loss

 recognize the tendency for loss or gain based on the electronic

configuration or the position in the periodic table ( for the first twenty

elements )
4
 state that ions are formed by the gain or loss of the electrons

SYLLABUS REQUIREMENTS - BONDING

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 Define anion and cation

 Recognize that charge is equal to protons minus electrons

 Identify the number of protons , electrons in and the electronic

configuration of an ion . ( first twenty elements only )

 Write symbols for ions and molecules

 Identify the two main types of bonding as ionic / electrovalent and
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covalent

SYLLABUS REQUIREMENTS - BONDING

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•Explain metallic bonding

•State the differences between ionic, covalent and metallic bonding

•Identify the types of bonding present in substances based on their

properties

•Predict the properties of substances based on the bonding present

•Draw diagrams to illustrate the types of bonding

•Predict the types of bonding between elements 6

SYLLABUS REQUIREMENTS
7 STRUCTURE

SYLLABUS REQUIREMENTS - STRUCTURE

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Students should be able to

1.Define and give examples of ionic crystals, simple

molecular structures and giant molecular crystals

2.Explain the term allotropy

3.Relate structures of sodium chloride, diamond and

graphite to their properties
8
4.Distinguish between ionic and molecular solids

REVIEW: ATOMS AND THE PERIODIC
TABLE
9

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10

These columns are known as GROUPS are also known as
GROUPS FAMILIES
GROUPS IN THE PERIODIC TABLE

10 11 12 13 14 15 16 17 18
1 2 3 4 5 6 7 8 9
There are 18
GROUPS

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Elements within a group have similar All have the same number of
physical and chemical properties electrons in their outermost or
valence shells

Example Na
(2,8,1) and
K (2,8,8,1)
are both in
Group 1

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The rows are known as PERIODS. There are 9 periods
MAIN PERIODS IN PERIODIC TABLE

1
2
3
4
5
6
7

8

9

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ELEMENTS AND BONDING
14 Introduction to ionic, covalent and metallic bonding

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PURE AND IMPURE SUBSTANCES
15 A review

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Matter can be sub-divided into PURE and IMPURE SUBSTANCES or MIXTURES.

PURE substances can be sub-divided into ELEMENTS and COMPOUNDS
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IMPURE substances or MIXTURES can be sub-divided into HOMOGENOUS and
HETEROGENOUS

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Can be separated into

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Can be
Can be separated
separated into
into

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Source: www.mghs.sa.edu.au/Internet/Faculties/Science/Year10/Pics/elementsAndCompounds.gif

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ELEMENTS VERSUS COMPOUNDS
An element.... A compound......

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 consists of only one kind  consists of atoms of two or
more different elements bound
of atom together,
 cannot be broken down  can be broken down into a
into a simpler type of simpler type of matter
(elements) by chemical means
matter by either physical (but not by physical means),
or chemical means  has properties that are different
 can exist as either atoms from its component elements,
(e.g. argon) or molecules  always contains the same ratio
of its component atoms Prep
(e.g., nitrogen). ared
by
JGL

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AN ELEMENT
Consists of only one kind of atom

Ar
Ar
can exist as either atoms (e.g.
argon)

or molecules (e.g., nitrogen).

N N cannot be broken down into a
simpler type of matter by either
physical or chemical means

N N If you try to break apart an atom or
molecule, you get an ATOMIC
BOMB

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H H A COMPOUND
O consists of atoms of two or more different
elements bound together

always contains the same ratio of its
component atoms
H H
O
Water (formula H2O)

H H O O H H For every water molecule, there are 2
O Hydrogen atoms for every 1 Oxygen atom

H H has properties that are different from its
O component elements

H H
O For example, hydrogen and oxygen are gases
but water is a liquid

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EXAMPLES OF ELEMENTS AND COMPOUNDS

Elements

Compounds

Source: www.physicalgeography.net/fundamentals/images/compounds_molecules.jpg

WHY DO COMPOUNDS FORM IN THE FIRST
PLACE?

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Scientists found that elements in Group 8 were very non-
reactive.
They also noticed that those in Groups 1,2,6 and 7 were
extremely reactive.
They also noticed that metallic substances had several
properties that were very different from other elements.
They could not at first understand why.
Eventually they discovered that it had to do with
ELECTRON CONFIGURATIONS
and
STABILITY
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ELECTRON CONFIGURATION AND STABILITY

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Scientists’ research showed that
in compounds, elements will
combine so that the valence or
outermost electrons will have
the same electron
configuration as the nearest
noble gas
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(in Group 8)

HOW CAN ELEMENTS COMBINE TO ACHIEVE THIS?
An element can gain electrons from the

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element it combines with to have the same

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electron configuration as the nearest noble
gas.
Once an atom gains one or more
electrons, it becomes a negatively charged
particle known as an ANION

An element can lose
electrons to another An element can share
element to have the valence electrons with
same electron another element to
configuration as the
nearest noble gas. have the same
Once an atom loses one
There electron configuration
are three as the nearest noble
or more electrons, it gas.
forms a positively (3) ways
charged particle known
as a CATION. 24

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YOU MAY WELL

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BE ASKING
WHAT DOES
THIS MEAN? 25

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LET’S TAKE A LOOK

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AT SOME EXAMPLES
TO UNDERSTAND
THIS CONCEPT MORE
FULLY 26

Sodium’s atomic number Neon’s atomic number is
is Z=11. Its electron Z=10. Its electron
Let’s take sodium as an example

configuration is therefore configuration is 2,8. It is
2,8,1 the nearest noble gas to
sodium.

Sodium will combine with another element so that it
can change its electron configuration from 2,8,1 to
2,8. To do this, it must lose 1 electron and give it to
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the element with which it combines.
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Chlorine’s atomic number Argon’s atomic number
is Z=17. Its electron is Z=18. Its electron
Let’s take chlorine as an example

configuration is therefore configuration is 2,8,8. It
2,8,7 is the nearest noble gas
to chlorine.

Chlorine will combine with another element so that it
can change its electron configuration from 2,8,7 to
2,8,8. To do this, it must gain 1 electron from the
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element with which it combines.
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SODIUM AND
CHLORINE UNDERGO
WHAT IS KNOWN AS
IONIC BONDING.
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IN IONIC BONDING……

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 An element can lose or gain electrons to another
element within the same compound to have the
same electron configuration as the nearest noble
gas.

 These are known as IONS.

 The electrostatic attraction between these ions is
known as an IONIC bond

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In order to form

IONIC BONDING OF SODIUM CHLORIDE
the compound
sodium
chloride, there are

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three (3) steps.

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First, the sodium
atom loses one
electron to form a
positive sodium
ion. (cation)

Then the chlorine
atom accepts the
electron from the
sodium atom to
form a negative
chloride ion
(anion).

Then the sodium
cation and
chloride anion
become attracted
to each due to
their different
charges, forming
an ionic bond 31
Source: www.revisionworld.co.uk

Carbon’s atomic number Neon’s atomic number is
is Z=6. Its electron Z=10. Its electron
Let’s take carbon as an example

configuration is therefore configuration is 2,8. It is
2,4 the nearest noble gas to
carbon.

Carbon will combine with another element so that it can change
its electron configuration from 2,4 to 2,8. To do this, it must share
4 electrons with the element with which it combines as it is equally
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COVALENT BONDING

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 An element can share valence electrons with
another element to have the same electron
configuration as the nearest noble gas.

 This sharing of valence electrons is known as
COVALENT bonding.

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Covalent bond
is the sharing
of two
COVALENT BONDING

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electrons
between the 2

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atoms

Two hydrogen atoms can share their valence
electrons to attain the same electron
configuration of the nearest Noble gas
configuration, Helium 34

The valence (outermost)
electrons are loosely held by
METALLIC BONDING

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the metal ions, so much so
that they move away from

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The electrons are free to move from
the atom to form a positively
one positively charged ION to the
charged ION.
next (i.e. They are DELOCALISED)
However, metals like in covalent
and are shared (just behave
bonding among the various metallic
differently.
positively charged ions

Metallic bonding is similar to
both covalent and ionic
bonding
The number of electrons = the
number of protons.
Source: www.daviddarling.info/images/metallic_bond.jpg

The metal is therefore 35
Syllabus requirement met:
electrically NEUTRAL Explain metallic bonding

IN SUMMARY……

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Ionic
bonding

3 types
of
bonding
Metallic Covalent
bonding bonding
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COMPARE AND CONTRAST TYPES OF BONDING

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Similarities Differences

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 Metallic and ionic  Covalent bonding shares
bonding involve electrons rather than
electrostatic attractions having electrostatic
between positive and charges.
negatively charged
particles.  Ionic bonding will form
 Metallic bonding shares compounds whereas
electrons among the ions covalent bonding can
in a similar manner to form a compound or
how electrons are shared element and metallic
in covalent bonding. bonding is strictly found
in elements
Syllabus requirement met: 37
State the differences between ionic, covalent and
metallic bonding

REPRESENTING IONS AND
MOLECULES
38 Ionic notation, chemical formulae

Mass Number
RECALL BASIC ATOMIC NOTATION

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= number of protons +

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number of neutrons

A Atomic number
= Number of
protons

Z
X Element symbol
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Ionic charge

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WE CAN ALSO = number of protons -
number of electrons

