Bonding and Structure
How atoms bond together
The bonds between atoms always involve their outer
electrons
 Inert gases have ____ outer shells of electrons and
are very__________.
 When atoms bond together they share, pool or
transfer electrons to get full outer shells, making them
more ______ and less ________, like the inert gases.
 There are three types of strong chemical bonds: ionic,
covalent and ________.
 Mostly we show these bonds through drawing _____
structures, these are ‘dot and cross diagrams’.
metallic
full
reactive
Lewis
stable
unreactive
Ionic Bonding
Contents
 Describe the ionic bond as the electrostatic
attraction between oppositely charged ions.
 Describe how ions can be formed as a result of
electron transfer.
 Deduce which ions will be formed when elements
in group 1, 2 and 3 lose electrons.
 Deduce which ions will be formed when elements
in groups 5, 6 and 7 gain electrons.
 State that transition elements can form more
than one ion.
Ionic Bonding
 Ionic bonding occurs between metals and non-
metals and involves transfer of _________ from
metals to non-metals.
 Sodium chloride has ionic bonding.
 Remember sodium has ___ electrons, its electron
arrangement is _______.
 Chlorine has _____ electrons, its electron
arrangement is _________.
17 drawelectrons
11 electrons.
2,8,7
 Now
these atoms 2,8,1
showing all the
An electron is transferred. The single outer electron
of the sodium atom moves into the outer shell of the
chlorine atom
+
Na
-
Cl
+
2, 8 , 1
2, 8, 87
-
-
+
Cl
Na
2, 8
+
2, 8, 8
-
Each outer shell is now ____. Both sodium and chlorine have an inert gas
electron structure.
The two charged particles are called _____.
The sodium ion is ________charged, because it has _____an electron.
The chloride ion is ________charged, because it has _____an electron.
The two ions are attracted to each other and to other oppositely
charged ions in the sodium chloride compound by ____________forces
gained
full
electrostatic
positively
lost
ions
negatively
 The number of electrons lost or gained will depend on the
number of electrons in the outer shell of the atom.
2+
Mg
2-
O
 Mg has 2 electrons in its outer shell, it can therefore lose
2 electrons and become a 2+ ion
 Oxygen has 6 electrons in its outer shell, it can therefore
gain 2 electrons and become a 2- ion
+
Na
O
+
Na
2-
Look what happens when sodium
bonds with oxygen?
Sodium can only donate one
electron, but oxygen needs two
One oxygen atom needs to react
with two sodium atoms to balance
the charges.
In summary
 Ionic bonding is the result of electrostatic
attraction between oppositely charged ions.
 The attraction extends throughout the compound
making a structure called a lattice.
 The formula for sodium chloride is NaCl, because
we know that for every one sodium ion there is
one chloride ion.
 The formula for magnesium oxide is MgO, because
there is one magnesium for every oxide.
 The formula for sodium oxide however is Na2O
because two sodium ions are needed for each
oxide ion
Common ions formed
 Metals always form positive ions, by losing electrons
 Non-metals always form negative ions by gaining
electrons
Element
Ions formed
+1
+3
+2
-1
-3
-2
Group 1
Group 2
Group 3
Group 5
Group 6
Group 7
Transition elements
variable
Common ions formed in the
transition elements
Sc
+III
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
+I
+I
+I
+I
+I
+I
+I
+I
+II
+II
+II
+II
+II
+II
+II
+II
+III
+III
+III
+III
+III
+III
+III
+III
+IV
+IV
+IV
+IV
+IV
+IV
+IV
+V
+V
+V
+V
+V
+VI
+VI
+VI
+VII
Zn
+II
Contents
 State the formula of common polyatomic ions
formed by non metals in period 2 and 3.
 Describe the lattice structure of ionic compounds.
Other polyatomic complex
ions
Name
hydroxide
sulphate
nitrate
Formula
NO3MnO4OH-
carbonate
SO42-
phosphate
Cr2O72-
manganate (VII)
dichromate (VI)
PO43CO32-
Properties of ionic compounds
– melting point
 Ionic compounds are always solids. They have
giant structures and therefore high melting
points.
 This is because in order for them to melt an ionic
compound, energy must be supplied to break up
the lattice of ions.
Properties of ionic compounds
– conducting electricity
 Ionic compounds conduct electricity when molten
or dissolved in water (aqueous) but not when solid.
 This is because the ions that carry the current
are free to move in the liquid state but are not
free in the solid state.
-
+
+
-
+
-
Cathode -
+
-
-
+
Anode +
Cathode -
-
-
+
-
+
-
+
-
+
+
Anode +
LIQUID
SOLID
Properties of ionic compounds
– hardness
 Ionic compounds tend to be brittle and shatter easily
when given a sharp blow. This is because they form a
lattice of alternating positive and negative ions.
 A blow in the direction shown may move the ions and
produce contact between ions with like charges.
Covalent Bonding
Contents
 Describe the covalent bond as the electrostatic
attraction between a pair of electrons and
positively charged nuclei.
 Describe how the covalent bond is formed as a
result of electron sharing.
 Deduce the Lewis (dot and cross) structures of
molecules and ions for up to four electron pairs on
each atom.
 State and explain the relationship between the
number of bonds, bond length and bond strength.
