MODULE 1

Atomic Structure and the periodic

table

John Dalton proposed the atomic Theory in 1803

He theorized that:
● All atoms of the same element are identical
● Atoms cannot be broken down any further
● Atoms of different elements have different
masses
● Compounds are formed by a combination of two
or more different atoms
● Atoms cant be created nor destroyed in a
chemical reaction
● Chemical reactions are rearrangement of atoms

In modern day, some of his postulates were

disproven

● All atoms of the same element are not

identical, due to the existence of isotopes
● Atoms can be broken down into smaller parts,
known as protons, neutrons and electrons
within the nucleus of an atom.

Structure of the atom

Students must be able to identify:
 ● Properties of protons, neutrons and
electrons only; their relative masses
and charges and location
Protons and neutrons- in the nucleus
Electrons-in the orbitals

● Their behaviour in magnetic and electric

fields.

● Protons are positive, therefore they will

be attracted to the negative field
● Electrons are negative, therefore they
will be attracted to the positive field
● Neutrons are neutral(no charge) therefore
it will neither be attracted to the
negative nor the positive field

The angle of deflection is greater in

electrons than in protons because electrons
are lighter than protons, thus making it
easier to pull electrons towards a field.

Define the following terms:

(a)mass number - The mass number is the total
amount of protons and neutrons in the nucleus
of an atom.
(b)isotopes- isotopes are atoms of the same
element that contain different chemical
properties but have the same physical
properties. The amount of neutrons that they
have differ, but they all have the same amount
of protons.
(c)relative atomic and isotopic masses-
Relative atomic mass is the ratio of an
element's average atomic mass to one-twelfth
of the mass of carbon-12.

Explain the phenomenon of radioactivity-

When there are nuclei with too many neutrons
or too many protons, it is considered
unstable. When it tries to get rid of excess

or add the lacking neutrons or protons, that

process is called radioactive decay.

Identify three uses of isotopes

Iodine 131- treats hyperthyroidism
Carbon 14- carbon dating
Cobalt-60 is used medically for radiation
therapy

calculate the relative atomic mass of an

element, given isotopic masses and abundances;

To determine relative atomic mass, we simply

● multiply each isotopic mass by its
abundance
● add all the values together

● divide the total value by 100 percent.

 Effectively we are calculating the
weighted average of the masses of
isotopes, taking into account their
relative abundances.

🌈 Emission Spectra and Discrete Energy Levels

Atoms can absorb and emit energy in the form

of light electromagnetic radiation. When atoms
are excited (for example, by heating or an
electric current), their electrons absorb
energy and move to higher energy levels
(excited states). These excited states are
unstable so the electrons fall back to lower
energy levels, releasing energy as light.

This emitted light can be separated using a

spectroscope to produce an
Emission spectrum—a series of bright lines
at specific wavelengths.

What the Emission Spectrum Tells Us:

- The lines appear at specific wavelengths.
- This means that only certain amounts of
energy are emitted.
- Therefore, the energy levels in atoms must
be discrete (not continuous).

This is evidence that:

Electrons in atoms occupy specific energy
levels and transitions between these levels
result in the emission of photons of specific
energies.

---

Bohr Model and the Hydrogen Atom

The Bohr Model (1913) was developed by Niels

Bohr to explain the line spectrum of hydrogen.

● Electrons orbit the nucleus in specific

energy levels.
● When an electron jumps from a higher
energy level (n₂) to a lower one (n₁), a
photon is emitted.
● The energy of this photon corresponds to
the difference between the energy levels.
Hydrogen Emission Spectrum: Lyman & Balmer
Series

Hydrogen’s emission spectrum is made up of a

series of lines. Each series corresponds to
transitions to a specific energy level.

1. Lyman Series
- Transitions: to n = 1
- Region: Ultraviolet (UV)
- Example: Electron falls from n = 2, 3, 4...
to n = 1

2. Balmer Series
- Transitions: to n = 2
- Region: Visible light
- Example: Electron falls from n = 3, 4, 5...
to n = 2

These series support the idea of quantized

energy levels because the emitted photons have
specific frequencies (or wavelengths).
Atomic Orbitals and Quantum Numbers

Electrons occupy orbitals, described using

quantum numbers:

- Principal Quantum Number (n): Energy level

or shell (n = 1, 2, 3…).
- Sublevels: Each level has sublevels — s, p,
d, f.
- Orbital shapes:
- s orbital: spherical shape.
- p orbital: dumbbell shape (along x, y, z
axes).
- d orbitals: cloverleaf shapes (start from
n=3).

