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When Mendeleev first arranged the known elements into the periodic table in 1869, he did so on the basis of repeating chemical behaviour — ordering by atomic mass and leaving gaps where his pattern demanded an undiscovered element. The modern table, ordered by atomic number and partitioned by sub-shell filling, makes the underlying logic explicit: the chemistry of an element is a direct consequence of its electron configuration. Period 3 — sodium through argon — is the single most-examined row at A-Level, because it spans the full structural sweep from reactive metal to giant-covalent network to molecular non-metal to monatomic noble gas across only eight elements. This lesson previews the five quantitative trends you must master: atomic radius, first ionisation energy (IE), electronegativity, melting point, and the acid-base behaviour of the Period 3 oxides. Each trend is grounded in two competing factors — nuclear charge and shielding — and the anomalies (the Mg → Al and P → S IE drops; the dramatic melting-point fall at phosphorus) are exactly where the highest-tariff marks are earned.
Spec mapping (AQA 7405): This lesson anchors §3.2.1 (periodicity — classification of elements in s, p, d blocks; properties of Period 3 elements Na to Ar; the structures and melting points of the Period 3 elements; the trends in atomic radius, first ionisation energy and electronegativity). It builds directly on §3.1.1.3 (the patterns in successive ionisation energies, covered in atomic-structure L3 of this course series), and extends into §3.2.3 (Period 3 oxides — covered in detail in L3 of the present course). It is cross-referenced by L1 (Group 2 trends) and L2 (Group 7 trends) of this course. Refer to the official AQA specification document for exact wording.
Assessment objectives: AO1 recall items include the definitions of periodicity, first ionisation energy, atomic radius, and electronegativity. AO2 features dominate this lesson — sketching and explaining the trend in IE across Period 3, predicting structural changes and their effect on melting points, and writing balanced equations for oxide formation. AO3 reasoning appears in the high-tariff anomaly questions: rationalising the Mg → Al IE drop (sub-shell energy difference), the P → S IE drop (3p electron pairing), and predicting trends in oxide structure and acid-base behaviour from the bonding-character continuum.
Periodicity is the repeating pattern of physical and chemical properties across periods of the Periodic Table. Period 3 (Na to Ar) is the canonical worked example at A-Level because every property you study — atomic radius, ionisation energy, electronegativity, melting point, oxide structure, and oxide acidity — shows a sharp, well-defined trend across just eight elements. Period 2 (Li to Ne) shows the same patterns but with second-row anomalies (e.g. unusually small atomic radii) that obscure the underlying physics; Period 3 is where the textbook trends are cleanest.
The two underlying factors are nuclear charge (number of protons, Z) and shielding (the screening of outer electrons from the nucleus by inner electrons). Across a period the nuclear charge increases by one per element, while the shielding contribution from inner shells (1s²2s²2p⁶ in Period 3) remains essentially constant. The net effective nuclear charge experienced by the outer electrons (Z_eff) therefore rises, with predictable consequences for radius, IE, and electronegativity.
Atomic radius decreases across Period 3 from sodium to chlorine.
| Element | Na | Mg | Al | Si | P | S | Cl | Ar |
|---|---|---|---|---|---|---|---|---|
| Atomic radius / pm | 186 | 160 | 143 | 117 | 110 | 104 | 99 | 71 |
| Protons (Z) | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| Electron config | [Ne]3s¹ | [Ne]3s² | [Ne]3s²3p¹ | [Ne]3s²3p² | [Ne]3s²3p³ | [Ne]3s²3p⁴ | [Ne]3s²3p⁵ | [Ne]3s²3p⁶ |
Exam Tip: When asked to explain the trend in atomic radius, always refer to (1) increasing nuclear charge, (2) approximately constant shielding (same inner-shell electrons), and (3) greater attraction for outer electrons. Never write "more protons attract more electrons" without specifying that shielding remains roughly constant — examiners deduct marks for omitting the shielding clause.
The first ionisation energy (IE₁) is the energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous unipositive ions:
X(g) → X⁺(g) + e⁻
| Element | Na | Mg | Al | Si | P | S | Cl | Ar |
|---|---|---|---|---|---|---|---|---|
| 1st IE / kJ mol⁻¹ | 496 | 738 | 578 | 786 | 1012 | 1000 | 1251 | 1521 |
First ionisation energy generally increases across Period 3. The reasons mirror those for atomic radius:
This is the first of two carefully-examined departures from the upward trend.
This pattern (IE drop on entering a new sub-shell) is general: it also appears between Be → B (Period 2) and Ca → Ga (Period 4).
The second anomaly is smaller but examined just as often.