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DISPLAY OTHER
INFORMATION n+/-
Number of atoms

X m
of element X in
molecule or ion

Element symbol
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Mass Number Ionic charge
ALTOGETHER of protons +
= number

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= number of protons -
number of neutrons number of electrons

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A n+/-
Atomic

X
number Number of atoms
= Number of of element X in
protons molecule or ion

Z m Element symbol
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Mass Number
EXAMPLE: SODIUM ION

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Element symbol Na 23 1+
Ionic charge
Charge = +1

Atomic Number Z = 11

11
Na
Mass number A = 23 Element symbol

Atomic number Z

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Mass Number
EXAMPLE: HYDROGEN MOLECULE

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Number of

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Element symbol H 2 atoms of
hydrogen in a

Charge = 0

Atomic Number Z = 1
H molecule of
hydrogen

1 2
Mass number A = 2
Element symbol

Number of Hydrogen Atomic number Z
atoms in a molecule of
hydrogen = 2
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ION FORMATION
44

IONS DEFINED

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An ion is an atom or molecule
where the total number of
electrons is not equal to the
total number of protons, giving
it a net positive or negative
electrical charge.
Syllabus objective met: 45
State that ions are formed by the gain or
loss of the electrons

REVIEW – ELECTRON CONFIGURATIONS

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 What is an electron configuration?
Definition: Electron configuration is the arrangement of electrons in
an atom, molecule or other body.

 How do we represent electron configurations?
By using Bohr-Rutherford diagrams

Or electron configuration notation

2,8,1
11 p
10 n

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REMEMBER – “CONTRAST” MEANS “LOOK AT THE DIFFERENCES”

LET’S CONTRAST –F AND NE

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Fluorine Neon

 Element symbol F  Element symbol Ne
 Group 17  Group 18
 Atomic Number Z = 9  Atomic Number Z = 10
 Mass number A = 19  Mass number A = 20
 Electron configuration: 2,7  Electron configuration: 2,8
 Bohr-Rutherford diagram  Bohr-Rutherford diagram

9p 10 p
10 n 10 n

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LET’S CONTRAST – NA & NE

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Sodium Neon

 Element symbol Na  Element symbol Ne
 Atomic Number Z = 11  Atomic Number Z = 10
 Mass number A = 23  Mass number A = 20
 Electron configuration: 2,8,1  Electron configuration: 2,8
 Bohr-Rutherford diagram  Bohr-Rutherford diagram

11 p 10 p
12 n 10 n

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Remember –
“Compare” means
“Look at

COMPARE AND CONTRAST ALL 3 ELEMENTS

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Similarities Differences

 F and Ne have the  Different atomic numbers
(Z) and therefore protons
same number of
 Different mass numbers (A)
electron shells and therefore different
neutrons
Scientists found that when
 F needs to gain 1 electron
elements from Group 1 and
to have the same number of
Group 7 combined, they lost or
electrons as Ne
gained an electron to have the
same number of electrons as  Na needs to lose 1 electron
the nearest Noble Gas. to have the same number of
electrons as Ne
i.e. F and Na form ions that Syllabus requirement met: 49
are ISO-ELECTRONIC with state that atoms like to achieve a stable
Ne state by electron gain or loss

IN GENERAL

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Groups 1, 2 and 3 Groups 15, 16 and 17

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 To become ISO-  To become ISO-ELECTRONIC
ELECTRONIC with the with the nearest Noble Gas
nearest Noble Gas (either (either within the same Period
within the same Period or the or the Period just above)
Period just above) 1. Group 15 elements gain 3 e-
1. Group 1 elements lose 1 e- 2. Group 16 elements gain 2 e-
2. Group 2 elements lose 2 e- 3. Group 17 elements gain 1 e-
3. Group 3 elements lose 3 e-
 This only happens when
combining or reacting with
 This only happens when another element(s) from
combining or reacting with Groups 1,2 or 3
another element(s) from
Groups 15,16 or 17

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CATION FORMATION
51

CATION FORMATION

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Let’s take an unknown Let’s assume that this
element X that has an atom of X loses 1
atomic number Z = 10 electron.
and a electrical charge
of zero. Now
# of protons (p) = 10
For an atom of X, # of electrons (e) = 10 -1 = 9
# of protons (p) = 10 Charge on ion = 10 – 9 = +1
# of electrons (e)= 10
Charge of atoms = 10 – 10 =0

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IF WE THINK OF IT LIKE AN EQUATION,

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For an atom of X For an positive ion of X

+ 10 p + 10 p
- 10 e - 9e
0 +1

In general, a positive ion formed by the loss of one or more
electrons is known as a CATION.
Syllabus objectives met:
Define cation
Recognize that charge is equal to protons minus electrons 53

Group 1 elements have 1 This is the same electron
valence (outermost) electron configuration as Ne, the noble
gas just above Na (Period 2,
THE PERIODIC TABLE
If Na loses 1 electron, its electron configuration Na ion and Ne
Group 8). i.e.
becomes (2,8) are ISO-ELECTRONIC

If K loses 1
electron, its electron
configuration This is the same electron
Example Na configuration as Ar, the noble
becomes (2,8,8)
(2,8,1) and gas just above K (Period
K (2,8,8,1) 2, Group 8). i.e. K ion and Ar
are both in are ISO-ELECTRONIC
Group 1

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CATION FORMED

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 If Na loses 1 p = +11
electron, its electron e- = - 10
configuration moves Charge = +1
from (2,8,1) to (2,8).
 Since the number of
protons remains the It therefore becomes a
same (p=11) but the cation with an electrical
number of electrons charge of +1
change (e- = 10), it is
no longer electrically
neutral
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CATION FORMED

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 If K loses 1 electron, its p = +19
electron configuration e- = - 18
moves from (2,8,8,1) to Charge = +1
(2,8,8).
neutral
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IN GENERAL

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Li, Na and K are in group
All group 1 1
elements
So
lose 1 e- to Li loses 1 e- to form Li+
form a
Na loses 1 e- to form Na+
cation with
an electrical K loses 1 e- to form K+

charge of +1 57

valence (outermost) electrons configuration as Ne, the noble
gas just above Mg (Period
T HE PERIODIC TABLE
If Mg loses 2 electrons, its electron configuration 2, Group 8). i.e. Mg ion and
becomes (2,8) Ne are ISO-ELECTRONIC

If Ca loses 2
electrons, its
Example Mg electron This is the same electron
(2,8,2) and configuration configuration as Ar, the noble
Ca(2,8,8,2) becomes (2,8,8) gas just above Ca (Period
are both in 2, Group 8). i.e. Ca ion and
Group 2 Ar are ISO-ELECTRONIC

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CATION FORMED

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 If Mg loses 2 p = +12
electrons, its electron e- = - 10
from (2,8,2) to (2,8).
neutral
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CATION FORMED

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 If Ca loses 2 p = +20
from (2,8,8,2) to
(2,8,8).
It therefore becomes a
protons remains the cation with an electrical
same (p=20) but the charge of +2
number of electrons
neutral 60

IN GENERAL

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Be, Mg and Ca are in
All group 2 group 2
elements
So
lose 2 e- to Be loses 2 e- to form Be2+
form a
Mg loses 2 e- to form Mg2+
cation with
an electrical Ca loses 2 e- to form Ca2+

charge of +2 61

valence (outermost) electrons configuration as He, the noble
gas just above B (Period
THE PERIODIC TABLE 1, Group 8). i.e. B ion and He
If B loses 3 electrons, its electron configuration
becomes (2) are ISO-ELECTRONIC

If Al loses 3
electrons, its
Example electron
B(2,3) and configuration This is the same electron
Al(2,8,3) are becomes (2,8) configuration as Ne, the
both in noble gas just above Al
Group 3 (Period 2, Group 8). i.e. Al
ion and Ar are ISO-
ELECTRONIC
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Recognize the tendency for loss of electrons based on the electronic configuration
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CATION FORMED

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 If B loses 2 electrons, p = +5
its electron e- =-2
configuration moves Charge =+3
from (2,3) to (2).
change (e- = 2), it is no
longer electrically
neutral
63

CATION FORMED

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 If Al loses 3 p = +13
from (2,8,3) to (2,8).
neutral
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IN GENERAL

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B and Al are in group 13
All group 13
elements So
B loses 3 e- to form B3+
lose 3 e- to
form a Al loses 3 e- to form Al3+

cation with
an electrical
charge of +3 65

RECALL:
METALS Group 1 to 13 and periods 8 and 9 are METALS

Metals

10 11 12 13 14
1 2 3 4 5 6 7 8 9

Metals
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RECALL

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Basic Math equation Ionic half-equation

For the basic mathematical We could write
equation: Na – 1e- → Na+
X–3=0
If we treat the “→” as
In order to solve for x , you
“=“, we could take across
take the number 3 across to
the 1e- to the RHS and
the right hand side (RHS) of
the equation and change change the sign as well
the sign.
Na → Na+ +1e-
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X = +3

SO FOR GROUP 1 ELEMENTS

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Statement Ionic half-equation

Li loses 1 e- to form Li+
Or Li – 1e- → Li+ Li → Li+ +1e-

Na loses 1 e- to form Na+
Or Na – 1e- → Na+ Na → Na+ +1e-

K loses 1 e- to form K+
K → K+ +1e-
Or K – 1e- → K+
68


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Be loses 2 e- to form Be2+
Or Be – 2e- → Be2+ Be → Be2+ + 2e-

Mg loses 2 e- to form Mg2+
Or Mg – 2e- → Mg2+ Mg → Mg2+ +2e-

Ca loses 2 e- to form Ca2+
Ca → Ca2+ +2e-
Or Ca – 1e- → Ca2+
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B loses 3 e- to form B3+
Or B – 3e- → B3+ B → B3+ + 3e-

Al loses 3 e- to form Al3+
Or Al – 3e- → Al3+ Al → Al3+ + 3e-

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IN GENERAL METALS LOSE ELECTRONS TO

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FORM POSITIVE IONS

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Group 1 metals form mono-positive cations
For example
In words: sodium loses 1 electron to form sodium cation
In chemical equation form: Na → Na+ + 1e-
Group 2 metals form di-positive cations.
For example
In words: Magnesium loses 2 electrons to form magnesium cation
In chemical equation form: Mg → Mg2+ + 2e-
Group 13 metals form tri-positive cations
In words: Aluminium loses 3 electrons to form aluminium cation
In chemical equation form: Al → Al3+ + 3e- 71

WHAT ABOUT TRANSITION METALS?

