IBDP CHEMISTRY
Chapter 4
Chemical Bonding
Types of Bonding
attract electrons towards itself . It is measured on the
Electronegativity is the ability of an element to ……………………………………………………………………………………………
fluorine, F ) to a minimum of 0.7 (……………………………………
francium, Fr ).
Pauling’s Scale, from a maximum of 4.0 (…………………………………
H
He
2.2
Li
Be
Na
Mg
K
Ca
1.0
0.9
0.8
Rb
B
1.6
2.0
Al
1.3
1.0
Sc
1.5
Ti
1.6
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Mo
Tc
Ru
Rh
Pd
Ag
Cd
W
Re
Os
Pt
Au
Hg
0.8
1.0
1.2
Y
Zr
Nb
Cs
Ba
La
Hf
Ta
Fr
Ra
Ac
0.8
0.7
Sr
1.4
0.9
0.9
1.3
1.1
1.3
1.6
1.5
1.7
2.2
1.7
1.6
2.1
1.9
1.8
2.2
2.2
1.9
2.3
Ir
2.2
1.9
2.2
2.2
1.9
1.9
2.4
1.6
1.7
1.9
C
2.6
F
Ne
4.0
1.9
2.2
2.6
3.2
Cl
Ar
Ga
Ge
As
Se
Br
Kr
In
Sn
Sb
Te
2.1
2.7
I
Xe
Tl
Pb
Bi
Po
At
Rn
1.8
1.8
2.0
2.0
1.8
P
O
3.4
1.6
1.8
Si
N
3.0
2.2
2.0
1.9
S
2.6
3.0
2.0
2.2
2.6
1.1
The Periodic Table of Elements, showing the electronegativity of the various elements.
The difference in electronegativity between elements determines the type of bonding formed.
•
ionic bonding .
large difference in electronegativity results in ……………………………………………
•
electronegative .), result in …………………………………………………
covalent bonding .
small difference in electronegativity (both ……………………………………………
note: AlCl3, AlBr3 and AlI3 are covalent while AlF3 and Al2O3 are ionic.
•
metallic bonding .
electropositive .), result in …………………………………………………
small difference in electronegativity (both ……………………………………………
Metallic Bonding
Metallic bonding occurs between elements which are generally
delocalisation
………………………………………
electropositive ,
…………………………………………
resulting in a
delocalised electrons , held
cations in a ‘sea’ of …………………………………………………………
of electrons. This results in a lattice of ……………………
electrostatic forces of attraction .
together by …………………………………………………………………………………………………
Strength of a Metallic Lattice
 Number of delocalised electrons and cationic charge
e.g. Mg vs Na: Mg has a greater number of
delocalised electrons
……………………………………………………………
cationic charge ,
and a greater ……………………………………………
stronger electrostatic attractions between the metal ions and delocalised electrons.
resulting in …………………………
 Size of metal ion
smaller metallic radius and
e.g. K vs Na: Both K and Na delocalise one electron per atom, but Na has a ………………………………………………………………
stronger electrostatic attractions between the metal
higher charge density than K, resulting in …………………………
a ………………………………………………………………
ions and delocalised electrons.
Chemical Bonding
2
© IBDP HL Chemistry
Ionic Bonding
a large difference in electronegativity , resulting in a
Ionic bonding occurs between elements with ………………………………………………………………………………………………………………
transfer of electrons from the electropositive element to the electronegative element. This results in
…………………………
electrostatic forces of attraction .
oppositely charged ions held together by ………………………………………………………………………………………………
Strength of an Ionic Lattice
 Magnitude of charge
greater magnitudes of charge than Na+ and Cl– respectively,
e.g. MgO vs NaCl: Mg2+ and O2– ions have ……………………………………………………………………………………
stronger electrostatic attractions between the oppositely charged ions in MgO.
resulting in …………………………
 Size of ions
smaller ionic radii than K+ and Cl– respectively, resulting in
e.g. NaF vs KCl: Na+ and F– ions have …………………………………………………
stronger
…………………………
electrostatic attractions between the oppositely charged ions in NaF.
The two factors above are collectively referred to as theoretical lattice energy, which assumes the compound
to be completely ionic. It can be represented as the following:
L.E.theoretical 
defined as the energy
required to break one mole
of ionic compound.
q+× q−
r+ + r−
However, some ionic compounds may have some degree of covalent character. This causes deviation from
strengthen or ……………………………
weaken the lattice.
theoretical lattice energy, which could either …………………………………
 Covalent character – intermediate bonding
This is most significant when the
small and highly charged ;
• cation is ……………………………………………………………………………
large anion and hence ……………………………………………………………………………………………………………
easily polarised / distorted / skewed .
• anion is …………………
EXPECTATION
3+
Al
The partial sharing of
electrons strengthens the
attractive forces through a
better orbital overlap.
