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Published by Farahiyah Omar, 2019-07-21 20:41:45

overview

overview

LEWIS DOT SYMBOLS

Consist of the symbol of an element and one dot for each
valence electron in an atom of the element

❑ Dot ☞ valence e-

EXAMPLE: ·· ·· ·· ·
· N· Cl
· Na ··
·

Octet Rule : An atom other than H tends to form bonds

(by losing or gaining or sharing e-) until it is surrounded
by eight valence e-

Exception to Octet Rule Stability Electron Configuration of
Ions

Exception to Octet Rule

Incomplete octet Expanded octet Odd electron

Central atoms have less Central atoms have more Contain an unpaired
than eight e- around them than eight e- around them e-

Stability Electron
Configuration of Ions

Noble gas Pseudonoble gas Half–filled
configuration configuration orbitals

Atoms may lose or gain enough The (n–1)d10 configuration Some transition metal
e- so as to forms stable ion with of a p–block metal atom atoms form cations that
that empties its outer level
octet (or duplet) configuration have e- configuration
☞ ns2 np6 associated with half–filled

EXAMPLE: 1s2 2s2 2p6 EXAMPLE: [Kr]4d10 d orbital (d5)

EXAMPLE: 3d5

INCOMPLETE
OCTET

EXPANDED ODD–ELECTRON
OCTET
MOLECULES

Writing Lewis Structure

Formal
Charges

Predicting the most plausible RESONANCE
Lewis structure
STRUCTURE
Higher priority Two or more Lewis structures for
a single molecule that cannot be
• All zero formal charge represented accurately by only
• Small formal charge one Lewis structure
• Negative formal charges are placed
EXAMPLE: ozone (O3)
on the more electronegative atoms
·· ·· ·· ·· ·· ·· ·· ··
OO O ·O· O O
·· ·· ··

I II

Chemical Bonds

Ionic Bond Covalent Bond Coordinate
Covalent Bond
Metal with non–metal
Electron transfer Non–metal with non–metal

Electron sharing

A shared electron pair is considered

to be localized between the two atoms

Metal loses valence electrons Formed when one of the atoms
- becomes cation, donates both e-
non–metal gains electrons
- becomes anion Also called covalent dative bond

Bond Length

Distance between nuclei of two
covalently bonded atoms in a molecule

Bond length: single > double > triple

Molecular VSEPR Theory Linear
Shape & Basic Molecular Trigonal Planar
Polarity Tetrahedral
Shape Trigonal bipyramidal
Octahedral
Molecular Shape V-shaped
Trigonal pyramidal
Bond Angle T-Shaped
Dipole Moment See-saw
Molecular & Bond Square pyramidal

Polarity Square planar

Dipole moment (μ)
Resultant Dipole moment
(μ)

VSEPR THEORY Valence Shell Electron Pair Repulsion
-Each group of valence electrons around a central atom
is located as far away as possible from the others in order to

minimize repulsion

To predict the molecular shape from the Lewis structure

The followings count as one e- group: Bonding pair

e- GROUP ARRANGEMENT Lone pair
Lone e-
e- Groups Arrangement
2 Linear Determined by number of e- groups
3 Trigonal planar around the central atom
4 Tetrahedral
5 Trigonal bipyramidal
6 Octahedral

e- GROUP REPULSION

Number Number Number E- Group Molecular Example
of e- of of Lone Arrangemen Shape and

Group Bonding Pair t Bond Angle
Pairs
0 linear linear
22 180o 180o

CB C
le l

3 3 0 trigonal trigonal F

planar planar 120o
120o
B
FF

2 1 trigonal V–shaped ••

planar (bent) Sn

< 120o Cl 95o Cl

Number Number Number E- Group Molecular Example
of e- of of Lone Arrangement Shape and

Group Bonding Pair Bond Angle
Pairs

4 4 0 tetrahedral Tetrahedra H 109.5o

l

109.5o C H
H

H

3 1 tetrahedral trigonal ••

pyramidal N
HH
107.3o
(< 109.5o) H 107.3o

2 2 tetrahedral V–shaped •• ••

(bent)

104.5o O
(< 109.5o) H 104.5o H

Number Number Number E- Group Molecular Example
of e- of of Lone Arrangement Shape and
Group Bond Angle
Bonding Pair
5 Pairs
0 trigonal trigonal C
5 1 bipyramidal bipyramidal
120o, 90o C l 90o
4 trigonal 120o l
see saw P C
3 C 90l o
C
l lF

86.8o

bipyramidal < 120o, < 90o F S

101.5o •
••
F

F

2 trigonal T–shaped F 86.2o
Br F
bipyramidal < 90o •• • • •
• ••
2 3 trigonal Linear F
I
bipyramidal 180o
180o

I

I

Number Number Number E- Group Molecular Example
of e- of of Lone Arrangement Shape and
Group 90o F F
Bonding Pair Bond Angle F
6 Pairs F S
0 octahedral octahedral F F
6 F F
90o 81.9o F

