IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

By the end of this chapter you should be able to describe the modern periodic table; deduce the type of an element and its properties from its position; explain why atomic radius decreases across a period; name and locate the blocks; relate the electronic configuration to the group number; define atomic radius, ionization energy, electron affinity and electronegativity and compare the last two; locate metals and non-metals; relate atomic radius to ionization energy and electron affinity; explain the variation of basic and acidic character with atomic number; calculate oxidation numbers; identify redox reactions; and write oxidation and reduction half-reactions.
Lessons in this chapter:

The modern periodic table and its blocks
Atomic radius
Ionization energy, electron affinity and electronegativity
Metallic and non-metallic character; acidic and basic oxides
Oxidation numbers
Oxidation–reduction reactions and balancing ionic equations

🌳 Concept tree

Chapter 3 — The Periodic Table and Classification of Elements
├── Lesson 1 The modern periodic table

│ ├── 1.1 The basis of arrangement

│ ├── 1.2 Description of the table

│ ├── 1.3 The four blocks of the periodic table

│ ├── 1.4 Types of elements

│ └── 1.5 Using the configuration to find position
Atomic radius

│ ├── 2.1 Definition

│ ├── 2.2 Variation across a period (left to right)

│ ├── 2.3 Variation down a group

│ └── 2.4 Ionic radius compared with atomic radius
├── Lesson 3 Ionization energy, electron affinity and electronegativity
│ ├── 3.1 Ionization energy (ionization potential)

│ ├── 3.2 Electron affinity

│ ├── 3.3 Electronegativity

│ ├── 3.4 Electron affinity versus electronegativity
│ └── 3.5 Summary of the trends (representative elements)
├── Lesson 4 Metallic and non-metallic character; acidic and basic properties

│ ├── 4.1 Metals and non-metals

│ ├── 4.2 Trends of metallic character

│ ├── 4.3 Oxides: acidic, basic, amphoteric and neutral
│ └── 4.4 Acidic and basic character in the periodic table
Oxidation numbers

│ ├── 5.1 Concept

│ ├── 5.2 Ionic compounds

│ ├── 5.3 Covalent compounds

│ ├── 5.4 Rules

│ └── 5.5 Worked examples

└── Lesson 6→Oxidation–reduction (redox) reactions

├── 6.1 Definitions

├── 6.2 Behaviour of an agent according to its oxidation number
├── 6.3 Identifying redox reactions by oxidation numbers
├── 6.4 Ionic equations, spectator ions and half-reactions
├── 6.5 Balancing ionic redox equations (half-reaction method)

├── 6.6 Special types of redox reactions

└── 6.7 Final summary

Chapter 3 — The Periodic Table and Classification of Elements · 1 / 22

1.1 The basis of arrangement

As knowledge of the atom developed, the true energy levels and sublevels were discovered.
By applying the Aufbau principle, the elements could be arranged so that each element has one electron more than the element before it.
The modern periodic table is therefore built on two bases:

the elements are arranged in ascending order of atomic number;
the elements are arranged according to the Aufbau principle (the order of filling of sublevels).

The order of filling sublevels coincides with the arrangement of the periods:

Period	Sublevels filled (in order of increasing energy)	Number of elements
1	1s	2
2	2s < 2p	8
3	3s < 3p	8
4	4s < 3d < 4p	18
5	5s < 4d < 5p	18
6	6s < 4f < 5d < 6p	32
7	7s < 5f < 6d < 7p	32 (as far as discovered)

Chapter 3 — The Periodic Table and Classification of Elements · 2 / 22

1.2 Description of the table

The table consists of 18 vertical groups and 7 horizontal periods.
Elements are arranged in ascending atomic number; each element has one electron more than its predecessor in the period.
Each period starts by filling a new energy level with one electron and then fills the sublevels in turn, ending with a noble gas.
In a vertical group, the elements have similar electron configuration of the outer energy level, but different principal quantum number n.
The period number equals the number of the outermost main level (n) of the element.
Electronic distribution of the first 30 elements (number of electrons in each sublevel):

Z	Element	Configuration
1–2	H, He	1s¹, 1s²
3–10	Li … Ne	[He] 2s¹ … 2s² 2p⁶
11–18	Na … Ar	[Ne] 3s¹ … 3s² 3p⁶
19	K	[Ar] 4s¹
20	Ca	[Ar] 4s²
21	Sc	[Ar] 4s² 3d¹
22	Ti	[Ar] 4s² 3d²
23	V	[Ar] 4s² 3d³
24	Cr	[Ar] 4s¹ 3d⁵
25	Mn	[Ar] 4s² 3d⁵
26	Fe	[Ar] 4s² 3d⁶
27	Co	[Ar] 4s² 3d⁷
28	Ni	[Ar] 4s² 3d⁸
29	Cu	[Ar] 4s¹ 3d¹⁰
30	Zn	[Ar] 4s² 3d¹⁰

(Chromium and copper are exceptions because half-filled (d⁵) and completely filled (d¹⁰) d sublevels are extra stable.)

Chapter 3 — The Periodic Table and Classification of Elements · 3 / 22

1.3 The four blocks of the periodic table

The table is divided into four main regions or blocks
named after the sublevel which receives the last electron.

s-blockon the left.

The outermost electrons occupy the s sublevelgroup 1A (ns¹) and group 2A (ns²), where n is the period number and the number of the last energy level.

Comparison aspect	Main idea	Keywords
p-block	on the right	right
The outermost electrons occupy the p sublevel	groups 3A (ns² np¹) 4A (np²) 5A (np³) 6A (np⁴) 7A (np⁵) and the zero group (noble gases np⁶) where p is saturated	groupszerogroupnoble
•	(Helium is 1s².)	Helium

in the middle.
The d sublevel is filled successively; as d holds up to 10 electrons, the block has ten vertical columns: seven belong to the B groups (3B to 7B, then 1B and 2B in the table’s numbering) and three columns together form the eighth group (Fe, Co, Ni and their analogues).

placed separately at the bottom so the table does not become too long.
The f sublevel (up to 14 electrons) is filled successively; it has two series:

Lanthanidesthe 4f sublevel is filled, 14 elements; the outer sublevel of all is 6s².