Covalent bonding
 Covalent bonds form between __________atoms.
 The atoms _____ some of their outer electrons
so that each has a full outer shell of electrons.
 A covalent bond is a shared _____ of electrons.
 We can represent one pair of shared electrons in
a covalent bond by a line, Cl – Cl.
 Many non-metal elements exist as ________
molecules, covalently bonded to each other
diatomic
non-metal
pair
share
Forming molecules
 A small group of covalently bonded atoms is called
a molecule.
 For example chlorine is a gas which is made of
molecules, chlorine has 17 electrons and an
electron arrangement 2,8,7.
 Two chlorine atoms make a molecule.
 The two atoms share one pair of electrons
 Each atom now has a full outer shell.
 The molecule does not contain charged particles
because no electrons have been transferred from
one atom to another.
Cl
Cl
Cl
Cl
 The Cl – Cl bond is formed from the sharing of two
electrons; one from each of the chlorine atoms to
form the diatomic, covalent molecule, Cl2
Methane
 Methane gas is a covalent compound of carbon and
hydrogen.
 Carbon has 6 electrons with electron arrangement
2,4 and hydrogen has just one electron.
 In order for carbon to get a full outer shell,
there are four hydrogen atoms to every carbon
atom.
H
H
Four hydrogen's
are needed for
each carbon
atom
C
H
Hydrogen has just 1
electron in its outer
shell.
H
H
Carbon has the
electron
arrangement 2,4
Double Covalent Bonds
 In a double bond, four electrons are shared.
 For example, the two atoms in an oxygen molecule
share two pairs of electrons, so that the oxygen
atoms have a double bond between them.
O
O
Activity:
 Draw the triple bond between two nitrogen atoms
Properties of simple
covalent molecules
 Substances composed of molecules are gases,
liquids or solids with low melting points. (The
strong covalent bonds are only between the atoms
in the molecules, not between the molecules
themselves).
 They are poor conductors of electricity (because
there are no charged particles to carry the
current).
 If they dissolve in water, the solutions are poor
conductors of electricity. (Again there are no
charged particles).
Dative Covalent Bonding
 A single covalent bond consists of a pair of electrons
shared between two atoms.
 In most covalent bonds, each atom provides one of the
electrons, but in some bonds, one atom provides both the
electrons.
 This is called dative covalent bonding, it is also called
coordinate bonding.
 In a dative bond;
 The atom that receives the electrons is electron
deficient.
 The atom that is donating the electrons has a pair of
electrons that is not being used in a bond, called a lone
pair.
The ammonium ion
 For example, ammonia, NH3, has a lone pair of
electrons. Draw the Lewis structure of ammonia.
H
Lone pair
Ammonium ion
+
H
H
N
H
H+
H
N
H
H
 In the ammonium ion, NH4+, the nitrogen uses its lone
pair of electrons to form a dative bond to an H+ ion (a
‘bare’ proton with no electrons at all and therefore
electron deficient).
 Dative covalent bonds are represented by an
arrow,
. The arrow points towards the atom
that is receiving the electron pair. But this is only
to show how the bond was made.
 The ammonium ion is completely symmetrical and
all the bonds have exactly the same strength and
length.
 Dative bonds have exactly the same strength and
length as ordinary covalent bonds between the
same pair of atoms.
 The ammonium ion has covalently bonded atoms,
but is a charged particle. Ions like this are called
complex ions
Activity:
 Draw Lewis structures showing all electrons for
the following covalent compounds:
 Water, H2O
 Carbon dioxide, CO2
 Fluorine, F2
 Ethane, C2H6
 Ammonia, NH3
Shapes of covalent
molecules
Contents
 Predict the shape and bond angles for species
with four, three and two negative charge centers
on the central atom using the valence shell
electron pair repulsion theory (VSEPR).
 Predict whether or not a molecule is polar from
its molecular shape and bond polarities.
The shapes of covalent molecules
 Molecules are three dimensional and they come in
many different shapes.
 We can predict the shape of a simple covalent
molecule – for example, one consisting of a central
atom surrounded by a number of other atoms – by
using the ideas that:
 A group of electrons around an atom will repel all
other electron groups.
 The groups of electrons will therefore take up
positions as far apart as possible.
 This is called the Valence Shell Electron Pair
Repulsion theory or VSEPR (the valence shell is
another name for outer shell).
 A group of electrons may be:
 A set of two shared in a single bond
 A set of four in a double bond
 An unshared (lone) pair
 The shape of a molecule depends on the number
of groups of electrons that surround the central
atoms.
 To work out the shape of any molecule you must
first draw a Lewis, ‘dot and cross’, diagram.
Two groups of electrons
 If there are two groups of electrons around the atom,
the molecule will be linear.
 The furthest away from each other the two groups
can get from each other is 180o.
 Carbon dioxide has this shape, because it has two
groups of electrons. Each group has four electrons in
a double bond between carbon and oxygen.
180o
O
C
O
O = C = O
Three groups of electrons
 If there are three groups of electrons around the
central atom, they will be 120o apart.
 The molecule is flat and the shape is trigonal planar.