5 1 octahedral square F
I
pyramidal
< 90o

F ••
••
4 2 octahedral square ••
planar F F

90o 90o Xe

FF

BOND POLARITY - Atoms with different electronegativities form
polar bonds

- Depicted as a polar arrow:

MOLECULAR - Net imbalanced of charge
POLARITY - e- rich regions (δ–) and e- poor regions (δ+)

DIPOLE - Quantitative measure of molecular polarity
MOMENT (μ) μ=Qxr

RESULTANT DIPOLE - Determined by molecular shape and

MOMENT (μ) bond polarity
μ > 0 ☞ polar, μ = 0 ☞ nonpolar

A molecule will be nonpolar if ……

The bonds are nonpolar ●● ●● ●● ●● ●●●F●●●

CI CI
●● ●●

No lone pair in the central atom and all ●●●●F●● B ●F●●●●●
the surrounding atoms are the same ••
H
A molecule in which the central atom has N
lone pair e- will usually be polar with few H
exceptions
H
●●●X●●●
●●●●X●● •• ●●
•• •• A EXCEPTION•• ●●●●X●● A
●X●●●
●●X ● •• ●●X●●●●

●●

Orbital Overlap Describe the formation of sigma (σ)
& and pi (π) bonds from overlapping
orbitals
Hybridization
Draw and explain the formation of
hybrid orbitals for central atom

sp, sp2, sp3, sp3d & sp3d2

Draw orbitals overlap and label
sigma (σ) and pi (π) bonds of a
molecule

VALENCE BOND THEORY A theoretical model that explain covalent bond
(VB)
Covalent bonds - formed by sharing electrons from overlapping

atomic orbitals
- Due to the overlapping, electrons are localized in

the bond region
- Orbital overlap ↑, Bond strength(stability) ↑

SIGMA (σ ) PI (π)
BOND
BOND
❑ Resulting from end–to–end overlap
❑ Has two regions of e- density
❑ Allow free rotation ☞ one above and one below
❑ Has highest e- density along the bond axis
the
❑ All single bonds are σ bond ❑ π boσn–dborensdtraicxtiss

rotation

EXAMPLE:

π

2p 2p

HYBRIDIZATION Process of mixing of two or more orbitals to form a
new set of equivalent hybrid orbitals

TYPES OF HYBRID • sp2 hybrid orbital
ORBITALS Combination of one s orbital
and two p orbital
• sp hybrid orbital
s
Combination of one s orbital and one p orbital orbital

2 2p two sp p three sp2

▪ Eslectron gxroup = 2 orbitals orbitals orbitals
▪ Electron group = 3

▪ Electron group arrangement = Linear, ▪ Electron group arrangement = trigonal

180 º planar,

120 º

180º 120 º sp2 hybrid
orbital

Al

sp hybrid B sp2 hybrid sp2 hybrid
orbital sep hybrid orbital orbital

orbital

• sp3 hybrid orbital • sp3d hybrid orbital

Combination of one s orbital Combination of one s orbital, three p
and three p orbital orbital and one d orbital

= five equivalent sp3d hybrid
▪ Eolerbctitraoln group = 5
▪ Electron group arrangement =

trigonal bipyramidal
120 º , 90 º

one s orbital four sp3d hybrid orbital
+ 90 º
equivalent
three p sp3 orbitals
orbital
sp3d
▪ Electron group = 4 hybrid P sp3d
orbital hybrid orbital
▪ Electron group arrangement = tetrahedral, 120 º

109.5 º

sp3 hybrid

109.5 º orbital

sp3 hybrid C sp3 hybrid
orbital
orbital

• sp3d 2 hybrid orbital Type of e- e- group
hybrid group arrangement
Combination of one s orbital, three p orbitals
orbital and two d orbital Linear
sp 2 Trigonal
= 6 equivalent sp3d hybrid planar
▪oErlbeictatrlon group = 6 sp2 3 Tetrahedral
▪ Electron group arrangement =
octahedral, 90 º Trigonal
bipyramidal
sp3d2 hybrid octahedral
orbital

sp3d2 hybrid S sp3 4
orbital sp3d 5
sp3d2 hybrid
orbital

sp3d2 6

DETERMINING HYBRIDS ORBITALS

Molecular Lewis Molecular shape &
Formula Structure e- group

arrangement

Use VSEPR Lone pair Hybrid
theory to around orbitals
central atom
determine the No lone pair
molecular around central
shape
atom

Molecular shape = e- group arrangement

FORMATION OF π BOND σσ

in Ethene, C2H4 C σ C
σ
• Besides sp2 hybrid orbitals, each carbon atom σ
also has a p unhybridised orbital and oriented
at 90º angle to the plane of the sp2 orbitals. Unhybridized
• One lobe of this p orbital is above the plane, p orbitals
and the other lobe is below the plane of sp2
hybrid orbitals π bond
• The p orbitals from the adjacent carbon atoms
undergo sideways overlapping to form pi ( π ) H sp2 C sp2 sp2Csspp22 H
bond. sp2
HH

FORMATION OF π BOND 2p

in Ethyne, C2H2 z

• Each carbon atom still has two unhybridized 2py
2p orbitals (py and pz), oriented at right
angles to each other and the axis of the sp
hybrid orbitals.