REALITY
3+
2–
Al
S
2–
S
However, it weakens the
existing ionic bond as the
charge density of each ions
are now lower.
large electron cloud of the anion
is polarised by the cation, resulting
in ‘partial sharing’ of electrons.
e.g. MgO (m.p. 2900 °C) vs Al2O3 (m.p. 2100 °C): Al3+ in Al2O3 has a greater charge than Mg2+ in MgO and is
expected to have a higher melting point. However, the
polarises the electron cloud
………………………………………………………………………………
very high charge density
………………………………………………………………………
of Al3+
partial covalent character , causing the
of O2–, which results in a …………………………………………………………………………
weaker than expected.
attractive forces to be ………………………
Chemical Bonding
3
© IBDP HL Chemistry
Covalent Bonding
Covalent bonding occurs between elements with
small or no difference in electronegativity ,
………………………………………………………………………………………………………………………
sharing of electrons. The resulting covalent bond formed is an electrostatic attraction between
resulting in a ………………………
the two nuclei and the shared pair of electrons
……………………………………………………………………………………………………………………………………….
Sigma () vs Pi () Bonds
Sigma () Bonds are
head-on overlaps
……………………………………………………
s or p ) or hybridised orbitals
between atomic orbitals (……………………
sp, sp2 and sp3 ) resulting in electron density concentrated between the nuclei of the bonding atoms.
(…………………………………………
-bond between two ‘s’ orbitals
-bond between an ‘s’ orbital and a ‘p’ orbital
-bond between two ‘p’ orbitals
-bond between two ‘hybrid’ orbitals
side-way overlaps between atomic orbitals resulting in electron density above and
Pi () Bonds are ……………………………………………………
p orbitals can form -bonds.
below the plane of the nuclei of the bonding atoms. Only …………………………………
-bond between
two ‘p’ orbitals
Bond Order
bond order = 1
bond order = 2
bond order = 3
A single bond, e.g. X–X is always a
sigma () bond.
A double bond, e.g. X=X is always one
sigma () bond and one pi () bond.
A triple bond, e.g. XX is always one
sigma () bond and two pi () bonds.
•
sigma bond .
The first bond formed between any two atoms is always a ……………………………………
•
stronger than a pi bond.
Sigma bond is ……………………………
•
Pi bond is broken first before sigma bond is broken.
Chemical Bonding
4
© IBDP HL Chemistry
Strength of a Covalent Bond
 Bond Order
one σ and one π bond , while a C–C bond consists of
e.g. C=C vs C–C: A C=C bond consists of ……………………………………………………………
one σ bond only .
………………………………………………
additional π bond in C=C, hence C=C is
More energy is required to overcome the ……………………………………………………
stronger.
 Bond Length
smaller atomic radius than silicon, hence the C–C bond is ………………………
shorter than
e.g. C–C vs Si–Si: Carbon has a ………………………
better extent of effective orbital overlap , requiring more
the Si–Si bond. The shorter bond has a ……………………………………………………….………….………….………………………………………
energy to overcome.
 Bond Polarity (Ionic Character) – intermediate bonding
Occurs due to small differences in electronegativity between the bonding atoms, resulting in unequal
polar bond.
sharing of the bonding electrons and formation of a ……………………
note: bonds formed between carbon, phosphorus, sulfur and hydrogen are considered non-polar due to
extremely small difference in electronegativities.
weaken the covalent bond.
strengthen or ……………………………
Similarly, the small degree of ionic character can either …………………………………
EXPECTATION
C
The additional
attractive forces
between the charged
dipoles strengthens the
covalent bond.
REALITY
+ C
O
O
−
However, the decrease
in extent of effective
orbital overlap weakens
the covalent bond.
The more electronegative atom pulls
electrons towards itself, resulting in
partial charges (dipoles) formed.
polar . The polar C=O bond
non-polar , while C=O is …………………
e.g. C=C (614 kJ mol-1) vs C=O (804 kJ mol-1): C=C is ……………………………
is strengthened by the
electrostatic attraction between oppositely charged dipoles ,
………………………………………………………………………………………………………………………………………………………………………
which
requires more energy to overcome.
polar . The distortion of
non-polar , while N–H is …………………
e.g. C–H (414 kJ mol-1) vs N–H (391 kJ mol-1): C–H is ……………………………
extent of effective orbital overlap , requiring less energy
the electron cloud in N–H decreases the ………………………………………………………………………………………………………
to overcome.
