Chemical bonds are formed by the sharing or transfer of valence electrons between atoms. Valence electrons play an important role in bond formation as atoms seek to achieve stable electronic configurations like noble gases. There are two main types of bonds: 1) Ionic bonds are formed by the transfer of electrons from metals to nonmetals, resulting in positively charged cations and negatively charged anions that are attracted to each other. 2) Covalent bonds are formed by the sharing of electron pairs between nonmetals. Atoms share electrons to achieve stable octet configurations. Single, double, and triple covalent bonds are distinguished by the number of electron pairs shared. Lewis structures use dots or crosses to represent valence

1. CHAPTER 3
CHEMICAL BONDING

2.  Chemical bond is the force that
holds two atoms together in a
molecule or compound
 Valence electrons play an
important role in the formation of
chemical bonds

3. LEWIS STRUCTURE
A Lewis symbol consists of:
 the symbol of an element
 dots or cross is used to represent the
valence electrons in an atom of the
element.

4. EXAMPLE
THE LEWIS SYMBOL OF ATOM
Group 1 2 13 14 15 16 17 18
Valence
electron
1 2 3 4 5 6 7 8
Lewisdot
symbol

5.  Elements in the same group have the
same valence electronic configurations
 similar Lewis symbols.

6. OCTET RULE
 Octet rule states that atoms tend to form
bonds to obtain 8 electrons in the valence
shell
 Atoms combine to achieve stability
 to have the same electronic
configuration as a noble gas

7.  Atoms achieve noble gas configuration
through:
i) transferring electrons
ii) sharing electron
 Bond formation involve transferring or
sharing of only valence electrons

8. Electronic Configuration of
Cations and Anions
1)Noble gas configuration
Group 1, 2 and 13 elements donate valence
electrons to form cations with noble gas
configurations
Example:
Na : 1s22s22p63s1
Na+ : 1s22s22p6 (isoelectronic with Ne)
Ca : 1s22s22p63s23p64s2
Ca2+ : 1s22s22p63s23p6 (isoelectronic with Ar)

9.  Group 15, 16 and 17 elements accept electrons
to form anions with noble gas configurations
Example:
O : 1s22s22p4
O2 : 1s22s22p6
(isoelectronic with neon)
Cl : 1s22s22p63s23p5
Cl : 1s22s22p63s23p6
(isoelectronic with Ar)

10. 2) Pseudonoble gas configuration
 d block elements donate electrons from
4s orbitals to form cations with
pseudonoble gas configuration.
Example:
Zn : 1s22s22p63s23p64s23d10
Zn2+ : 1s22s22p63s23p63d10
(pseudonoble gas configuration )

11. 3) Stability of the half-filled orbitals
 d block element can also donate electrons to
achieve the stability of half-filled orbitals
Example:
Mn : 1s22s22p63s23p64s23d5
Mn2+ : 1s22s22p63s23p63d5
(stability of half-filled 3d orbital )
Fe : 1s22s22p63s23p64s23d6
Fe3+ : 1s22s22p63s23p63d5
(stability of half-filled 3d orbital)

12. FORMATION OF THE BONDS
USING LEWIS SYMBOLS
i.Ionic bond
ii.Covalent bond

13. IONIC BOND
 Ionic bond (electrovalent bond) is
an electrostatic attraction between
positively and negatively charged
ions.
 Ionic compounds are formed when
electrons are transferred between
atoms (metal to nonmetal) to give
electrically charged particles that
attract each other .

14. Example 1: NaCl
Sodium, an electropositive metal, tends to
remove its valence electron to obtain noble gas
electronic configuration (Ne)
Chlorine, an electronegative element, tend to
accept electron from Na to obtain noble gas
electronic configuration (Ar)

15.  The electrostatic forces between Na+ and
Cl- produce ionic bond
 These two processes occur simultaneously
+

16. Example 2: CaCl2
Ca: 1s2 2s2 2p6 3s2 3p6 4s2
(Has two electrons in its outer shell)
Cl: 1s2 2s2 2p6 3s2 3p5
(Has seven outer electrons)

17. Calcium Chloride
If Ca atom transfer 2 electrons, one to each
chlorine atom, it become a Ca2+ ion with the
stable configuration of noble gas.
At the same time each chlorine atom to achieve
noble gas configuration gained one electron
becomes a Cl- ion to achieve noble gas
configuration.
The electrostatic attraction formed ionic bond
between the ions.

