Bonding and structure - ionic compounds, covalent compounds and metals. Relationship between intermolecular forces and physical properties. Allotropes.

1. STRUCTURE AND BONDING
Prepared by Janadi Gonzalez-Lord

2. 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

3. Prepared by JGL 8/21/2009
SYLLABUS REQUIREMENTS
3 BONDING

4. SYLLABUS REQUIREMENTS – IONIC BONDING
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8/21/2009
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

5. SYLLABUS REQUIREMENTS - BONDING
Prepared by JGL
8/21/2009
The students should be able to :
 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
5
covalent

6. SYLLABUS REQUIREMENTS - BONDING
Prepared by JGL
8/21/2009
The students should be able to :
•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

7. Prepared by JGL 8/21/2009
SYLLABUS REQUIREMENTS
7 STRUCTURE

8. SYLLABUS REQUIREMENTS - STRUCTURE
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8/21/2009
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

9. Prepared by JGL 8/21/2009
REVIEW: ATOMS AND THE PERIODIC
TABLE
9

10. Prepared by JGL
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10

11. 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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12. 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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13. 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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14. Prepared by JGL 8/21/2009
ELEMENTS AND BONDING
14 Introduction to ionic, covalent and metallic bonding

15. 3/30/2010
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PURE AND IMPURE SUBSTANCES
15 A review

16. 3/30/2010
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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
16
IMPURE substances or MIXTURES can be sub-divided into HOMOGENOUS and
HETEROGENOUS

17. 3/30/2010
Can be separated into
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Can be
Can be separated
separated into
into
17
Source: www.mghs.sa.edu.au/Internet/Faculties/Science/Year10/Pics/elementsAndCompounds.gif

18. 3/30/2010
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

19. 19
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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

20. 20
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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

21. 21
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EXAMPLES OF ELEMENTS AND COMPOUNDS
Elements
Compounds
Source: www.physicalgeography.net/fundamentals/images/compounds_molecules.jpg

22. 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
22

23. 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
23
(in Group 8)

24. 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

25. Prepared by JGL
YOU MAY WELL
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BE ASKING
WHAT DOES
THIS MEAN? 25

26. Prepared by JGL
LET’S TAKE A LOOK
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AT SOME EXAMPLES
TO UNDERSTAND
THIS CONCEPT MORE
FULLY 26

27. 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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28. 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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29. Prepared by JGL
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SODIUM AND
CHLORINE UNDERGO
WHAT IS KNOWN AS
IONIC BONDING.
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30. 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
30

31. 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

32. 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
difficult to lose or gain four electrons. Prepared by JGL
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33. 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.
33

34. 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

35. 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

36. IN SUMMARY……
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Ionic
bonding
3 types
of
bonding
Metallic Covalent
bonding bonding
36

37. 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

38. Prepared by JGL 8/21/2009
REPRESENTING IONS AND
MOLECULES
38 Ionic notation, chemical formulae

39. 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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40. 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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41. 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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42. 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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43. 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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44. Prepared by JGL 8/21/2009
ION FORMATION
44

45. 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

46. 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
46

47. 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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48. LET’S CONTRAST – NA & NE
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Sodium Neon
 Element symbol Na  Element symbol Ne
 Group 1  Group 18
 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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49. 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

50. 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
50

51. Prepared by JGL 8/21/2009
CATION FORMATION
51

52. 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
52

53. 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

54. 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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55. 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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56. 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).
 Since the number of
protons remains the It therefore becomes a
same (p=19) but the cation with an electrical
number of electrons charge of +1
change (e- = 18), it is
no longer electrically
neutral
56

57. 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

58. Group 2 elements have 2 This is the same electron
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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59. CATION FORMED
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 If Mg loses 2 p = +12
electrons, its electron e- = - 10
configuration moves Charge = +2
from (2,8,2) to (2,8).
 Since the number of
protons remains the It therefore becomes a
same (p=12) but the cation with an electrical
number of electrons charge of +2
change (e- = 10), it is
no longer electrically
neutral
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60. CATION FORMED
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 If Ca loses 2 p = +20
electrons, its electron e- = - 18
configuration moves Charge = +2
from (2,8,8,2) to
(2,8,8).
 Since the number of
It therefore becomes a
protons remains the cation with an electrical
same (p=20) but the charge of +2
number of electrons
change (e- = 18), it is
no longer electrically
neutral 60

61. 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

62. Group 13 elements have 3 This is the same electron
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
Syllabus requirements met:
Recognize the tendency for loss of electrons based on the electronic configuration
or the position in the periodic table ( for metals) Prepared by JGL
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63. 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).
 Since the number of
protons remains the It therefore becomes a
same (p=5) but the cation with an electrical
number of electrons charge of +3
change (e- = 2), it is no
longer electrically
neutral
63

64. CATION FORMED
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 If Al loses 3 p = +13
electrons, its electron e- = - 10
configuration moves Charge = +3
from (2,8,3) to (2,8).
 Since the number of
protons remains the It therefore becomes a
same (p=13) but the cation with an electrical
number of electrons charge of +3
change (e- = 10), it is
no longer electrically
neutral
64