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Recall: Transition metals are Groups 3 to 12
They also form cations
However, they ccan form cations with multiple valences.
For e
In chemical equation form: Na → Na+ + 1e-
Group 2 metals form di-positive cations.
For example
In words: Magnesium loses 2 electrons to form magnesium cation
In chemical equation form: Mg → Mg2+ + 2e-
Group 13 metals form tri-positive cations
In words: Aluminium loses 3 electrons to form aluminium cation
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In chemical equation form: Al → Al3+ + 3e-

SUMMARY – CATION FORMATION

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Cations are positive
ions.

The atoms lose
electrons to achieve the They are formed by loss
electron configuration of of electrons.
the nearest noble gas

Metals (Groups 1 to 13)
form cations
73

ANION FORMATION
74

This is the same electron
Group 15 elements have 5
configuration as Ne, the noble
valence (outermost) electrons
gas in the same period as N
T HE PERIODIC TABLE (Period 3, Group 8). i.e. N ion
If N gains 3 electrons, its electron configuration and Ne are ISO-
becomes (2,8) ELECTRONIC

If P gains 3
electrons, its
Example electron This is the same electron
N(2,5) and configuration configuration as Ar, the
P(2,8,5) are becomes (2,8,8) noble gas in the same period
both in as P (Period 3, Group 8). i.e.
Group 5 P ion and Ar are ISO-
ELECTRONIC

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gas in the same period as O
T HE PERIODIC TABLE (Period 2, Group 8). i.e. O ion
If O gains 2 electrons, its electron configuration and Ne are ISO-

If S gains 2
electrons, its
Example electron
O(2,6) and configuration This is the same electron
S(2,8,6) are becomes (2,8,8) configuration as Ar, the noble
both in gas in the same period as S
Group 6 (Period 3, Group 8). i.e. S
ion and Ar are ISO-
ELECTRONIC

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gas in the same period as O
T HE PERIODIC TABLE (Period 2, Group 8). i.e. F ion
If F gains 1 electron, its electron configuration and Ne are ISO-

If Cl gains 1
electron, its electron
Example
configuration
F(2,7) and This is the same electron
becomes (2,8,8)
Cl (2,8,7) configuration as Ar, the noble
are both in gas in the same period as Cl
Group 7 (Period 3, Group 8). i.e. Cl
ion and Ar are ISO-
ELECTRONIC
Syllabus requirements met:
Recognize the tendency for loss or gain based on the electronic configuration or
the position in the periodic table ( for the first twenty elements) Prepared by JGL
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RECALL:
NON-METALS Groups 14 to 18 are Non-metals

Non-
Metals

14 15 16 17 18

Atoms and The Periodic Table Prepared
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IN GENERAL, NON-METALS GAIN

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ELECTRONS TO FORM NEGATIVE IONS

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Group 15 non-metals form tri-negative anions
For example
In words: nitrogen gains 3 electrons to form nitride anion
In chemical equation form: N + 3e-→ N3-

Group 16 non-metals form di-negative anions
For example
In words: oxygen gains 2 electrons to form oxide anion
In chemical equation form: O + 2e-→ O2-

Group 17 non-metals form mono-negative anions
For example
In words: fluorine gains 1 electron to form fluoride anion
In chemical equation form: F + 1e-→ F- 79

ANION FORMATION

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Let’s take an unknown Let’s assume that this
element Y that has an atom of Y gains 2
atomic number Z = 11 electron.
and a electrical charge
of zero. Now
# of protons (p) = 11
For an atom of Y, # of electrons (e) = 11 +2 = 13
# of protons (p) = 11 Charge on ion = 11 – 13 = -2
# of electrons (e)= 11
Charge of atoms = 11 – 11 =0

80

IF WE THINK OF IT LIKE AN EQUATION,

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For an atom of Y For a negative ion of Y

+ 11 p + 11 p
- 11 e - 13 e
0 -2

In general, a negative ion formed by the gain of one or
more electrons is known as a ANION.
Syllabus objectives met:
Define anion
Recognize that charge is equal to protons minus electrons 81
State that ions are formed by the gain or loss of the electrons

SUMMARY – ANION FORMATION

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Anions are negative
ions

Non-metals lose
electrons to attain the Non-metals gain
electron configuration of electrons to form anions
the nearest noble gas.

Non-metals are
elements in Groups 14
to 18 82

IONIC BOND FORMATION
83

RECALL...

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Cations Anions

 Are positive ions  Are negative ions

 Are formed from metals  Are formed from non-
(groups 1-13) metals (groups 14-18)
 Are formed when  Are formed when non-
metals lose electrons metals gain electrons
The ionic bondattain the electron
to attain the electron to is the
configuration of the configuration of the
electrostatic attraction gas
nearest noble gas nearest noble
between anion and cation 84

HOW TO RECOGNIZE AN IONIC COMPOUND

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Ionic
Metal Non-metal
compound

85

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Metals Non-

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Group 1 to 13 are METALS
Metals
Groups 14 to
18 are Non-
Metal Non-metal metals Ionic
compound

Periods 8 and 9 are METALS
Metals

86

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LET US EXAMINE THE BONDING BETWEEN
MAGNESIUM AND FLUORINE

87

Fluorine
Magnesium

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Metals Non-

Noble Gases Group 8

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Metals
Neon

88

USING THE PERIODIC TABLE...

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Magnesium Flourine

 Period 3  Period 2
 Metal  Non-metal
 Forms cation  Forms anion
 Z = 12  Z=9
 Electron configuration  Electron configuration
(2,8,2) (2,7)
 Nearest noble gas Ne  Nearest noble gas Ne
 Electron configuration of  Electron configuration of
Ne = (2,8) Ne = (2,8)
89

FORMATION OF IONIC BOND

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Cation formation Anion formation

2 electrons need to be 1 electron needs to be
lost to go from (2,8,2) gained to go from (2,7)
to (2,8) to (2,8)
Mg → Mg2+ + 2e- F + 1e- → F-
# of e- lost # of e- gained
The Law of conservation of matter states that matter can neither created
nor destroyed.
Therefore

# of e- lost HOW? # of e- gained 90

THE ONLY WAY THIS CAN HAPPEN….

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Is if Then

Another Fluorine atom
(F) accepts the 2nd
electron
F + 1e- → F-
Mg → Mg2+ + 2e- And
F + 1e- → F-

2 Fluorine ions are needed to bond to each
Magnesium ion.
91

IN TERMS OF IONIC EQUATIONS

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Cation equation Anion equation

Mg → Mg2+ + 2e- F + 1e- → F-
F + 1e- → F-

2F + 2e- → 2F-

92

ADDING THE 2 EQUATIONS

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Total ionic equation Ionic equation

Mg → Mg2+ + 2e- Mg2+ + 2F- → MgF2
2F + 2e- → 2F-

Mg + 2F → Mg2+ + 2F-

93

THEREFORE THE CHEMICAL FORMULAE

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FOR IONIC COMPOUND FORMED BETWEEN
MAGNESIUM AND FLUORINE IS
MGF2

A chemical formulae
expresses the ratio of
atoms of elements within a
compound.
94

IONIC BOND

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FORMATION PROCESS

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CATION formation IONIC Compound
formation
Mg → Mg2+ + 2e- Mg2+ + 2F- → MgF2

ANION formation Electrostatic
2F + 2e- → 2F- attraction

95

IONIC BONDING EXERCISES

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 Demonstrate the type 1. Sodium and fluorine
of bond formed 2. Magnesium and
between the two sets oxygen
of elements below in 3. Lithium and Sulphur
1. Diagrammatic form
2. Using atomic notation
3. Using ionic equations

96

IONIC BONDING BETWEEN SODIUM now p= 11 and e-=10.
There are AND
FLUORINE – protons, p=11 CATION FORMATIONcharge of +1 and
The number of Since each p has a
STEP 1:

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each e- has a charge of -1, the overall

8/21/2009
(because Z=11) charge becomes +11-10=+1.
 Using Bohr-Rutherford diagrams atom
The nearest noble gas to Na is therefore forms a positive ion
Na
neon, Ne. The atomic number for Ne 1+
The number of or cation of monopositive charge +1
is Z=10. Its electronic configuration
neutrons, n=12 is therefore 2,8
(n = A – Z)

11 p - 1 e- 11 p + 1 e-
12 n 12 n
To achieve this electronic
Na atom has electronic configuration, Na must lose 1
configuration
e-
of 2,8,1 because there are 2 e- in the
1st shell, 8 2,8,1 the 2nd shell and 1 e- in
e- in 2,8
Using atomic notation,
the 3rd or outermost (valence) shell Now the electronic configuration
becomes 2,8 which is iso-electronic
Na Na + + 1e -
with Ne.