18. IONIC BOND
(FORMED BY TRANSFER OF
ELECTRONS)
CALCIUM CHLORIDE
+ + 2

19. EXAMPLE 3: LIF
+

20. LEWIS STRUCTURE AND FORMATION OF IONIC
COMPOUNDS
+ +
1) CaCl2
2
2) MgO
+

21. 3) CaBr2
+ +

22.  Ionic bond is very strong, therefore ionic
compounds:
1. Have very high melting and boiling
points
2. Hard and brittle
3. Can conduct electricity when they
are in molten form or aqueous
solution because of the mobile ions

23. LEARNING CHECK
By using Lewis structure, show how the
ionic bond is formed in the compounds
below.
( a ) KF
( b ) BaO
( c ) Na2O

24. COVALENT BOND
Definition of covalent bond
i. Chemical bond in which two or more electrons
are shared by two atoms.
ii.The electrostatic force between the electrons
being shared the nuclei of the atoms.
Why should two atoms share electrons?
To gain stability by having noble gas
configuration (octet)

25. F F
+
7e- 7e-
F F
8e- 8e-
F F
F F
Lewis structure of F2
lone pairs
lone pairs
lone pairs
lone pairs
single covalent bond
single covalent bond
Example

26. COVALENT COMPOUNDS:
 Compounds may have these
covalent bonds:
i. Single bond
ii. Double bond
iii. Triple bond.

27. 8e-
H H
O
+ + O
H H
O H
H
or
2e- 2e-
Lewis structure of water
single covalent bonds

28. Double bond – two atoms share
two pairs of electrons
O C O or O C O
8e- 8e-
8e-
double bonds
double bonds

29. +
Triple bond – two atoms share
three pairs of electrons
N N
8e-
8e-
N N
or
triple bond
triple bond

30. LEWIS STRUCTURE OF
COVALENT SPECIES
i.Single bond
i.Double bond
i.Triple bond

31. STEPS IN WRITING LEWIS STRUCTURES
1. Count total number of valence e- of atoms involved.
2. Add 1 for each negative charge. Subtract 1 for each
positive charge.
3. Draw skeletal structure of the compound. Put least
electronegative element in the center.
4. Complete an octet for all atoms except hydrogen
5. If structure contains too many electrons, form double
and triple bonds on central atom as needed.

32. LEARNING CHECK
Draw the Lewis structure for each of the
following compounds:
i. HF
ii. CH4
iii. CHCl3
iv. NH3
v. H2O

33. Total no. of valence
electrons
H : 1e
F : 7e
Total : 8e

34. Number of electrons
C : 4e
4H : 4e
Total : 8e

35.  Count electrons:
C : 4e
H : 1e
3Cl : 21e
Total: 26 e
Center atom: C

36. Count number of
electrons
N : 5e
3H : 3e
Total : 8e

38. Ionic Compound Covalent Compound
Boiling Point High Low
Volatility Non – volatile Volatile (can change to vapour
when heated)
Solubility
Usually soluble in water and
polar solvents but
insoluble in organic
Usually soluble in organic
solvents such as
benzene but insoluble in water
Electrical
Conductivity
Conducts electricity in the
molten state or aqueous
Solution
Does not conduct electricity in
any state.

39. BOND LENGTH
Compare the bond length between single,
double and triple bond
Bond length :
The distance between nuclei of the atoms involves in the
bond
C C C C C C
1.54 Å 1.34 Å 1.20 Å
As the number of bonds between the carbon increase,
the bond length decreases because C are held more
closely and tightly together
As the number of bonds between two atoms increases,
the bond grows shorter and stronger

41. Formal charge
Formal charge is the charge on a
certain atom in a Lewis structure
Formal Charge = The number of valence electron – the number of electron at the atom

42. The sum of formal charge on each atom
should equal:
i. zero for a molecule
ii.the charge on the ion for a polyatomic
ion
Formal charge is used to find the most
stable Lewis structure

43. A stable structure has:
i. Formal charge on each atom
closest to zero
ii. Formal –ve charge should be on
a more electronegative atom and
formal +ve charge should be on
a more electropositive atom

44. 1) Draw all the possible Lewis structure
of COCl2.
2) Predict the most plausible structure.
EXAMPLE

45. SOLUTION
The most plausible structure is (2)
Formal charge is determined before
completing a Lewis structure to predict
the most stable structure because
formal charge closest to zero.
1) 2)

46. LEARNING CHECK
 Draw the possible Lewis
structures for HNO2.
 Determine the most plausible
Lewis structures for HNO2.

