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7/23/2019 Chemistry Form 6 Sem 1 03a
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PRE-UNIVERSITY
SEMESTER 1CHAPTER 3
CHEMICAL
BONDING
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Chemical Bonding can be generally divide to 5 main group
Electrovalent bonding (ionic) Covalent bonding
Metallic bonding
Hydrogen bonding
Van der Waals bonding
To represent the types of bonding, a Lewis diagram (dot-and-
cross) is used. Each dot or cross represent one electron in valence
shell and its a more convenient way in showing electrovalent. For both ionic & covalent bonding, octet rule must be fulfill where
tendency of atoms to achieve noble gas configuration. Table 6.2
show some cation/anion with difference number of valence
electron.
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Electrovalent bond (ionic bond)
Formed by transfering 1 or more e- from outer orbital to another.The atom donate electron is name as cation and the atom whoreceive electron is name as anion. The bond form whenelectrostatic attraction occur between 2 opposite charge ions.
Formation of ionic compound involving a metal with low IE and anon-metal with high EA. Example for lithium fluoride (LiF). Theelectronic structure of the lithium and fluorine are :
Lithium (Li) = 1s2 2s1
Fluorine (F) = 1s2 2s2 2p5
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Practice : Draw the Lewis dot and cross diagram for these ionic compound
Sodium chloride Magnesium fluoride
Aluminium oxide
Na
+
Cl
_
Mg
2+
F
_
2
Al
3+
2
O
2-
3
Na+
Cl- Mg
2+F
-
2
Al 3+ O 2-32
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Covalent Bonding : Sharing of Electron
Covalent bond is bond that formed in between atoms by sharingelectron from its atoms in order to achieve a stable electronic
configuration of ns2 np6 for atoms involve. (hydrogen achieve 1s2)
Some non-metallic elements exist naturally as diatomic molecules
like hydrogen, and halogens groups.
From example above, we can see that in covalent bond, molecules
may form single bond, double bond or triple bond in order to
achieve stable valence electrons. Though, there are somemolecules with the exceptions of achieving stable valence electrons.
Hydrogen molecule Chlorine molecule Oxygen molecule Nitrogen molecule
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Electron deficient compounds compounds which the molecule
(especially the center atom) does not achieve octet electron arrangement.
Examples of these molecules are BeCl2 ; BF3 and AlCl3.
Beryllium dichloride Boron trifluoride Aluminium trichloride
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Electron rich compounds compounds which have more than 8 electrons
at center atom of molecules, such as PCl5, SF6 and ICl5.
However, not all compounds can have more or less than 8 electrons in the
center of the atom. There are certain limitation towards the application of
the expansion of center atom
Phosphorous pentachloride Sulphur hexachloride Iodine pentachloride
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For example, nitrogen (N) and phosphorous (P) are both from Group
but phosphorous can exist as PCl3 and PCl5 while nitrogen can
only have NCl3 but not NCl5. This is because
.....................
..................................
Same things occur when it come to hydrolysis of CCl4 and SiCl4. SiCl4
can undergoes hydrolysis with water according to the equation
.
while CCl4 cannot. Despite the factors that they are from the same group
(Group ), CCl4 cannot undergoes hydrolysis as
...
15
nitrogen which only have 2 shell, do not have empty d-orbital available,
but phosphorous contain d-orbital to fill in more electron
SiCl4 + 2 H2O SiO2 + 4 HCl
14
carbon which only have 2 shell, do not have empty d-orbital available, so
water cannot form coordinative with carbon hence cannot undergoes
hydrolysis.
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Examples : Draw the Lewis structure for the following molecules.
CO2 HCN CH3COOH
C2H2 NH3 CO32-
SO4
2- C3H
6
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6.2.1 Dative bond
Now, try drawing the Lewis structure for these molecules : SO2
, SO3
,
NO3- or CO.
Dative bond is formed when an atom that has lone pair electrons
which can donate to molecule/ion that has empty unhybridise
orbital.
Following are a few applications of dative bond in covalent molecules
SO2 SO3 NO3- CO
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1. Dative bond in helping molecule to achieve octet.
NH4+
BF3.NH3
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Dative bond in forming dimer ~ 2 monomer combine forming a dimer.
Forming Al2Cl6 Forming polymer of BeCl2
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3. Dative bond in formation of complex ion.