Chemical Bonding
5
© IBDP HL Chemistry
Solid State – Four Structures
Volatility
Forces
Particles
All substances can be placed into one of the following categories of structure:
Metallic Lattice
Ionic Lattice
lattice of positive
lattice of
ions, in a ‘sea’ of
positive and
delocalised electrons
negative ions
electrostatic
electrostatic
attractions between
Giant Covalent
Simple Molecular
lattice of
lattice of
atoms
discrete molecules
covalent bonds
covalent bonds
attractions between
within the molecule,
(electrostatic attractions
positive ions and
……………………………………
positive ions and
……………………………………
between two nuclei and
intermolecular
…………………………………………
delocalised
electrons
………………………………………………………
negative ions
……………………………………
shared pair of electrons)
Depends on ESAs:
Depends on ESAs:
Depends on CBs:
1. charge of cation &
no. of electrons
2. size of cation
molecules
1. charge of ions
1. bond order
2. size (radii) of ions
2. bond length
L.E. 
product of charges
sum of radii
forces between
……………………
Depends on IMFs:
To be learnt in
a later chapter.
3. ionic character
Electrical Conductivity
3. covalent character
Able to conduct
Unable to conduct
Unable to conduct
Unable to conduct
electricity due to
electricity when solid
electricity.
electricity.
mobile electrons .
………………………………………………
ions are held in
as ……………………………………………
fixed positions .
……………………………………………
(Exception: Graphite -
(Exception: Acids or
(The greater the no of
Able to conduct
Only three electrons in
bases which can
delocalised electrons,
electricity when molten carbon are used in
dissociate in water to
the greater the
or aqueous due to
bonding; the fourth
produce mobile ions in
conductivity.)
mobile ions.
electron is delocalised
aqueous solution.)
and is hence mobile.)
Chemical Bonding
6
© IBDP HL Chemistry
Exception to Octet rule
Expansion of octet
An element may expand octet if it is from the
because it
third period and higher
……………………………………………………………………
possesses energetically-accessible d-orbitals
……………………………………………………………………………………………………………………………………
in the Periodic Table. This is
to hold the additional electron. For
example, phosphorus can form two chlorides – PCl3 and PCl5.
phosphorus trichloride (PCl3)
phosphorus pentachloride (PCl5)
××
Cl
×
Cl
P × Cl
×
Cl × P× × Cl
××
Cl
Cl
Cl
Incomplete octet configuration
odd number of valence electrons and exist as radical.
Some elements do not fulfil octet rule and have an ……………………………………………………………………………………………
For example, nitrogen in NO and NO2.
nitrogen monoxide (NO)
×
×
×
×
×
×
×
×
nitrogen dioxide (NO2)
×
×
even number of valence electrons in the valence shell.
Some elements have ………………………………………………………………………………………………
For example, boron in BF3, aluminium in AlCl3 and beryllium in BeCl2.
boron trifluoride (BF3)
×
Chemical Bonding
×
×
aluminium chloride (AlCl3)
×
7
×
×
beryllium chloride (BeCl2)
Cl × Be× Cl
© IBDP HL Chemistry
Dative Bonding
A dative bond (also known as a co-ordinate bond) is a type of covalent bond in which one atom contributes
two electrons to a bond, while the other atom contributes none. This is in comparison to a regular covalent
bond in which each atom contributes one electron.
One example of such bonding can be seen in carbon monoxide:
dot-and-cross diagram
Lewis diagram
A dative bond (also known as a ‘coordinate
bond’) is a covalent bond in which the shared
pair of electrons originate from the same atom.
Dative bonds occur when one atom has
an energetically-accessible empty orbital ,
……………………………………………………………………………………………………………………
while the other
a lone pair of electrons available for donation . Dative bonds are always sigma () bonds,
atom has ……………………………………………………………………………………………………………………………………
and are identical in properties (including bond strength) to normal covalent bonds.
Lewis acid .
Lewis base while the electron-pair acceptor is known as ………………………………
The electron-pair donor is known as ………………………………
+
AlCl3 has not achieve octet
and hence has energetically
accessible 3p empty orbital.
(Lewis acid)
⎯→
NH3 has a lone pair electrons
available for donation.
(Lewis base)
dative bond formed between
AlCl3 and NH3
AlCl3 is able to form a dimer through the formation of dative bond between two AlCl3 molecules. This process
is known as dimerisation.
Al atom of one molecule of AlCl3 accepts one lone pair of electrons from Cl atom of another molecule of AlCl3.
This is due to the availability of empty 3p orbital on the Al atom which can accept two electrons from Cl.
Chemical Bonding
8
© IBDP HL Chemistry
Constructing Lewis Diagrams
Polyatomic anions
①
Arrange the atoms.
The atom(s) that can form the most bonds
will be the central atom.
②
Allocate the negative charge(s).
Place negative charge(s) on more
electronegative elements.
③
Fill in the bonds.
Draw bonds from each atom to its
neighboring atoms. Use dative bonds as a
‘last resort’.
④
Fill in the lone pairs.
Convert to dot-and-cross diagram if
necessary.
Example: ICl4–
Polyatomic cations
①
Arrange the atoms.
The atom(s) that can form the most bonds
will be the central atom.
②
Allocate the positive charge(s).