47. 47
SHAPE OF A MOLECULE
 Basic shapes are based on the repulsion
between the bonding pairs.
 Tips to determine the molecular shape :
Step 1
Draw Lewis structure of the molecule
Step 2
Consider the number of bonding pairs
Step 3
Place bonding pairs as far as possible to
minimize repulsion.

48. 48
A. Molecules with 2 bonding pairs
Example: BeCl2
Lewis structure
 shape
Linear
180°
Be : 2e
2Cl : 14e
Total : 16 e
Cl ..
:
..
Cl
Be
..
:
..

49. 49
B. Molecules with 3 bonding-pairs
 Example: BCl3
Lewis structure
B: 3e
3Cl : 21e
Total: 24e
B ..
:
..
Cl
Cl
Cl
..
:
..
..
:
..
 Repulsive forces
between pairs are the
same
120°
Trigonal planar

50. 50
C. Molecules with 4 bonding pairs
 Example: CH4
Lewis structure
C
H
H
H
H
 Equal repulsion
between bonding pairs
– equal angle
109.5°
Tetrahedral

51. 51
D. Molecules with 5 bonding pairs
 Example: PCl5
 Lewis structure
P
Cl
Cl
Cl
Cl
Cl
..
:
..
..
:
..
..
:
..
..
:
..  Shape:
120°
Trigonal bipyramidal
90°

52. 52
Class of
molecules
Number
of
bonding
pairs
Number
of lone
pairs
Shape
AB2 2 0
Linear
180°
2 electron pairs in the valence shell
of central atom

53. Class of
molecules
Number
of
bonding
pairs
Number
of lone
pairs
Shape
AB3 3 0
trigonal planar
3 electron pairs in the valence shell
of central atom
120°

54. Class of
molecules
Number of
bonding
pairs
Number
of lone
pairs
Shape
AB4 4 0
Tetrahedral
4 electron pairs in the valence shell
of central atom
109.5o

55. 55
Class of
molecules
Number
of
bonding
pairs
Number
of lone
pairs
Shape
AB5 5 0
Trigonal pyramidal
5 electron pairs in the valence shell
of central atom
120°
90°

56. ORBITAL OVERLAP AND
HYBRIDIZATION
56
1. Formation Covalent Bond
2. Formation Hybrid orbitals
3. Orbital Overlapping

57. 57
FORMATION OF COVALENT BOND
• Valence bond theory - Covalent bond
is formed when two neighbouring
atomic half-filled orbitals overlap.
• Two types of covalent bonds are
a) sigma bond (σ)
b) pi bond ()

58. 58
+
a)  bond
• formed when orbitals overlap along its
internuclear axis (end to end overlapping)
Example:
i. overlapping s orbitals
H H H H
 bond

59. 59
ii. Overlapping of s and p orbitals
H +
Px orbital
x x
H
 bond

60. 60
• iii. Overlapping of p orbitals
x
x
+ x
 bond

61. 61
b)  bond
• Formed when two p-orbitals of the same
orientation overlap sideways
y y
+
y y
 bond

62. 62
y y y
+
y
 bond

63. 63
Formation of bonds in a molecule
• Covalent bonds may form by:
a) overlapping of pure orbitals
b) overlapping of hybrid orbitals

64. 64
Overlapping of pure orbitals
• Example :
i. O2
ii. N2

65. 65
O2
Consider the ground state configuration:
O : 1s2 2s2 2p4
1s 2s 2p
Two unpaired electrons to be used in
bonding.
Overlapping occurs between
the p-orbitals of each atom
σ


y
x
O
y
O

66. 66
y
O
y
x
O
σ



67. 67
Formation Hybrid orbitals
• Overlapping of hybrid orbitals and the pure orbitals
occur when different type of atoms are involved in
the bonding.
• Hybridization of orbitals:
mixing of two or more atomic orbitals to form a new
set of hybrid orbitals
• The purpose of hybridisation is to produce new
orbitals which have equivalent energy
• Number of hybrid orbitals is equal to number of
pure atomic orbitals used in the hybridization
process.