Molecule / ion form dative bond (also known as coordinative bond) by
donating lone pair electron, which act as a .. in the formation ofcomplex ions. For example
hexaaquacopper (II) ion ;
[Cu(H2
O)6
]2+tetraamminenickel (II) ion ;
[Ni(NH3
)4
2+]
Hexacyanoferrate (III) ion ;
[Fe(CN)6
]3-
ligand
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6.2.3 Resonance ~ a molecule/polyatomic ion in which two or more
plausible Lewis structure can be written but the actual structure cannot be
written at allSulphur dioxide, SO2
Ethanoate ion, CH3COO
Nitrogen dioxide, NO2
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Sulphur trioxide, SO3
Carbonate ion, CO32-
Since the resonance structure cannot be determined as it does not have
a permanent structure so it is expressed as a combined of resonance
structure known as resonance hybrid
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Resonance hybrid
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Covalent Bonds : Overlapping of Orbitals
2 ways in explaining how covalent bond are attached : Valence bond theory
Valence-shell electron-pair repulsion theory (VSEPR)
Here we can explain and predict what type of molecular bond and
shape will form through the bonding formation but it does not
explain the stability of covalent bond.
For valence bond theory, it used atomic orbital overlapping that
result the formation of a new molecular orbital embracing bothnuclei. The strength of covalent bond is proportional to the area
where the atomic orbital overlap. Larger the area overlap, stronger
the covalent bond.
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Hybrid Atomic Orbitals
3 basic types of hybrid orbital
sp3 hybrid orbital (tetrahedral arrangement)
sp2 hybrid orbital (trigonal planar arrangement)
sp hybrid orbital (linear arrangement)
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6.3.2 sp3 hybridisation
The term sp3 gives an impression of the hybridisation involved _____ s
orbital and _____ p orbitals
Examples of molecules which give sp3 hybridisation are
For example, in methane, CH4, since carbon is in Group _____so the
valance electron of C is _______
Methane silicon tetrachloride
sulphate ion Perchlorate ion
1 3
CH4 SiCl4
SO42- ClO4
-
14
2s2 2p2
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State of
moleculesOrbital diagram Illustration / Explanation
Ground state
_____ _____ _____
2p
____
2s
Excited state ____ ____ ____ ____
2s 2p
Hybridisatio
n state
_____ _____ _____ _____
sp3
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109.50
tetrahedral
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6.3.3 sp2 hybridisation
The term sp2 gives an impression of the hybridisation involved _____ s
orbital and _____ p orbitals
Examples of molecules which give sp2 hybridisation are
Since boron is Group ______ element so the electron valance of B
is _________
Sulphur trioxide Boron trifluoride
Nitrate ion Carbonate ion
1 2
SO3 BF3
NO3- CO3
2-
13
2s2
2p1
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State of
moleculesOrbital diagram Illustration / Explanation
Ground state
_____ _____ _____
2p
____
2s
Excited state____ ____ ____ ____
2s 2p
Hybridisatio
n state
_____ _____ _____ ____
sp2 pz
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Formation of sp2 Hybrid Orbitals
Shape of molecule
Trigonal planar
Angle between bond
pair
120o
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6.3.4 sp hybridisation
The term sp gives an impression of the hybridisation involved _____ s
orbital and _____ p orbitals
Examples of molecules which give sp hybridisation are
Lets use beryllium chloride as example.