Place positive charge(s) on more
electropositive elements.
③
Fill in the bonds.
Draw bonds from each atom to its
neighboring atoms. Use dative bonds as a
‘last resort’.
④
Fill in the lone pairs.
Convert to dot-and-cross diagram if
necessary.
Chemical Bonding
Example: NO2+
9
© IBDP HL Chemistry
Hybridisation of Orbitals
Before an atom can form covalent bonds, its orbitals (e.g. s or p orbitals) must first be …………………………………
hybridised to form
degenerate
…………………………………
orbitals, i.e. orbitals of the same energy.
The Valence Shell Electron Pair Repulsion (VSEPR) Theory determines the arrangement of hybrid orbitals,
which are used to form …………………………
σ bonds . Unhybridised p orbitals are used to form …………………………
π bonds .
energy
energy
arrangement
hybridisation
np
ns
nsp3
energy level of s-orbital
is lower than p-orbital
sp3: tetrahedral
with no p orbitals
An ns orbital hybridised with three np orbitals,
to form four degenerate orbitals, i.e. sp3.
energy
energy
unhybridised
p orbital
np
hybridisation
nsp2
ns
arrangement
sp2: trigonal planar
with one p orbital
An ns orbital hybridised with two np orbitals,
to form three degenerate orbitals, i.e. sp2, with one unhybridised np orbital.
energy
energy
arrangement
unhybridised
p orbital
np
hybridisation
nsp
ns
sp: linear
with two p orbitals
An ns orbital hybridised with on np orbitals,
to form two degenerate orbitals, i.e. sp, with two unhybridised np orbital.
deformed dumb-bell shape.
Hybrid orbitals, i.e. sp, sp2 or sp3, have a ……………………………………………………………
A greater extent of hybridisation translates to a greater number of p-orbitals involved in the hybridisation,
which results in:
Chemical
Bondingp
• greater
10 level, weaker electrostatic attraction of
© IBDP
HL Chemistry
character, higher orbital energy
shared
pair of
electrons to the positive nuclei and weaker sigma bond formed between the hybrid orbitals.
Hybrid Orbitals
Hybrid orbitals are formed when the s and p (and sometimes d) sub-shells combine to form a new, hybrid
sub-shell. These hybrid orbitals are used to form -bonds or contain lone pairs.
6
electron domains
s

p
  
d
6 orbitals hybridised to form 6 -bonds/lone pairs
sp3d2
     
d
octahedral
sp3d2 orbital
*not in syllabus
5
electron domains
s

p
  
d
5 orbitals hybridised to form 5 -bonds/lone pairs
sp3d
    
d
trigonal bipyramidal
sp3d orbital
*not in syllabus
4
electron domains
s

p
 
4 orbitals hybridised to form 4 -bonds/lone pairs
sp3
   
3
electron domains
s

tetrahedral
sp3 orbital
p

3 orbitals hybridised to form 3 -bonds
sp2
  
2
electron domains
or
Chemical Bonding
s

p
trigonal planar
sp2 orbital
p
2 orbitals hybridised to form 2 -bonds
sp
 
linear
sp orbital
p
11
© IBDP HL Chemistry
VSEPR Theory
The Valence Shell Electron Pair Repulsion (VSEPR) Theory determines the arrangement of hybrid orbitals,
deduce the shapes of molecules or molecular ions.
hence ………………………………………………………
VSEPR: First Postulate
minimise repulsion between them.
This states that electron domains will occupy a position such as to …………………………………………………
In other words, electron pairs will “spread out” as much as possible.
6 domains
sp3d2
5 domains
sp3d
4 domains
sp3
3 domains
sp2
2 domains
sp
octahedral
(90)
trigonal bipyramidal
(90 and 120)
tetrahedral
(109.5)
trigonal planar
(120)
linear
(180)
VSEPR: Second Postulate
lone pair-lone pair repulsion is the greatest, followed by lone pair-bond pair
This states that ………………………………………………………………………………
repulsion, bond pair-bond pair repulsion, and then electron pair-single electron repulsion.
LP-LP repulsion > …………………
LP-BP repulsion > …………………
BP-BP repulsion > EP–single electron repulsion
i.e. …………………
This means that the greater the number of lone pairs a central atom has, the smaller the bond angle.
Example 1
methane, CH4
ammonia, NH3
water, H2O
electron domains
4 domains (4 BP, 0 LP)
4 domains (3 BP, 1 LP)
4 domains (2 BP, 2 LP)
109.5
107
104.5
molecular geometry
bond angle
All three molecules have the same electron geometry (sp3, tetrahedral),
but the increase in the number of lone pairs caused a decrease in the bond angle.
However, if there is a single electron (e.g. NO2), the bond angle will be larger instead.