The same pattern recurs between N → O (Period 2) and As → Se (Period 4): a half-filled p sub-shell is energetically favoured because spin-pair repulsion is minimised, so the next element loses one electron unusually easily.
Exam Tip: These two dips are examined frequently. For the Mg → Al dip, emphasise sub-shell energy levels (3p > 3s) and slight extra shielding from the 3s² pair. For the P → S dip, emphasise spin-pair repulsion in the doubly-occupied 3p orbital. Always quote the electron configurations explicitly — examiners reward this.
Q: Explain why the first ionisation energy of sodium (496 kJ mol⁻¹) is much lower than that of neon (2081 kJ mol⁻¹).
A: Sodium's outer electron occupies the 3s sub-shell, which is much further from the nucleus than neon's outer 2p electrons. Sodium has an additional inner shell (the n = 2 shell, with eight electrons) providing substantial extra shielding. Although sodium has one more proton (Z = 11 vs Z = 10 for Ne), the combined effect of greater radial distance and significantly increased shielding far outweighs the small extra nuclear charge, so sodium's first ionisation energy is much lower. This drop in IE on starting a new shell defines the boundary between periods and is the most dramatic IE change anywhere in the table.
Electronegativity (χ) is the ability of an atom to attract the bonding pair of electrons in a covalent bond towards itself. The Pauling scale is conventional in A-Level work, with fluorine assigned the maximum value of 3.98.
| Element | Na | Mg | Al | Si | P | S | Cl |
|---|---|---|---|---|---|---|---|
| Electronegativity (Pauling) | 0.93 | 1.31 | 1.61 | 1.90 | 2.19 | 2.58 | 3.16 |
Electronegativity increases smoothly across Period 3 from sodium to chlorine.
The increase is fairly linear across Period 3, in contrast to the IE trend which has two anomalies. This is because electronegativity is an averaged property over many bonding environments — the small spin-pair-repulsion effects that show up in IE are averaged out.
Common Misconception: Electronegativity is not the same as electron affinity. Electronegativity is a relative, dimensionless scale describing an atom's bonding behaviour in a molecule; electron affinity is the enthalpy change (kJ mol⁻¹) when a gaseous atom gains an electron to form a gaseous anion. The two correlate but measure different things.
Melting points across Period 3 show a distinctive zig-zag pattern that reflects discrete changes in bonding and structure, not a smooth electronic trend. This is the most-mistaught topic at A-Level — the explanation must always start with structure.
| Element | Na | Mg | Al | Si | P₄ | S₈ | Cl₂ | Ar |
|---|---|---|---|---|---|---|---|---|
| Structure | Giant metallic | Giant metallic | Giant metallic | Giant covalent | Simple molecular | Simple molecular | Simple molecular | Monatomic |
| Melting point / °C | 98 | 650 | 660 | 1410 | 44 | 113 | −101 | −189 |
Melting point increases sharply Na → Mg → Al (98 → 650 → 660 °C) because:
Silicon has by far the highest melting point in Period 3. It crystallises in a diamond-cubic structure: each Si atom is covalently bonded to four neighbours in a tetrahedral arrangement, with bond lengths of 235 pm. To melt silicon, you must break a substantial fraction of these strong covalent bonds (Si–Si bond energy 226 kJ mol⁻¹), which requires very high temperature. This single structural change — from metallic to giant covalent — accounts for almost the entire vertical spread in the Period 3 melting-point graph.
These elements exist as discrete molecules in the solid state, held together by weak London (dispersion) forces between molecules. Melting depends only on overcoming these intermolecular forces — the covalent bonds within each molecule remain intact. London-force strength scales with the number of electrons per molecule (more electrons → larger instantaneous dipoles → stronger forces):
The drop from Si (1410 °C) to P₄ (44 °C) is the largest single step in the Period 3 melting-point pattern and is the canonical A-Level question.
Argon exists as discrete monatomic atoms with no chemical bonding. Only very weak London forces between isolated atoms (18 electrons each) hold the solid together, giving the lowest melting point in Period 3.
Exam Tip: The single most important step in answering any melting-point question is to identify the structure type first (metallic, giant covalent, simple molecular, monatomic) and then discuss the forces broken on melting. A common mistake is trying to explain the trend using only one factor (e.g. nuclear charge) when the real explanation requires discussing the change of bonding type. Marks are routinely awarded for explicitly naming the structure.
Q: Explain why silicon has a much higher melting point than phosphorus.