Since beryllium is Group ______ element so the electron valance of
Be is ___________
Carbon dioxide Beryllium chloride
Cyanic acid Ethyne
1 1
CO2 BeCl2
HCN C2H2
2
2s2
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State of
molecules
Orbital diagram Illustration / Explanation
Ground state_____ _____ _____
2p
____2s
Excited state ____ ____ ____ ____2s 2p
Hybridisatio
n state
_____ _____ ___ ___
sp py pz
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Formation of sp Hybrid Orbitals
Shape of molecule
Linear
Angle between bond pair
180o
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6.4 Hybridisation in organic molecules
In this subtopic, were going to witness how is the formation of the bonding
that exist in some organic molecules. The 3 organic molecules which willbe discussed in this sub-topic are :
methane, CH4 ethene, C2H4
ethyne, C2
H2
All of the molecules above has carbon in it
Carbon is a group _____ element. It has the electronic configuration of
______________
The orbital diagram Ground state of carbon : _____ _____ _____ _____
2s 2p
14
2s2 2p2
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Methane, CH4 Type of hybridisation :
Excited state of carbon : _____ _____ _____ _____2s 2p
Hybridised state : _____ _____ _____ _____
sp3
Molecular shape :
Angle between the bonding pair :
..109.50
tetrahedral
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Ethene, C2H4 Type of hybridisation :
Excited state of C : _____ _____ _____ _____
2s 2p
Hybridised state : _____ _____ _____ _____
sp2 pz
Molecular shape
Angle between bond
pair bond pair
sp2
Trigonal planar
120o
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Ethyne, C2H2 Type of hybridisation :
Excited state of C : _____ _____ _____ _____2s 2p
Hybridised state : _____ _____ _____ _____
sp py pz
Molecular shape
Angle between bond
pair bond pair
sp
Linear
180o
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As a conclusion, the formation of double bond
(C=C) is due to ______sigma bond () and_____pibond ()
While the formation of triple bond (CC) is due
to ______sigma bond () and _____pibond
()
oneone
one two
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3.5 Hybridisation in water, H2O and ammonia, NH3
The hybridisation of ammonia is similar to that in methane (sp3
hybridisation). Nitrogen, N which has the electron valence as. where the ground state can be stated in the orbital
diagram below
Ground state : ____ ____ ____ ____
2s 2p
Excited state : ____ ____ ____ ____
2s 2p
Hybridised state : ____ ____ ____ ____
sp3
*Compare the angle between the bonding pair of NH to NH inammonia and CH to CH in methane
Angle between HNH < Angle between HCH
Shape :
S h h b idi i f ( 3 h b idi i ) O O
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Same goes to the hybridisation of water (sp3 hybridisation). Oxygen, O,
which has the electron valance as .., where the ground
state can be stated in the orbital diagram belowGround state : ____ ____ ____ ____
2s 2p
Excited state : ____ ____ ____ ____
2s 2p
Hybridised state : ____ ____ ____ ____
sp3
*Compare the angle between the bonding pair
of HOH in water and HCH in methane
Angle between HOH Angle between HCH
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From the 2 examples above, we can tell how the lone pair electrons
affecting the angle between the bonding pair and bonding pair. In
ammonia, not only that there is the repulsion between bondingpair and bonding pairbut theres also the repulsion between
bonding pair and lone pair.
Since the angle between the bonding pair and bonding pairdecrease, theres a probability that its due to the effect of stronger
repulsion between the bonding pair and lone pair electron. This
statement is supported as in the repulsion between the HOH in
water is smaller than in ammonia, NH3. as a conclusion, we canconclude that
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bonding-pair vs. bonding
pair repulsion
lone-pair vs. lone pair
repulsion
lone-pair vs. bonding
pair repulsion> >
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Valence Shell Electron Pair Repulsion (VSEPR) Theory
~ state that the electron-pair repulsion stated that electron pairs
around central atom repel each other
3 main rules
Bonding pairs and lone pairs of electrons arrange themselves to be as far
apart as possible. The order of repulsion strength of lone pair and bond pair are
lone-pair & lone-pair > lone-pair & bond-pair > bond-pair & bond-pair
Double / triple bond are considered as 1 bonding pair when predicting the
shape of molecules or ions
Diagram below shows the type of bonding and the molecular
shape predicted.