Actinidesthe 5f sublevel is filled, 14 elements, all radioactive (unstable nuclei).

Chapter 3 — The Periodic Table and Classification of Elements · 4 / 22

1.4 Types of elements

The elements of the modern table are classified into four types:

Comparison aspect	Main idea	Keywords
Noble (inert) gases	last vertical column of the p-block (group zero/18). Configuration ends with np⁶ (helium 1s²). All sublevels of the period are filled so they are completely stable and form compounds with extreme difficulty	lastverticalcolumnblock
Representative elements	the s-block and the p-block except the zero group. All energy levels are full except the outermost and they tend to attain the configuration ns² np⁶ by losing gaining or sharing electrons	blockexceptzerogroup
Main (d-block) transition elements	the d sublevel is filled successively	sublevelfilledsuccessively
Inner transition elements	the f-block where f is filled successively	blockfilledsuccessively

The sixth period contains all four types (noble gas Rn is in the 6th period; Cs and Ba, representative; Hf to Hg, d-block; and the lanthanides), so it has four types.

Chapter 3 — The Periodic Table and Classification of Elements · 5 / 22

1.5 Using the configuration to find position

Comparison aspect	Main idea	Keywords
Period	= number of the outermost main level (largest n)	numberoutermostmainlevel
Group of s and p block elements	(A groups) = number of electrons in the outermost level. Examples: 1A → ns¹ 2A → ns² 3A → ns² np¹ … 7A → ns² np⁵. In every case the number of valence electrons equals the group number	groupsnumberelectronsoutermost
Group of d block elements	= number of electrons in (n−1)d plus ns: Sc (3) group 3B Ti 4B V 5B Cr 6B Mn 7B Ni (8 10) group VIII Cu (11) 1B Zn (12) 2B	numberelectronsplusVIII
•	The block is the sublevel that receives the last electron	blocksublevelreceiveslast
Worked example 1.1.	Locate ₁₇Cl: 1s² 2s² 2p⁶ 3s² 3p⁵. Outer level n = 3 → period 3. Outer configuration ns² np⁵ → group 7A (halogens) p-block representative non-metal	Locate₁₇ClOuterlevel

Comparison aspect	Main idea	Keywords
Worked example 1.2.	Locate ₂₀Ca: [Ar] 4s² → period 4 group 2A s-block representative element (an alkaline earth metal)	Locate₂₀Caperiodgroup
Worked example 1.3.	Locate ₂₆Fe: [Ar] 4s² 3d⁶ → period 4 d-block (main transition element) group VIII (8 outer electrons: 2 + 6)	Locate₂₆Feperiodblock
Worked example 1.4.	An element has the configuration [Kr] 5s² 4d¹⁰ 5p⁴. Period 5 (n = 5) p-block group 6A (ns² np⁴) representative it is ₅₂Te (36 + 2 + 10 + 4 = 52)	elementconfiguration4d¹⁰Period

✍️ Worked examples

Comparison aspect	Main idea	Keywords
Worked example 1.5.	An element is in period 6 group 1A. Its outer configuration is 6s¹ and its atomic number is 55 (caesium) s-block	elementperiodgroupouter
•	Worked example 1.6 — match elements to types (like the chapter exercises)	Workedexamplematchelements
₈₆Rn	[Xe] 4f¹⁴ 5d¹⁰ 6s² 6p⁶ — noble gas p-block	4f¹⁴5d¹⁰nobleblock
₅₅Cs	6s¹ — representative element in the s-block	representativeelementblock
₂₃V	4s² 3d³ — transition metal in the fourth period	transitionmetalfourthperiod
₆₄Gd	4f⁷ 5d¹ 6s² — lanthanide inner transition	lanthanideinnertransition
₇₆Os	4f¹⁴ 5d⁶ 6s² — main transition metal in period 6	4f¹⁴maintransitionmetal

Chapter 3 — The Periodic Table and Classification of Elements · 6 / 22

Lesson 2: Atomic radius

We study the trends of properties in the representative elements (s and p blocks).

Because the wave theory shows it is impossible to fix the exact position of an electron, the atomic radius cannot be defined as "distance from the nucleus to the farthest electron".
Instead:
Atomic radius (covalent radius) is half the distance between the centres of the nuclei of two identical atoms bonded in a diatomic molecule.
It is measured in picometres (pm); 1 pm = 10⁻¹² m.
In ionic crystals (such as NaCl) the size of each cation or anion is described by its ionic radius, which depends on the number of electrons lost or gained.
Example in a Cl₂ molecule the distance between the two nuclei is 198 pm, so the atomic radius of chlorine = 198/2 = 99 pm.

Atomic radii in pm:

1A	2A	3A	4A	5A	6A	7A
H 37
Li 152	Be 113	B 88	C 77	N 70	O 66	F 64
Na 186	Mg 160	Al 143	Si 117	P 110	S 104	Cl 99
K 227	Ca 197	Ga 122	Ge 122	As 121	Se 117	Br 114
Rb 247	Sr 215	In 163	Sn 140	Sb 141	Te 143	I 133
Cs 265	Ba 217	Tl 170	Pb 175	Bi 155	Po 167	At 140

2.2 Variation across a period (left to right)

The atomic radius decreases from group 1A to group 7A. Reason: the effective nuclear charge (Z_eff) increases gradually.

Z_eff is the actual nuclear charge felt by an electron in the atom. It is always less than the real nuclear charge (number of protons) because the inner (core) electrons shield part of the nuclear attraction from the outer electrons.
Across a period the number of protons increases by one at each step while the electrons are added to the same main level, so the shielding by inner electrons does not change much. Z_eff rises, the valence electrons are pulled closer, and the atom contracts.
Hence the largest atom in a period is the alkali metal (1A) and the smallest is the halogen (7A).
Example period 3: Na (186) > Mg (160) > Al (143) > Si (117) > P (110) > S (104) > Cl (99) pm.