Sodium has mass number A=23 and atomic number Z=11.
97
The atomic notation for sodium atom is therefore 2311Na.
The number of electrons, e-=11 (because the number of protons is equal to the number
of electrons in an atom)

There are now p= 9 and e-=10.
I ONIC BONDING BETWEEN SODIUM AND Since each p has a charge of +1 and

FLUORINE TheTEP of ANION FORMATIONof -1, the overall
–S
number 2:
each e- has a charge

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protons, p=9 charge becomes +9-10=-1.
Na atom therefore forms a negative ion or

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(because Z=9)
 Using Bohr-Rutherford diagramsof mono-negative charge -1
anion
The nearest noble gas to F is 1-
The number of neon, Ne. The atomic number for Ne
neutrons, n=10 is Z=10. Its electronic configuration
(n = A – Z) is therefore 2,8
9p
+ 1 e-
10 n
9p
10 n
To achieve this electronic
configuration, F must gain 1 e-.
F atom has electronic configuration of Law of
According to the
2,8
2,7 conservation of matter, matter can
2,7 because there are 2 e- in the 1 st
Now the electronic configuration
shell and 7 e- in the 2ndneither be created nor destroyed. Te
Using atomic notation, or outermost
electron must therefore come 2,8 which is iso-electronic
becomes from -
(valence) shell
F + 1e -
the Na atom. with Ne. F
Fluorine has mass number A=19 and atomic number Z=9.
98
The atomic notation for fluorine atom is therefore 199F.
The number of electrons, e-=9 (because the number of protons is equal to the number of
electrons in an atom)

IONIC BONDING BETWEEN SODIUM AND FLUORINE –
STEP 3: IONIC BOND FORMATION

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 Using Bohr-Rutherford diagrams
1+ 1-

11
p + 9p
10 n
12
n
IONIC bond
2,8
2,8
Using atomic notation,

Na+ + F- NaF
The fluorine anion is attracted to the sodium cation as opposites attract.
The electrostatic (“electro” meaning “electrically” or “coming from electrons” and “static”
99
meaning “not moving”) attraction between the anion and cation is the IONIC bond.

IN SUMMARY, FOR SODIUM AND FLUORINE

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Na Na+ + 1 e- F + 1 e- F-

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9p
10 n
11 p
12 n

1-
1+

9p
11 p 10 n
12 n

100

Na+ + F- NaF

SIMILARLY, FOR MAGNESIUM AND OXYGEN,

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Mg Mg2+ + 2 e- O + 2 e- O2-

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8p
8n
12 p
12 n

1-
1+

9p
11 p 10 n
12 n

101

Mg2+ + O2- MgO

SIMILARLY, FOR LITHIUM AND SULPHUR,

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Li Li+ + 1 e- S + 2 e- S2-

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16p
16n
3p 3p
4n 4n

1-
1+ 1+

3p 9p
3p 10 n
4n 4n

102

2Li+ + S2- Li2S

REPRESENTING IONIC BONDING
103 Lewis structures (Dot and cross diagrams), chemical
formulae, ionic equations, ionic notation

EXAMPLE: ALUMINIUM OXIDE

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CATION information ANION information

Aluminium  Oxygen
Symbol Al  Symbol O
Group 3  Group 16
Period 3  Period 2
Metal  Non-metal
Forms cation  Forms anion
Z = 13  Z=8
Electron configuration (2,8,3)  Electron configuration (2,6)
Nearest noble gas Ne  Nearest noble gas Ne
Electron configuration of Ne =  Electron configuration of Ne
(2,8) = (2,8)
Needs to lose 3 electrons to  Needs to gain 2 electrons to
achieve electron configuration of achieve electron 104
Ne configuration of Ne

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HOW DO WE REPRESENT
THIS INFORMATION MORE
SIMPLY?

105

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An ionic equation
expresses what
happens at an
ionic level
106

CATION FORMATION: ALUMINIUM

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In words:

An aluminium atom
loses three electrons
to form
aluminium cation 107

CATION FORMATION: ALUMINIUM

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Aluminium atom is neutrally
charged. This is represented
Ionic equations must

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In by the element symbol ions in them Aluminium cation is
ionic equation form: have Al
alone. represented by the element
symbol and the charge on the
ion. In this case it is 3+

Al → Al3+ + 3 e -

There is no =.
Instead there is an RHS = Right hand
arrow. This means side
LHS = “changes into”. It os The RHS is electrically
Left called an equation neutral because the
Hand because the LHS is number of electrons (3-)
Side equivalent to RHS. cancels out the positive
108
charge (3+)

The 3
The charge is electrons that
represented by the cations
superscript. For loses is
CATION FORMATION

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aluminium represented
cation, it is 3+ here

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Using Bohr-Rutherford diagrams
3+

12 p 12 p 3 e-
12 n 12 n

The brackets indicate
that this is part of a
The outermost or larger entity. A cation
valence electrons are no usually has an anion to
longer on the outermost balance it off 109
shell electrically.

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ANOTHER WAY TO REPRESENT BONDING IS
LEWIS (DOT AND CROSS) DIAGRAMS

110

WHAT ARE LEWIS (DOT AND CROSS)
DIAGRAMS?

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 A Lewis structure is a simplified Bohr-Rutherford
diagram

 Since chemical reactions take place among the
valence and outermost electrons only
 Only these electrons are represented in the
diagram.

111

COMPARISON OF BOHR-RUTHERFORD
DIAGRAM TO LEWIS STRUCTURE

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Bohr-Rutherford diagram –
Lewis structure – Al atom
Al atom

13 p
14 n
Al
The 3 valence
112
electrons are
represented here.

COMPARISON OF BOHR-RUTHERFORD
DIAGRAM TO LEWIS STRUCTURE

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Bohr-Rutherford diagram Lewis structure
Al cation Al cation

3+ 3+

13 p
12 n
Al
No valence
electrons are
represented here all
113
have been lost to
form cation.

FOR OXYGEN

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Bohr-Rutherford diagram Lewis structure
O anion O anion
All 8 valence
2- electrons are 2-
represented here to
form anion.

13 p
12 n
O
114

ACCORDING TO THE LAW OF CONSERVATION OF
MATTER, MATTER CAN NEITHER BE CREATED NOR

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DESTROYED.

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Since Al needs to give up And since O needs to gain
3e- only 2e-

Al → Al3+ + 3e- O + 2e- → O2-

There is a need to balance the number of e- on
both sides so that the Law is not violated!

The only way to do this is to find the lowest
common multiple for the number of electrons 115
for the above.

EXCUSE ME?? WHAT IS THE LOWEST
COMMON MULTIPLE?

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 The Lowest Common Multiple (LCM) is the lowest
number that is divisible among the members of a
set of two or more numbers.
For example,
 For a set of numbers (2,3,6), 6 is the LCM as 6 is
the lowest number that can be divided by 2, 3 and 6
 For a set of numbers (2,4,6), 12 is the LCM as 12
is the lowest number that can be divided by 2, 4
and 6
 For a set of numbers (9,3,6), 18 is the LCM as 18 is
the lowest number that can be divided by 9, 3 and 6
116

SO FOR ALUMINIUM AND OXYGEN

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In order to balance this side, I would need to multiply by a
factor/number that would give me the LCM.

For this ionic half- For this ionic half-
equation, the factor would equation, the factor would
be 2 since be 3 since
2 x 3e- = 6 e- 3 x 2e- = 6 e-
Al → Al3+ + 3e- O + 2e- → O2-
x2 2Al → 2Al3+ + 6e- x3 3O + 6e- → 3O2-

The set of numbers in this case would be (3,2)
The LCM would therefore be 6 as 6 is the lowest 117
number that can be divided by both 2 and 3.

SO FOR ALUMINIUM AND OXYGEN

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Since the number of electrons is balanced on both
sides, the net effect is zero

2Al → 2Al3+ + 6e-
+
3O + 6e- → 3O2-
2Al+ 3O → 2Al3+ + 3O2-
Since it requires 3 oxygen anions to bond with 2 aluminium
cations, the ratio of Al:O is 2:3.