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Class
No of
surround
atoms
No of
lone pair
electron
Molecular shapeDiagram of the molecular
shape
Example of
molecules
AB2 2 0 Linear
BeCl2CO2
HCN
AB3 3 0Trigonal
Planar
CO32-
AlCl3BF3
AB4 4 0 Tetrahedral
CH4SiCl4SO4
2-
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Class
No of
surround
atoms
No of
lone pair
electron
Molecular shapeDiagram of the molecular
shape
Example of
molecules
AB5 5 0Trigonal
bipyramidal
PCl5BiCl5
AB6 6 0 Octahedral SF6
TeCl6
VSEPR
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Class
# of atomsbonded to
central atom
# lonepairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB3 3 0
trigonal
planar
trigonal
planar
AB2E 2 1trigonal
planarbent
10.1
VSEPR
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Class
# of atomsbonded to
central atom
# lonepairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB3E 3 1
AB4 4 0 tetrahedral tetrahedral
tetrahedraltrigonal
pyramidal
10.1
VSEPR
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Class
# of atomsbonded to
central atom
# lonepairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
AB4 4 0 tetrahedral tetrahedral
10.1
AB3E 3 1 tetrahedraltrigonal
pyramidal
AB2E2 2 2 tetrahedral bent
H
O
H
VSEPR
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Class
# of atomsbonded to
central atom
# lonepairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
10.1
AB5 5 0
trigonal
bipyramidal
trigonal
bipyramidal
AB4E 4 1trigonal
bipyramidal
see - saw
VSEPR
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Class
# of atomsbonded to
central atom
# lonepairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
10.1
AB5
5 0trigonal
bipyramidal
trigonal
bipyramidal
AB4E 4 1trigonal
bipyramidal
distorted
tetrahedron
AB3E2 3 2 trigonalbipyramidalT-shaped
ClF
F
F
VSEPR
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Class
# of atomsbonded to
central atom
# lonepairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
10.1
AB5
5 0trigonal
bipyramidal
trigonal
bipyramidal
AB4E 4 1trigonal
bipyramidal
distorted
tetrahedron
AB3E2 3 2 trigonalbipyramidalT-shaped
AB2E3 2 3trigonal
bipyramidallinear
I
I
I
VSEPR
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Class
# of atomsbonded to
central atom
# lonepairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
10.1
AB6
6 0 octahedraloctahedral
AB5E 5 1 octahedralsquare
pyramidal
Br
F F
FF
F
VSEPR
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Class
# of atomsbonded to
central atom
# lonepairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
10.1
AB6
6 0 octahedraloctahedral
AB5E 5 1 octahedralsquare
pyramidal
AB4E2 4 2 octahedral
square
planar
Xe
F F
FF
5 GENERAL STEPS TAKEN WHEN WRITING LEWIS
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5 GENERAL STEPS TAKEN WHEN WRITING LEWIS
STRUCTURE FOR MOLECULES AND IONS
Calculate the total number of valence electrons from all atoms
Arrange all the atoms surrounding the central atom by using a pair of
electron per bond
Assign the remaining electrons to the terminal atoms so that eachterminal atom has 8 electrons (H = 2 e-)
Place any left-over electron on the central atom.
@ Form multiple bonds if there are not enough electrons to give the
central atom an octet of electrons.
i) PCl3 ii) SF6
1 P 1 (5) 5 l t 1 S => 1 (6) = 6 electrons
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1 P => 1 (5) = 5 electrons
3 Cl => 3 (7) = 21 electrons
Total = 26 electrons
Step 2 : place 1 bond from surround to
center atom
e- used = 3 (2) = 6e- remained = 26
Step 3 : placed each surround atom
with 6 e-e- used = 3 (6) = 18
e- remained = 2
Step 4 : place remained e- at center of
atom
1 S => 1 (6) = 6 electrons
6 F => 6 (7) = 42 electrons
Total = 48 electrons
Step 2 : place 1 bond from surround to
center atom
e- used = 6(2) = 12
e- remained = 36
Step 3 : placed each surround atom
with 6 e-e- used = 6 (6) = 36
e- remained = 0
b) SO42- c) POCl3
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d) SF6 e) I3-
4 Surround Atom + 0 Lone pair e-
Arrangement : tetrahedral
Shape : tetrahedral
4 Surround Atom + 0 Lone pair e-
Arrangement : tetrahedral
Shape : tetrahedral
6 Surround Atom + 0 Lone pair e-
Arrangement : octahedral
Shape : octahedral
2 Surround Atom + 3 Lone pair e-
Arrangement : trigonal bipyramidal
Shape : linear
f) ICl3 g) SbCl52-
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i) PCl5 j) CO32-
3 Surround Atom + 2 Lone pair e-
Arrangement : trigonal bipyramidal
Shape : T shape
5 Surround Atom + 1 Lone pair e-Arrangement : octahedral
Shape : square pyramidal
5 Surround Atom + 0 Lone pair e-
Arrangement : trigonal bipyramidal
Shape : trigonal bipyramidal
3 Surround Atom + 0 Lone pair e-
Arrangement : trigonal planar
Shape : trigonal planar
6.6 Electronegativity and Polar Molecules
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Electronegativity are measurement of ability of an atom in molecules to
attract a pair of electron For 2 identical atoms, since they have same electronegativity so they
have no difference in electronegativity. These molecules are called polar
molecules
While if 2 not identical form a covalent bond, the bonding electrons willattracted more strongly by more electronegative element. We can
indicate the polarity of hydrogen chloride molecules in 2 ways.
H Cl The separation of charge (between + and ) in a poplar molecule is
called dipole
When 2 electrical charges of opposite sign are separated by small
distance, dipole moment is established
+
Molecules that are polarhave large dipole moments.