2.3 Variation down a group

The atomic radius increases with atomic number. Reasons:

the number of energy levels increases;
the filled inner levels shield the outer electrons from the nucleus;
the repulsion between the electrons increases.

Example→group 1A: Li 152 < Na 186 < K 227 < Rb 247 < Cs 265 pm.

Species	Na	Na⁺	Fe	Fe²⁺	Fe³⁺	Cl	Cl⁻
Radius (pm)	186	102	126	78	64.5	99	181
Protons	11	11	26	26	26	17	17
Electrons	11	10	26	24	23	17	18

Comparison aspect	Main idea	Keywords
Metal cation is smaller than its atom	(Na⁺ < Na Fe²⁺ < Fe). Protons are the same but there are fewer electrons so the effective nuclear charge per electron increases and the outer level is lost (for Na the 3s electron is removed). The larger the positive charge the smaller the ion (Fe³⁺ < Fe²⁺ < Fe)	Protonssamefewerelectrons
Non-metal anion is larger than its atom	(Cl⁻ > Cl). The protons are the same but there is one more electron so repulsion between electrons increases and the nuclear attraction per electron decreases (as in tug of war: the same pull shared by more electrons)	protonssameelectronrepulsion

✍️ Worked examples

Worked example 2.1. Arrange in decreasing radius: Mg, Na, Cl, S. All in period 3: Na (186) > Mg (160) > S (104) > Cl (99).
Worked example 2.2. Which is larger, Br or Br⁻? Br⁻, because it has one electron more than its (same) 35 protons.
Worked example 2.3.
Compare the ions S²⁻, Cl⁻, K⁺ and Ca²⁺ (all have 18 electrons) with 16, 17, 19 and 20 protons.
More protons attract the same number of electrons more strongly, so the radius order is S²⁻ > Cl⁻ > K⁺ > Ca²⁺.
Worked example 2.4 (periodic table figure type question).
In the table element X is in period 2, group 1A; element A in period 3, group 1A; element B in period 3, group 7A.
Radii A > X (extra shell); A > B (decrease across period).

Chapter 3 — The Periodic Table and Classification of Elements · 7 / 22

3.1 Ionization energy (ionization potential)

If an atom gains a moderate amount of energy its electrons are excited to higher levels.
If the energy is large enough the weakest-bound electron is ejected and the atom becomes a positive ion.
Ionization energy (potential) = the amount of energy required to remove one mole of electrons from one mole of isolated gaseous atoms.
Because one, two or three electrons may be removed there are first, second, third … ionization energies:

M(g) → M⁺(g) + e⁻ (first ionization energy, IE₁)

M⁺(g) → M²⁺(g) + e⁻ (second ionization energy, IE₂), and so on.
Unit kJ/mol. All are endothermic (ΔH positive).
First ionization energies (kJ/mol), selected H 1312; Li 520, Be 900, B 801, C 1086, N 1402, O 1314, F 1681, Ne 2081; Na 496, Mg 738, Al 578, Si 787, P 1012, S 1000, Cl 1251, Ar 1521; K 419; Cs 376.
Trend in a period:

generally increases from left to right because the effective nuclear charge increases and the atomic radius decreases, so the nucleus attracts the valence electrons more strongly and more energy is needed to remove one.
Ionization energy is inversely related to atomic radius.
(Small irregularities occur at 3A and 6A because of the stability of full and half-filled sublevels, e.g.

N > O.)

Trend in a group decreases from top to bottom, because the number of shells increases, the atomic radius increases and the attraction of the nucleus for the valence electrons falls, so less energy is required to remove it.
Noble gases have very high first ionization energies because of the stability of their saturated outer level.

Successive ionization energies.

IE₂ > IE₁ because the electron is removed from a positive ion in which the effective nuclear charge per electron is greater.
For group 2A elements IE₃ is much bigger than IE₂, because the third electron must be taken from a saturated inner level.
For magnesium:

Mg(g) → Mg⁺(g) + e⁻ ΔH = +737 kJ/mol

Mg⁺(g) → Mg²⁺(g) + e⁻ ΔH = +1450 kJ/mol

Mg²⁺(g) → Mg³⁺(g) + e⁻ ΔH = +7730 kJ/mol (a sudden large jump: the electron comes from the stable 2p⁶ level)

✍️ Worked examples

Worked example 3.1.
From IE values 496, 4562, 6910 kJ/mol for an element, find its group.
A very big jump after the first value means one valence electron: group 1A (sodium).
Worked example 3.2.
Which has the larger first ionization energy Cl or Na?
Cl, since it is at the right side of the same period (smaller radius and greater Z_eff).
Worked example 3.3. Which has the smaller IE: Rb or K? Rb, because it is lower in the group (larger radius).
Worked example 3.4 — energy to ionise a sample (using the ΔH). The energy needed to remove one electron from every atom in 0.5 mol of gaseous magnesium atoms = 0.5 × 737 = 368.5 kJ.

Chapter 3 — The Periodic Table and Classification of Elements · 8 / 22

3.2 Electron affinity

Removing an electron from a gaseous atom needs energy.
When one mole of gaseous atoms gains one mole of electrons the energy change may be release or absorption:
Electron affinity = the amount of energy change (released or absorbed) when one mole of isolated gaseous atoms gains one mole of electrons.
Example Br(g) + e⁻ → Br⁻(g) ΔH = −325 kJ/mol (energy released, so the sign is negative).
Electron affinity values (kJ/mol), s and p blocks:

Li −59.6	Be >0	B −26.7	C −122	N +7	O −141	F −328	Ne +29
Na −52.9	Mg >0	Al −42.5	Si −134	P −72.0	S −200	Cl −349	Ar +35
K −48.4	Ca −2.4	Ga −28.9	Ge −119	As −78.2	Se −195	Br −325	Kr +39
Rb −46.9	Sr −5.0	In −28.9	Sn −107	Sb −103	Te −190	I −295	Xe +41
Cs −45.5	Ba −14	Tl −19.2	Pb −35.2	Bi −91.3	Po −183.3	At −270	Rn +41