A: Silicon has a giant covalent structure in which each atom is covalently bonded to four neighbours in a tetrahedral lattice. To melt silicon, many strong covalent bonds (Si–Si, 226 kJ mol⁻¹) must be broken, requiring very high temperatures (1410 °C). Phosphorus exists as discrete P₄ molecules in which atoms are covalently bonded internally but only weak London (dispersion) forces operate between molecules. Melting phosphorus requires only that the weak intermolecular forces be overcome — the strong P–P covalent bonds within each tetrahedron remain intact. Hence the dramatic drop in melting point from Si to P₄.
The oxides of Period 3 elements show a parallel structural progression (giant ionic → amphoteric → giant covalent → molecular covalent) that maps onto a clear acid-base trend (basic → amphoteric → acidic). This is the direct extension into L3 of the present course (Period 3 oxides — full treatment) and is anchored in §3.2.3 of the AQA specification. A concise overview is given here to make the periodicity story complete.
| Oxide | Na₂O | MgO | Al₂O₃ | SiO₂ | P₄O₁₀ | SO₃ | Cl₂O₇ |
|---|---|---|---|---|---|---|---|
| Bonding | Ionic | Ionic | Ionic with covalent character | Giant covalent | Molecular covalent | Molecular covalent | Molecular covalent |
| Structure | Giant ionic | Giant ionic | Giant ionic (polarised) | Giant covalent | Molecular | Molecular | Molecular |
| Mp / °C | 1132 | 2852 | 2072 | 1610 | 340 | 17 | −9 |
| pH in water | 13–14 | 9–10 | insoluble | insoluble | 1–2 | 0–1 | 0–1 |
| Acid-base | Strongly basic | Basic | Amphoteric | Weakly acidic | Acidic | Strongly acidic | Strongly acidic |
The trend tracks the change from metallic (Na, Mg, Al) to non-metallic (Si, P, S, Cl) elements:
Q: Predict whether Al₂O₃ is acidic, basic, or amphoteric, and write equations for any reactions with HCl(aq) and NaOH(aq).
A: Al₂O₃ sits at the boundary between metallic and non-metallic oxides. Aluminium has intermediate electronegativity, and the high charge density of Al³⁺ polarises O²⁻ substantially. Al₂O₃ is therefore amphoteric:
The aluminate ion [Al(OH)₄]⁻ is the conjugate base. Amphoteric behaviour is also shown by Al(OH)₃, Zn(OH)₂, BeO, and PbO — a useful periodic-trend pattern.
A standard School Science Review-style demonstration links periodicity directly to a measurable laboratory observation. The procedure is straightforward and routinely tested in Section A of Paper 3.
Procedure (qualitative):
Expected observations:
Safety: Na₂O reacts violently with water and should be added in very small quantities to a large excess of water behind a safety screen. P₄O₁₀ is hygroscopic and corrosive — handle in a fume cupboard. Always wear eye protection.
This practical maps the metallic-to-non-metallic transition onto a single pH ruler and is a memorable way to reinforce the underlying structural argument.
| Property | Trend across Period 3 | Key explanation |
|---|---|---|
| Atomic radius | Decreases Na → Cl | Increasing nuclear charge, constant shielding |
| 1st ionisation energy | Generally increases (dips at Al and S) | Increasing Z_eff; sub-shell energy gap (Mg → Al); spin-pair repulsion (P → S) |
| Electronegativity | Increases Na → Cl | Increasing nuclear charge, decreasing radius |
| Melting point | Rises across metals (Na, Mg, Al), peaks at Si (giant covalent), falls sharply to P₄, peaks again at S₈, drops to Cl₂ and Ar | Change of structure type dominates over electronic factors |
| Oxide acidity | Basic → amphoteric → acidic | Bonding character changes ionic → covalent across the period |
Question 1. [13 marks total]
(a) Sketch a graph of first ionisation energy (y-axis) against atomic number (x-axis) for the elements Na to Ar. Mark numerical values on the axes and label the two anomalies. [3 marks]
(b) Explain, using electron configurations, why the first ionisation energy of aluminium is lower than that of magnesium. [3 marks]
(c) Silicon has the highest melting point in Period 3. Explain this observation in terms of structure and bonding, and compare with the melting point of phosphorus. [3 marks]
(d) A student dissolves separate samples of Na₂O, P₄O₁₀, and MgO in distilled water, measuring the pH of each resulting solution. Predict the approximate pH of each solution and explain your reasoning in terms of the bonding character of the oxides and the species formed in water. [4 marks]
(a) IE graph across Period 3 [3 marks, AO2]
Accept any reasonable scale on the IE axis. Penalise once for a graph showing the dips but with no numbers, or with numbers but no dips.
(b) Mg → Al IE drop [3 marks, AO2]
Do not award marks for a vague "Al has a 3p electron" without the energy/shielding argument.