68. 68
10.4
Hybridization
• Hybrid orbitals have different shapes
from original atomic orbitals
• Types of hybridisation reflects the
shape/geometry of a molecule
• Only the central atoms will be involved
in hybridisation

69. 69
Hybridization of orbitals
i. sp
ii. sp2
iii. sp3

70. 70
sp3 hybridization
• one s orbital and three p orbitals
are mixed to form four sp3 hybrid
orbitals
• the geometry of the four hybrid
orbitals is tetrahedral with the
angle of 109.5o .

71. 71
sp3 hybrid
• Mixing of s and three p orbitals
sp3
sp3

72. 72
Example:
1) CH4
• Lewis structure :
• Valence orbital diagram ;
H :
C ground state :
C excited :
C hybrid :
• Orbital Overlap :
• Molecular Geometry :

73. 73
Example : Methane, CH4
Ground state : C : 1s2 2s2 2p2
C
H H
H
H
Lewis Structure
1s 2s 2p
Excitation: to have 4
unpaired electrons
Excited state :
1s 2s 2p
sp3 hybrid
shape: tetrahedral
sp3
sp3
sp3
sp3
H
H
H
H
C

74. 74
Fig. 10.8
sp3-Hybridized C atom in CH4
sp3
sp3
sp3
sp3
1s
1s
1s

75. 75
Example 2 :
NH3
Lewis structure :
Valence orbital diagram ;
H :
N ground state :
N excited :
N hybrid :
Orbital Overlap :
Molecular Geometry :

76. 76
Fig. 10.9
sp3
sp3
sp3
sp3
1s
1s
1s

77. 77
Example:
3) H2O
• Lewis structure :
• Valence orbital diagram;
O ground state :
O hybrid :
• Orbitals overlap:

78. 78
sp2 hybridization
• one s orbital and two p orbitals are
mixed to form three sp2 hybrid
orbitals
• the geometry of the three hybrid
orbitals is trigonal planar with the
angle of 120o .

79. 79
s
px py
sp2
sp2 sp2
one s orbital + two p orbitals three sp2 orbitals

80. 80
Example:
1) BF3
• Lewis structure :
• Valence orbital diagram;
F :
B ground state :
B excited :
B hybrid :
• Orbital overlap:

81. 81
Example: BF3
sp2
sp2
sp2



F : 1s22s22p5
Shape : trigonal planar
Pure p orbital

82. 82
Example:
2) C2H4
• Lewis structure :
• Valence orbital diagram;
C ground state :
C excited :
C hybrid :
• Orbital overlap:

83. 83
 bond
 bonds

84. 84
10.5

85. 85
sp hybridization
• one s orbital and one p orbital are
mixed to form two sp hybrid orbitals
• the geometry of the two hybrid
orbitals is linear with the angle of
180o

86. 86
Formation of sp Hybrid Orbitals
Types of hybrid orbitals
sp sp Produces linear shape

87. 87
Example:
1) BeCl2
• Lewis structure :
• Valence orbital diagram;
Cl :
Be ground state :
Be excited :
Be hybrid :
• Orbital overlap:

88. 88

89. 89
Example:
2) C2H2
• Lewis structure :
• Valence orbital diagram;
C ground state :
C excited :
C hybrid :
• Orbital overlap:

90. 90

91. No of Lone Pairs
+
No of Bonded Atoms Hybridization Examples
2
3
4
sp
sp2
sp3
BeCl2
BF3
CH4, NH3, H2O
How do I predict the hybridization of the central atom?
Count the number of lone pairs AND the number
of atoms bonded to the central atom

92. 92
Intermolecular Forces
 Intermolecular forces are the
attractive forces between
molecules

93. 93
Effects of intermolecular forces
on physical properties
 Have effects on these physical
properties:
a) boiling point
b) melting point
c) solubility
d) density
e) electrical conductivity

94. 94
Intermolecular Forces
Van der Waal
Forces
Hydrogen
Bond
Between
covalent
molecules
Between
covalent
molecules
with H
covalently
bonded to
F, O or N

95. 95
van der Waal Forces
 Forces that act between covalent
molecules
 Three types of interaction:
i. Dipole-dipole attractive
forces
- act between polar molecules
ii. London Dispersion forces
-act between non-polar
molecules

96. •Polar molecules occur when there is an electronegativity difference between the bonded atoms.
•Nonpolar molecules occur when electrons are shared equal between atoms of a diatomic
molecule or when polar bonds in a larger molecule cancel each other out.