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Molecules that are non polar have zero dipole moment.
Still, for some molecules, even there are different in electronegativity
but it doesnt mean that these molecules there are polar molecules.
When the surrounding atom are symmetrically surrounded by
identical (same) atom, they are non-polar Example of molecules which are non polar
Dipole Moments and Polar Molecules
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p
H F
electron rich
regionelectron poor
region
+
= Q x rQ is the charge
r is the distance between charges
1 D = 3.36 x 10-30
C m
Which of the following molecules have a dipole moment?
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Which of the following molecules have a dipole moment?
H2O, CO
2, SO
2, and CH
4
O
dipole moment
polar molecule
S
CO O
no dipole moment
nonpolar molecule
dipole moment
polar molecule
C
H
H
HH
no dipole moment
nonpolar molecule
Nitrogen dioxide, NO2 Methane, CH4 Ethene, C2H4 Benzene, C6H6
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Boron trifluoride, BF3 Cyanide acid, HCN Sulphur dioxide, SO2 Sulphur trioxide, SO3
Ammonia, NH3 Ammonium ion, NH4+ Ethane, C2H6 Chloroethane, C2H5Cl
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Cyclohexane, C6H12 Chlorocyclohexane,
C6H11Cl
Carbon dioxide, CO2 Carbonate ion, CO32-
Phosphorous trichloride,
PCl3
Phosphorous
pentachloride, PCl5
cisbut-2-ene transbut-2-ene
A simple experiment which can be
used to determine either a
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used to determine either a
molecule is polar or non polar isillustrated below
By using the liquid form of the
compound, it is flow out slowly
from burette while a negativecharged rod is bring close to the
flow of the liquid.
If the liquid is deflected to the
direction of negative charged, thisliquid is
If it remain undeflected, this liquid
is .
polar
non-polar
From the example above, classified which compounds can be deflected
and which cannot
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and which cannot
Compound which can be deflected by
charged rod
Compound which cannot be deflected
by charged rod
Nitrogen dioxide,
Cyanide acid,
Sulphur dioxide,
Ammonia,
Chloroethane,Chlorocyclohexane,
Phosphorous trichloride
Cis-but-2-ene
Methane, ethene, benzene,
Sulphur trioxide,
Ammonium ion,
Ethane, cyclohexane,
Carbon dioxide,Carbonate ion,
Phosphorous pentachloride,
Trans-but-2-ene
Electronegativity and Type of Chemical Bond.
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Actually, the type of bond that would form can be tell by using the
difference of electronegativity (EN). More larger the difference,the more tendency of electron form low EN move an electron to
higher EN atom and ionic compound is formed.
The relationship between the ionic character and the difference inthe electronegativity of the bonded atom is shown on next slide (or
page 220).
The presence of dipoles gives ionic character to polar covalent
molecules. When the polarity of the covalent molecule increases,the ionic character also increase.
An ionic bond is formed if the cation has a small ionic radius
anion has a large ionic radius both cation & anion carries a lowelectrical charge.
Polarisation ~ the distortion of the charge cloud of the negative ion
by a neighbouring positive ion.
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Fig. 9.18
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3.6.1 Covalency Properties in Ionic Molecules
From the graph above the dotted line represent the arbitrary
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From the graph above, the dotted line represent the arbitrary
line between ionic and covalent characteristic of a molecule.To be more specific, there more likely an ionic compound may
have high covalent characteristic (exemplified by LiI), or
conversely covalent compound having high ionic characteristic(exemplified by HF).
The covalent characteristic of a molecule is dependent on the
ability of a cation to polarise an anion. Polarisation indicates
the ability of a cation to attract the electron density of an anion
when put next to the cation involved. When a cation is able to
pull the electron density of the anion closer to it, as if the anion
wanted to share electron with cation, hence increase thecovalency of the molecule
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Highly ionic compound
Large cationic size
Small anionic size
Highly covalent compound
small cationic size
large anionic size
The covalency properties of a molecule is dependent on the
cation and anion where they can be explained qualitatively via
Polarisation power of cation
Polarisability of anion
3.6.1.1 Polarisation Power of Cation
Polarisation Power of Cation measure the ability of a cation to
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Polarisation Power of Cation measure the ability of a cation to
polarise the electron cloud of the anion. 2 factors determining the polarisation power of cation
Charge of cation Size of cation
Greater the charge of ion, higher theeffective nuclear charge of cation, hence it
will be able to attract the neighboring
electron density of anion. This will caused
the polarization power of cation increase,hence increase the covalent characteristic
of cation.
