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Chemical Bonding Notes Asteria Education©
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1. Electronegativity
Electronegativity generally:
Increases across the period (increase in nuclear charge strengthens attractive force
between the shared pair of electrons and the nucleus)
Decreases down the group (The increase in shielding effect down the group due to
increasing number of shells outweighs the increase in nuclear charge effect due to increasing
number of protons. This results in the shared pair of electrons being less attracted to the
nucleus decreasing the net attractive force between the shared pair of electrons and the
nucleus)
The three most electronegative atoms are N, O and F (Top right of the periodic table)
Knowing the electronegativity of an atom is important as it is used to assign oxidation number,
to predict the nature of intramolecular bonds and to find out the bond polarity
2. Type of intramolecular bond formed
Type of intramolecular bond formed depends on the atoms’ electronegativity:
If two atoms of similar electronegativity form a bond, it results in a covalent bond by sharing
electrons
If two atoms of large difference in electronegativity form a bond, it results in an ionic bond
formed by transferring electrons
If two atoms of intermediate difference in electronegativity form a bond it results in a bond
with intermediate character (partially ionic and covalent)
Covalent compound with ionic character (for polar molecules such as HCl) Due to the
electronegativity difference between the bonded atoms, there is polarisation of the covalent
bond and it causes an unequal sharing of electrons (polar covalent bonds)
Ionic compound with covalent character (e.g Al2O3 is an ionic compound but with covalent
nature) Due to the polarisation of anion electron cloud by the cation it causes an overlap of
electron clouds (sharing of electrons)
3. Factors affecting the extent of polarisation of anion electron cloud:
3.1. Charge density of cation
The higher the charge density, the stronger the polarising power, the greater the extent
of polarisation of anion electron cloud
Therefore, to have a strong polarising power, a cation should have high ionic charge and
small ionic radius
Electronegativity is the ability of an atom to attract the shared pair of elections in a covalent
bond.
Charge density is directly proportional to |𝑞+|
𝑟+
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3.2. Size of anion electron cloud
The larger the size of anion electron cloud, the easier it is to polarise the electron cloud.
(Ease of polarizability increases)
4. Intramolecular Bonding
4.1. Covalent bond
Electron pairs that are involved in bonding are called bond pairs while those not involved in
bonding are called lone pairs
Double bond, triple bond and dative bond are all considered be 1 bonding domain
Bond order is the number of electron pairs shared between two bonded atoms. Single bond
(bond order 1). Double bond (bond order 2) Triple bond (bond order 3)
4.1.2. Dative covalent bonds (Coordinate bonds)
A dative bond is a special kind of covalent bond that only differs from a normal covalent bond
in the origin of the bonding electrons. (Both electrons come from the same atom). Other
than that, it cannot be distinguished from a normal covalent bond as it is equally strong.
(This means you treat it as a normal covalent bond in terms of bond strength and chemical
reactions)
A dative bond is represented by an arrow pointing towards the atom that is accepting the lone
pair of electrons (not contributing anything). In the dot and cross diagram, both electrons are
coming from nitrogen.
4.1.3. Drawing dot and cross diagram of covalent compounds:
Step 1: Identify the central atom
Note: Central atoms are normally group 13 14 15 as there are more valence electrons for
bonding and more vacant orbitals, Group 1 and 2 are not likely due to their low
electronegativity hence, tend to form ionic bonds with non-metals. For period 3, group 16 17
can be a central atom as well as they can expand their octet, due to energetically accessible
vacant d orbitals
Step 2: Assign charges
Note: Polarising power is only used to describe cations while ease of polarizability is used
to describe anions.
Covalent bond is a chemical bond that involves the sharing of electron pairs between
atoms.
A Dative bond is a covalent bond in which both electrons of the shared pair come from
one of the two bonded atoms.
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When there are charges, assign the positive charges to the least electronegative atom and
the negative charges to the most electronegative atom.
Step 3: Fulfil the octet configuration for the peripheral atoms
Draw in the number of bonds that the peripheral atoms (whatever that’s not the central atoms)
need in order to achieve octet. For example, Group 16 would require 2 bonds to achieve octet
and we would draw 2 bonds to the group 16 atom
Step 4: Checking of octet configuration for the central atom
Count the number of electrons surrounding the central atom (including the lone pairs), check
whether there are more than 8 electrons around the central atom. If it is and is from period 2,
consider dative bonding. Dative bonding is normally formed from the atom with an excess of
electrons to other atoms. If it is from period 3 and above, it is the correct answer.