Notes:

Comparison aspect	Main idea	Keywords
•	The trend of electron affinity is irregular in periods and groups unlike atomic radius and ionization energy	trendelectronaffinityirregular
•	The most negative values (largest energy release) belong to the halogens (7A) because by gaining one electron they reach the configuration of the nearest noble gas	mostnegativevalueslargest
Be and Mg	have positive values (absorb energy) because the added electron must enter the higher-energy p sublevel while the s sublevel is complete	positivevaluesabsorbenergy
Nitrogen	has a positive value: its half-filled 2p³ is stable an added electron has to pair with one already present which causes repulsion and reduces stability	positivevaluehalffilled
Noble gases	have positive values because of the stability of their saturated configuration (the electron would have to enter a new higher level)	positivevaluesbecausestability

Chapter 3 — The Periodic Table and Classification of Elements · 9 / 22

3.3 Electronegativity

Electronegativity→is the ability of an atom to attract the electrons of the chemical bond towards itself.

It is expressed by a relative numberthe larger the number, the greater the ability to attract bonding electrons.

Examples (Pauling scale)F 4.0, O 3.5, N 3.0, Cl 3.2, C 2.5, H 2.1, Na 0.9, Cs 0.7.

Comparison aspect	Main idea	Keywords
Trends	increases across a period (increasing atomic number decreasing radius) decreases down a group	increasesacrossperiodincreasing
Fluorine	is the most electronegative element	mostelectronegativeelement
•	The difference in electronegativity between two elements decides the type of bond between them (studied in Chapter 4 of the book)	differenceelectronegativitybetweenelements

Chapter 3 — The Periodic Table and Classification of Elements · 10 / 22

3.4 Electron affinity versus electronegativity

Comparison aspect	Electron affinity	Electronegativity
Meaning	Energy change when a free gaseous atom gains an electron	Tendency of an atom in a bond to attract the bonding electrons
State of the atom	Isolated atom in elementary state	Atom combined with other atoms
Expressed by	Energy (kJ/mol)	Relative number without units
Regularity	Irregular trend	Regular trend

Chapter 3 — The Periodic Table and Classification of Elements · 11 / 22

3.5 Summary of the trends (representative elements)

Comparison aspect	Main idea	Keywords
Across a period (left → right)	atomic radius ↓ ionization energy ↑ electronegativity ↑	atomicradiusionizationenergy
Down a group	atomic radius ↑ ionization energy ↓ electronegativity ↓	atomicradiusionizationenergy
•	So the element with the largest ionization energy and electronegativity is at the top right (F with He having the highest IE) and the largest atoms are at the bottom left (Cs)	elementlargestionizationenergy

✍️ Worked examples

Worked example 3.5. Arrange F, Cl, Br, I in increasing electronegativity: I < Br < Cl < F.
Worked example 3.6.
Why is the electron affinity of nitrogen positive while that of carbon is negative?
C (2p²) can add an electron into an empty 2p orbital to reach the stable 2p³; N (2p³ half-filled) can only add by pairing electrons, which costs energy.
Worked example 3.7 (figure-type question).
In the first four periods an element D lies in group 1A period 3 and F in group 7A period 3, X in group 1A period 2.
Ionization energy F > D, and X > D.
The order of increasing IE D < X < F.

Chapter 3 — The Periodic Table and Classification of Elements · 12 / 22

Lesson 4: Metallic and non-metallic character; acidic and basic properties

4.1 Metals and non-metals

Berzelius (early 19th century) was the first to classify elements into metals and non-metals, before anything was known about the atom.
The classification is still used, although there is no sharp boundary.
Metals

Their valence shell generally has less than half its capacity of electrons.
They lose valence electrons to reach the noble-gas configuration, forming positive ions; hence they are electropositive.
They conduct electricity because their few valence electrons move easily from place to place in the metal.
They have large atomic radii and therefore small ionization energies.

Non-metals

Their valence shell generally has more than half its capacity of electrons.
They gain a few electrons to reach the noble-gas configuration, forming negative ions; they are electronegative.
They are electrical insulators because the valence electrons are bound strongly to the nucleus and cannot move.
They have small atomic radii and high ionization energies.

Metalloids

(semimetals) B, Si, Ge, As, Sb, Te, At lie on a diagonal line between metals and non-metals.
They have the appearance of metals but most properties of non-metals; their electronegativity is intermediate; their electrical conductivity is lower than that of metals but much higher than non-metals.
They are used in transistors and semiconductors.
Position all metals lie to the left of the metalloids in the table; non-metals lie to the right.

Comparison aspect	Main idea	Keywords
Across a period	(left to right): the first group contains the strongest metals metallic character decreases gradually until the metalloids then non-metallic character increases and is highest in group 7A	leftrightfirstgroup
Down a group	metallic character increases	metalliccharacterincreases
•	So the strongest metals are at the bottom left (caesium is the most metallic) and the strongest non-metals at the top right (fluorine the strongest non-metal)	strongestmetalsbottomleft
Worked example 4.1.	Classify Si K. Metals: K non-metal: Cl metalloids: Si	ClassifyMetalsmetalmetalloids
•	Worked example 4.2	Workedexample

Why is Al a metal but Cl a non-metal?
Al (3 valence electrons, large radius, low ionization energy) loses electrons; Cl (7 valence electrons, small radius, high electron affinity) gains one.