Boron trifluoride is an example of this:
F
F
F
B
B
Outer electrons shown only
F
F
120o
F
Four groups of electrons
 If there are four groups of electrons, they are
furthest apart when they are arranged so that
they point to the four corners of a tetrahedron
(a triangular based pyramid) The angles here are
109.5o.
 Methane, CH4, is an example:
H
H
C
H
H
C
H
Outer electrons shown only
H
109.5o
H
H
Ammonium ion
 The ammonium ion is also shaped like a
tetrahedron.
 It has four groups of electrons surrounding the
nitrogen atom.
+
H
109.5o
+
H
H
Outer electrons
shown only
N
H
H
N
H
H
H
Molecules with lone pairs
 Some molecules have unshared (lone) pairs of
electrons. These are electrons that are not part
of a bond.
 The lone pairs affect the shape of the molecule.
 Water and ammonia are good examples of this
effect.
 There is an increase in the repulsion between the
following groups:
 Bonding pair – bonding pair
 Lone pair – bonding pair
 Lone pair – lone pair
Repulsion increases
Ammonia, NH3
 Ammonia has four groups of electrons and one of the
groups is a lone pair.
 With four groups, the ammonia
molecule, like the water molecule,
has a shape based on a
tetrahedron. In this case there
are only three ‘arms’.
 The lone pair squeezes the three
shared pairs together and the
bond angles are approximately
107o. The shape is described as
pyramidal
H
N
H
H
Outer electrons shown only
N
H
H
H
107o
Water, H2O
H
O
H
104.5o
H
Outer electrons shown only
O
H
 Look at the Lewis diagram . There are four groups
of electrons around the oxygen so the shape is
based on a tetrahedron.
 However, two of the ‘arms’ of the tetrahedron are
lone pairs that are not part of a bond.
 This results in a ‘V’ shaped or angular molecule.
Lone pairs explained
 The angles of a perfect tetrahedron are all 109.5o but lone





pairs affect these angles.
The shared pairs of electrons are attracted towards the
oxygen nucleus and also the hydrogen nucleus.
However, lone pairs are attracted only by the oxygen
nucleus and are therefore pulled closer to it than shared
pairs.
The lone pairs therefore repel more effectively than
shared pairs, and ‘squeeze’ the hydrogen together,
reducing the H-O-H angle.
An approximate rule of thumb is 2o per lone pair.
The actual bond angle in water is about 104.5o.
Contents
 Predict the shape and bond angles for species
with five and six negative charge centres using
the VSPER theory.
Shapes of five and six
negative charge centres
 VSEPR theory can be
extended to cover five
and six pairs of
electrons or negative
charge centres.
 The electron pairs will
arrange themselves
around the central atom
so that they are as
mutually repulsive as
possible.
 Five pairs will give a
trigonal bipyramid shape
with bond angles of 90˚,
120 ˚, 180˚ e.g.
phosphorus pentachloride
PCl5
 Six pairs will give an
octahedral shape with
bond angles of 90˚ and
180˚ e.g. sulphur
hexafluoride SF6
Activity
 Use the orbit molecular building system to create
a trigonal bipyramid shape and an octahedral
shape.
Cl
Cl
F
Cl
P
Cl
Cl
Trigonal
bypyramid, PCl5
F
F
S
F
F
F
Octahedral, SF6
Expanding the Octet
 The presence of five or six pairs of electrons around
the central atom implies that the octet has been
expanded. This cannot happen with the second period
elements such as nitrogen, oxygen or fluorine, but can
happen with the third period elements such as
phosphorus, sulphur and chlorine.
 This is because, in the third period, the elements have
3d orbitals, which are close enough in energy to the
3p orbitals that they are available to be utilised.
 Nitrogen, for example, forms only one chloride, NCl3,
whereas phosphorus can form two chlorides, PCl3 and
PCl5.
Non-bonding pairs
 When there are non-bonding pairs of electrons then the




same rules apply, in that non-bonding pairs exert a
greater repulsion than bonding pairs.
Consider the structure of xenon tetrafluoride (XeF4).
There are six pairs of electrons around the central
xenon atom. Four of them are bonding pairs and two are
non-bonding pairs.
This could result in two possible structures. In the first
structure the bond angle between the non-bonding pairs
is 90˚, whereas in the second structure it is 180˚,
attempt to draw these two structures. Which structure
will exist? What is the name of this shape?
The actual shape of xenon tetrafluoride is therefore
square planar
Bond Length, Strength
and Resonance Hybrids
Contents
 State and explain the relationship between the
number of bonds, bond length and bond strength.
 Describe resonance hybrid structures.
 Draw Lewis diagrams for resonance hybrid
structures
Bond length and strength
 The more pairs of electrons that are shared
between two atoms in a bond then the _______
the bond and the _________the bond length.
 In almost all cases single bonds are _______ and
__________ than double bonds, and triple bonds
are even ________ and __________ than double
bonds.
shorter stronger
weaker stronger shorter longer
Resonance hybrids
 Consider the Lewis structure for the carbonate
ion, CO32-. The carbon atom has four outer
electrons of its own, and each oxygen has six
outer electrons.
 Together with the two extra electrons this gives
a total of 24 valence electrons: that is 12 pairs.
 The structure shows that one of the carbon-tooxygen bonds is a double bond, whereas the other
two are single bonds.