Step 5: Drawing of remaining lone pairs
Draw in all the lone pairs on the peripheral and central atoms.
In drawing of ‘dot and cross’ diagrams for covalent compounds, we prioritise to fulfil the octet
configuration of the peripheral atoms.
Example 1: PCl3
Step 1: Identify the central atom
Phosphorus is the central atom as Phosphorus is from group 15 and in period 3 while Chlorine
is from group 17. Hence, Phosphorus is a better choice than Chlorine.
Step 2: Assign charges
No Charges
Step 3: Fulfil the octet configuration for the peripheral atoms
Chlorine is from group 17 and hence only requires the formation of one bond to achieve octet
configuration.
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Step 4: Checking of octet configuration for the central atom
Phosphorus is from group 15 and hence would have 5 valence electrons. Three electrons are
used for bonding with Chlorine and hence 2 electrons are left and not used for bonding. We
draw them in as lone pairs. Now, Phosphorus has 8 electrons around it and has achieved
octet configuration.
Step 5: Drawing of remaining lone pairs
Example 2: O3 (With dative bonding)
Step 1: Identify the central atom
Oxygen is the only choice here
Step 2: Assign charges
No charges involved
Step 3: Fulfil the octet configuration for the peripheral atoms
The rest of the oxygen atoms must be the peripheral atoms and since oxygen is from group
16, it requires two bonds to be formed in order to achieve octet
Step 4: Checking of octet configuration for the central atom
Since oxygen is from group 16 and has 6 valence electrons, for the central oxygen, four are
used for bonding with the other oxygens and hence 2 remains as unbonded lone pairs.
However, we realise that the central oxygen has 10 valence electrons around it and is from
period 2. (Unable to expand octet) We start to consider dative bonding from the central oxygen
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atom to the other oxygen atom. (Dative bonding is from atoms with extra electrons to other
atoms)
The one receiving the dative bond contributes 0 electrons and receives 2 electrons posting a
net gain of +2 electrons, hence for the peripheral oxygen, one dative bond is equivalent to
forming 2 bonds and is able to achieve octet.
Step 5: Drawing of remaining lone pairs
4.2. Exceptions to Octet Rule for covalently-bonded species:
4.2.1. Electron deficient molecules
The central atom has an incomplete octet, electron deficient molecules usually contain group
2 or group 13 atoms (such as Be and Al) as central atom. (Remember the priority when
drawing dot and cross diagram for covalent compounds you try to fill up the peripherals to
octet configuration first.)
Practice 1: AlCl3 and BeCl2
4.2.2. Odd electron molecules (Free radicals)
A few molecules contain a central atom with an odd number of valence electrons, so they
cannot possibly have all their electrons in pairs, these are called free radicals as they contain
a lone electron (single, unpaired) electron, which makes it highly reactive as it will try to find
another radical in order to form an octet.
Example 3: NO2
Step 1: Identify the central atom
Nitrogen is the best choice here as nitrogen is from group 15 while oxygen is from group 16.
Step 2: Assign charges
No charges to be assigned
Step 3: Fulfil the octet configuration for the peripheral atoms
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Step 4: Checking of octet configuration for the central atom
Nitrogen is from group 15 and has 5 valence electrons, 4 are used for bonding with the
peripheral oxygens and hence 1 valence electron is left. However, there are 9 valence
electrons surrounding nitrogen and nitrogen is from period 2 and is unable to expand its octet.
Now, we consider dative bonding from nitrogen to the peripheral atoms to reduce the excess
electron.
Now, nitrogen has 7 valence electrons around it and it still has not achieved octet
configuration, but there is no other better way for nitrogen and hence we leave the answer as
it is. (It is better to have less electrons than octet than to exceed octet when in period 2)
Step 5: Drawing of remaining lone pairs
Practice 2: ClO2
4.2.3. Expanded Octet
Central atoms from period 3 and beyond can expand their octets. This is due to the fact that
atoms from these periods can accommodate additional electrons using energetically
accessible empty d orbitals in addition to s and p orbitals,
Period 2 elements cannot expand their octets as the 2nd shell does not have any d orbitals so
they need to use the higher energy orbitals in the 3rd shell which is not energetically accessible.