65. 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

66. 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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67. 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-
67
X = +3

68. 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

69. SO FOR GROUP 2 ELEMENTS
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Statement Ionic half-equation
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+
69

70. SO FOR GROUP 13 ELEMENTS
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Statement Ionic half-equation
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-
70

71. 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

72. 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-

73. 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

74. Prepared by JGL 8/21/2009
ANION FORMATION
74

75. 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
Prepared by JGL
75 8/21/2009

76. This is the same electron
Group 16 elements have 6
configuration as Ne, the noble
valence (outermost) electrons
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-
becomes (2,8) ELECTRONIC
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
Prepared by JGL
76 8/21/2009

77. This is the same electron
Group 17 elements have 7
configuration as Ne, the noble
valence (outermost) electrons
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-
becomes (2,8) ELECTRONIC
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
77 8/21/2009

78. RECALL:
NON-METALS Groups 14 to 18 are Non-metals
Non-
Metals
14 15 16 17 18
Atoms and The Periodic Table Prepared
78 by JGL
3/30/2010

79. IN GENERAL, NON-METALS GAIN
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ELECTRONS TO FORM NEGATIVE IONS
8/21/2009
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

80. ANION FORMATION
Prepared by JGL
8/21/2009
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

81. IF WE THINK OF IT LIKE AN EQUATION,
Prepared by JGL
8/21/2009
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

82. SUMMARY – ANION FORMATION
Prepared by JGL
8/21/2009
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

83. Prepared by JGL 8/21/2009
IONIC BOND FORMATION
83

84. RECALL...
Prepared by JGL
8/21/2009
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

85. HOW TO RECOGNIZE AN IONIC COMPOUND
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8/21/2009
Ionic
Metal Non-metal
compound
85

86. Prepared by JGL
Metals Non-
8/21/2009
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

87. Prepared by JGL
8/21/2009
LET US EXAMINE THE BONDING BETWEEN
MAGNESIUM AND FLUORINE
87

88. Fluorine
Magnesium
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Metals Non-
Noble Gases Group 8
8/21/2009
Group 1 to 13 are METALS
Metals
Neon
88

89. USING THE PERIODIC TABLE...
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8/21/2009
Magnesium Flourine
 Group 2  Group 17
 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

90. FORMATION OF IONIC BOND
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8/21/2009
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

91. THE ONLY WAY THIS CAN HAPPEN….
Prepared by JGL
8/21/2009
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

92. IN TERMS OF IONIC EQUATIONS
Prepared by JGL
8/21/2009
Cation equation Anion equation
Mg → Mg2+ + 2e- F + 1e- → F-
F + 1e- → F-
2F + 2e- → 2F-
92

93. ADDING THE 2 EQUATIONS
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8/21/2009
Total ionic equation Ionic equation
Mg → Mg2+ + 2e- Mg2+ + 2F- → MgF2
2F + 2e- → 2F-
Mg + 2F → Mg2+ + 2F-
93

94. THEREFORE THE CHEMICAL FORMULAE
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8/21/2009
FOR IONIC COMPOUND FORMED BETWEEN
MAGNESIUM AND FLUORINE IS
MGF2
A chemical formulae
expresses the ratio of
atoms of elements within a
compound.
94

95. IONIC BOND
Prepared by JGL
FORMATION PROCESS
8/21/2009
CATION formation IONIC Compound
formation
Mg → Mg2+ + 2e- Mg2+ + 2F- → MgF2
ANION formation Electrostatic
2F + 2e- → 2F- attraction
95

96. IONIC BONDING EXERCISES
Prepared by JGL
8/21/2009
 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

97. 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:
Prepared by JGL
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)

98. 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
8/21/2009
(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)

99. IONIC BONDING BETWEEN SODIUM AND FLUORINE –
STEP 3: IONIC BOND FORMATION
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8/21/2009
 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.

100. IN SUMMARY, FOR SODIUM AND FLUORINE
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Na Na+ + 1 e- F + 1 e- F-
8/21/2009
9p
10 n
11 p
12 n
1-
1+
9p
11 p 10 n
12 n
100
Na+ + F- NaF

101. SIMILARLY, FOR MAGNESIUM AND OXYGEN,
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Mg Mg2+ + 2 e- O + 2 e- O2-
8/21/2009
8p
8n
12 p
12 n
1-
1+
9p
11 p 10 n
12 n
101
Mg2+ + O2- MgO

102. 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

103. Prepared by JGL 8/21/2009
REPRESENTING IONIC BONDING
103 Lewis structures (Dot and cross diagrams), chemical
formulae, ionic equations, ionic notation

104. 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

105. Prepared by JGL
8/21/2009
HOW DO WE REPRESENT
THIS INFORMATION MORE
SIMPLY?
105

106. Prepared by JGL
8/21/2009
An ionic equation
expresses what
happens at an
ionic level
106

107. CATION FORMATION: ALUMINIUM
Prepared by JGL
8/21/2009
In words:
An aluminium atom
loses three electrons
to form
aluminium cation 107

108. CATION FORMATION: ALUMINIUM
Prepared by JGL
Aluminium atom is neutrally
charged. This is represented
Ionic equations must
8/21/2009
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+)

109. The 3
The charge is electrons that
represented by the cations
superscript. For loses is
CATION FORMATION
Prepared by JGL
aluminium represented
cation, it is 3+ here
8/21/2009
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.