97. 97
+
-
Cl
H
+ -
Cl
H
Dipole-dipole forces
(permanent dipole forces)
 Exist in polar covalent compounds
 Polar molecules have permanent dipole
due to the uneven electron distributions
Example:
Chlorine is more
electronegative,
thus it has higher
electron density
Dipole-dipole forces; the
partially positive end attracts
the partially negative end

98. 98
London Dispersion Forces
 attractive forces that exist between
non-polar molecules
 result from the temporary
(instantaneous) polarization of
molecules
 The temporary dipole molecules will be
attracted to each other and these
attractions is known as the London
Forces or London Dispersion forces

99. 99
The formation of London forces
 At any instant, electron distributions
in one molecule may be
unsymmetrical.
 The end having higher electron
density is partially negative and the
other is partially positive.
 An instant dipole moment that exists
in a molecule induces the
neighboring molecule to be polar.

100. 100
Example:
London forces in Br2
Br Br
Electrons in a molecule
move randomly about the
nucleus
Br Br
At any instant, the
electron density might
be higher on one side
-
+
Br Br
The temporary dipole
molecule induce the
neighboring atom to
be partially polar
Temporary
dipole molecule
-
+
London forces

101. 101
Factors that influence the strength
of the van der Waals forces.
 The molecular size/molecular mass
Molecules with higher molar mass have
stronger van der Waals forces as they tend to
have more electrons involved in the London
forces.
Example:
CH4 has lower boiling point than C2H6
 Note:
However if two molecules have similar
molecular mass, the dipole-dipole interaction
will be more dominant.
Example: H2S has higher boiling point than
CH3CH3

102. 102
Hydrogen intermolecular bond
 Dipole-dipole interaction that acts between
a Hydrogen atom that is covalently bonded
to a highly electronegative atom ; F, O ,N in
one molecule and F,O or N of another
molecule.
Example:
+ -
F
H
+ -
F
H
Hydrogen
intermolecular bond

103. 103
Other examples:
NH3 liquid
N
..
N
..
Covalent bond
Hydrogen
intermolecular bond
H2O
O
O O
O
Hydrogen
intermolecular bond
Hydrogen
intermolecular bond

104. 104
in
CH3OH
C
O
Consider ethanol, CH3OH
Hydrogen bond
CH3OH and methane
C
C
O
Not a hydrogen
bond
H is not
bonded to
either F, O
or N

105. 105
Example: H2O
O H
H
O H
H
O
H
H
O H
H
O H
H
___ covalent bond
----- hydrogen bond

106. 106
Properties of compounds with
Hydrogen intermolecular forces
 Have relatively high boiling point than
compounds having dipole-dipole forces
or London forces
 the Hydrogen bond is the strongest
attraction force compared to the dipole-
dipole or the London forces.
Boiling point

107. 107
Solubility
A. Dissolve in polar solvent
 The molecules that posses
Hydrogen bonds are highly polar.
 They may form interaction with
any polar molecules that act as
solvent.
B. Dissolve in any solvent that can
form Hydrogen bonds

108. 108
Example
 NH3 dissolves in water because it can form
Hydrogen intermolecular bond with water.
N
..
O
N
..
Hydrogen bond

109. 109
The effect of Hydrogen bond on water
molecules
 The density of water is relatively high
compared to other molecules with similar
molar mass.
Reason:
Hydrogen intermolecular bonds are
stronger than the dipole-dipole or the
London forces. Thus the water molecules
are drawn closer to one another and
occupy a smaller volume.

110. Ice (solid H2O) has lower density
compared to its liquid. Refer to the
structure of ice
Density

111. 111
Hydrogen bond takes
one of the tetrahedral
orientation and occupy
some space
Ice form tetrahedral arrangement

112. 112
 H2O(l) is denser than H2O(s) because
 the hydrogen bond in ice arrange the
H2O molecules in open hexagonal
crystal
 H2O molecules in water have higher
kinetic energy and can overcome
the hydrogen bond
 V-shaped water molecules slide
between each other.