Example 2
nitrite ion, NO2–
nitrate ion, NO3-
nitrogen dioxide, NO2
electron domains
3 domains (2 BP, 1 LP)
3 domains (3 BP)
3 domains (2 BP, 1 SE)
below 120
120
above 120
molecular geometry
bond angle
The presence of a lone pair (as opposed to a bond pair) decreases the bond angle,
while the presence of a single electron increases the bond angle.
Chemical Bonding
12
© IBDP HL Chemistry
VSEPR: Third Postulate
electronegative the central atom, the …………………………
greater the bond pair-bond pair
This states that the more ……………………………………………
repulsion. Note that the bond pair-bond pair repulsion will still be less than lone pair-lone pair repulsion (i.e.
the second postulate still holds), but to a smaller extent.
Example 3
NH3
PH3
AsH3
electron domains
4 domains (3 BP, 1 LP)
4 domains (3 BP, 1 LP)
4 domains (3 BP, 1 LP)
107
93.5
91.8
molecular geometry
bond angle
The more electronegative central atom ‘pulls’ the bonding pairs closer towards the central
atom, increasing the bond pair-bond pair repulsion and increasing the bond angle.
electronegative the
For molecules with the same central atoms but different terminal atoms, the more ……………………………………………
weaker the bond pair-bond pair repulsion.
terminal atom, the …………………………
Example 3
NH3
NCl3
NF3
electron domains
4 domains (3 BP, 1 LP)
4 domains (3 BP, 1 LP)
4 domains (3 BP, 1 LP)
107
106
102
molecular geometry
bond angle
The more electronegative terminal atom ‘pulls’ the bonding pairs closer towards the
terminal atom, decreasing the electron density around the central atom, thus decreasing the
bond pair-bond pair repulsion and decreasing the bond angle.
Chemical Bonding
13
© IBDP HL Chemistry
Shapes of Molecules
Electron geometry refers to the
orientation in space of the electron pairs
(lone pairs and bond pairs) present.
Six Electron Domains
Molecular geometry refers to the shape of
a molecule. Since lone pairs are not used in
bonding, the lone pairs are not taken into
consideration when describing its shape.
octahedral electron geometry
sp3d2 hybrid orbital
A central atom with six electron domains will result in the following shapes (molecular geometry):
Example
Draw Shapes
Shape Name
octahedral
square
pyramidal
square planar
T-shaped
90°
< 90°
90°
< 90°
Bond Angle
Five Electron Domains
trigonal bipyramidal electron geometry
sp3d hybrid orbital
A central atom with five electron domains will result in the following shapes:
Example
Draw Shapes
Shape Name
Bond Angle
Chemical Bonding
trigonal
bipyramidal
see-saw
T-shaped
linear
90°, 120°
< 90°, < 120°
< 90°
180°
14
© IBDP HL Chemistry
Four Electron Domains
tetrahedral electron geometry
sp3 hybrid orbital
A central atom with four electron domains will result in the following shapes:
Example
Draw Shapes
Shape Name
Bond Angle
tetrahedral
trigonal pyramidal
bent
109.5°
107°
104.5°
Three or Two Electron Domains
trigonal planar electron geometry
sp2 hybrid orbital
linear electron geometry
sp hybrid orbital
A central atom with three or two electron domains will result in the following shapes:
Example
Draw Shapes
Shape Name
Bond Angle
Chemical Bonding
trigonal planar
bent
linear
120°
< 120°
180°
15
© IBDP HL Chemistry
Delocalisation of Electrons
Delocalisation is defined as the sharing of electrons between
three or more adjacent atoms .
…………………………………………………………………………………………
with
parallel to each other . This usually occurs between conjugated -bonds
overlapping p-orbitals that are …………………………………………………………………
and atoms with lone pairs of electrons (in p-orbitals).
The -bond indicates two
adjacent p-orbitals here.
The three -bond in the benzene ring are
conjugated, with six p-orbitals adjacent to each
other, hence the electrons are delocalised and
shared amongst all six carbon atoms.
The lone pair indicates
another p-orbital here.
Since the three p-orbitals are adjacent,
the electrons are delocalised and shared
amongst all three atoms.
Other common examples of delocalised electrons are:
amide linkage
carbonate ion
ethenamine
butadiene
Allotropes of Carbon
diamond (C)
Macromolecular
………………………………………………
structure;
graphite (C)
Macromolecular
………………………………………………
structure;
fullerene (C60)
Simple molecular
………………………………………………
structure;
carbon atoms arranged
carbon atoms covalently bonded
60 carbon atoms covalently
tetrahedrally into an entire
into planar sheets of hexagonal
bonded into a ‘soccer-ball’
network of covalent bonds.
rings.
configuration.
sp3 hybridised;
Atoms are ……………
sp2 hybridised;
Atoms are ……………
Atoms are ……………
sp2 hybridised;
109.5° ;
bond angle …………………………
no delocalised electrons are
available to conduct electricity.