Comparison aspect	Main idea	Keywords
Non-metal oxides	dissolve in water to give acids and are called acidic oxides. They react with alkalis to give a salt and water. CO₂(g) + H₂O(l) → H₂CO₃(aq) SO₃(g) + H₂O(l) → H₂SO₄(aq) CO₂(g) + 2NaOH(aq) → Na₂CO₃(aq) + H₂O(l)	dissolvewatergiveacids
Metal oxides	are generally basic oxides. Some dissolve in water forming alkalis (alkaline oxides) they react with acids to give a salt and water. Na₂O(s) + H₂O(l) → 2NaOH(aq) K₂O(s) + H₂O(l) → 2KOH(aq) Na₂O(s) + 2HCl(aq) → 2NaCl(aq) + H₂O(l) MgO(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂O(l)	generallybasicSomedissolve
Amphoteric oxides	(Al₂O₃ ZnO Sb₂O₃ SnO) react both with acids and with alkalis: Al₂O₃(s) + 6HCl(aq) → 2AlCl₃(aq) + 3H₂O(l) Al₂O₃(s) + 2NaOH(aq) → 2NaAlO₂(s) + H₂O(l) (sodium aluminate) ZnO(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂O(l) ZnO(s) + 2NaOH(aq) → Na₂ZnO₂(aq) + H₂O(l) (sodium zincate)	Al₂O₃Sb₂O₃reactboth
Neutral oxides	(CO N₂O) react neither with acids nor with alkalis and do not dissolve readily in water	reactneitheracidsalkalis

4.4 Acidic and basic character in the periodic table

Comparison aspect	Main idea	Keywords
Across a period	as atomic number increases the basic character of the oxide decreases and the acidic character increases. Period 3: Na₂O (strongly basic) MgO (basic) Al₂O₃ (amphoteric) SiO₂ (weakly acidic) P₄O₁₀ (acidic) SO₃ (strongly acidic) Cl₂O₇ (very strongly acidic). P₄O₁₀ + 6H₂O → 4H₃PO₄ Cl₂O₇ + H₂O → 2HClO₄	atomicnumberincreasesbasic
Down a group	(e.g. group 1): basic character of the hydroxides/oxides increases as atomic number increases (LiOH < NaOH < KOH < RbOH < CsOH)	basiccharacterhydroxidesoxides
•	For the hydrides of group 7A the acidic strength increases down the group: HF < HCl < HBr < HI. As the radius of the halogen increases the attraction of the halogen to the hydrogen atom decreases so hydrogen is ionised more easily	hydridesgroupacidicstrength

Comparison aspect	Main idea	Keywords
Worked example 4.3.	Write balanced equations showing that sodium oxide is basic and sulphur trioxide is acidic. Na₂O + 2HCl → 2NaCl + H₂O SO₃ + 2NaOH → Na₂SO₄ + H₂O	Writebalancedequationsshowing
Worked example 4.4.	Show that tin(II) oxide is amphoteric. SnO + 2HCl → SnCl₂ + H₂O SnO + 2NaOH → Na₂SnO₂ + H₂O	Showoxideamphoteric2HCl
Worked example 4.5.	Classify the oxides CaO SO₂ ZnO Li₂O. CaO basic SO₂ acidic ZnO amphoteric NO neutral Li₂O basic	ClassifyoxidesLi₂Obasic
Worked example 4.6.	Which is more acidic HI or HF in water? HI since the H–I bond is longer and weaker	acidicwatersincebond

Chapter 3 — The Periodic Table and Classification of Elements · 13 / 22

Lesson 5: Oxidation numbers

5.1 Concept

Before the nature of chemical bonding was understood, valency expressed an atom’s ability to combine with a number of other atoms.
Today the concept of oxidation number is used instead.
Oxidation number is a number that expresses the electric charge (positive or negative) that the atom or ion carries (or appears to carry) in the compound, whether ionic or covalent.

5.2 Ionic compounds

The oxidation number of an ion or group equals its charge, with its sign: K⁺Br⁻ (K +1, Br −1), Na⁺Cl⁻, Mg²⁺O²⁻, Ca²⁺CO₃²⁻, Cu²⁺SO₄²⁻.

5.3 Covalent compounds

The atoms do not form ions, but the shared electrons are displaced towards the more electronegative atom; this atom is assigned a negative number and the less electronegative one a positive number.

In molecules of an element (S₈, P₄, O₃, Cl₂, H₂, N₂, O₂) the electronegativities are equal, so there is no displacement and each atom has oxidation number zero.
In molecules of two different atoms the bonding electrons are counted with the more electronegative atom. HCl: H +1, Cl −1. H₂O: H +1, O −2. CO₂: C +4, O −2.

The sum of oxidation numbers of all atoms in a neutral compound is zero; in an ion it equals the charge of the ion.
Group 1A elements are always +1; group 2A always +2; group 3A (Al) +3; fluorine is always −1. Start with these.
is −2 in most compounds, except: peroxides (H₂O₂, Na₂O₂) −1; superoxides (KO₂) −½; oxygen difluoride (OF₂) +2.
is +1 in most compounds, except metal hydrides (NaH, CaH₂, AlH₃) where it is −1.
The oxidation number is calculated for one atom (or ion) at a time; some compounds contain atoms of the same element with different oxidation numbers (NH₄NO₂: N is −3 in NH₄⁺ and +3 in NO₂⁻).
The highest oxidation state of a representative element does not exceed its group number (A groups).
For a polyatomic group the oxidation number is the charge of the group: NO₃⁻ (−1), CO₃²⁻ (−2), SO₄²⁻ (−2), NH₄⁺ (+1).

Illustrations→Na₂O: Na +1, O −2. Na₂O₂: O −1. KO₂: K +1, O −½. CaH₂: Ca +2, H −1. AlH₃: Al +3, H −1.