Practice 3: SO3
4.2.3 Dative Bonding Dimers
AlCl3 is an electron deficient molecule. However, AlCl3 can achieve the octet configuration
through the formation of its dative bonding dimer, Al2Cl6.
Equation: 2AlCl3 → Al2Cl
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4.3. Valence Shell Electron Pair Repulsion (VSEPR) Theory
It is based on:
1, Electron pairs in the valence shell of a central atom are located as far apart from one another
as possible to minimise repulsion.
2. The repulsion between electron pairs increase in the order:
BP-BP < BP – LP < LP – LP
BP – Bond pair LP – Lone Pair
Total no. of domains
Bonding domains
Electron domains
Geometry Examples Angle
2
2
0
Linear
BeCl2
180
3
3
0
Trigonal Planar
BF3
120
2
1
V-Shaped (bent)
SnCl2
118
4
4
0
Tetrahedral
CH4
109.5
3
1
Trigonal Pyramidal
NH3
107
2
2
V-Shaped (bent)
H2O
105
5
5
0
Trigonal Bipyramidal
PCl5
120 and 90
4
1
See-Saw
SF4
<120 and <90
3
2
T-Shaped
ClF3
<90
2
3
Linear
XeF2
180
6
6
0
Octahedral
SF6
90
5
1
Square Pyramidal
BrF5
<90
4
2
Square Planar
XeF4
90
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4.3.1 Bond angle of Odd Electron Molecules
A single unpaired electron still occupies an orbital hence, this electron is treated similar to a
lone pair and is considered as an electron domain when determining the shape of molecules.
However, there is only one electron in the orbital, the space occupied and repulsion between
the different domains will be much lesser thus it causes the bond angle to be larger than
expected. (Less repulsion)
Examples: NO2, NO3-
5. Sigma and Pi Bonds
Pi bonds can only occur between 2 p orbitals
In the case of a more bond that is double or triple, the first bond would be Sigma while the
rest will be Pi
Note: Sigma bond is stronger than a pi bond as the degree of overlap is greater. Collinear
overlap has a greater degree of orbital overlap than collateral overlap
A Sigma bond is a covalent bond resulting from head-to-head (collinear) overlap of
atomic orbitals.
A Pi bond is a covalent bond resulting from sideways overlap (Collateral overlap) of
atomic orbitals.
Sigma bonds can occur between 2 s orbitals, 1 s and 1 p
orbital, 2 p orbitals.
All sigma bonds have direct overlap with each other.
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5.1 Factors affecting Bond Strength
Covalent bonds with shorter bond lengths are stronger as the electrons are more attracted to
the nuclei than those with a longer bond length resulting in a stronger bond. Bond length is
inversely proportional to bond strength
Atoms from 1st and 2nd period form stronger bonds than those of other periods as 1s and 2s
orbitals are small and close to nucleus hence, the electrons experience a stronger attraction
to both binding nuclei. Exceptions: (O-O is weaker than S-S and F-F is weaker than Cl-Cl, due
to their small size and electrons the two atoms face repulsion)
Polar bonds are stronger than non-polar bonds due to the addition electrostatic forces of
attractions from the partial chargers
Multiple bonds are stronger than single bond
5.2 How to determine whether a molecule is polar
1st: Use VSEPR to determine the shape of the molecule
2nd: Determine the electronegativity of each of the molecule, draw bond dipole arrows
3rd Find the net dipole
4th no net dipole = non-polar got net dipole = polar
Try Drawing: H2O, CCl4, CF2Cl2
6. Ionic Bonding
Ionic bonding is non-directional as each positive ion is attached to all surrounding negative
ions and each negative ion is similarly attracted to surrounding positive ions.
MgCl2 ≡ Mg2+ + 2 Cl- Mg3N2 ≡ 3 Mg2+ + 2 N3-
Na2O2 ≡ 2Na+ + 2O- Na2O ≡ Na+ + O2-
Strength of ionic bonding is directly proportion to the magnitude of lattice energy
The strength of ionic bond increases with greater ionic charger and smaller ionic radius
a. Amount/Degree of orbital overlap between two bonded atoms. The larger the degree
of overlap the stronger the bond.
b. How strongly the electrons in the overlap region are attracted to the nuclei of the atoms:
the stronger the attraction, the stronger the bond.