Relative Energies of Orbitals

-Orbital energy increases in this order:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p
- Even though 3d is in the third shell, it has
higher energy than 4s, so 4s is filled before
3d.

---

Electronic Configuration

Configurations follow the Aufbau principle

(fill lower-energy orbitals first), Pauli
exclusion principle(no more than two
electrons can occupy an orbital) and Hund’s
rule(When electrons are filling orbitals of
the same energy, they prefer to enter empty
orbitals first. These electrons all have the
same spin)

• S orbitals – 1 orbital per shell – holds 2

electrons total
• P orbitals – 3 orbitals per shell – holds 6
electrons total
• D orbitals – 5 orbitals per shell – holds 10
electrons total

Examples:
- H : 1s¹
- O : 1s² 2s² 2p⁴
- Fe: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
- Zn: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰

Ions:
- Na:
⁶ 1s² 2s² 2p⁶ (lost 1 electron from 3s)
- Fe²⁺: [Ar] 3d⁶ (lost 2 from 4s)

Factors Affecting First Ionisation Energy

Ionisation Energy (IE) = energy needed to

remove 1 electron from 1 mole of gaseous
atoms.

Factors:
- Nuclear charge(larger nuclear charge=
increase in ionization energy)
- Atomic radius(larger atomic radius= decrease
in ionization energy (electrons farther from
nucleus)
- Shielding- increases ionization energy
(inner electrons block attraction)

Ionisation Energy Data & Sub-shells (Use

Period 3)

Period 3 elements (Na to Ar)

- General trend:IE increases across the period

(due to increasing nuclear charge and
decreasing atomic radius).
- But there are dips in IE at:
- Mg → Al: because Al starts filling the 3p
sub-shell (which is higher in energy).
- P → S: due to electron pairing in 3p
orbital (repulsion makes it easier to remove
an electron).

Using Successive Ionisation Energies to Deduce

Configuration

Successive IEs are the energies required to

remove electrons one after the other.

- A large jump in IE shows removal of an

electron from a lower energy level (closer to
the nucleus).
- For example,Mg:
- 1st IE = 738 kJ/mol
- 2nd IE = 1451 kJ/mol
- 3rd IE = 7733 kJ/mol → big jump = electron
being removed from full 2p sub-shell.
So Mg has 2 electrons in the outer shell →
configuration: 1s² 2s² 2p⁶ 3s²

Forces of Attraction

Types of Forces of Attraction Between

Particles

1. Ionic Bonds: attraction between oppositely

charged ions (e.g., Na⁺ and Cl⁻ in NaCl).
2. Covalent Bonds: Sharing of electron pairs
between atoms (e.g., H₂, O₂, CH₄).
3. Metallic Bonds: Lattice of positive metal
ions surrounded by a 'sea' of delocalised
electrons (e.g., Cu, Fe).
4. Hydrogen Bonds: Attraction between a
hydrogen atom (bonded to N, O, or F) and a
lone pair on another N, O, or F atom (e.g., in
water).
5. Van der Waals Forces:
- Permanent-permanent dipole: Attraction
between molecules with permanent dipoles
(e.g., HCl).
- Induced-induced dipole (London forces):
Temporary attractions due to fluctuating
electron clouds, present in all molecules
(e.g., I₂).
- Permanent-induced dipole: A molecule with
a permanent dipole induces a dipole in another
(e.g., HCl and Ar).

---
States of Matter and Forces of Attraction

-Solids: Strongest intermolecular forces;

fixed shape and volume.
-Liquids: Weaker forces than solids; fixed
volume but variable shape.
-Gases: Very weak forces; no fixed shape or
volume.

Stronger forces= higher melting/boiling points

Weaker forces = lower melting/boiling points

---
Physical Properties Related to Bonding

-Melting/Boiling Points: Higher in

ionic/metallic compounds due to strong forces;
lower in covalent and molecular solids.
- Solubility:
 - Ionic compounds dissolve in polar solvents
(e.g., NaCl in water).
- Non-polar substances dissolve in non-polar
solvents (e.g., I₂ in hexane).
- Electrical Conductivity:
- Ionic compounds: conduct when molten or in
solution (ions are free to move).
- Covalent compounds: generally poor
conductors unless they form ions.
---

Formation of Bonds

Ionic Bonds
- Formed by transfer of electrons.
- Example: Na (metal) loses 1e⁻ → Na⁺; Cl (non-metal) gains
1e⁻ → Cl⁻ → NaCl.