 If this structure is correct we would expect one
of the carbon-to-oxygen bonds to be shorter than
the other two.
 In fact all the carbon-to-oxygen bonds have the
same length, which is intermediate between a
carbon-to-oxygen single bond and a carbon-tooxygen double bond.
 This can be explained by stating that the true
structure of the carbonate ion lies somewhere
between the three possible extreme structures,
which are called resonance hybrids.
2-
O
C
O
O
Electrons
gained from
another
species, eg 2
Na
Outer electrons shown only
The resonance hybrids of the
Carbonate ion
2-
O
O
C
O
2-
2-
O
C
C
O
O
O
O
 The true structure of the carbonate ion lies
somewhere in between these three extremes.
O
Resonance Hybrid
structures
 Activity:
 Other compounds or ions for which resonance
hybrid structures can be drawn include; sulphur
dioxide,SO2; ozone, O3; the nitrate ion, NO3-; the
nitrite ion, NO22-; and benzene, C6H6.
 Look up these structures in your text books on page
67, figure 15, and copy them into your notes.
 NB you may be asked to draw these structures in an
exam.
Contents
 Describe σ and π bonds.
 Explain hybridization in terms of the mixing of
atomic orbitals to form new orbitals for bonding.
 Identify and explain the relationships between
Lewis structures, molecular shapes and types of
hybridization (sp, sp2 and sp3).
 Describe the delocalization of π electrons and
explain how this can account for the structures of
some species.
Dual Nature of the electron
 Remember the dual nature of the electron - ‘wave
particle duality’. They possess properties associated
with both particles and waves.
similarly,
two
waves
are
exactly
out of phase
 This
means
that
theythat
show
the
same properties
of
combine
waves.
Sodestructively.
when two waves are in phase meet, they
combine constructively;
Bonding molecular orbitals.
 A similar situation arises when two atomic orbitals
combine.
 When two atomic orbitals from different atoms
combine constructively the electron density
between the two atoms increases, resulting in a
molecular orbital with a lower energy than either
of the two atomic orbitals.
 The molecular orbital is of a lower energy, and
electrons tend to fill lower energy levels first,
this is known as bonding molecular orbital.
Anti-bonding orbitals
 Similarly, if the two atomic orbitals combine
destructively the electron density between the
two atoms decreases, resulting in a molecular
orbital with higher energy than either of the two
atomic orbitals.
 In this case the electrons will tend to stay in
their individual atomic orbitals, and the molecular
orbital formed is known as an anti-bonding orbital
Hydrogen and helium
 The model below explains why hydrogen forms a
diatomic molecule when the two 1s atomic orbitals from
each atom combine, whereas helium is monatomic.
Hydrogen 1s
Helium 2s
energy
Antibonding
molecular orbital
Atomic
orbital
Antibonding
molecular orbital
Atomic
orbital
molecular
orbital
1s
orbital
Atomic
orbital
No overall change in
With lower net energy
energy so helium
the hydrogen diatomic
atoms do not form
molecule forms.
helium molecules.
Atomic
orbital
molecular
orbital
Shapes of orbitals
Do you remember the shapes of the different
orbitals?
Try and draw them, then click on the screen to see
if you were right.
s
px
py
Recall that all of them are 3-dimensional and thus
occupy a region along axes. This is particularly
important in the formation of bonds.
pz
Sigma σ, and Pi π bonds
 When covalent bonds form there is overlap of the
valence electron orbitals.
 The overlap can happen in two way:
 Head-on (along the plane of axis)
 Side –on
 Any overlapping between an s orbital with another
s orbital or a p orbital is head on, this forms a
sigma, σ bond.
 Side on overlapping can happen between two like p
orbitals (these are p orbitals in the same plane),
this form a pi, π bond.
Sigma σ bonds
Sigma bonds (σ), occur
when orbitals combine
head-on.
axis
Two s orbitals
combining
axis
An s orbital combining with a p orbital
axis
Two p orbitals combining
Pi, π bonds
z
Pi bonds form when like p
orbitals (py with py or pz with
pz) align with each other side
by side.
z
x
x
y
y
These form
electron dense
areas above and
below the plane of
the nuclei, these
are the π bonds.
 The head–on nature of the sigma bond means that if a p
orbital of one atom bonds head on with a p orbital of
another atom then no more sigma bonds can form between
those two atoms because it is not possible for more than
one set of p orbitals to meet head-on, thus only pi bonds
can form.
y axis
π
π
σ
z axis
x axis
Hybridisation
 The molecular orbital theory has to be adapted to explain




the bonding (and shape) of tetrahedral molecules such as
methane.
Carbon has the electron configuration 1s2 2s2 2p2.
Thus it is impossible for it to retain this configuration and
form four equal bonds pointing to the corners of a
tetrahedron when it combines with the 1s1 electrons from
the four hydrogen atoms.
For a start it has only two unpaired electrons, so it might
reasonably be expected to have a combining power of 2.
In addition, the two p electrons are in orbitals that are at
90˚ to each other, and so will not form bond angles of
109.5˚ when they combine with the s orbitals on the
hydrogen atoms, so the orbitals have to merge together.