< 120° ; delocalised
120° ; delocalised bond angle …………………………
bond angle …………………………
electrons can conduct electricity
along each ‘sheet’
……………………………………………………
of graphite.
electrons are present but are
unable to conduct electricity
…………………………………………………………………………………
across molecules.
Chemical Bonding
16
© IBDP HL Chemistry
Resonance
more than one possible position for a double bond in a molecule. A resonance
Occurs when there is ………………………………………………………………………………………………………………………………………………
structure has
two or more Lewis diagrams to represent the same molecule ,
…………………………………………………………………………………………………………………………………………………………………………
and cannot be
described fully with one Lewis diagram alone.
While delocalisation may not result in resonance, resonance is a result of delocalisation.
Example 1: Benzene, C6H6
Hence in benzene, the six C–C bonds have equal bond energies, with a strength inbetween that of a single bond
1.5 .
and of a double bond, i.e. a bond order of ……………
Example 2: Carbonate Ions, CO32–
In carbonate ions, the three C–O bonds have an equal bond energies, with a strength in between that of a
11/3 .
single bond and of a double bond, i.e. a bond order of ……………
Chemical Bonding
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Intermolecular Forces
electrostatic attractions between opposite dipoles . There are three types
Intermolecular forces are ……………………………………………………………………………………………………………………………………………
of intermolecular forces (i.e. London dispersion forces, dipole-dipole forces and hydrogen bonding) and a
molecule may concurrently have more than one type.
hydrogen bonding > ……………………………………………
dipole-dipole > ………………………………………………
London forces .
Generally strength of …………………………………………………………
Instantaneous dipole-induced dipole (id-id) forces
Instantaneous dipoles or temporary dipoles occurs in both
polar
……………………
non-polar molecules and are
and ………………………………
constant random motion of electrons within the molecule. The uneven distribution of
created from the ……………………………………………………………………
electrons can make one side of the atom more negatively charged than the other, thus creating a temporary
dipole, even on a non-polar molecule.
attracts the electron cloud of a neighbouring molecule and creates an
The temporary dipole formed ………………………………………………………………………………
induced dipole ,
…………………………………………
electrostatic forces of attraction between these dipoles.
resulting in ……………………………………………………………………………………………………
This random motion of electrons can also distort an existing polarised electron cloud of a polar molecule
to a greater extent ,
…………………………………………………………
thereby increasing the charge density of the dipoles and strengthening the
electrostatic attractions between the charged dipoles.
London dispersion forces .
In IB syllabus, id-id forces are referred to as ………………………………………………………………………………
electrostatic attractions between temporary dipoles
The strength of London forces depends on:
 Size of atom or molecule
larger and ………………………………………………
more polarisable electron cloud, hence more energy is required
e.g. Br2 vs Cl2: Br2 has a ……………………
to overcome the stronger London forces between Br2 molecules.
 Shape of molecule
more elongated and
e.g. CH3CH2CH2CH3 (butane) vs CH3CH(CH3)CH3 (methylpropane): Butane has a ………………………………………………
hence more polarisable electron cloud. Thus, more energy is required to overcome the stronger and
more extensive
………………………………………………
Chemical Bonding
London forces between butane molecules.
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Permanent dipole-permanent dipole (pd-pd) forces
polar molecules have additional pd-pd forces. In IB syllabus,
All molecules have London forces, but only ………………………………………………
pd-pd forces are referred to as ………………………………………………………………
dipole-dipole forces .
electrostatic attractions between permanent dipoles
show a net dipole .
For a molecule to be polar, it must have polar bonds that do not cancel out, i.e. ……………………………………………………
CO2 is linear, and
the dipoles cancel out
SO2 is bent, and the
dipoles do not cancel out
The strength of dipole-dipole forces depends on:
 Polarity of the molecule (see e.g in hydrogen bonding)
stronger the dipole-dipole forces.
In general, the greater the polarity of the molecule, the ……………………………
Hydrogen bonding
+ hydrogen atom that is
strongest form of pd-pd forces formed between a ……………………………………………………
Hydrogen bonding is the ………………………………
– lone pair of electrons on an F, O or N. Hence, a molecule can
directly bonded to an F, O or N and a ………………………………………………………………………
only have dipole-dipole forces OR hydrogen bonding, but never BOTH.
The strength of hydrogen bonding depends on:
 Extent of hydrogen bonding
two hydrogen bonds per molecule while NH3 forms an average
e.g. H2O vs NH3: H2O forms an average of ………………
one hydrogen bond per molecule, hence more energy is required to overcome the more extensive
of ………………
hydrogen bonds between H2O molecules.
 Type of hydrogen bond (polarity of the hydrogen bond)
more electronegative than nitrogen and hence more energy is required to
e.g. HF vs NH3: Fluorine is ………………………………………………………………
overcome the stronger hydrogen bonds between HF molecules.