Ionic bonding is the strong electrostatic attraction between positively charged cation and
negatively charged anion.
Lattice energy is directly proportional to |𝑞+𝑞−|
𝑟+ +𝑟−
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Ionic compounds are hard because there are strong electrostatic forces of attraction between
the positively charged cations and negatively charged anions which require a large amount of
energy to overcome. However, they are brittle as layers of ions can slide pass each other and
same charges come opposite each other this cause strong repulsions between the planes and
the ionic lattice shatters
7. Metallic Bond
Strength of metallic bond is dependent on the number of valence electron in the metal element.
The larger the number of valence electron the stronger the metallic bond.
Example: Al metal has a higher melting boiling point than Na since each Al has 3 valence
electrons while Na has only 1 valence electron
8. Intermolecular Bonds
4 Types of Intermolecular bonding you need to know
PD-PD, ID-ID, Hydrogen Bonding, Ion-Dipole Interaction
8.1. Permanent dipole- Dipole forces
ONLY applicable for polar molecules. This is due to electronegativity difference between two
atoms of the molecule resulting in one atom of the molecule being electron deficient while the
other atom is electron rich.
The positive end (electron deficient) of one molecule attracts the negative end (electron rich)
of another molecule’s dipole. This attraction is known as permanent dipole-dipole forces
The strength of PD-PD forces increases with increasing polarity (increasing
electronegativity difference)
8.2. Dispersion forces (Instantaneous dipole – Induced dipole)
Dispersion / ID-ID forces arise due to instantaneous temporary dipole that induces a
corresponding dipole in a neighbouring molecule. There is attraction between the
instantaneous temporary dipole and the induced dipole. This attraction is called dispersion
forces.
At another instant, the molecule may change its dipole through another movement of electrons
thus resulting in inducing another set of dipoles in a neighbouring molecule, this new set of
dipoles will continue to induce other molecules.
Dispersion forces are present in BOTH non-polar and polar molecules
In non-polar: IDID is the only type of intermolecular force present
In Polar: both IDID and PDPD exist
-For Large Molecules: dispersion forces can outweigh the PD-PD/Hydrogen Bonding
-For Small Molecules, PD-PD/Hydrogen Bonding are stronger than ID-ID
Metallic bond is an electrostatic attraction between the positive metal cations and the
negative electrons
Sea of delocalised electrons
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8.3. Factors affecting the strength of dispersion forces:
8.3.1. No. of electrons
The higher the number of electrons, the larger the electron cloud thus it becomes more
polarisable and the strength of dispersion forces become stronger
8.3.2. Molecular shape
For non-polar molecules of same Mr, the Strength of dispersion forces depend ont eh shape
as greater area of contact between molecules allow dispersion to act at more points and thus
stronger attractions
8.4. Hydrogen bonding
8.4.1. Conditions for hydrogen bonding:
Things to take note when drawing hydrogen bonding:
1. Lone Pairs of electrons on electronegative atom (NOF)
2. Delta Charges
3. Label and draw the dotted line
Hydrogen bonding is an extra strong PDPD force (10 times stronger)
8.4.2 Hydrogen bonding Dimer
Some molecules exist as dimer under certain conditions due to hydrogen bonding. For
instance, Mr of ethanoic acid is found to be 120 instead of 60 at temperature above its boiling
point and when dissolved in non-polar organic solvent.
Note: Ethanoic acid is the only one you need to know
1. One molecule must have at least one lone pair of electrons on a highly electronegative
atom (NOF)
2. One molecule must have an H atom covalently bonded to a highly electronegative
atoms (NOF)
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8.4.3 Comparing BP of NH3, H2O and HF
Differences in the strength of hydrogen bonds
Higher BP of H2O compared to HF and NH3
H2O has two H atoms attached to an electronegative atom (e.g O) and two lone pairs of
electrons on O atom to form hydrogen bonding. Hence, on average each H2O molecule
forms two hydrogen bond.
HF and NH3 can form on average one hydrogen bond per molecule since HF is limited to
having 1 H atom, while NH3 is limited to only having one lone pair of electrons on N.
Therefore, H2O can form a more extensive intermolecular hydrogen bonding.
Higher BP of HF compared to NH3
F is more electronegative than NH3, Dipole moment of HF molecule is larger than NH3, HF
forms a stronger hydrogen bonding than NH.