3
sp
CH , exhibits
Hybridisation Methane,
sp hybridisation in its
3
σ
σ
All 4 hybrid
orbitals
arenow
used
Carbon
can
Carbon
can
now
Click
on
the
Click
on the
But
this
can
only
Click
on
the
in
forming
σ
bonds
form
4
equal
form
4 bonds
screen
totosee
how
screen
see
how
produce
2 With
bonds
screen
to
see
how
thus
the
bonds,
e.g.
but
they
are
not
anthe
electron
is
electrons
and
we
know
in
the
orbitals
hybridisation
is
hydrogen
to
equal
in energy.
promoted
to a
3 as
arrange
methane,
CHit4 is
combine
to
form
called
sp
form
CH
4
higher
energy
themselves
in a made
the carbon
hybrid
orbitals
up of 1s and
level
during
carbon atom
forms 4 equal
3p
bonding
bonds
4
bonding but what does
this mean?
σ
σ
Click on the
screen to see
how carbon
can form 4
equal bonds
with
hydrogen
2
sp
Hybridisation
σ
Click
Click on
on the
the
screen
screen to
to see
see
the
howfilling
the of
three
electrons
of the
orbitals
arrange leaving
one
themselves
p electron.
in a
carbon atom
σ
Ethene, C2H4, exhibits
sp2 hybridisation in its
bonding but what does
this mean?
σ
π
on the
When
bonding
forms
σ bonds
can
OnlyDuring
3ethene
σ bonds
are only 3 Click
to see
form,
an(remember
electron
isa σ bond screen
only forms
formed
using 3 of
when
orbitals
overlap
and 1salonghow
the the
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This
the promoted
hybrid
orbitals
orbitals
and
leaves
phybrid
orbital
on each of
the carbon
and the
2pathus
this is
electrons
2
atoms
orbitals
.sp
The
lobes
are overlap above
and
called
below
formed
the plane to form a πarrange
bond
hybridisation
themselves
sp Hybridisation
σ
Click on the
screen to see
what happens in
ethyne.
σ
Ethyne, C2H2, exhibits
sp hybridisation in its
bonding but what does
this mean?
π
π
What do you think will happen in
ethyne, when only 2 σ bonds can
form on each carbon (one with
hydrogen and one with carbon), but
breaking the C – C bond requires
more energy than just one σ bond.
Shapes of molecules and ions.
 We have seen that the shapes and bond angles of
simple molecules and ions can be determined using
VSEPR theory. They can also be determined using
hybridisation theory.
 If the hybridisation is known, then the shape and
bond angles can be predicted; similarly, if the shape
and bond angles are known, the type of hybridisation
can be deduced.
 Hybridisation can also be used to explain shapes of
molecules with 5 and 6 charge centres. This involves
mixing s, p and d orbitals to form, for example, d2sp3
hybrid orbitals for octahedral shapes.
Hybridisation, shapes and bond
angles.
Shape
Examples
Bond angle
Type of
hybridisation
Linear
N2, C2H2, HCN
180˚
sp
Trigonal planar
C2H4, BF3, SO3
120˚
sp2
Tetrahedral
CH4, NH4+, BF4-
109.5˚
sp3
Bent (parent shape
trigonal planar)
SO2, CH3COCH3
~120˚
sp2
Trigonal pyramid
(parent shape
tetrahedral)
NH3, PCl3
107˚
sp3
Bent (parent shape
tetrahedral)
H2O
105˚
sp3
Delocalisation of the pi electrons
and resonance hybrids
 Earlier we used the idea of resonance structures to
explain certain structures e.g. the carbonate ion.
 Another example of a resonance structure is the benzene
molecule. For many years chemists had a problem in
determining the structural formula of benzene, which was
known to have the molecular structure, C6H6.
 This problem was solved by Friedrich Kekule, who first
showed that benzene could be written in a cyclic or ring
form using alternate double and single bonds between the
carbon atoms.
 Later this was described using two resonance structures.
Can you draw the two resonance structures of benzene.
Resonance hybridisation theory
v molecular orbital theory.
 Resonance hybridisation theory can explain many of
the properties of benzene, and in particular the fact
that all the carbon-to-carbon bond lengths are equal
and lie between a C-C single bond and C=C double
bond, and the fact that benzene is more stable
energetically and less reactive chemically expected.
 The Molecular orbital theory can also explain the
structure of benzene and account for even more of its
properties.
Molecular orbital theory
and benzene
 Each atom in benzene is sp2 hybridised so that it
forms three sigma (σ) bonds, one to a hydrogen atom
and one each to the adjacent carbon atoms to form a
planar hexagonal ring.
 The remaining electron on each carbon atom is in a p
orbital vertically above and below the plane of the
ring.
 Instead of pairs of p orbitals combining together to
give alternate double and single bonds, all six p
orbitals combine to form one ring – shaped molecular
orbital containing six delocalised pi electrons.
Pi delocalisation in benzene
The pi electrons above
and below the plane
become delocalised to
form a ring of
delocalised electrons
 This is more stable by about 150 kJmol-1 this is known as the
delocalisation enthalpy or resonance energy.
 Like the resonance hybrid model it explains why benzene does
not readily undergo addition reactions, because this extra
amount of energy would need to be put in to break down the
delocalised pi bond. But unlike the resonance hybrid model it
explains why benzene does react with electrophiles.