5.5 Worked examples

Comparison aspect	Main idea	Keywords
Worked example 5.1.	Oxidation number of Cr in K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 → 2x = 12 → x = +6	OxidationnumberK₂Cr₂O₇
Worked example 5.2.	Mn in KMnO₄: +1 + x + 4(−2) = 0 → x = +7. Mn in MnO₂: x − 4 = 0 → +4. Mn in MnCl₂: x − 2 = 0 → +2	KMnO₄MnO₂MnCl₂
Worked example 5.3.	S in H₂SO₄: 2 + x − 8 = 0 → +6. S in Na₂S₂O₃: 2 + 2x − 6 = 0 → x = +2. S in SO₃²⁻: x − 6 = −2 → +4. S in H₂S: 2 + x = 0 → −2	H₂SO₄Na₂S₂O₃SO₃²
Worked example 5.4.	N in NH₄⁺: x + 4 = +1 → −3. N in NO₃⁻: x − 6 = −1 → +5. N in HNO₂: 1 + x − 4 = 0 → +3. N in N₂O: 2x − 2 = 0 → +1. N in NO₂: x − 4 = 0 → +4. N in N₂H₄: 2x + 4 = 0 → −2	HNO₂N₂H₄

Worked example 5.5.Cl in NaClO: +1 + x − 2 = 0 → +1; in KClO₃: +1 + x − 6 = 0 → +5; in HClO₄: 1 + x − 8 = 0 → +7.

Worked example 5.6.C in CH₄: x + 4 = 0 → −4; in CH₃OH: x + 3 + (−2) + 1 = 0 → −2; in HCOOH: 2 + x − 4 = 0 → +2; in CO: +2; in CO₂: +4.

Worked example 5.7.O in H₂O₂: 2(+1) + 2x = 0 → −1; in OF₂: x + 2(−1) = 0 → +2; in KO₂: +1 + 2x = 0 → −½.

Worked example 5.8 — average number.Fe in Fe₃O₄: 3x − 8 = 0 → x = +8/3 (average of 2 Fe³⁺ and 1 Fe²⁺).

✍️ Worked examples

Worked example 5.9 — from the group number.
The highest oxidation number of chlorine (7A) is +7; of nitrogen (5A) is +5; of sulphur (6A) is +6.
The lowest is (group number − 8) Cl −1, N −3, S −2.

Chapter 3 — The Periodic Table and Classification of Elements · 14 / 22

6.1 Definitions

Oxidation the process in which an atom/ion loses electrons, resulting in an increase in its oxidation number (positive charge increases).
Reduction the process in which an atom/ion gains electrons, resulting in a decrease in its oxidation number.
The reducing agent is the reactant that loses electrons (it is oxidised); the oxidizing agent is the reactant that gains electrons (it is reduced).
Oxidation and reduction always occur together; the number of electrons lost by the reducing agent equals the number gained by the oxidizing agent.
A reaction in which no oxidation numbers change is not a redox reaction (e.g. neutralisation NaOH + HNO₃ → NaNO₃ + H₂O; precipitation AgNO₃ + NaCl → AgCl + NaNO₃; CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂).

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6.2 Behaviour of an agent according to its oxidation number

Central atom at its maximum oxidation number → oxidizing agent only: K₂Cr₂O₇ (Cr +6), HNO₃ (N +5), H₂SO₄ (S +6), KMnO₄ (Mn +7).
Central atom at its minimum oxidation number → reducing agent only: NH₃ (N −3), CH₄ (C −4), H₂S (S −2), H₂.
Central atom at an intermediate oxidation number → can be either, depending on the reaction: HNO₂ (N +3), SO₂ (S +4), H₂O₂ (O −1).

Chapter 3 — The Periodic Table and Classification of Elements · 16 / 22

6.3 Identifying redox reactions by oxidation numbers

Comparison aspect	Main idea	Keywords
Worked example 6.1.	Zn + CuSO₄ → ZnSO₄ + Cu. Zn: 0 → +2 (oxidation Zn is the reducing agent). Cu: +2 → 0 (reduction CuSO₄/Cu²⁺ is the oxidizing agent)	CuSO₄ZnSO₄oxidationreducing
Worked example 6.2.	CuO + H₂ → Cu + H₂O. Cu +2 → 0 (reduced CuO is the oxidizing agent). H 0 → +1 (oxidised H₂ is the reducing agent)	reducedoxidizingagentoxidised
Worked example 6.3.	Fe₂O₃ + 3CO → 2Fe + 3CO₂. Fe +3 → 0 reduction (Fe₂O₃ oxidizing agent) C +2 → +4 oxidation (CO reducing agent)	Fe₂O₃3CO₂reductionoxidizing
Worked example 6.4 — a non-redox reaction.	CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂: Ca +2 C +4 O −2 H +1 Cl −1 on both sides. Nothing changes: not redox	CaCO₃2HClCaCl₂both
Worked example 6.5 — oxidation or reduction?	NO₂ → N₂O₄: N +4 → +4 neither. Cr₂O₇²⁻ → Cr₂O₃: Cr +6 → +3 reduction. MnO₄⁻ → MnO₂: Mn +7 → +4 reduction. ClO₃⁻ → ClO₄⁻: Cl +5 → +7 oxidation. FeCl₂ → FeCl₃: Fe +2 → +3 oxidation. O₂ → O₃: no change in oxidation number (not redox)	N₂O₄neitherCr₂O₇²Cr₂O₃

Chapter 3 — The Periodic Table and Classification of Elements · 17 / 22

6.4 Ionic equations, spectator ions and half-reactions

Consider the redox reaction 2NaI(aq) + Cl₂(g) → 2NaCl(aq) + I₂(s).