As hydrogen bonding and PDPD origins from the dipole moment of molecules, the disparity
between the electronegativity of atoms causes stronger intermolecular forces of attraction.
8.4.4. Intramolecular Hydrogen Bonding
As melting and boiling requires the breaking down of intermolecular forces of attractions,
molecules that can form intramolecular hydrogen bonding experience a lower than expected
MP and BP as less intermolecular hydrogen bonding can be formed.
1: Able to form intramolecular hydrogen bonds only but not intermolecular hydrogen bonds
2-nitrophenol
The same molecule must have one lone pair on a highly electronegative atom as well as H
atom attached to a highly electronegative atom. The lone pair and the electron deficient H
atom must be in proximity such that the hydrogen bond can be formed.
2: Able to form intermolecular hydrogen bonds only but not intramolecular hydrogen bonds
4-nitrophenol
4-nitrophenol has a higher MP and BP as compared to 2-nitrophenol as it can form a more
extensive intermolecular hydrogen bonding than 2-nitrophenol
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8.4.5. Density of ice and water due to hydrogen bonding
The density of ice is lower than liquid water this causes the ponds and lakes to freeze
downwards creating an insulation for the water below and thus marine life can still survive in
winter. In solid ice, the intermolecular hydrogen bonds become stronger and thus can hold the
ice in a lattice structure whereby one oxygen atom is surrounded by 4 hydrogen atoms at a
tetrahedral shape, 109.5 o
8.5. Ion-Dipole Interaction (not due to hydrogen bonding and is much stronger)
Ion-dipole are electrostatic forces of attraction between and ion and a polar molecule.
Note that the molecule can be any polar molecule such as water or even ethanol, as long as
the molecule is polar.
Steps to take when drawing ion-dipole interaction
9. Relative strength of van der Waals forces and hydrogen bonding
9.1. For two molecules small with similar Mr
Compare the types of intermolecular forces
Different intermolecular forces
Strength: Hydrogen Bonding > Permanent Dipole Permanent Dipole > Instantaneous Dipole-
Induced Dipole
9.2. Hydrogen bonds
Compare the number of average number of hydrogen bonds per molecule formed, the more
number of hydrogen bonds formed per molecule the higher the MP and BP as a more
extensive hydrogen bonding is formed.
Compare electronegativity difference of bonds, the bigger the electronegative difference of the
bond, the stronger the hydrogen bond. (Remember: Hydrogen bond is a kind of PDPD
bond)
9.3. PD-PD (Polar molecules)
The PD-PD forces are dependent on the electronegativity difference of the atoms in the bond.
The larger the electronegativity difference, the stronger the bond.
9.4. Dispersion forces (non-polar molecules)
Compare the shapes of molecules, the larger the area of contact between molecules, the
stronger the dispersion force. Spherical shapes experience less extensive dispersion force
due to a smaller area of contact.
1. Draw the dotted lines to show ion-dipole interaction
2. Label Delta Charges and Ion-dipole interaction
3. For polyatomic anions such as SO42- as well NO3
-, the full structure of the ion has to be
drawn out.
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9.5. For two molecules with significantly different Mr
9.5.1. between two polar molecules
Compare the strength of dispersion forces, the molecule with the larger size (Mr) (more
electrons) is more polarisable and thus more extensive and stronger dispersion force.
(Remember dispersion forces exist in all molecules)
9.5.2. Small polar vs large non-polar
The dispersion forces between the large molecules may be stronger than the PD-PD forces
between molecules of a smaller Mr.
10. Bonding and physical properties of simple molecules
BP and MP depend on the strength of intermolecular forces between molecules
However, MP also considers the packing of particles in solid
Electrical conductivity depends on mobile charge carriers such as in the case of metal, the
sea of delocalised electrons, as well as the mobile ions in ionic bonding.
11, Solubility
For a solute to dissolve in solvent, there must be sufficient energy released from the solute-
solvent interaction to overcome the attractive forces between the solute-solute interactions as
well as the solvent-solvent interactions. Energy is released when new attractive forces are
formed due to solute-solvent interactions since there is bond formation.
A substance is soluble when the energy released in solute-solvent interactions can
overcome the attractive force between the solute-solute interactions and attractive force
solvent-solvent interactions.