 These are attracted to the delocalised pi electrons above and
below the plane of the ring to bring about substitution
reactions.
Ozone
 The delocalisation of pi electrons can always be used as
an alternative explanation to resonance hybrids. For
example, in ozone the two resonance hybrids are formed
depending on which two out of the three oxygen atoms
combine to form the double bond.
 Using molecular orbital theory the p atomic orbitals on
all three of the oxygen atoms combine together to form
a “banana” shaped molecular orbital containing two
delocalised pi electrons.
Activity:
 Other examples of the delocalisation of pi electrons
include; the nitrate ion, NO3-; the nitrite ion, NO2-;
the carbonate ion, CO32-; and the ethanoate ion,
CH3COO-.
 Draw the two resonance structures of the ethanoate
ion, and then draw a diagram to show how the p
orbitals on the carbon atom and two oxygen atoms
can combine to form delocalised pi bonding.
Electronegativity and the
nature of bonds
Contents
 Predict whether a compound of two elements
would be ionic from the position of the elements
in the periodic table or from their electronegativity values.
 Predict whether a compound of two elements
would be covalent from the position of the
elements in the periodic table or from their
electronegativity values.
 Predict the relative polarity of bonds from
electronegativity values.
 Compare and explain the properties of substances
resulting from different types of bonding.
The nature of the bond.
 We can predict whether a bond between two elements will be




ionic or covalent by looking at the ____________ in
electronegativity values.
Elements in groups 1, 2 and 3 tend to have ______
electronegativity values.
Elements in Groups 5, 6 and 7 tend to have ______
electronegativity values.
The __________ the difference in electronegativity values,
then the more likely it is that the bond will be _________.
Generally, the difference in electronegativity value between the
two elements needs to be about 1.8 or ________ than for ionic
bonding to occur.
greater
difference
low
ionic
high
greater

Copy out the table and fill in the data to determine
the type of bonding present in these structures
Mg
Electronegativity
Difference
Formula
Type of bonding
Cl
1.2
3.0
Be
Cl
1.5
3.0
Al
O
1.5
3.5
Al
Cl
1.5
3.0
1.8
1.5
2.0
MgCl2
BeCl2
Al2O3
Al2Cl6
ionic
covalent
ionic
covalent
1.5
Exceptions to the rule
Activity:
 Look up and record neatly in your notes the
electronegativity values for lead and bromine.
 Calculate the difference between these two
values.
 What type of bonding would you expect this
compound to display, explain your answer?
 Do the same for boron and fluorine.
 Lead bromide is ionic and yet the difference in
electronegativity values is only 1.0, whereas boron
trifluoride is not ionic and yet the difference in
electronegativity values is 2.0.
 Many ionic compounds show some covalent character,
and many covalent compounds show some ionic
character, so often it is not a case of ionic or covalent
but somewhere in between.
 Generally if a compound conducts electricity when
molten or in aqueous solution then ionic bonding
predominates and if a compound is a poor conductor of
electricity when molten or in aqueous solution then
covalent bonding predominates
Polarity of bonds
 When a single covalent bond is formed between
two atoms of the same element, for example a
chlorine molecule, Cl2, the electron pair will on
average be shared equally between the two atoms,
and the bond will be non-polar.
Cl
Cl
Electron pair shared
equally
 However, if two different atoms are bonded
together covalently, for example a hydrogen
chloride molecule, HCl, then the nuclei of the
different atoms will exert different attractive
forces on the electron pair, and it will not be
shared equally.
 The atom that attracts the electron pair more
strongly will then be slightly negatively charged,
δ-,compared with the other atom, which will be
slightly positively charged, δ+. +
δ
δ This results in a polar bond
H
Electron pair pulled
towards the chlorine
Cl
 If the polar molecule is placed between two
electrically charged plates, then it is said to have
a dipole moment, because the δ- end of the
molecule will be attracted to the positive plate
and the δ+ end of the molecule will be attracted to
the negative plate.
 The bigger the difference in electronegativities,
then the more polar the bond and the greater the
dipole moment.
 Calculate the difference in electronegativity
between Carbon and Hydrogen. Now try and
explain why methane CH4 is not a polar molecule.
Intermolecular forces
Contents
 Describe the types of intermolecular forces
(attractions between molecules that have
temporary dipoles, permanent dipoles or hydrogen
bonding) and explain how they arise from the
structural features of molecules.
 Describe and explain how intermolecular forces
affect the boiling points of substances.
Types of intermolecular
forces.
 So far you have learnt about the strong
intramolecular forces which hold atoms to each
other to form molecules, these affect the energy
in chemical reactions, but there are other forces
that control the physical properties of the
molecules. These are intermolecular forces, there
are three main types of intermolecular force:
 Dipole – dipole
 van der Waals
 Hydrogen bonding
Polarity of bonds
 To understand dipole – dipole forces you need to
understand how some molecules are polar.
 When a single covalent bond is formed between
two atoms of the same element, e.g. a chlorine
molecule, Cl2, the electron pair, on average, will be
shared equally between the two atoms, and the
bond will be non-polar.
 However, if two different atoms are bonded
together covalently, e.g. a hydrogen chloride
molecule, HCl, then the nuclei of the different
atoms will exert different attractive forces on
the electron pair, and it will not be shared equally.