Smaller the size of cation, closer the
neighboring anion to the nucleus of cation,
hence easier for the cation to polarise the
anion and result an increment in the
polarization power of cation, and increase
the covalent characteristic of cation.
Both factors can be explained in another term called as charge density where
Charge Density = Charge / Ionic Radius
From the equation above, Charge Density will have a greater value, provided that cation has
a high charge and small cationic radius.
Greater the charge density, higher the polarization power, greater the covalent characteristic
of the cation.
3.6.1.2 Polarisability of Anion
Polarisability of an anion ~ ability of the anion to allow the electron density
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y y y
to be polarised by cation. 2 factors determining the polarisability of an anion
Unlike cation, anion does not have a term that combined both factors of
charge and ionic radius. However, information of polarisability of anion
enable the prediction of the covalent characteristic of a molecule, since in
order to form a covalent bond, it depend on both polarisation power of
cation and polarisability of the anion
Charge of anion Size of anion
Greater the charge of anion, lower theeffective nuclear charge of anion. This will
weakened the electrostatic attraction forces
between nucleus and the outermost electron in
anion, and increase the polarisability of theanion, hence increase the covalent
characteristic of anion
Larger the size of anion, further the
outermost electron from the nucleus of
the anion, easier for the cation to
polarise the anion, and cause the
polarisability to increase, hence increase
the covalent characteristic of anion.
3.6.2 Prediction of Chemical Bond :Fajans Rule
In 1923, Kazimierz Fajans formulated an easy guidance to predict
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j y g p
whether a chemical bond will be covalent or ionic, and depend on thecharge on the cation and the relative sizes of the cation and anion. They
can be summarized in the following table
Based on these guidance, the bonding of a few compounds shallbe discussed to understand the application of Fajans Rule in the
chemical bonding
Ionic compound Low positive charge Large cation Small anion
Covalent compound High positive charge Small cation Large anion
Lithium halide (LiX)
Lithium ion, Li+ (1s2) has a small size due to only 1 shell present in
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, ( ) y p
its ion. But since it has a low charge, so its charge density is not toohigh. That is why, all lithium halide are ionic compound. The
covalency of lithium halide varies from a highly ioniccharacteristic to
highly covalency, depending on the polarisability of the anion next
to Li+
When a group of halide, F ; Cl; Br; I is put close to Li+, the
covalency of lithium halide increase when going down to Group
17 halide. LiF is highly ionic, since the fluoride ion has small ionicsize and low charge, hence has low polarisability. Ionic size
increase with the increasing shell when going down to Group 17
halide, hence increase the polarisability, which allowed lithium ion
to polarise the anions electron density, hence increase thecovalency
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Aluminium halide (AlX3) and aluminium oxide (Al2O3)
Aluminium ion (Al3+) has high charge density, due to its high charge unit
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( ) g g y, g g
and its small ionic radius. So, depending on the anion, aluminium has ahigh tendency to form covalent compound. For example, when going
down to Group 17 halide, aluminium fluoride (AlF3) forms ionic
compound (since F- has a low polarisability), while aluminium trichloride
(AlCl3), aluminium tribromide (AlBr3) and aluminium iodide (AlI3) formcovalent compound (since chloride, bromide and iodide have high
polarisability). This explained why aluminium fluoride has a high melting
point (10400C), while aluminium trichloride and tribromide are 1920C and
780
C respectively. As for aluminium oxide (Al2O3), it is an ionic compound with high
covalent characteristic, as aluminium ion has high covalent characteristic
due to its high charge density. This explained the high melting point of
Al2O3 (20500C) yet it is insoluble in water. It also explained theamphoteric properties of aluminium oxide where aluminium oxide can act
as an acid (covalent characteristic), as well as a base (ionic
characteristic).
Metallic Bonding
The properties of metals cannot be explained in terms of the ionic
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The properties of metals cannot be explained in terms of the ionic
/ covalent bond. In ionic / covalent compound, electron are notfree to move under the influence of applied potential (charge)
difference. Therefore, ionic solid and covalent compound are
insulator.
In metal, electron are delocalised and metal atoms are effectively
ionised.