Covalent Bonds
- Formed by sharing electrons.
- Orbital overlap results in:
- Sigma (σ) bonds: head-on overlap (e.g., H–
H).
- Pi (π) bonds: sideways overlap (in
double/triple bonds).

Metallic Bonds
- Metal cations in a lattice + delocalised
electrons.
- Explains conductivity and malleability of
metals.

Electronegativity:
- Difference in electronegativity determines
bond polarity.
- Greater difference → more ionic character.
---

Coordinate(Dative Covalent) Bonds

- A lone pair is donated from one atom to

another.
- Example: NH₃ donates lone pair to BF₃ → NH₃→BF₃.
- Dot-cross diagrams illustrate this clearly.

---

Shapes and Bond Angles of Molecules (VSEPR

Theory)

| Geometry | Example | Bond Angle |

| Linear | BeCl₂ | 180° |

| Trigonal | BF₃ | 120° |
| Tetrahedral | CH₄, NH₄⁺ | 109.5° |
| Pyramidal | NH₃ | ~107° |
| Non-linear | H₂O | ~104.5° |
| Octahedral | SF₆ | 90° (6-
sided) |

Shapes of Organic Molecules

- Ethane (C₂H₆): Tetrahedral (sp³)

 - Ethene (C₂H₄): Trigonal planar (sp²)
- Benzene (C₆H₆): Planar hexagonal ring,
resonance stabilised (sp²)

---

Lattice Structures and Physical Properties

Simple molecular
I₂
Weak van der Waals, Low m.p./b.p., non-conductive

Hydrogen bonded
Ice
Hydrogen bonding, Higher m.p. than other simple
molecular

Giant covalent
SiO₂
Strong covalent bonds, High m.p., hard, brittle

Ionic
NaCl
Lattice of ions,High m.p., soluble in water

Metallic
Cu
Metal cations + e⁻ cloud, Conducts electricity,
malleable

Giant atomic
Diamond, graphite
Strong covalent bonds, Very hard (diamond),
conductive (graphite)
 Mole Concept

3.1 Apply Avogadro’s Law

Law: Equal volumes of gases at the same

temperature and pressure contain equal
numbers of molecules.
Demo:
At RTP (room temperature and pressure), 1 mole of
any gas = 24 dm³
E.g.:
Find volume of 0.5 mol of oxygen at RTP:
Volume=0.5 mol×24 dm3/mol=12 dm3

3.2 Define the Mole

 The mole is the amount of substance that
contains 6.022 × 10²³ particles (Avogadro’s
number).
3.3 Define Molar Mass

Molar mass = mass of 1 mole of a substance

(g/mol)
E.g.:
Molar mass of CO₂ = 12.01 (C) + 2 × 16.00 (O) =
44.01 g/mol

3.4 Write Balanced Molecular and Ionic Equations

🧪 Molecular:

HCl(aq)+NaOH(aq)→NaCl(aq)+H2O(l)
🧪 Ionic:
H+(aq)+OH−(aq)→H2O(l)

3.5 Perform Calculations Using the Mole Concept

🧮 Mass → Moles:
Moles=Mass/Molar Mass

🧮 Solution:
Moles=Volume (dm³)×Concentration

🧮 Gas Volume:
At RTP,
Moles=Volume/24

3.6 Apply Mole Concept to Equations

🧪 Example:

N2+3H2→2NH3
If 2 mol of N₂ reacts completely, it needs 6 mol
of H₂ and produces 2 mol of NH₃

3.7 Empirical and Molecular Formulae

🧮 Empirical Formula (from % or mass):

1. Convert to moles
2. Divide by smallest
3. Use ratios

E.g.:
75% C, 25% H
→ 75/12 = 6.25 mol C
→ 25/1 = 25 mol H
→ Divide: C = 6.25/6.25 = 1, H = 25/6.25 = 4
→ Empirical formula = CH₄
🧮 Molecular Formula:
Molecular formula=n×Empirical

Find n using molar mass.

3.8 Perform Titrimetric Analyses

🧪 Types:
● Acid/base: HCl vs NaOH
● Redox:

○ KMnO₄ (purple → colorless)

○ K₂Cr₂O₇ (orange → green)

 ○ Thiosulfate/Iodine (blue-black to
colorless)
○ H₂O₂ (oxidizing agent)

3.9 Use Titration Results for Calculations

🧮 (a) Mole Ratios – from balanced equation

🧮 (b) Molar Concentration:
C=n×V

Find moles from one solution, use ratio to find

unknown.