 The atom that attracts the electron pair more
strongly (thus more electronegative), will pull the
electron pair toward itself making it slightly
negative, this is denoted by the symbol, δ Thus the less electronegative atom will become
slightly positive, δ+.
 The molecule is polar, if it is placed between two
electrically charged plates, the δ- end of the
molecule will be attracted to the positive plate
and the δ+ end will be attracted to the negative
plate and it is said to have a dipole moment.
 The bigger the difference in electronegativities,
then the more polar the bond and the greater the
dipole moment.
Polar molecules
 The polarity of a molecule depends on the polarity of
the bonds in the molecule and the shape of the
molecule.
 There is a degree of polarity between Carbon and
oxygen making the bond polar;
δ+
δC
O
 This molecule has polar bonds and is asymmetrical so
the molecule is polar.
 But carbon dioxide is not polar... can you suggest why.
δδ+
δO
C
O
Distinguishing between polar
and non-polar molecules
 We can distinguish between polar and non-polar
molecules by:
 Comparing boiling points; molecules with higher
boiling points but with similar molecular mass and
number of electrons will be polar.
 Comparing the effect of a charged rod on a stream
of different liquids; a charged rod will only have an
effect on polar molecules.
 Microwave radiation will only effect a polar liquid; a
microwave oven works by causing the polar water
molecules to line up with the microwave radiation.
Activity:
 Using chapter 10 of your course companion find
out the displayed formulae for the following
compounds and decide if you think they are polar
molecules.
 Cyclohexane
 Dichlorohexane
 Ethanol
 Hexane
 Propanone
 Water
 Watch the video clip on the following slide for the
answers
Electrostatic deflection of different
covalent molecules showing their polarity.
Click on the box to start movie
A piece of perspex
was charged by
rubbing it with nylon.
Different covalent
liquids were allowed
to run out of a
burette.
Their polarity was
measured by the
amount the liquid was
deflected when the
charged perspex was
brought close to it.
van der Waals forces
 For non-polar molecules there are no permanent
electrostatic forces of attraction between them.
 The forces that do exist are called van der Waals’
forces.
 The stronger the van der Waals’ forces, then the
higher is the boiling point, as more energy is
required to overcome the attraction between the
molecules and separate them.
Factors affecting the
strength of van der Waals’
 Look up and compare the boiling points of the
following non-polar molecules:
Fluorine
Chlorine
Bromine
Iodine
Methane, CH4
Ethane, C2H6
Propane, C3H8
Butane, C4H10
 Now compare the structures of the halogens and
the alkanes separately.
 Can you work out the possible factors affecting
the strength of the van der Waals’ ?
Hydrogen Bonding
 When hydrogen is bonded directly to one of the small,
highly electronegative atoms fluorine, oxygen or
nitrogen, then the polarity of the covalent bond is
very high.
 In addition, as the electron pair is drawn away from
the hydrogen atom, all that remains is the proton in
the nucleus, as there are no outer electrons.
 The negative electronegative atom of another
molecule is thus attracted by a very strong dipoledipole attraction..
 This type of very strong dipole-dipole attraction is
given its own name – hydrogen bonding.
Hydrogen bonding in water
δ-
δ-
O
O
H
δ+
H
δ
δ
O
+
H
δ-
δ+
H
O
H
δ+
H
H
-
δ+
δ+
δ+
H
O
H
-
O
H
O
H
δ+
δ
δδ+
H
δ-
δ+
H
δ+
δ+
δδ+
O
H
δ+
H
δ+
H
δ+
Strength of hydrogen-bonding
 The strength of hydrogen bonding can be
demonstrated by the hydrides of the elements of
groups 5, 6 and 7.
 Each of the first members of their respective series
– ammonia, NH3, water, H2O and hydrogen fluoride, HF
– have a much higher boiling point than the other
members of the group.
 This is particularly noticeable with water, which is a
liquid at room temperature with a boiling point of
100oC at atmospheric pressure compared with all the
other group 6 hydrides, which are gases at room
temperature and atmospheric pressure.
Hydrogen bonding for Group IV, V,
VI and VII hydrides.
 Activity:
 Using the boiling
points for the
hydrides opposite,
plot a graph of boiling
point (on the y axis)
against period number
(on the x axis).
 Describe and explain
the shape of the
graph in terms of
types of
intermolecular forces.
Hydrogen bonding for Group IV, V,
VI and VII hydrides.
The unique properties of water
 Water is in fact an almost unique liquid, because when it
freezes it expands, nearly all other liquids contract in
volume when they freeze.
 The structure of ice is very open. Each oxygen atom is
bonded to four hydrogen atoms in a giant tetrahedral
arrangement. Two of these bonds are the strong
covalent O-H bonds in the water molecule; the other
two are weaker and longer hydrogen bonds between the
2δ- charge on each oxygen atom and the δ+ charge on
each of the two hydrogen atoms from other water
molecules.
 When ice melts, the molecules can move closer
together, water has its maximum density at 4oC
Properties of giant covalent
structures
Contents
 Describe and compare the structure and bonding
in the three allotropes of carbon (diamond,
graphite and C-60 Fullerene)
 Describe the structure of and bonding in silicon
and silicon dioxide.
