Chapter 3 — The Periodic Table and Classification of Elements

• 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:

1. the elements are arranged in ascending order of atomic number;
2. 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:

• 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):

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

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

• 1.
• s-block: on the left.
• The outermost electrons occupy the s sublevel: group 1A (ns¹) and group 2A (ns²), where n is the period number and the number of the last energy level.

• 2.
• p-block: on the 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.
• (Helium is 1s².)

• 3.
• d-block: 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).

• 4.
• f-block: 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:

• Lanthanides: the 4f sublevel is filled, 14 elements; the outer sublevel of all is 6s².
• Actinides: the 5f sublevel is filled, 14 elements, all radioactive (unstable nuclei).

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

1. 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.
2. 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.
3. Main (d-block) transition elements: the d sublevel is filled successively.
4. Inner transition elements: the f-block, where f is filled successively.

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.

• Period = number of the outermost main level (largest n).
• 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.
• 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; Fe, Co, Ni (8, 9, 10) group VIII; Cu (11) 1B; Zn (12) 2B.
• The block is the sublevel that receives the last electron.

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.

Worked example 1.2. Locate ₂₀Ca: [Ar] 4s² → period 4, group 2A, s-block, representative element (an alkaline earth metal).

Worked example 1.3. Locate ₂₆Fe: [Ar] 4s² 3d⁶ → period 4, d-block (main transition element), group VIII (8 outer electrons: 2 + 6).

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).

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:

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.

The atomic radius increases with atomic number. Reasons:

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

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

• 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).
• 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).

• 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)

• 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:

Notes:

1. The trend of electron affinity is irregular in periods and groups, unlike atomic radius and ionization energy.
2. 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.
3. 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.
4. 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.
5. Noble gases have positive values because of the stability of their saturated configuration (the electron would have to enter a new, higher level).

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

• It is expressed by a relative number: the 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.

• Trends: increases across a period (increasing atomic number, decreasing radius); decreases down a group.
• Fluorine is the most electronegative element.
• The difference in electronegativity between two elements decides the type of bond between them (studied in Chapter 4 of the book).

Across a period (left → right): atomic radius ↓, ionization energy ↑, electronegativity ↑. Down a group: atomic radius ↑, ionization energy ↓, electronegativity ↓. 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).

• 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

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

Non-metals

1. Their valence shell generally has more than half its capacity of electrons.
2. They gain a few electrons to reach the noble-gas configuration, forming negative ions; they are electronegative.
3. They are electrical insulators because the valence electrons are bound strongly to the nucleus and cannot move.
4. 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.

• 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.
• Down a group: metallic character increases.
• 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).

• 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)
• 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)
• 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)
• Neutral oxides (CO, NO, N₂O) react neither with acids nor with alkalis and do not dissolve readily in water.

• 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.
• Down a group (e.g. group 1): basic character of the hydroxides/oxides increases as atomic number increases (LiOH < NaOH < KOH < RbOH < CsOH).
• 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.

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.

Worked example 4.4. Show that tin(II) oxide is amphoteric. SnO + 2HCl → SnCl₂ + H₂O; SnO + 2NaOH → Na₂SnO₂ + H₂O.

Worked example 4.5. Classify the oxides CaO, SO₂, ZnO, NO, Li₂O. CaO basic; SO₂ acidic; ZnO amphoteric; NO neutral; Li₂O basic.

Worked example 4.6. Which is more acidic, HI or HF in water? HI, since the H–I bond is longer and weaker.

• 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.

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⁻.

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.

1. The sum of oxidation numbers of all atoms in a neutral compound is zero; in an ion it equals the charge of the ion.
2. Group 1A elements are always +1; group 2A always +2; group 3A (Al) +3; fluorine is always −1. Start with these.
3. Oxygen is −2 in most compounds, except: peroxides (H₂O₂, Na₂O₂) −1; superoxides (KO₂) −½; oxygen difluoride (OF₂) +2.
4. Hydrogen is +1 in most compounds, except metal hydrides (NaH, CaH₂, AlH₃) where it is −1.
5. 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⁻).
6. The highest oxidation state of a representative element does not exceed its group number (A groups).
7. 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.

Worked example 5.1. Oxidation number of Cr in K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 → 2x = 12 → x = +6.

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.

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.

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.

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⁺).

• 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).

• 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).

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).

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).

Worked example 6.3. Fe₂O₃ + 3CO → 2Fe + 3CO₂. Fe +3 → 0 reduction (Fe₂O₃ oxidizing agent); C +2 → +4 oxidation (CO reducing agent).

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.

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).

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

• 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.
• Iodide I⁻ (−1) → I₂ (0): oxidation. Chlorine Cl₂ (0) → Cl⁻ (−1): reduction.
• 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⁻.

Steps:

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

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 ✓.

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 ✓.

Worked example 6.11. Zn + Fe³⁺ → Zn²⁺ + Fe²⁺. Oxidation: Zn → 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 ✓.

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

• 2H₂O₂ → 2H₂O + O₂: O in H₂O₂ is −1; it becomes −2 in H₂O (reduction) and 0 in O₂ (oxidation).
• 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 which the element has an intermediate state.

• 2H₂S + SO₂ → 3S + 2H₂O: S −2 (oxidised) and S +4 (reduced) both give S⁰.
• 5I⁻ + IO₃⁻ + 6H⁺ → 3I₂ + 3H₂O: I −1 and I +5 give I 0. (Check: atoms I 6, O 3, H 6; charge 0 both sides ✓.)

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).

Worked example 6.15. SO₂ + Cl₂ + 2H₂O → H₂SO₄ + 2HCl: S +4 → +6 oxidised (SO₂ reducing agent); Cl 0 → −1 reduced (Cl₂ oxidizing agent).

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).

• 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.