Chemistry Form 6 Sem 1 03a

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


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


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


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


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


Arrangement : tetrahedral

Shape : tetrahedral


Arrangement : octahedral

Shape : octahedral


Arrangement : trigonal bipyramidal

Shape : linear

f) ICl3 g) SbCl52-


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i) PCl5 j) CO32-



Shape : T shape

5 Surround Atom + 1 Lone pair e-Arrangement : octahedral

Shape : square pyramidal



Shape : trigonal bipyramidal


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

Chemistry Form 6 Sem 1 03a

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Transcript of Chemistry Form 6 Sem 1 03a