Chemical Bonding
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Solubility of Substances
To determine the solubility of a particular substance, we have to consider the type of attractive forces that can
be formed between the solute particles and the solvent.
ion-dipole interactions
hydrogen bonds
A substance is soluble if the formation of attractive forces between the solute and solvent
sufficient energy
…………………………………………………
releases
……………………………
overcome existing attractive forces in the solute and
(not necessarily greater) to ……………………………
solvent respectively.
Ionic Substances
generally soluble in polar solvents such as water.
❖ Ionic substances are …………………………………………………
energy released from the formation of
❖ If an ionic compound is soluble in water, it is because the ………………………………………………
ion-dipole interactions
………………………………………………………………
sufficient to overcome the ………………………………
ionic bonds within the
(hydration energy) is …………………………………………………………………
ionic solid (lattice energy) and the hydrogen bonds between the water molecules.
❖ The magnitude of the lattice energy and hydration energy depends on charge and ionic radii:
lattice energy  q+q–
r++r–
(ionic bonds)
hydration energy 
(ion-dipole interactions)
Q
r
Simple Molecular Substances
❖ Generally, polar substances are more soluble in polar solvents while non-polar substances are more
soluble in non-polar solvents.
hydrogen bonds formed
aqueous, polar solvents as the ……………………………………………
❖ Hydrogen-bonded molecules are soluble in …………………………………………………………………
releases sufficient energy to overcome the initial respective
between the solute and water molecules …………………………………………………………………………
hydrogen bonds
………………………………………………
in the solvent and solute molecules.
❖ Non-polar molecules are soluble in
organic, non-polar solvents
………………………………………………………………………………
between the solute and solvent molecules
London forces formed
as the …………………………………………
releases sufficient energy
…………………………………………………………………………
to overcome the initial
London forces in the solvent and solute molecules.
respective …………………………………………
Chemical Bonding
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Effects of Hydrogen Bonding
Intramolecular Hydrogen Bonding
❖ Intramolecular hydrogen bonding arises when the molecule contains at least two functional groups
-OH, -COOH, -NH2 ) with one another. In addition, these
capable of forming hydrogen bonds (e.g. ………………………………………………………………
close proximity in order to come together to form the hydrogen bond.
two groups must be in ………………………………………………
decrease the extent of intermolecular hydrogen
❖ The ability to form intramolecular hydrogen bonds will ……………………………
decreases .
bonds, hence the boiling point of a substance ……………………………………
intramolecular
e.g. propanedioic acid vs ethanedioic acid: The ability of propanedioic acid to form …………………………………………
hydrogen bonds
…………………………………………………
decreases the
extent of intermolecular hydrogen bonds ,
………………………………………………………………………………………………………………………
hence requiring
less energy to overcome.
propanedioic acid (m.p. 135 C)
ethanedioic acid (m.p. 190 C)
Hydrogen Bonding Causing Dimerisation
❖ Hydrogen bonds are significantly stronger than all other forms of dipole-dipole attractions, and may cause
carboxylic acids .
dimerisation of …………………………………………………
carboxylic acids can dimerise
due to hydrogen bonds
aluminium chloride can
dimerise due to dative bonds
organic solvent or in …………………
pure form. In aqueous state,
This occurs when the carboxylic acid is dissolved in ……………………………………………
hydrogen bonds with water molecules instead.
the carboxylic acid forms ………………………………………………
Hydrogen Bonding in the Bifluoride Ion
hydrogen bond within it.
❖ The bifluoride ion, HF2–, is an unusual ion as it contains a ……………………………………………
The hydrogen bond in this ion is particularly strong, with a bond strength of 162 kJ mol –1, making it
stronger than some covalent bonds even!
Chemical Bonding
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Hydrogen Bonding in Ice
open-cage arrangement, with each
❖ When liquid water freezes to become ice, the molecules snap into an ………………………………
water molecule forming the maximum possible number of four hydrogen bonds.
empty space between molecules, explaining the unique property
This arrangement leaves significant ………………………
of ice being
less dense
………………………………
hard and brittle
………………………………………………
than water. Also, the directional nature of the hydrogen bonds explains the
texture of ice.
104.5
The lone pairs of electrons are used for hydrogen bonding, hence increasing the bond angle from …………………
109.5(when solid).
(when liquid or gaseous) to …………………
Chemical Bonding
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Formal Charge vs Oxidation State
Formal charge indicates the number of missing or additional electrons on an atom, assuming that electrons
shared equally , i.e. all atoms are of equal electronegativity.
in all bonds are ……………………….…………………………
Oxidation state indicates the number of missing or additional electrons on an atom, assuming that electrons
completely transferred to the more electronegative atom
in all bonds are ………………………………………………………………………………………………………………………………………………………………………
i.e. ionic bonding.