The chemical formulae is therefore Al2O3 118

2Al3+ + 3O2-
USING BOHR-RUTHERFORD DIAGRAMS

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2-

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Al3+ 3+
Al3+ 3+

18p
8n

13 p
13 p 14 n
14 n

O2-
2- 2-

8p
8p
8n
8n

119

O2- O2-

USING LEWIS (DOT-AND-CROSS) DIAGRAMS

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2 -

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[Al] 3+ O [Al] 3+
2- 2-

O O

2Al3+ + 3O2-

120

A SIMPLER METHOD FOR DETERMINING CHEMICAL

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FORMULAE IS BY CROSSING CHARGES

8/21/2009
To cross charges,
1. Write the symbol for first the cation and then the
anion
2. Take the number of the charge of the anion and
bring it to the bottom right hand corner of the
cation.
3. Do the same for the cation.
4. If the charges are equal, then leave the formula
as a 1:1 ratio
5. If the charges are unequal but are both even
numbers then divide by the LCM.
6. If the charges are unequal but both are uneven
121
numbers, then leave as is.

DETERMINING CHEMICAL FORMULA OF IONIC

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COMPOUNDS

8/21/2009
From the previous examples, when sodium and
chlorine react:
 one sodium atom gives 1 electron to a chlorine
atom.
 The loss of 1 electron transforms sodium atom into
sodium cation
 The gain of 1 electron from sodium atom transforms
chlorine atom into chlorine anion
 An ionic compound forms between sodium cations
and chlorine cations
 The ratio of sodium ion to chlorine ions involved in 122
ion formation is 1:1

DETERMINING CHEMICAL FORMULAE OF IONIC

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COMPOUNDS. EXAMPLE SODIUM CHLORIDE

8/21/2009
Step 1: Form cation • Na → Na+ + 1e-
Step 2: Form anion • Cl + 1e- → Cl-
Step 3: Write
chemical symbols
for cation and anion
• Na 1+ + Cl1-

Step 4: Cross
charges of anion
and cation
• Na Cl 123

IN CLASS EXERCISE

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Determine the chemical formulae of the
Answer
following:

 Magnesium and sulphur  MgS
 Potassium and nitrogen  K3N

 Aluminum and chlorine  AlCl3

 Calcium and phosphorous  Ca3P2

 Sodium and oxygen  Na2O

 Beryllium and Fluorine  BeF2

 Boron and nitrogen  BN

 Magnesium and bromine  MgBr2
124

IONIC NOMENCLATURE
125 IUPAC naming rules for ionic compounds

(International Union of Pure and Applied Chemistry)

IUPAC RULES FOR IONIC BONDING

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RULE #1:
1. The name of the metal comes first followed by the non-
metal
2. Use lowercase letters throughout

RULE #2:
The metal’s name remains unchanged

RULE #3:
1. The non-metal’s name changes to end with suffix – ide.
2. For oxygen, sulphur and nitrogen, remove “ygen”, “ur” and
“ogen” respectively and replace with “ide”
3. For all others, remove the last 3 letters and replace with
“ide”
4. It does not matter the number of non-metal ions in the 126
compound – they are not included in the name

(International Union of Pure and Applied Chemistry)


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 Using the first 3 rules stated in the previous
slide, name the ionic compounds formed assuming
that the elements undergo ionic bonding
1. Oxygen, magnesium
2. Aluminum, nitrogen magnesium oxide
3. Chlorine, sodium aluminum nitride
4. Fluorine, potassium sodium chloride
5. Sulphur, calcium potassium fluoride
calcium sulphide

127


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RULE#4:
For transition metals which can have more that one type of
charge (oxidation state or valency), the valence number
of the metal must follow the name using roman
numerals in brackets
 Name the ionic compounds formed given the following
information:
1. Sn+ , oxygen tin(I) oxide
2. Iodine, Pb+ lead (I) iodide
3. Cu2+, chlorine copper (II) chloride
4. Fluorine, Ag+ silver (I) fluoride
5. Mn7+, oxygen manganese (VII) oxide
6. Sulphur, Fe3+ iron (III) sulphide 128

COVALENT BONDING
129 Covalent bonding, polar covalent bonds, Lewis
structures

Recall that

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Metals Non-

8/21/2009
Metals
Groups 14 to
18 are Non-
Metal Non-metal metals Ionic
compound

Periods 8 and 9 are METALS
Metals

130

RECALL:
NON-METALS Groups 14 to 18 are Non-metals

What happens Non-
when elements Metals
that are non-
metals want to 14 15 16 17 18

combine?
Atoms and The Periodic Table Prepared
131 by JGL
3/30/2010

COVALENT BONDING

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 If we look at group 14 elements , there are 4
valence electrons.
 The question arises – what is the best way to
achieve 8 valence electrons?
 Should 4 electrons be lost or gained?

 In this case, the choice to achieve a stable octet is
by sharing electrons
 This is known as

Definition: A covalent bond is a bond formed between 2 atoms in
which a pair of electrons are shared so that both atoms can
132
achieve the electron configuration of the nearest noble gas.

COVALENT BONDING

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7
8 8
6
7

5 1
6
2
4 5
3
3 4

What happens is that the electron shells overlap and
the electrons are counted as if they belong to both
nuclei. 133

EXAMPLE: CARBON AND OXYGEN

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Carbon information Oxygen information

Carbon  Oxygen
Symbol C  Symbol O
Group 4  Group 16
Period 2
 Period 2
Non-Metal
 Non-metal
Z = 6
Electron configuration (2,4)
 Z=8
Nearest noble gas Ne  Electron configuration (2,6)
Electron configuration of Ne =  Nearest noble gas Ne
(2,8)  Electron configuration of Ne
Needs to share 4 electrons to = (2,8)
achieve electron configuration of  Needs to share 2 electrons to
Ne achieve electron 134
configuration of Ne


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Oxygen requires 2 e-. Oxygen requires 2 e-.

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It shares 2 e- with the carbon It shares 2 e- with the carbon atom.
atom. 1 2 5 6

35 1
3 8p
6p
8p 6n
8n 8n
2 4
46

C atom 7 8
8 7
O atom O atom
However, carbon requires 4 e-. 135
Even after it shares 2 e- with an oxygen atoms, it still requires 2 more e- in
order to achieve its stable octet.
It does so by sharing e- with another oxygen atom


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Because the ratio of C atoms to O atoms is 1:2

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the chemical formula is CO2

6p
8p 6n
8p
8n 8n

C atom

O atom O atom
The covalent bonding is represented as shown above. 136

Note that there are 4 covalent bonds because there are 4 pairs of shared electrons

USING LEWIS STRUTURES

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The
covalent
O C O bonding is
represented
here

137

USING LEWIS STRUTURES

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Instead of

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drawing the
O C O “
Note that each
“ represents a “dots” and
pair of electrons.
“crosses”, a
covalent
bond can be
OC O shown using
a“ ” 138

IN CLASS EXERCISE

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 For the pairs of  Sulphur and sulphur
elements listed, draw  Nitrogen and nitrogen
the following:  Oxygen and oxygen
 Bohr-Rutherford
 Hydrogen and hydrogen
diagram showing
 Nitrogen and hydrogen
bonding between
elements  Nitrogen and phosphorous

 Lewis structure
showing bonding
between elements
 Chemical formula
139

SULPHUR AND SULPHUR
S S

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S S 16p
16 n
16p
16 n

S2 140

NITROGEN AND NITROGEN
N N

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N N 7p
7n
7p
7n

N2 141

OXYGEN AND OXYGEN
O O

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O O

O2
8p
8p 8n
8n
142

HYDROGEN AND HYDROGEN

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H H H H
1p 1p
0n 0n

H2 143

NITROGEN AND HYDROGEN

HN H

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H 1p
0n
7p
7n
1p
0n

1p

HN H 0n

H NH3 144

NITROGEN AND PHOSPHORUS

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PN
15p
7p

NP
16 n
7n

P N 145

DATIVE OR COORDINATE COVALENT
BONDING
146

 Sometimes one of the elements involved in the
bonding will give up or share both of its electrons

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 This type of covalent bond is called a

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DATIVE OR COORDINATE
COVALENT BOND
This usually occurs with
elements that have lone pairs
of electrons 147

WHAT IS A LONE PAIR OF ELECTRONS?

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A pair of valence
electrons that is
not directly
involved in
bonding 148

HOW MANY LONE PAIRS DO THE FOLLOWING
HAVE?

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Atoms…. Ions

 Oxygen  Oxygen ion in
 Fluorine
magnesium oxide
 Fluorine in sodium
 Nitrogen
fluoride
 Phosphorous  Nitrogen in calcium
 Aluminium nitride
 Sodium  Phosphorous in sodium
phosphide
 Aluminium in
aluminium chloride 149

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SOLUTIONS
3 2

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1
3 2 2

2
O 3 1
F 3 1
N
Lewis structure for Lewis structure for Lewis structure for
oxygen atom fluorine atom nitrogen atom

2 2 1 0

1
P 1
Al Na
Lewis structure for Lewis structure for 150
Lewis structure for
Phosphorous atom fluorine atom sodium atom

SOLUTIONS
This pair of
3 electrons is not

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included as they
2- involved in
are 2+

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1
the ionic bond

2
O Mg3 This pair of
electrons is not
Lewis structure for
magnesium oxide 3 included as they
2- involved in
are 1+
1 the ionic bond

2
F Na 3
Lewis structure for 151
sodium fluoride

SOLUTIONS
This pair of
electrons is not This pair of

4 electrons is not

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included as they
Thisare involved in
2- isof
pair
2- of as they
included
This are involved in
pair

8/21/2009
electrons ionic bond4
1 the not electrons is not bond
included as they the ionic
included as they

N
are involved in

2
the ionic bond

3
N are involved in
the ionic bond

2+ 2+ 2+

Ca Ca Ca
Lewis structure for calcium nitride 152

LET US LOOK

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AT
CARBON
MONOXIDE 153


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Oxygen requires 2 e-.