113. 113
 The boiling points of these substances are
affected by:
a) the number of hydrogen bonds per
molecule
b) the strength of H intermolecular
forces which directly depends on
the polarity of the hydrogen bond
Example:
Explain the trend of boiling points given below:
The order of the increase in boiling point is:
H2O > HF > NH3 > CH4
Boiling points of substance with
Hydrogen intermolecular bonds

114. 114
 by looking at the polarity of the bond, we have
(Order of polarity: HF > H2O > NH3)
but H2O has the highest boiling point.
For H2O, the number of hydrogen bonds per molecule
affects the boiling point.
 Each water molecule can form 4 hydrogen bonds with
other water molecules. More energy is required to
break the 4 Hydrogen bonds.
 HF has higher boiling point than NH3 because F is
more electronegative than Nitrogen.
 CH4 is the lowest - it is a non polar compound and
has weak van der Waals forces acting between
molecules.
Answer:

115. 115
Effects of intermolecular forces on
physical properties
1)Boiling point
 For molecules with similar size, the order
of intermolecular strength:
Hydrogen bond > dipole-dipole forces >
London dispersion forces
 Strength of intermolecular forces ↑
 boiling point ↑

116. 116
Why boiling point H2O > HF
and HF > NH3?
 Fluorine is more electronegative than
oxygen, therefore stronger hydrogen
bonding is expected to exist in HF liquid than
in H2O.
 However, the boiling point of H2O is higher
than HF because each H2O molecules has
4 hydrogen bonds.

117. 117
 On the other hand, H-F has only 2 hydrogen
bonds.
 Therefore the hydrogen bonds are stronger
in H2O rather than in H-F.

118. 118
Boiling point HF > NH3
 Fluorine is more electronegative than
nitrogen ,thus the hydrogen bonding in H-F
is stronger than H-N.

119. Metallic bond
 An electrostatic force between
positive charge metallic ions and the
sea of electrons.
 Bonding electrons are delocalized
over the entire crystal which can be
imagined as an array of the ions
immersed in a sea of delocalized
valence electron.
119

120. 120
Metallic bonds
e e e e
e e e e
e e e e
Positive
ions are
immersed in
the sea of
electrons
Free
moving
electrons

121. 121
Electrostatic force in a metal
Metallic Bond (Electron-sea Model)
 Metals form giant metallic structure
 Each positive ion is attracted to the
‘sea of electrons’.
 These atoms are closely held by the
strong electrostatic forces acting
between the positive ions and the ‘sea
of electrons’.
 These free moving electrons are
responsible for the high melting point
of metals and the electrical
conductivity.

122. 122
 metals have high melting point
 high energy is required to overcome these
strong electrostatic forces between the
positive ions and the electron sea in the
metallic bond
Physical properties of metals
+ + + + +
+ + + + +
+ +
+ + +
e e e e e e
e e e e e e
e e e e e e
Metallic bonds formed
by the electrostatic
forces exist between
positive ions and the
free moving electrons

123. 123
 The strength of the metallic bond
increases with the number of valence
electrons and the size of ions.
 The smaller the size of positive ions the
greater is the attractive force acting
between the ions and the valence
electrons
The strength of the metallic
bonds

124. 124
Boiling points in metals
+1 +1 +1 +1
+1 +1
+1 +1 +1 +1
+1 +1
+1 +1 +1 +1
+1 +1
e e
e e e e
e e e e e e
+2 +2 +2 +2 +2 +2 +2 +2
+2 +2 +2 +2 +2 +2 +2 +2
+2 +2 +2 +2 +2 +2 +2 +2
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Na Mg
Has one valence electron
the electrostatic force acting
between positive ions and
free moving electrons form
metallic bonds
Has 2 valence electrons
Stronger metallic bond due to the
size of Mg being smaller than Na
and the strong electrostatic force
between +2 ions and the two
valence electrons,
Mg has higher boiling point than Na

125. 125
Example:
Explain the difference in the
boiling point of the two metals
given:
Magnesium – 11300 oC
Aluminum – 24500 oC

126. 126
 The cationic size of Al is smaller
compared to magnesium and its
charge is higher (+3).
 Mg has two valence electrons
 Al has three valence electrons
involved in the metallic bonding.
 The strength of metallic bond in
Aluminium is greater than that of
Magnesium
 Al has higher boiling point
Answer

127.  The strength of metallic bond is
directly proportional to the boiling
point.
 The stronger metallic bond,the
higher the boiling point.
127