• These p orbitals undergo sideways overlapping
to form a pair of π bond.

2pz π bond 2pz

σ sp 2py 2py

sp sp σ

π bond

FORMATION OF π BOND

Benzene, C6H6 sp2 C C sp2

Csp2 C
sp2 C
sp2
Unhybridised
C 2 pz orbitals

sp2

• Benzene molecule consists of a flat, hexagonal π bonds
ring of six carbon atoms with bond angle 120º
Unhybridised
• The six carbon atoms is sp2 hybridised and form 2 pz orbitals
σ bonds with another and six hydrogen atom

• Each carbon atom in the benzene ring still has an
unhybridised 2pz orbital which contains single electron.
The 2pz orbitals are oriented perpendicular to the plane
of the benzene ring.

• Each 2pz orbital undergoes sideways overlapping with two
neighbouring 2pz orbitals to form π bonds. Thus, three
π bonds are formed.

Intermolecular Dipole-dipole
Forces London Forces

Van Der Waals Forces
Hydrogen Bond

Intermolecular Factors That Molar Mass
Forces Influence VDWF Molecular Shape

Effects of Hydrogen Boiling Point
Bond Solubility
Density

Intermolecular Forces Vs Explain the
Vapor pressure relationship
&

Vapor pressure Vs Boiling
point

TYPES OF Hydrogen Bonding
INTERMOLECULAR FORCES
- Force between partially positive H of in
polar bond such with N–H, O–H or F –H
and partially negative lone pair e- on
N, O, F of another molecule

Van der Waals Dipole-Dipole
Force

-Attractive force between
polar molecules
-The positive pole of one molecule attract
the negative pole of another

London Dispersion
Force

-Results from instantaneous dipole –
induced dipole
-Also called dispersion force

FACTORS THAT INFLUENCE THE STRENGTH
OF VDWF

Molecular shape Polarizability (size / molar
mass)

For nonpolar substance with the •Molar mass ↑
same molar mass: • Size ↑
• Number atoms ↑
-Surface area of m- olecules ↑ • Number of e- ↑
-Contact between molecules ↑ • Polarizability ↑

-London force ↑

• LF ↑

SUMMARY

LF exists between all particles DD force exists only in polar molecules
•Molar mass ↑ Strength ↑
•Also depend on molecular shape •Adds to the effect of LF

Effects of Hydrogen
Bond

Boiling Point Solubility

Different no.of HB Same no. of HB Capability of any
No. of HB↑ Bp↑ compounds that can form

H-bonding with water
molecules

“like dissolves like”

Shape Molar mass Density
Molar mass ↑ Bp ↑
Branch -Ice is less denser than water.
-surface area ↓ Straight Chain
-bp ↓ -surface area ↑ - Geometric arrangement of H bond
-bp ↑ in water, ice has an open,
hexagonally shaped crystal structure.

Intermolecular Forces vs Vapor pressure
&

Vapor pressure vs Boiling point

-The weaker intermolecular forces, the

more volatile the liquid (the higher its

vapor pressure)
☞ the lower boiling point

INTERMOLECULAR BOILING POINT ↑
FORCES ↑
α

VAPOR

PRESSURE ↓

Metallic Properties of Malleability
Bond metal Ductility
Electrical Conductivity
Formation of Thermal Conductivity
metallic Bond
Electron sea Model

Factors affecting Strength of MB
Metallic Bond Boiling Point

Effect of Metallic
Bond

Metallic Bond

Properties of metal Formation of Metallic bond Factors affecting
Metallic bond
All metal atoms in the sample contribute their
valence e- to form an “electron sea” that is Number of valence e-:
delocalized throughout the substance
• Valence e- ↑ delocalized e-
(mobile e-)↑

•Attraction between
nucleus–delocalized e- ↑

• Strength metallic bond ↑

metal conduct heat and Metal are meleable and Size of atoms:
electricity well in both the ductile because:
solid and liquid states •Size ↓
because: • When a piece of metal deformed by •Attraction between
a hammer, the metal ions (nuclei) slide
metals atoms have past each other through the e- sea to nucleus–delocalized e- ↑
delocalized e- new lattice positions •Strength metallic bond ↑

Most solids have moderate mp and • So, the metal ions (nuclei) do not repel Effect of Metallic bond
higher bp due to: each other

•the attraction between moveable
cations (nuclei) and mobile e- need not
be broken during melting

•Boiling requires each cation (nuclei) and
the mobile e- to break away from the

others
☞ the metallic bond is strong enough
to resist separation of the atom


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