Giant covalent structures
 Carbon exhibits allotropy. This means that it can
exist in more than one physical form.
 The three main allotropes of pure carbon are:
 Diamond
 Graphite
 Buckminster Fullerene
 In all three allotropes the carbon atoms are
bonded covalently, but in diamond and graphite,
instead of small simple molecules, the covalent
bonds link across the carbon atoms to form a
single large (or giant) molecule.
Diamond
 In diamond each carbon is bonded equally to four
other carbon atoms to form a giant tetrahedral
structure.
 All the C-C bond lengths are equal, and there is no
plane of weakness through the structure, so
diamond is an extremely hard substance.
 All the outer electrons around each carbon atom
are localized to form the four bonds to other
carbon atoms, so diamond does not conduct
electricity, because there are no delocalized
electrons.
Diamond
All bonds are
equal length and
the bond angle is
perfectly
tetrahedral
109.5o
Graphite
 Each carbon atom forms strong covalent bonds to three




other carbon atoms in a trigonal planar structure so
that the carbon atoms link up to form hexagonal rings.
The C – C bonds in the ring are in fact stronger and
shorter than the C – C bond in diamond.
The forces of attraction between the layers are very
weak, because they are formed by delocalised electrons
that can move between the layers.
This means that graphite is one of the few non-metals
to be a good conductor of electricity
This also means that the layers can easily slide over
each other, so graphite feels waxy to the touch and is a
good lubricant.
Graphite
Three bonds are equal length and
in fact shorter than those in
diamond,their bond angles are
trigonal planar 120o
The red lines represent forces of attaction between the
layers, these are delocalised electrons that are free to move
Buckminster Fullerene
 In 1996 Robert Curl (1933 - ), Harold Kroto (1939 - ) and
Richard Smalley (1943 – 2005) were jointly awarded the
Nobel prize in Chemistry for their discovery of
fullerenes.
 The basic fullerene is C60 in which 60 carbon atoms are
joined in a combination of hexagonal and pentagonal rings
to form a sphere.
 Since their discovery, more than 1000 new compounds
involving fullerenes have been made.
 Some contain metals (e.g. Lanthanum) trapped inside a
fullerene cage; others consist of long tubes that can be
closed or open at one end. These are called nanotubes,
because they have an extremely small diameter, in the
order of one nanometer (1 x 10-9m)
Activity:
 Watch the Horizon video ‘Molecules have
sunglasses’, this gives an idea of the struggles
throughout the discovery of the fullerenes.
 It is about 50 minutes long.
Metallic bonding
Contents
 Describe the metallic bond as the electrostatic
attraction between a lattice of positive ions and
delocalized electrons.
 Explain the electrical conductivity and malleability
of metals.
Metallic bonding
 When atoms of metals bond together in the solid
state, one or more of their valence electrons becomes
detached from each atom to become delocalised.
 These valence electrons are no longer associated with
a particular atom, but are free to move throughout
the metallic structure.
 The bonding in metals thus consists of the attraction
between these delocalized valence electrons and the
remaining positive metal ions (cations).
 It is sometimes said that metals are made up of an
array of cations in a ‘sea’ of moblie electrons.
Physical properties of
metals
 The physical properties of metals include:
 Good conductors of electricity
 Good thermal conductors
 Generally high melting points but varies with the
number of valence electrons and size of the cation.
 Malleable – the ability to be beaten into shape
without breaking
 Ductile – it can be drawn into a wire
Conductors of electricity and heat
 It is because the valence electrons are no longer
located on a particular atom, but are free to move
throughout the structure, that metals are such
excellent conductors of electricity.
 If there are impurities in the metal, then this can
hinder the movement of electrons and increase
the electrical resistance: this explains why copper
needs to be refined or purified before it is used
for electrical wiring.
 The movement of electrons through metals also
enables the transmission of kinetic energy, so
metals are also good conductors of heat.
Melting points
 The melting point of metals is related to the
strength of the attractive forces holding the
cations in the ‘sea’ of delocalised electrons. This
depends on the number of valence electrons
delocalized from each atom, the size of the
cations and the way in which the cations are
packed together.
 In general, the melting point decrease as the size
of the cation increases, which explains why
melting points decrease down Group 1 (alkali
metals).
Malleablity and ductility
 Many metals are malleable and ductile.
 Both of these properties can be explained by the
cations being able to slide past each other to
arrange the overall shape of the solid.
 Because the electrons are delocalised this can
happen without significant change in the bonding
forces.
Properties related to
bonding
Contents
 Compare and explain the properties of substances
resulting from different types of bonding.
Physical properties
Type of
force
Mp / bp
Solubility in
water
Solubilty in
organic solvent
Electrical
conductor
van der
Waals
low
insoluble
soluble
Non conductor
Dipoledipole
Higher
than v d W
miscible
miscible
Non-conductor
Hydrogen
bonding
Higher
than d-d
miscible
Less soluble
Non-conductor
Ionic
high
Mostly
soluble
insoluble
Conducts when
molten or in
solution
Metallic
high
insoluble
insoluble
Good conductor
Giant
covalent
high
insoluble
insoluble
Non-conductor
except graphite