Metallic bond ~ electrostatic attraction between the positively
charged metal ion and the electron delocalised. Because of this, electron now can freely move from cathode to
anode when a metal is subjected to an electrical potential. The
mobile electron can also conduct heat by carrying the kinetic
energy from a hot part of the metal to a cold part. This electron
delocalised can also use to explain the electrical and thermal
conductivities of metal
The Band Theory : Overlapping of Orbital
The number of molecular orbitals produced is equal to the number
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p q
of atomic orbitals that overlap.
In a metal, the number of atomic orbitals that overlap is very large.
Thus the number of molecular orbital produced is also very large.
The energy separations between these metal orbitals areextremely small. So, we may regard the orbital as merging
together to form a continuous band of allowed energy state. This
collection of very closed molecular orbital energy levels is called
an energy band. This theory for metal is called band theory
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Electrical Conductors
Molecular orbital model == 2 group of energy level.
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Molecular orbital model 2 group of energy level.
Lower energy level valence band form from overlap of outer most
orbital containing valence electron of each atom.
Higher energy level conduction band energy level filled with mobile
electron
But there are some case where valence band can also serve as
conduction band (caused by the movement of delocalised
molecular orbital)
Electrical conductivities decrease when temperature increase vibration of the lattice of ion impedes the free movement of
electron in conduction band.
conduction band
valence band
Insulator
Difference between conductors, semi-conductors, and insulator
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depend on the energy gap between the 2 bands. Conductor 2 bands overlaps so conduction band always partly
filled.
Insulator gap between the band is large and no electron exist inthe conduction band. E.g. insulator diamond
When 2s and 2p orbital of C is combine to form 2 energy bands,
valence band is filled with electron.
In insulator, the energy gap between the band is large. Under
normal condition, few electrons in valence band can jump across
to conduction band. If electron cannot reach conduction band
across the gaps, the electrical conduction cannot take place.
Semiconductor
Theres still energy gaps between 2 bands in semiconductor, but it
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is smaller than insulator. In semiconductor, some electrons have sufficient energy to jump
across the energy gaps and electron can move freely in
conduction band thus enable electrical conduction. Still, the electrical activity is not as good as metal (conductor)
Increasing temperature can help to improve the conductivity
because electron gain thermal energy and are able to reach
conduction band. It can also improve its effectiveness by adding small amount of
substance. This adding is what we called doping. It can help to
increase electrons to fill in valence band.
Example of doping is Si dope P (n-type). Si dope Ge (p-type)
Depend on the needs, this process can help to create the various
type of semiconductor in electronic characteristic.
7.1 Van der Waals forces
Van Der Waals forces are the intermolecular forces formed
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between covalently bond molecules which exist as simplemolecules.
There are 2 types of Van Der Waals forces namely
Permanent Dipole Permanent dipole forces Temporary dipole induced dipole forces
7.1.1 Dipole-dipole attraction forces
1. Polar molecule possessed dipole moment. Each of the polar
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molecules have an overall magnitude. For example in hydrogenchloride
H Cl
+
2. The dipole inside polar molecules is permanent and the forces
between the molecule form as the positive end of dipole will attract to
the negative end of another molecules dipole.
3. This kind of forced are called permanent dipole-dipole forces.
4. The strength of the attraction depends on two factors : dipole moment
and relative molecular mass
5. Higher the dipole moment the more polar the molecule
stronger the Van Der Waals forces
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6. Comparisons were made between 4 molecules that have nearlyequaled of molecular mass, but with different dipole moment
7. Methyl cyanide exhibit the highest boiling point among the 3molecules as it has the highest dipole moment among these
molecules, which makes the attraction between the dipole-dipole
attraction become stronger, and required a higher temperature to
break the attraction forces among CH3CN-----CH3CN.
Compounds RMM DM Boiling point (C)
Propane , CH3CH2CH3 44 0.1 - 18.0
Methyl methoxide, CH3OCH3 44 1.3 4.0
Chloromethane 50.5 1.9 6.0
Methyl cyanide, CH3CN 41 3.9 56.0
8. Another factor which influence the strength of permanent dipole-
dipole forces, are the factor of relative molecular mass.
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9. Higher the mass, stronger the forces of attraction ( Van Der Waalsforces ), higher the boiling point or melting point of the substance
RMMMelting
point (C)
Boiling
point (C)
Hydrogen chloride, H Cl 36.5 - 114 - 85
Hydrogen bromide, H Br 81.0 - 87 - 66
Hydrogen iodide, H I 128 - 51 - 35
7.1.2 Temporary dipole induce dipole forces
Non-polar molecules have a dipole moment = 0. Basically,
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they wont have any attraction between the molecules asthere are no significant poles with charge in the molecule,so how they interact ??!!!