Comparison aspect	Main idea	Keywords
The overall ionic equation	2Na⁺ + 2I⁻ + Cl₂ → 2Na⁺ + 2Cl⁻ + I₂
Na⁺	is a spectator ion — its oxidation number does not change and it does not take part actively. Removing it gives the net ionic equation: 2I⁻ + Cl₂ → 2Cl⁻ + I₂	spectatoroxidationnumberdoes
Iodide I⁻ (−1) → I₂ (0)	oxidation. Chlorine Cl₂ (0) → Cl⁻ (−1): reduction	oxidationChlorinereduction
•	Writing each separately and adding electrons to keep the charge balanced: Oxidation half-reaction: 2I⁻ → I₂ + 2e⁻ Reduction half-reaction: Cl₂ + 2e⁻ → 2Cl⁻ Worked example 6.6 — metal with copper sulphate. Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s). Overall ionic: Zn + Cu²⁺ + SO₄²⁻ → Zn²⁺ + SO₄²⁻ + Cu. Spectator: SO₄²⁻. Net: Zn + Cu²⁺ → Zn²⁺ + Cu. Half-reactions: Zn → Zn²⁺ + 2e⁻ (oxidation) Cu²⁺ + 2e⁻ → Cu (reduction). Worked example 6.7 — sodium with water. 2Na + 2H₂O → 2NaOH + H₂. Ionic: 2Na + 2H₂O → 2Na⁺ + 2OH⁻ + H₂. Oxidation: Na → Na⁺ + e⁻ (×2). Reduction: 2H₂O + 2e⁻ → H₂ + 2OH⁻. Na is the reducing agent water (H +1) is the oxidizing agent. OH⁻ is not changed in oxidation number (it appears as product only). Worked example 6.8 — calcium with water. Ca + 2H₂O → Ca(OH)₂ + H₂. Net ionic: Ca + 2H₂O → Ca²⁺ + 2OH⁻ + H₂. Oxidation: Ca → Ca²⁺ + 2e⁻. Reduction: 2H₂O + 2e⁻ → H₂ + 2OH⁻	Writingseparatelyaddingelectrons

Chapter 3 — The Periodic Table and Classification of Elements · 18 / 22

6.5 Balancing ionic redox equations (half-reaction method)

Steps:

Separate the oxidation and the reduction half-reactions.
Balance atoms/ions in each half-reaction.
Apply conservation of charge by adding electrons (lost on the right of the oxidation, gained on the left of the reduction).
Multiply to make the electrons lost equal to the electrons gained.
Add the two half-reactions (electrons cancel).

Comparison aspect	Main idea	Keywords
Worked example 6.9 (pattern of the book).	Cr(s) + H⁺ → Cr³⁺ + H₂
Oxidation	Cr → Cr³⁺ + 3e⁻. Reduction: 2H⁺ + 2e⁻ → H₂. Multiply the first by 2 and the second by 3: 2Cr → 2Cr³⁺ + 6e⁻ 6H⁺ + 6e⁻ → 3H₂. Add: 2Cr + 6H⁺ → 2Cr³⁺ + 3H₂. Charge check: left +6 right +6 ✓	ReductionMultiplyfirstsecond
Worked example 6.10.	Al + Ni²⁺ → Al³⁺ + Ni
Oxidation	Al → Al³⁺ + 3e⁻ (×2). Reduction: Ni²⁺ + 2e⁻ → Ni (×3). Add: 2Al + 3Ni²⁺ → 2Al³⁺ + 3Ni. Charge: +6 = +6 ✓	Reduction3Ni²2Al³Charge

Worked example 6.11.Zn + Fe³⁺ → Zn²⁺ + Fe²⁺.

OxidationZn → Zn²⁺ + 2e⁻. Reduction: Fe³⁺ + e⁻ → Fe²⁺ (×2). Add: Zn + 2Fe³⁺ → Zn²⁺ + 2Fe²⁺. Charge: +6 = +2 + 4 ✓.

Worked example 6.12.Br⁻ + Cl₂ → Cl⁻ + Br₂: 2Br⁻ → Br₂ + 2e⁻; Cl₂ + 2e⁻ → 2Cl⁻. Add: 2Br⁻ + Cl₂ → Br₂ + 2Cl⁻.

Worked example 6.13 — balancing with oxidation numbers.Sn²⁺ + Fe³⁺ → Sn⁴⁺ + Fe²⁺. Sn rises by 2; Fe falls by 1. Two Fe³⁺ per Sn²⁺: Sn²⁺ + 2Fe³⁺ → Sn⁴⁺ + 2Fe²⁺. Total charge: +8 = +8 ✓.

Chapter 3 — The Periodic Table and Classification of Elements · 19 / 22

6.6 Special types of redox reactions

Disproportionation→oxidation and reduction occur for the same element in one reactant, which is both oxidised and reduced.

Comparison aspect	Main idea	Keywords
H₂O₂ → 2H₂O + O₂	O in H₂O₂ is −1 it becomes −2 in H₂O (reduction) and 0 in O₂ (oxidation)	becomesreductionoxidation
Cl₂ + 2NaOH → NaCl + NaClO + H₂O	Cl 0 → −1 and 0 → +1
Comproportionation	(synproportionation) — two compounds or ions containing a common element in two different oxidation states react to give one product in the element has an intermediate state	synproportionationcompoundsionscontaining

Comparison aspect	Main idea	Keywords
H₂S + SO₂ → 3S + 2H₂O	S −2 (oxidised) and S +4 (reduced) both give S⁰	oxidisedreducedbothgive
I⁻ + IO₃⁻ + 6H⁺ → 3I₂ + 3H₂O	I −1 and I +5 give I 0. (Check: atoms I 6 O 3 H 6 charge 0 both sides ✓.)	giveCheckatomscharge
Worked example 6.14.	Identify the oxidation reduction oxidizing agent and reducing agent in MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O (a pattern the book uses). Mn: +4 → +2 reduction MnO₂ oxidizing agent. Cl: −1 → 0 oxidation HCl reducing agent (only two of the four Cl⁻ change)	Identifyoxidationreductionoxidizing
Worked example 6.15.	SO₂ + Cl₂ + 2H₂O → H₂SO₄ + 2HCl: S +4 → +6 oxidised (SO₂ reducing agent) Cl 0 → −1 reduced (Cl₂ oxidizing agent)	2H₂OH₂SO₄2HCloxidised
Worked example 6.16.	Which of the following is not redox: (a) 2Mg + O₂ → 2MgO (redox) (b) NaOH + HCl → NaCl + H₂O (not redox) (c) H₂ + Cl₂ → 2HCl (redox)? (b)	followingredox2MgONaOH

Chapter 3 — The Periodic Table and Classification of Elements · 20 / 22

6.7 Final summary

Oxidation number rises → oxidation (loss of electrons) → reducing agent.
Oxidation number falls → reduction (gain of electrons) → oxidizing agent.
Balance redox equations so that electrons lost = electrons gained and charge is conserved.