Formal Charge
Oxidation State
bonding electrons are ‘split’ equally
between the two atoms
bonding electrons are given to the
more electronegative atom
The oxygen atom has
6 valence electrons.
The carbon atom has
4 valence electrons.
normally,
oxygen is
supposed
to have 6
valence
electrons
normally,
carbon is
supposed
to have 4
valence
electrons
Formal Charge = 0
Formal Charge = 0
The oxygen atom has
8 valence electrons.
The carbon atom has
0 valence electrons.
normally,
oxygen is
supposed
to have 6
valence
electrons
normally,
carbon is
supposed
to have 4
valence
electrons
Oxidation State = –2
Oxidation State = +4
Comparing Lewis Diagrams
We can use formal charge to compare two or more electron-dot (Lewis) diagrams.
Structure A
Structure B
Formal Charge = –1 ⎯→
Formal Charge = 0 ⎯→
Formal Charge = +2 ⎯→
Formal Charge = +1 ⎯→
less formal charges in the molecule.
B is more stable as there are ………………………………………………………………
Structure ……………
Structure A
–1
+1
Structure B
0
0
+1
–1
-1 charge reside on the more electronegative atom .
B is more stable as the ………………………………………………………………………………………………………………………………………………………
Structure ……………
Chemical Bonding
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The Ozone Layer
ultra-violet radiation from the sun, hence
The ozone layer plays a crucial role in absorbing 98% of the ………………………………………………………………
overexposure to uv rays, which cause skin cancer and cataract .
protecting us from ……………………………………………………………………………………………………………………………………….…………………………………………
Ultra-violet rays from the sun can broadly be classified into three categories:
ultra-violet A (uvA)
315 nm to 400 nm
least harmful; not absorbed by ozone layer
ultra-violet B (uvB)
280 nm to 315 nm
absorbed by ozone layer
ultra-violet C (uvC)
100 nm to 280 nm
absorbed by ozone layer and atmosphere
Absorption of UV Rays
In our atmosphere, ozone (O3) and oxygen (O2) are being continually
equilibrium
……………………………….……
interconverted ,
………………………………………………
establishing an
which maintains the ozone layer.
The conversion of O3 into O2
absorbs uvB and uvC radiation:
𝑢𝑣B /𝑢𝑣C
O3 →
O3 + O →
O2 + O
2 O2
2
3
resonance in ozone results
in a bond order of 1.5
oxygen gas has a doublebond, i.e. bond order of 2
The conversion of O2 into O3
absorbs uvC radiation:
𝑢𝑣C
O2 →
2O
O2 + O →
O3
Bond Strength & Wavelength
The exact wavelength of ultra-violet radiation that is absorbed depends on the bond energy of the covalent
bond broken, i.e. bond order 1.5 for ozone, and bond order 2 for oxygen.
Recall this formula from Atomic Structure.
From the Data Booklet:
h = 6.63  10–34 J s
c = 3.00  108 m s–1
L = 6.02  1023 mol–1
This value is typically provided in kJ mol–1,
hence a multiplication by 1000 is necessary.
Chemical Bonding
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Depletion of Ozone Layer
In the presence of
chlorine radicals
…………..…………………………………
(from chlorofluorocarbons) or
oxides of nitrogen ,
………………………………………………………
ozone
molecules can be decomposed into oxygen molecules.
Chlorofluorocarbons (CFCs)
CFCs are released into the atmosphere through their use
as a propellant in aerosol cans and in
……………………………………………………………………………………………………………
refridgerators .
……………………………………………
Since 1996, CFCs have been banned around the world but we are still suffering from its effects today.
CFCs are able to deplete ozone as follows:
CxFyClz →
𝑢𝑣
•Cl + O3 →
•OCl + O3 →
CxFyClz–1 + •Cl
•OCl + O2
•Cl + 2 O2
Overall Equation:
2 O3 →
•C𝑙
3 O2
number of chlorine atoms present in the
The ozone depletion potential (ODP) of CFCs scales with the ……………………………………………………………………………
molecule. Ultraviolet radiation provides sufficient energy to cause the breakage of C−Cl bonds but not C−F
bonds.
Oxides of Nitrogen (NO, NO2)
Oxides of nitrogen are released into the atmosphere through
lightning activity and the operation
…………………………………………………………………………………………………
of an internal combustion engine .
………………………………………………………………………………….…………
We use catalytic converters to reduce emissions of oxides of nitrogen from car exhausts.
Oxides of nitrogen are able to deplete ozone as follows:
Chemical Bonding
•NO + O3 →
•NO2 + O3 →
•NO2 + O2
•NO + 2 O2
•NO2 + O3 →
•NO + O3 →
•NO + 2 O2
•NO2 + O2
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Overall Equation:
2 O3 →
•NO
3 O2
Overall Equation:
•NO2
2 O3 →
3 O2
© IBDP HL Chemistry
Chemical Bonding
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