For non-polar molecules, they may have a chance to form
asymmetrical structure, as the distribution of electron withinthe molecule are not even, giving the atom a temporarydipole moment.
During the formation of temporary dipole moment, induction
process takes place where the distribution of electron areuneven and give the atom which are temporary rich ofelectron to form dipole. These dipoles also known asinduce dipole.
When induced dipole is formed , a temporary interactionbetween the molecules formed and produces weak forcesamong them.
This theory is introduced by Frite London in 1930. It is known as
London dispersion forces.
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In (a) the non-polar molecule which does not have a dipole within
the molecule begin to fluctuate and thus forming a temporarydipole as in (b). Thus the forces of attraction will formed between
the temporary dipole and this forces is named as London Forces
7.2 Effect of the intermolecular forces ( Van der waals ) on the physical
properties of the molecules
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H vapourisation give a quantitative measurement of strength ofattractive forces present in liquid. So, H vapourisation , the boiling
point , the intermolecular forces among its molecules.
When a molecule increase in size, the number of electron also
increase, so the attraction between the electron valence and nucleusbecome less. This distortion of electron cloud can easily occur and
increase the polarisability of the negative ion.
This can be relating with the dispersion forces among molecules
thereforeH vapourisation , e.g. : Value of boiling point of halogen gas
increase. ( from F2 I2)
In hydrocarbon, boiling point increase with relative molecular mass
(RMM). Molecule with higher RMM will have a higher boiling point. The effect of branched chain in hydrocarbon will also affect the boiling
point of hydrocarbon involved
Structure RMM Boiling point
(C)
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This is due to a larger surface area in a straight chain of
hydrocarbon, and allows greater forces between the molecules
giving larger Van der Waals forces compare to branch chain
hydrocarbon
2,2dimethyl
propane72 4
2-methylbutane 72 18
npentane CH3 CH2 CH2 CH2 CH3 72 36
7.3 Hydrogen Bonding
Hydrogen bond is a special dipoledipole interaction between H
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atom with otheratom with high electronegativity. ( N, O, F )
It is extra stable than normal Van der waals forces and required a
high energy to break the bond. This explained why the boiling point
of NH3, H2O and HF are higher than other hydrogen compound
from each of their particular group.
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Hydrogen bond can also be used to explain the different of boiling
point of some organic compound. In the diagram above, the trend of
th d i th d i t f N O d F it f
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the compound in the same group deviates for N, O and F, as it formhydrogen bond among themselves.
Hydrogen bond can be compared among NH3 , H2O and HF. HF has
a higher boiling point than NH3 due to higher electronegativity of
fluorine compare to nitrogen. So the dipole moment of HF is greaterthan NH, which results greater hydrogen bond. Though, O has a
lower electronegativity than F, but H2O has a greater boiling point
compare to HF because in between H2O ---- H2O molecules, they can
form 2 hydrogen bond between the molecule but between HF --- HFcan only form one hydrogen bond. So, the more the hydrogen
formed, greater the forces, higher the boiling point.
The factors of hydrogen bonding can also use to explain the solubility
of some organic compound in water, like example, ethane cannot
dissolve in water but ethanol can dissolve in water, due to the
hydrogen bonding.
Some of the molecules gain more stability by forming dimerwith its
molecules. E.g. : When ethanoic acid is brought to mass
t t f d t ti d it i k t / t 120 Thi
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spectrometer for detection and it gives a peak at m/e at 120. Thisindicates the shows that ethanoic acid (CH3COOH) has a RMM of
120, as CH3COOH , RMM = 60.
This indicate ethanoic acid exist as dimer where interaction of
hydrogen bonding between end of each functioning group COOH
occur.
There is another application of hydrogen bond, which is the
intermolecular forces and intramolecular forces. In 2-nitrophenol
and 4 nitrophenol the boiling point of the 2 compo nds can be
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and 4-nitrophenol, the boiling point of the 2 compounds can beexplain below :
Since 2-nitrophenol form strong hydrogen bond as intramolecular
forces, the interaction between 2-nitrophenol molecules are
weaker among each other, compare to 4-nitrophenol, which used
hydrogen bond as their intermolecular forces. With strongerhydrogen bond which act as the intermolecular forces, the boiling
point of 4-nitrophenol is expected to be higher than 2-nitrophenol