110. Prepared by JGL
8/21/2009
ANOTHER WAY TO REPRESENT BONDING IS
LEWIS (DOT AND CROSS) DIAGRAMS
110

111. WHAT ARE LEWIS (DOT AND CROSS)
DIAGRAMS?
Prepared by JGL
8/21/2009
 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

112. COMPARISON OF BOHR-RUTHERFORD
DIAGRAM TO LEWIS STRUCTURE
Prepared by JGL
8/21/2009
Bohr-Rutherford diagram –
Lewis structure – Al atom
Al atom
13 p
14 n
Al
The 3 valence
112
electrons are
represented here.

113. COMPARISON OF BOHR-RUTHERFORD
DIAGRAM TO LEWIS STRUCTURE
Prepared by JGL
8/21/2009
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.

114. FOR OXYGEN
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8/21/2009
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

115. ACCORDING TO THE LAW OF CONSERVATION OF
MATTER, MATTER CAN NEITHER BE CREATED NOR
Prepared by JGL
DESTROYED.
8/21/2009
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.

116. EXCUSE ME?? WHAT IS THE LOWEST
COMMON MULTIPLE?
Prepared by JGL
8/21/2009
 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

117. SO FOR ALUMINIUM AND OXYGEN
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8/21/2009
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.

118. SO FOR ALUMINIUM AND OXYGEN
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8/21/2009
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

119. 2Al3+ + 3O2-
USING BOHR-RUTHERFORD DIAGRAMS
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2-
8/21/2009
Al3+ 3+
Al3+ 3+
18p
8n
13 p
13 p 14 n
14 n
O2-
2- 2-
8p
8p
8n
8n
119
O2- O2-

120. USING LEWIS (DOT-AND-CROSS) DIAGRAMS
Prepared by JGL
2 -
8/21/2009
[Al] 3+ O [Al] 3+
2- 2-
O O
2Al3+ + 3O2-
120

121. 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.

122. 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

123. 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

124. IN CLASS EXERCISE
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8/21/2009
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

125. Prepared by JGL 8/21/2009
IONIC NOMENCLATURE
125 IUPAC naming rules for ionic compounds

126. (International Union of Pure and Applied Chemistry)
IUPAC RULES FOR IONIC BONDING
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8/21/2009
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

127. (International Union of Pure and Applied Chemistry)
IUPAC RULES FOR IONIC BONDING
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8/21/2009
 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

128. IUPAC RULES FOR IONIC BONDING
Prepared by JGL
8/21/2009
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

129. Prepared by JGL 8/21/2009
COVALENT BONDING
129 Covalent bonding, polar covalent bonds, Lewis
structures

130. Recall that
Prepared by JGL
Metals Non-
8/21/2009
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
130

131. 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

132. COVALENT BONDING
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8/21/2009
 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.

133. 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

134. EXAMPLE: CARBON AND OXYGEN
Prepared by JGL
8/21/2009
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

135. USING BOHR-RUTHERFORD DIAGRAMS
Prepared by JGL
Oxygen requires 2 e-. Oxygen requires 2 e-.
8/21/2009
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

136. USING BOHR-RUTHERFORD DIAGRAMS
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Because the ratio of C atoms to O atoms is 1:2
8/21/2009
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

137. USING LEWIS STRUTURES
Prepared by JGL
8/21/2009
The
covalent
O C O bonding is
represented
here
137

138. USING LEWIS STRUTURES
Prepared by JGL
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

139. IN CLASS EXERCISE
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8/21/2009
 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

140. SULPHUR AND SULPHUR
S S
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S S 16p
16 n
16p
16 n
S2 140

141. NITROGEN AND NITROGEN
N N
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N N 7p
7n
7p
7n
N2 141

142. OXYGEN AND OXYGEN
O O
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O O
O2
8p
8p 8n
8n
142

143. HYDROGEN AND HYDROGEN
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H H H H
1p 1p
0n 0n
H2 143

144. NITROGEN AND HYDROGEN
HN H
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H 1p
0n
7p
7n
1p
0n
1p
HN H 0n
H NH3 144

145. NITROGEN AND PHOSPHORUS
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8/21/2009
PN
15p
7p
NP
16 n
7n
P N 145

146. Prepared by JGL 8/21/2009
DATIVE OR COORDINATE COVALENT
BONDING
146

147.  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

148. WHAT IS A LONE PAIR OF ELECTRONS?
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A pair of valence
electrons that is
not directly
involved in
bonding 148

149. 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

150. Prepared by JGL
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