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However, carbon It shares 2 e- with the carbon atom.
requires 4 e-. 5 6

Even after it shares 2
e- with an oxygen
atoms, it still requires 2 1
more e- in order to 3 8p
6p
achieve its stable octet. 6n 8n
2 4
It does so by sharing a 7
lone pair of e- with the 8
same oxygen atom i.e.
oxygen forms a C atom
dative bond with
O atom
carbon
154

Note that each
“ “ represents
CARBON MONOXIDE
CO

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a pair of electrons.

8/21/2009
OC
8p 6p
8n 6n

Because there are three (3)
covalent bonds, carbon

OC monoxide is said to have a
triple bond
155

COVALENT NOMENCLATURE
156

IUPAC RULES

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Rule 1:
 The more electronegative element (that is – the
element that is closest in atomic number to
Fluorine) is written last.
 This element is given the suffix – ide.

Rule 2
 When there is only one atom in the molecule, it
remains as the element name

157

IUPAC RULES 1 mono

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2 di

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Rule 3:
 When there is more that one atom for a
given element in the molecule the
3 tri
element is given the relevant prefix as
follows: 4 tetra
 2 atoms – di
 3 atoms – tri
5 penta
 4 atoms – tetra 6 hexa
 5 atoms – penta
 6 atoms – hexa 7 hepta
 7 atoms – hepta
 8 atoms – octa
8 octa
 9 atoms – nona 9 nona
 10 atoms - deca 158

10 deca

IN CLASS EXERCISE

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Name the following
Solutions
covalent compounds

1. ICl7  Iodine heptachloride
2. N2O5  Dinitrogen pentaoxide
3. P2O3  Diphosphorous trioxide
4. SO2  Sulphur dioxide
5. SO3  Sulphur trioxide
 Nitrogen dichloride
6. NO2
 Hydrogen chloride
7. HCl
 Hydrogen iodide
8. HI
 Tetraphosphorous
9. P4O10 decoxide 159

ATOMIC RADIUS
160 Concepts – electron-shell shielding, electron-electron
repulsion, effective nuclear charge

WHAT IS ATOMIC RADIUS?
This is defined
Nucleusas the half the
Nucleus distance

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between the
nuclei of two

8/21/2009
atoms that are
not bound to
the same
molecule

Why not
measure the
radius of a
Atomic single atom?

radius
Because an
atom is so
small, it is
Edge of really hard to
Edge of
electron measure a
electron
cloud single atom. 161
cloud

FACTORS AFFECTING ATOMIC RADIUS

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Number
of
electrons

Number
Number
of
of
electron
protons
shells 162

EXPLAINING INCREASING ATOMIC RADIUS

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Less
More More
attraction Larger
shielding of electron-
High between Reduced electron
valence electron
number of protons of effective cloud i.e.
(outermost) repulsion
electron nucleus nuclear Larger
electrons among
shells and charge atomic
by inner inner
valence radius
electrons electrons
electrons.

163

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EXPLAINING DECREASING ATOMIC RADIUS

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Higher Less
Smaller
attraction shielding of
electron Increased
between valence Low number
cloud i.e. effective
protons of (outermost) of electron
Smaller nuclear
nucleus and electrons by shells
atomic charge
valence inner
radius
electrons. electrons

164

Atomic radius increases

ATOMIC RADII IN PERIODIC TABLE
down a group.

Increasing number of shells and
For every consecutive

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period, there is an increase
Increasing atomic radius

in the number of electron

8/21/2009
shells and inner electrons.

inner electrons
This means that there is
increased shielding of
valence electrons from
nucleus by inner electrons.

There is also increased
electron-electron repulsion
among inner electrons.

This reduces the effective
nuclear charge (that is the
attraction between the
positive nucleus and the
valence electrons)

Therefore , the atomic
radius increases.
Source: www.camsoft.co.kr/.../VFI_Atomic_Radii_sm.jpg

165

Atomic radius decreases

ATOMIC RADII IN PERIODIC TABLE
across a period

For every consecutive

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element within a given
period, there is an

8/21/2009
increase in the number of
protons and inner
electrons.

Although the number of
Decreasing atomic radius electrons is equivalent tl
the number of proton, the
distance between the
electron shell and the
nucleus remains the
Increasing number of protons with same.
same number of shells This means that there is
decreased shielding of
valence electrons from
nucleus by inner
electrons.

This results in a
Source: www.camsoft.co.kr/.../VFI_Atomic_Radii_sm.jpg contracting of the electron
cloud,, resulting in
decreasing atomic radius166

ELECTRONEGATIVITY
167 How atomic radius affects electronegativity

The more
protons in the
DEFINING ELECTRONEGATIVITY

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nucleus

8/21/2009
 Electronegativity is the
The smaller the atomic to
ability of an atom
radius
attract electrons to And the fewer
itself. electron shells
between the
 Dependent on atomic nucleus and
radius and effective the valence
nuclear charge (outermost)
electrons
 Developed by Linus
Pauling (Nobel Prize is the
The effective nuclear charge
protons’ ability to attract valence And the greater the
winner (outermost) electrons) nucleus’ ability to attract
electrons to itself
i.e. The higher the
168
electronegativity

ATOMIC RADII AND ELECTRONEGATIVITY
Fluorine (F) has the smallest
atomic radius and therefore the

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highest electronegativity

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Francium (Fr) has the
largest atomic radius and
therefore the lowest
electronegativity

169

Source: www.camsoft.co.kr/.../VFI_Atomic_Radii_sm.jpg

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The most

8/21/2009
electronegative
element is
fluorine (F)
electronegative is
francium (Fr)

Increasing
The least

electronegativity

170

INTERMOLECULAR AND
INTRAMOLECULAR FORCES
171 Ionic bond, covalent bond, dipole-dipole
interactions, ion-dipole, Wan der Waals, hydrogen
bonding

DIFFERENCES BETWEEN

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INTRA-molecular forces INTER-molecular forces

 “Intra” means “internal”  “Inter” means
 Think “intra-venous”
“between”
 Think “inter collegiate”
(IV) which means
“inside the vein” which means “between
colleges”
 Intramolecular forces
 Intermolecular forces
are those bonds inside are those bonds
the molecule or between one molecule
compound or ion and another
172

TYPES OF

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INTRA-molecular forces INTER-molecular forces

 Covalent bonds  Covalent bonds
 Ionic bonds  Ionic bonds

 Van Der Waals

 Dipole-dipole
interactions
 Hydrogen bonding

 Ion-dipole interactions

173

INTERMOLECULAR FORCE AND STATES OF

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MATTER

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Solid Liquid

For a compound to exist For a compound to exist
in the solid state at in the liquid state at
room temperature and room temperature and
pressure, pressure,
 the intermolecular
 the intermolecular
forces must be fairly
forces must be strong strong
 at the same time the  at the same time the
molecules/ions have molecules/ions have
low kinetic energy. medium kinetic energy.
174

COMPARATIVE STRENGTH OF INTERMOLECULAR

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FORCES

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Ionic Bond 300-600 kJ/mol
Decreasing bond strength

Covalent 200-400 kJ/mol

Hydrogen Bonding 20-40 kJ/mol

Ion-Dipole 10-20 kJ/mol

Dipole-Dipole 1-5 kJ/mol

Instantaneous Dipole/Induced Dipole 0.05-2 kJ/mol 175

VAN DER WAALS FORCES
176 Also known as London dispersion forces


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 Weakest of all  One instantaneous dipole
intermolecular forces. can induce another
 It is possible for two instantaneous dipole in an
adjacent neutral molecules adjacent molecule (or
to affect each other. atom).
 The nucleus of one  The forces between
molecule (or atom) attracts instantaneous dipoles are
the electrons of the called Van Der Waals
adjacent molecule (or forces.
atom).  Polarizability is the ease
 For an instant, the electron with which an electron
clouds become distorted. cloud can be deformed.
 In that instant a dipole is  The larger the molecule
formed (called an (the greater the number of
instantaneous dipole). electrons) the more
polarizable 177


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The other side
This results in a slight dipole in is slightly more

8/21/2009
the atom/molecule. negative in
However, as electrons are in comparison.
constant motion, this deipole is
instantaneuos/

As these electrons
are drawn nearer to
the adjacent atom’s This makes this side
nucleus, the electrons of the atom more
The protons from this
from the adjacent positive than the next
nucleus are attracted to 178
atom are repelled side
the electrons from the
adjacent atom slightly

Source: http://universe-review.ca/I12-14-vanderwaals2.gif

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179

Atomic number increases down a At RTP, Fluorine is a gas (F2(g))

Increasing density
group.

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The mass of the atom must At RTP, Chlorine is a gas (Cl2(g))
therefore increase down a group.