🧮 (c) Mass Concentration:

Mass conc.=Molar conc.×Molar mass
 Redox Reaction

✅ 4.1 – Explain redox reactions in terms of:

➤ Electron transfer

● Oxidation = loss of electrons

● Reduction = gain of electrons

🎯 OIL RIG – Oxidation Is Loss, Reduction Is

Gain
Example:

Zn→Zn2+ + 2e−(Oxidation)
Cu2+ + 2e− → Cu(Reduction)

➤ Changes in oxidation state

● Oxidation: increase in oxidation number

● Reduction: decrease in oxidation number

Example:

Fe2+→Fe3+(oxidation, +2 to +3)
Cl2→Cl−(reduction, 0 to -1)
Oxidation Reduction

• For an atom in elemental form (an element

standing alone with no charge)
the oxidation number is always zero.
Examples: S, N2, K
• For any monatomic ion (an ion consisting of only
one element) the oxidation
number equals the charge on the ion.
Examples: O2-

• Non-metals generally have negative oxidation

numbers.
• Group I metals in a compound are always +1.
• Group II metals in a compound are always +2.
• Fluorine in a compound is always -1.
• Hydrogen in a compound with metals is always -1.
• Hydrogen in a compound with non-metals is always
+1.
• Oxygen in a compound is generally -2 (UNLESS in
peroxides or with
fluorine, in which case it is -1).
• Halogens (Group VII) in a compound are generally
-1.
• The sum of the oxidation numbers of all atoms in
a neutral compound is zero.
• When a monatomic ion is attached to a polyatomic
ion, the known charge of
the polyatomic ion can be used to determine the
oxidation number of the
monatomic ion.
• The sum of the oxidation numbers in a polyatomic
ion equals the charge on
the ions.
✍️ 4.2 – Construct Half Equations (Acidic & Basic
Conditions)

➤ Acidic Conditions (add H⁺ and H₂O as needed)

Example: MnO₄⁻ → Mn²⁺ (in acid):

1. Balance Mn: Already balanced

2. Balance O with H₂O: MnO₄⁻ → Mn²⁺ + 4H₂O

3. Balance H with H⁺: Add 8H⁺
4. Balance charge with e⁻: Add 5e⁻ to left

MnO4− + 8H+ + 5e− → Mn2+ +4H2O

➤ Basic Conditions (first use acid steps, then neutralize H⁺ with

OH⁻)

Example: Cr(OH)₃ to CrO₄²⁻ in base:

1. Use acidic method

2. Add OH⁻ to both sides to cancel H⁺
3. Combine H⁺ and OH⁻ to make H₂O

🧪 4.3 – Deduce Balanced Redox Equations from Half

Equations

Steps:
1. Write oxidation and reduction half-
equations
2. Multiply to equalize electrons
3. Add equations and cancel

Example: Zn + Cu²⁺ → Zn²⁺ + Cu

Half-equations:

Zn→Zn2+ +2e−(Ox)
Cu2+ +2e− →Cu(Red)

✅ Add:
Zn+ Cu2+ →Zn2+ +Cu

🔁 4.4 – Order Elements by Oxidising/Reducing Ability

➤ Displacement Reactions:

More reactive elements will displace less reactive

ones.
Examples:
1. Zinc + Copper(II) Sulfate:

Zn(s)+CuSO4(aq)→ZnSO4(aq)+Cu(s)

Zinc displaces copper → Zn is a stronger reducing

agent

2. Chlorine Water + Potassium Iodide:

Cl2+2I− →2Cl− +I2

Chlorine displaces iodine → Cl₂ is a stronger

oxidizing agent

● Oxidising agent: Causes oxidation, gets

reduced
● Reducing agent: Causes reduction, gets
oxidized
 Kinetic Theory
– Basic Assumptions of the Kinetic Theory (Ideal Gases)

An ideal gas is a theoretical gas that perfectly

follows kinetic molecular theory.
Assumptions:
1. Gases consist of many small particles in
constant, random motion.
2. Gas particles have negligible volume
compared to the volume of the container.
3. No intermolecular forces between gas
particles.
4. Collisions are perfectly elastic (no
energy lost).
5. Average kinetic energy is proportional to
temperature in Kelvin.