Chapter 3 — The Periodic Table and Classification of Elements · 21 / 22

Part 22

📖

📌 Comprehensive Scientific Terminology & Definitions:

Comparison aspect	Main idea	Keywords
Modern Periodic Table	Systematic tabular arrangement of elements in increasing order of atomic numbers and the filling of subshells according to the Aufbau principle composed of 7 periods and 18 groups across four blocks: s and f	Systematictabulararrangementelements
Atomic Radius	Half the distance between the nuclei of two identical covalently bonded atoms. Decreases across a period due to increased effective nuclear charge (Z_eff) and increases down a group due to the addition of principal energy shells.	Halfdistancebetweennuclei
Ionization Energy (IE)	The minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state: X(g) + IE → X⁺(g) + e⁻	minimumrequiredremovemost
Electronegativity	A relative measure of the tendency of an atom in a covalent bond to attract the shared pair of electrons toward itself. Fluorine is the most electronegative element (Pauling value = 4.0)	relativemeasuretendencyatom
Electron Affinity (EA)	The energy change associated with the addition of an electron to a gaseous atom in its ground state to form a univalent negative ion: X(g) + e⁻ → X⁻(g) + EA	energychangeassociatedaddition

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

Comparison aspect	Main idea	Keywords
Effective Nuclear Charge	Zeff = Zactual – Shielding (S)	ZeffZactualShielding
Periodic Trends Across a Period (Left to Right)	Atomic Radius Decreases \| IE Electronegativity EA Increase	AtomicRadiusDecreasesElectronegativity
Periodic Trends Down a Group (Top to Bottom)	Atomic Radius Increases \| IE Electronegativity EA Decrease	AtomicRadiusIncreasesElectronegativity
Basic Oxides (Metallic)	React with acids to form salt + water (e.g Na₂O MgO CaO)	Reactacidsformsalt
Acidic Oxides (Nonmetallic)	Dissolve in water to form acids react with bases (e.g CO₂ SO₃ P₂O₅)	Dissolvewaterformacids
Amphoteric Oxides	Act as bases with acids and as acids with bases (e.g Al₂O₃ ZnO SnO PbO)	basesacidsAl₂O₃

🔍 In-Depth Explanations, Deductions & Core Concepts:

Comparison aspect	Main idea	Keywords
Why does atomic radius decrease across a period?	Because the nuclear charge increases by one proton with each consecutive element while electrons enter the same principal shell resulting in greater attractive pull on valence electrons without added shielding	Becausenuclearchargeincreases
Why is the 2nd ionization energy of sodium exceptionally high?	Because removing a second electron requires disrupting a stable filled noble-gas core (Ne octet 2p⁶) demanding an immense quantity of energy	Becauseremovingsecondelectron
Why is Al₂O₃ classified as amphoteric?	Because it reacts with strong acids (e.g HCl) forming aluminum chloride (basic behavior) and with strong bases (e.g NaOH) forming sodium meta-aluminate NaAlO₂ (acidic behavior)	Becausereactsstrongacids

Periodic Property	Across a Period (Left to Right)	Down a Group (Top to Bottom)	Fundamental Driving Cause
Atomic Radius	Decreases	Increases	Increased Z_eff across / Added electron shells down
Ionization Energy	Increases	Decreases	Stronger electrostatic pull across / Increased shielding down
Electronegativity	Increases	Decreases	Nucleus closer to bond pairs across / Shielded down
Metallic Character	Decreases (Nonmetallic increases)	Increases	Valence electrons lost more easily down group

💡 High-Yield Exam Traps & Key Takeaways:

Most electronegative elements F (4.0) > O (3.5) > N (3.0) ≈ Cl (3.0).
Largest atomic radius belongs to Cesium (Cs); smallest non-noble gas is Fluorine (F).

Chapter 3 — The Periodic Table and Classification of Elements · 22 / 22

Curriculum

Chemistry — Second Secondary

Chemistry — Second Secondary

Chemistry — Second Secondary 2026/2027

Chapter 3 — The Periodic Table and Classification of Elements

🌳 Concept tree

│ ├── 1.1 The basis of arrangement

│ ├── 1.2 Description of the table

│ ├── 1.3 The four blocks of the periodic table

│ ├── 1.4 Types of elements

│ ├── 2.1 Definition

│ ├── 2.3 Variation down a group

│ ├── 3.2 Electron affinity

│ ├── 3.3 Electronegativity

│ ├── 4.1 Metals and non-metals

│ ├── 4.2 Trends of metallic character

│ ├── 5.1 Concept

│ ├── 5.2 Ionic compounds

│ ├── 5.3 Covalent compounds

│ ├── 5.4 Rules

│ └── 5.5 Worked examples

├── 6.1 Definitions

├── 6.6 Special types of redox reactions

└── 6.7 Final summary

✍️ Worked examples

2.2 Variation across a period (left to right)

2.3 Variation down a group

✍️ Worked examples

M(g) → M+(g) + e− (first ionization energy, IE₁)

N > O.)

Mg(g) → Mg+(g) + e− ΔH = +737 kJ/mol

Mg+(g) → Mg2+(g) + e− ΔH = +1450 kJ/mol

✍️ Worked examples

✍️ Worked examples

4.1 Metals and non-metals

4.4 Acidic and basic character in the periodic table

5.1 Concept

5.2 Ionic compounds

5.3 Covalent compounds

5.5 Worked examples

✍️ Worked examples

📌 Comprehensive Scientific Terminology & Definitions:

⚖️ Governing Scientific Laws, Principles & Mathematical Formulas:

🔍 In-Depth Explanations, Deductions & Core Concepts:

💡 High-Yield Exam Traps & Key Takeaways:

Well done! You finished this lesson

M(g) → M⁺(g) + e⁻ (first ionization energy, IE₁)

Mg(g) → Mg⁺(g) + e⁻ ΔH = +737 kJ/mol

Mg⁺(g) → Mg²⁺(g) + e⁻ ΔH = +1450 kJ/mol