8/21/2009
Van Der Waals forces must also
increase down the group as
there are more temporary At RTP, Bromine is a liquid (Br2(l))
electrostatic attractions because
there are more electrons
the atoms are heavier and more At RTP, Iodine is a solid (I2(s))
likely to come in contact with each
other

180

DIPOLE-DIPOLE INTERACTIONS
181

WHAT IS A DIPOLE? A dipole- dipole

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 “di” = “2” interaction is an
 Batteries have a
positive pole and a
electrostatic
negative pole signified attraction and/or
by a “+” sign and a “-”
sign.
repulsion
 This is an example of a between 2
dipole.
dipoles within
molecules
182

DIPOLE-DIPOLE INTERACTIONS

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Occur when Dipole-dipole interactions

 The elements within a
molecule have very
different
electronegativities
 The atomic radii of the
elements are small

183
Source:
http://www.chem.unsw.edu.au/coursenotes/CHEM1/nonunipass/
HainesIMF/images/dipoledipole.jpg

The greek This means that the Cl Chlorine is closer
symbol ∂ atom’s side of the to fluorine than
(pronounced

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molecule is slightly more hydrogen is. It is
delta) means negative than the H therefore the more
“slightly”

8/21/2009
atom’s side of the electronegative
molecule element

The attraction
between the slightly
positive end of one
molecule and the
slightly negative end The pair of
of an adjacent This means that the H
atom’s side of the electrons within the
molecule is known as covalent bond is
a dipole-dipole molecule is more
positive than the Cl more attracted to
interaction the Cl atom then
atom’s side of the 184
atom the H atom

HYDROGEN BONDING
185

WHAT IS A HYDROGEN BOND?

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 This is an extreme form of dipole-dipole interactions
 It occurs in molecules where H atoms (hence the
term “hydrogen bond”) are bonded to a very
electronegative element (N, O and any of the
halogens – F, Cl, Br, I)

186

HOW IS A HYDROGEN BOND FORMED?

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The pair of e- in the covalent

8/21/2009
bond between H and O atom
This accounts
is drawn towards the more
elctronegative element O
for the high Let us use
bond strength
H only has 1 e-.
of athis e- is drawn
If hydrogen
water as
towards the O
bondonly 1 p is left.
atom, as
This makes this side
an
compared with
of the molecule highly
positive.
other dipole- example
dipole
187
interactions. Source:
http://www.ccs.k12.in.us/chsBS/kons/kons/wonderful_world_of_fi
les/image006.gif

ION-DIPOLE INTERACTIONS
188 Why ions dissolve in polar solvents

IONS
WHAT ARE ….

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POLAR SOLVENTS

This is a
SOLVENT
where the
molecules of
In every ionic the solvent
compound, there have a
are dipole.
negative ions
(anions)
and
positive ions 189
(cations)

WHEN AN IONIC COMPOUND IS ADDED TO A
POLAR SOLVENT……..

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Note that this is a physical and not a

8/21/2009
chemical change. The ions remain as
ions and the molecules of the polar
solvent have not changed. They are only
separated a little more than previously.

The molecules of the water However,salty –
This is why salt tastes it requires
polar solvent are salt is still there. It particles in
because the
many of the
is just
attracted to the ions to surround each ion
a different physical state than before.
to separate them and
overcome the forces of 190

attraction.

INTERMOLECULAR FORCE AND
PHYSICAL PROPERTIES
191 How the type of intermolecular force within a
compound determines that compound’s physical
properties

RECALL : COMPARATIVE STRENGTH OF

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INTERMOLECULAR FORCES

8/21/2009
Ionic Bond 300-600 kJ/mol
Decreasing bond strength

Covalent 200-400 kJ/mol

Hydrogen Bonding 20-40 kJ/mol

Ion-Dipole 10-20 kJ/mol

Dipole-Dipole 1-5 kJ/mol

Instantaneous Dipole/Induced Dipole 0.05-2 kJ/mol 192

THE STRONGER THE INTERMOLECULAR

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FORCE, THE MORE DENSE THAT STATE OF MATTER

8/21/2009
Type of Bond Most Examples
intermolecular strength likely
forces (kJ/mol) state of
matter
Van Der Waals 0.05-2.00 Gas (g) Neon Ne(g)
Oxygen O2(g)
Nitrogen N2(g)
Dipole-dipole 1.00-5.00 Gas (g); Hydrogen chloride HCl (l)
interactions Liquid (l)
Hydrogen 20-40 Liquid (l) Water H2O (l)
bonding
Covalent 200-400 Solid Diamond C(s)

Ionic 300-600 Solid Sodium chloride NaCl(s)
193

THE STRONGER THE INTERMOLECULAR

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FORCE, THE HIGHER THE MELTING/BOILING POINT

8/21/2009
The higher Bond strength of the bond, point
Type of
intermolecular
the
strength
Most likely state of
matter
Melting the
Boiling
point/

greater the (kJ/mol) (in the form of heat
forces energy
energy) required to separate theLow
Van Der Waals 0.05-2.00 Gas (g)
particles in a given physical state of
Increasing bond strength

Dipole-dipole 1.00-5.00 Gas (g); Liquid (l) Low
matter.
interactions
Hydrogen 20-40 Liquid (l) Medium
bonding
This means that the temperatures at
Covalent 200-400 Solid Very high
which solid turns to liquid (melting) and
Ionic 300-600 Solid Very high
liquid to gas (boiling) would be much
194
higher.

PROPERTIES OF IONIC COMPOUNDS
195

IONIC COMPOUNDS ….

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 Are solid at room temperature and pressure (RTP)
 Have high melting and boiling points

 Have crystalline structures

 Conduct electricity in the molten state or aqueous
state but not in the solid state
 Are soluble in polar solvents

 Are insoluble in organic or non-polar solvents

196

REVIEW: STATES OF MATTER

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 The state of matter of a  Solids are denser than
given substance is liquids
dependent upon the  Liquids are denser than
energy of the particles gases
as well as the strength  This is due to how
of the attraction
increasingly stronger
between the particles. the attraction between
 The stronger the the particles are as
attraction, the closer state changes from gas
the particles are and to liquid to solid
the denser the state of
matter 197

IONIC COMPOUNDS ARE SOLID ARE RTP

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 The ionic bond is very
strong.
 Every cation is ionically
bonded to an anion and
vice versa
 Please note that the
overall ratio of ions
remains the same
 The overall attraction
between the millions of
cations and anions
creates a fairly dense This is why ……
and strong structure
198

NEW CONCEPT – LATENT HEAT

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 To move from one state to
the next, enough energy
needs to be applied to the
substance to increase the
kinetic energy of the
particles to overcome the
forces of attraction to
change state.
 The energy required to
overcome these forces of
attraction to finally allow the
particles of a substance to
fully transfer from one state
to another is known as the
LATENT HEAT.
199

IONIC COMPOUNDS HAVE HIGH MELTING AND

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BOILING POINTS

8/21/2009
 Because the ionic bond is very strong.
 And the overall attraction between the millions of
cations and anions creates a fairly dense and
strong structure
 Quite a bit of energy is required to move from solid
phase to liquid phase
 Ionic compounds therefore have high latent heats.

This is why….
200

IONIC COMPOUNDS HAVE CRYSTALLINE

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STRUCTURES
This is why….

8/21/2009
 The ionic bond
formed between
cations and
anions form an
orderly regular
pattern and fairly
rigid angular
structure
 This allows light to
filter through and A crystal or crystalline solid is a solid
be reflected material, whose constituent atoms,
molecules, or ions are arranged in an
201
orderly repeating pattern extending in all
three spatial dimensions.

IONIC COMPOUNDS DISSOLVE IN POLAR

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SOLVENTS

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This is
why….

 Polar solvents are Because the ionic compounds have
those in which the negative particles (anions) and positive
particles have a particles (cations), they are attracted to
slightly positive the polar solvents.
and a slightly
However, it requires many of the particles
negative charge 202
at surround each ion to separate them
and overcome the forces of attraction.

IONIC COMPOUNDS CONDUCT ELECTRICITY IN THE

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MOLTEN STATE AND AQUEOUS STATE
This is why….

8/21/2009
 Once enough latent heat is
applied to transform an
ionic substance from the
solid to liquid states (i.e. the
substance melts), the ions
are free to move
 This also applies when it is
in the aqueous state

If a positive pole (ANODE) is placed in the ionic liquid, the anions will be
attracted to the anode.
If a negative pole (CATHODE) is placed in the ionic liquid, the cations will
be attracted to the cathode.
203
The movement of the charged particles in a complete circuit allows molten
or aqueous ionic compounds to conduct electricity

STRUCTURE OF SODIUM CHLORIDE (NACL)

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Sodium chloride (NaCl) or table salt has a 6:6 crystalline structure. This
means that every Na+ is in contact with 6 Cl- and that every Cl- is in
contact with 6 Na+.

204

PROPERTIES OF COVALENT
COMPOUNDS
205

COVALENT COMPOUNDS USUALLY….