5.2 – Real vs Ideal Gases

⚖️ Real Gases:

● Deviate from ideal behavior under:

○ High pressure
○ Low temperature
● Particles do attract each other
● Particles have finite volume

Ideal Gases:
● Follow kinetic theory assumptions
 ● Most accurate at:
○ Low pressure
○ High temperature

📊 Graphical Representation:
PV vs P graph:
● Ideal gas: straight line (constant PV)
● Real gas: dips at high pressure due to
intermolecular forces

5.3 / 11.5 – Boyle’s Law and Charles’ Law

📉 Boyle’s Law:

At constant temperature, pressure and

volume are inversely proportional.

P∝1/V or PV=k

📈 Charles’ Law:

At constant pressure, volume is directly

proportional to temperature (in K).

V∝T or VT=k

Graph:
Volume (V) vs Temperature (T in Kelvin) →
Straight line

5.4 Gas Law Calculations

🧮 (a) Boyle’s Law:
P1V1=P2V2

Example:
A gas at 1 atm occupies 4 L. What is its volume
at 2 atm?

1×4=2×V2
⇒V2=2L1

🧮 (b) Charles’ Law:

V1T1=V2T2

(Temperatures in Kelvin!)

🧮 (c) Ideal Gas Equation:

PV=nRT

Where:

● P = pressure (Pa)
● V = volume (m³)
● n = moles
● R = 8.31 J/mol·K
● T = temperature (K)

Example:
Calculate volume occupied by 1 mol of gas at 298
K and 1 atm (1 atm = 101325 Pa):
V=nRTP
=1×8.31×298101325
≈0.0245 m3=24.5 dm3

🔍 Relative Molar Mass from Ideal Gas Law:

Mr=mRT/PV

5.5 – States & Phase Changes

💧 (a) The Liquid State:

● Particles are close together but can move past

each other.
● Fixed volume, no fixed shape.

🧊 (b) Melting:

● Solid → Liquid

● Heat energy breaks some intermolecular forces.

● Temperature stays constant during melting.

💨 (c) Vaporisation:

● Liquid → Gas

● Molecules gain energy to overcome all

intermolecular forces.
● Happens by:
○ Evaporation (surface only)
○ Boiling (throughout liquid)
 Energetics

6.1 & 6.2: Energy Changes in Chemical Reactions

● Key Idea: Chemical reactions involve breaking

bonds (endothermic, ΔH is positive) and
forming bonds (exothermic, ΔH is negative).
● Example:

○ Breaking H–H bonds in H₂ → absorbs energy (+ΔH)

○ Forming O–H bonds in H₂O → releases energy (–ΔH)

Exothermic vs. Endothermic Reactions

● Exothermic: Releases energy (e.g.,

combustion), products have lower energy than
reactants.
● Endothermic: Absorbs energy (e.g.,
photosynthesis), products have higher energy
than reactants.

Energy Profile Diagrams:

🧪 6.4: Bond Energy

● Definition: Average energy needed to break one

mole of a type of bond in gaseous molecules.
● Unit: kJ mol⁻¹

Bond Energy, Strength & Reactivity

● Stronger bonds = higher bond energy = less

reactive (e.g., N≡N in nitrogen gas is very strong, so
N₂ is unreactive).
● Factors affecting bond energy:
○ Bond length
○ Electronegativity
○ Bond order (single, double, triple)
Types of Enthalpy Changes

● ΔH_f°: Enthalpy of formation

● ΔH_c°: Enthalpy of combustion
● ΔH_neut: Enthalpy of neutralization
● ΔH_rxn: Enthalpy of reaction
● ΔH_hydr: Enthalpy of hydration
● ΔH_soln: Enthalpy of solution
● ΔH_atom: Enthalpy of atomization
● Ionisation energy, Electron affinity, Lattice
energy

Lattice Energy

● More charge = stronger attraction = more

negative lattice energy
● Smaller ions = closer = stronger lattice
energy
● No calculation needed, just trends.

♻️ 6.8: Hess’s Law

“The total enthalpy change is the same,

regardless of the path taken.”
● Used to calculate enthalpy changes indirectly.
● Standard conditions: 298 K, 1 atm, 1 mol dm⁻³

🧮 6.9: Enthalpy Calculations

● Use experimental data or bond energies.

● Example: Bond energy method:

ΔH=∑(bonds broken)−∑(bonds formed)
●
● Born-Haber Cycles used for lattice energy and
ion formation.
● Practical examples:
○ Enthalpy of neutralization
○ Enthalpy of solution
○ Enthalpy of combustion