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 Are liquid or gases at room temperature and pressure
(RTP)
 Have low melting and boiling points

 Have non-crystalline structures

 Do not conduct electricity either in the molten state or
aqueous state
 Are insoluble in polar solvents

 Are soluble in organic or non-polar solvents

206

REVIEW: STATES OF MATTER

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 The state of matter of a  Solids are denser than
given substance is liquids
dependent upon the  Liquids are denser than
energy of the particles gases
as well as the strength  This is due to how
of the attraction
increasingly stronger
between the particles. the attraction between
 The stronger the the particles are as
attraction, the closer state changes from gas
the particles are and to liquid to solid
the denser the state of
matter 207

THE MAIN INTERMOLECULAR FORCES IN MANY

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COVALENT COMPOUNDS ARE
This occurs when

8/21/2009
Dipole-Dipole interactions
molecules are
Van Der Waals (London dispersion)

polar. The state of
matter would
mostly be liquid at
RTP.
This occurs when the
molecules are non- Examples include:
forces

polar . The state of H2O(l)
matter would mostly HCl(l)

be gas at RTP.

Examples include:
 Ne(g)
 O2(g)
208
 N2(g)

A hexagonal
ring of 6 carbon Each carbon is covalently bonded to 3 others.
E

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atoms is
XCEPTIONS TO THIS INCLUDE
formed.

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Each hexagonal Graphite is a solid at RTP
Graphite ring is joined to
because
form a sheet
3 1
5
6
4 3
1 2 2 Covalently
bonded
sheets of
carbon atoms
Graphite is made up of carbon increase
atoms .
Each sheet is loosely held density
209
by weak Van der Waa;ls
forces

A hexagonal
ring of 6 carbon Each carbon is covalently bonded to 4 others.
E

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atoms is
XCEPTIONS TO THIS INCLUDE
formed.

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Each hexagonal Graphite is a solid at RTP
Diamond ring is joined to
because
form a sheet
1
1

4 2
Covalently
bonded
3
sheets of
carbon atoms
increase
Each sheet is loosely held density
Diamond is made up of carbon
210
atoms . by weak Van der Waa;ls
forces


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Dipole-Dipole
interactions

Van Der Waals
(London dispersion)
forces

This means that the melting
points and boiling points
211
would be very low

EXCEPTIONS TO THIS INCLUDE

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Graphite Diamond

212


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Van Der Waals
Dipole-Dipole
(London dispersion) interactions
forces
A crystal or crystalline solid is a solid material, whose constituent
atoms, molecules, or ions are arranged in an orderly repeating pattern
extending in all three spatial dimensions

Van Der Waals forces and dipole-dipole interactions are
unstructured and fairly random.
There is therefore no ordered arrangement of the
molecules.
Therefore most covalent compounds have no crystal
213
structure.

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214

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Graphite

EXCEPTIONS TO THIS INCLUDE - GRAPHITE

EXCEPTIONS TO THIS INCLUDE

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Graphite Diamond

215

COVALENT COMPOUNDS HAVE MOLECULES THAT ARE
EITHER

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Polar

Non-polar

This means that there are no
free electrons or free positively
charged and/or free negatively
charged particles to conduct
electricity.
Therefore most covalent compounds do not conduct 216
electricity

There are no permanent dipoles to have electrostatic
attractions with polar solvents

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In covalent compounds with non- Polar

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polar molecules

Therefore most covalent compounds are
insoluble in polar solvents
217

In covalent
compounds with
There are Polar
polar molecules permanent dipoles

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which have

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electrostatic
attractions with
polar solvents

Examples include :
1. HX where X =F, Cl, Br or I
2. H2O (water)
3. NH3 (ammonia)

Therefore covalent compounds with highly polar 218

molecules are soluble in polar solvents

METALLIC BONDING
219

PROPERTIES OF METALLIC
COMPOUNDS
220

ALLOTROPES
221

THERMODYNAMICS OF ION
FORMATION
222 Ionization energy, electron Affinity, Lattice energy

BUT WHY DO IONS FORM?

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8/21/2009
 To understand why this Suppose you were very
happens, we need to hungry, but you had to
understand how energy walk to the end of the
works. school to get lunch.
Would you do it and why?
Similarly, element
 Yes, because I can fill my
s only bond or need (hunger) despite
combine with the walk required.
each other if there Would you walk 10 miles if
you did not have to?
is a benefit in  No....
terms of energy
223

FOR IONS TO FORM

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CATION ANION

 Electron(s) is/are  Electron(s) is/are
removed from valence added to valence
electron shell. electron shell
 This requires ENERGY  This releases ENERGY

Bond formation only takes place when the reaction is
energetically favourable.

This usually means that there is energy released (denoted
by a negative or minus(-) sign) at the end of the reaction.224

IONIZATION ENERGY AND ELECTRON AFFINITY

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Ionization energy (IE) Electron affinity (EA)

This is defined as This is defined as
The energy required to The energy released
remove an electron when an electron is
from an atom in its added to the valence or
gaseous state from its outermost electron
valence or outermost shell of an atom in its
electron shell . gaseous state.

For a given atom X, For a given atom X,
X(g) + IE → X+(g) + e- X(g) + e- → X-(g) + EA 225

LATTICE ENERGY

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8/21/2009
This is defined as

The energy released when a cation and anion bond
via an ionic bond is known as the LATTICE ENERGY
(LE)

For two ions, X+ and Y-, that combine to form an ionic
solid XY

X+(g) + Y-(g) → XY(s) + LE 226

EXAMPLE: FORMATION OF SODIUM CHLORIDE

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8/21/2009
To form sodium chloride

Sodium loses one electron
Na → Na+ + 1 e-

Chlorine gains one electron
Cl + 1 e- → Cl-

An ionic bond forms between sodium and chlorine to
form sodium chloride
227
Na+ + Cl- → NaCl

EXAMPLE: FORMATION OF SODIUM CHLORIDE -
RELATED ENERGY CHANGES

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Na → Na+ + 1 e- IE = + 495.8 kJ

8/21/2009
Cl + 1 e- → Cl- EA = - 352.4 kJ

IE + EA =+143.4 kJ

This is not energetically favourable. However, the LE
released compensates for this energy

Na+ + Cl- → NaCl LE = -787.3 kJ

TOTAL ENERGY = IE + EA + LE
= +143.4 – 787.3 kJ 228
= -643.9 kJ

IONIZATION ENERGY
229

FACTORS AFFECTING IONIZATION ENERGY

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8/21/2009
 Ionization Energy (IE) is the energy required to
remove an electron from an atom in its gaseous
state.
 Therefore anything that makes it harder to remove
an electron from the valence shell, will increase IE
 Conversely, anything that makes it easier to remove
an electron from the valence shell, will decrease IE
 Therefore there is need to have a low or small IE to
form a positive or cation.

230

BASKETBALL AND
If your opponent holds on Think if
tightly to the ball, it will
require more energy and you had to

Prepared by JGL
effort to take away the ball.
take away

8/21/2009
the ball
Similarly, the higher from

IONIZATION ENERGY
the IE, the harder to someone
remove the valence
electrons. during a
basketball
If he/she holds on loosely to
the ball, it will be easier and game.
require less energy to
remove the ball from his/her
hands.

Similarly, the lower the IE
, the easier it is to
remove the ball. 231

WHAT FACTORS INCREASE IONIZATION
ENERGY?

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8/21/2009
Atomic radius

Electro- Effective
nuclear
negativity Factors charge
affecting
Ionization
Energy

Electron
Number of
configuration valence
electrons 232

EFFECTIVE NUCLEAR CHARGE AND IONIZATION

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ENERGY The effective nuclear
charge is the protons’

8/21/2009
ability to attract valence
Atomic radius (outermost) electrons)

Electro- Effective
nuclear
affecting
Ionization The greater the
Energy effective nuclear
charge, the more
closely the
Number of valence electrons
Electron are held by the
electrons nucleus and the233
higher the IE

Atomic radius is
defined as the half
A

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the distance
TOMIC RADIUS AND IONIZATION ENERGY
between the nuclei

8/21/2009
of two atoms that
are not bound to
the same molecule Atomic radius The smaller the
atomic radius, the
closer the valence
electrons are to
Electro- Effective the nucleus and
nuclear the greater
negativity Factors charge attraction between
affecting
the two. It is would
Ionization therefore be
Energy harder to remove
the valence
electrons. This
Electron
Number of increases IE
electrons 234

ELECTRONEGATIVITY AND IONIZATION the

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The higher ENERGY
electronegativity the greater

8/21/2009
Electronegativity is the attraction of the electrons
the ability of an atom to the nucleus. This makes
Atomic radius
to attract electrons to removal of electrons more
itself. difficult and increases IE

Electro- Effective
nuclear
affecting
Ionization
Energy

Electron
Number of
electrons 235

ELECTRONEGATIVITY AND IONIZATION ENERGY

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8/21/2009
The closer the electron
configuration initially to that of
Atomic radius
the nearest noble gas, the
more likely the ion will form
and therefore the lower the IE

Electron Electro- Effective
nuclear
configuration is the
arrangement affecting
of electrons in Ionization
an atom, molecule or Energy
other body.

Electron
Number of
electrons 236

ELECTRONEGATIVITY AND IONIZATION ENERGY

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The more valence electrons, the

8/21/2009
more electrons will be required to be
removed in order to get the electron
Atomic radius
configuration of the nearest noble
gas configuration.

This means that it would require
Electro- Effective
more energy to remove all the
nuclear
negativity Factors and therefore the IE would
electrons
charge
increase.
affecting
The valence electrons are Ionization
the electrons in the Energy
outermost shell

Electron
Number of
electrons 237

Bonding - ionic covalent & metallic

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