(c) Si melting point [3 marks, AO2]
A comparative statement that names the bond/force type broken in each case is required for full marks.
(d) Oxide solutions [4 marks, AO2 + AO3]
Accept approximately correct pH ranges. The reasoning mark requires explicit bonding-character logic.
The three responses below cover the meaningful A-Level range: Grade C (the borderline-pass floor), Grade B (solid mark-scheme coverage), and Grade A* (top-band synthesis). The commentary after each response (editorial, not a real examiner report) names the marks earned and the specific moves that differentiate from adjacent bands.
(a) Graph rises from Na (≈ 500) up to Ar (≈ 1520) with two dips: one at Al (≈ 578) below Mg (≈ 738), and one at S (≈ 1000) below P (≈ 1012). Both dips labelled. Smooth curve through other points.
(b) Mg is [Ne]3s² and Al is [Ne]3s²3p¹. The outer electron in Al is in a 3p orbital, which is higher in energy than the 3s. The 3s electrons in Al also shield the 3p electron slightly. So removing the 3p electron from Al takes less energy than removing the 3s electron from Mg, even though Al has one more proton.
(c) Silicon has a giant covalent structure where every Si atom is bonded to four others by strong covalent bonds in a tetrahedral lattice. To melt it, you have to break many covalent bonds, which takes a lot of energy, so the melting point is very high (1410 °C). Phosphorus exists as P₄ molecules with strong covalent bonds inside the molecule but only weak London forces between molecules. Melting only breaks the weak forces, so the melting point of P₄ is only 44 °C.
(d) Na₂O is ionic and dissolves to give NaOH, so the solution is strongly alkaline (pH ≈ 13). MgO is also ionic but only slightly soluble, giving Mg(OH)₂, so the solution is weakly alkaline (pH ≈ 10). P₄O₁₀ is a molecular covalent oxide and reacts with water to give phosphoric acid H₃PO₄, so the solution is strongly acidic (pH ≈ 2).
Editorial commentary (Grade C): Solid coverage of all four parts with the correct overall structure-bonding framework. To progress to B, the student should quote balanced equations for (d), give explicit shielding arguments in (b), and use the words "London (dispersion) forces" and "intermolecular forces" precisely in (c). The pH predictions are correct but unjustified beyond a single sentence.
(a) Graph rises from Na (496) to Ar (1521) with a clear dip at Al (578, below Mg 738) and at S (1000, below P 1012). Both anomalies labelled. Axes correctly scaled.
(b) Mg: 1s²2s²2p⁶ 3s²; Al: 1s²2s²2p⁶ 3s²3p¹. The outermost electron in Al occupies a 3p orbital, which lies at slightly higher energy than the 3s. In addition, the filled 3s² pair provides a small amount of extra shielding for the 3p electron. Less energy is therefore needed to remove the 3p electron from Al than the 3s electron from Mg, despite Al having one more proton — the sub-shell energy difference outweighs the extra nuclear charge.
(c) Silicon has a giant covalent (macromolecular) structure: each Si atom is covalently bonded to four others in a tetrahedral diamond-cubic lattice, with bond energy 226 kJ mol⁻¹. Melting requires breaking many of these strong covalent bonds, giving a very high mp (1410 °C). Phosphorus exists as discrete P₄ molecules — strong P–P covalent bonds within each tetrahedron but only weak London (dispersion) forces between molecules. Melting overcomes only the weak intermolecular forces (the P–P bonds within P₄ remain intact), so mp = 44 °C — almost two orders of magnitude lower than Si.
(d) Na₂O is ionic; Na₂O + H₂O → 2 NaOH; the OH⁻ ion makes the solution strongly alkaline (pH ≈ 13–14). MgO is ionic but only sparingly soluble; gives Mg(OH)₂; pH ≈ 9–10. P₄O₁₀ is a molecular covalent oxide; P₄O₁₀ + 6 H₂O → 4 H₃PO₄; the H₃PO₄ dissociates as a weak triprotic acid giving pH ≈ 1–2. The pattern (basic → weakly basic → acidic) tracks the change from ionic to covalent bonding across Period 3.
Editorial commentary (Grade B): Now mark-scheme-rigorous: balanced equations in (d), explicit sub-shell + shielding argument in (b), and named force types in (c). To progress to A*, the response could connect the trend to ionic-covalent bonding-character continua, discuss why Al₂O₃ is amphoteric (the "missing oxide" in this question), and reference the polarising power of small high-charge cations.
(a) Graph: y-axis IE / kJ mol⁻¹ scaled 400–1600, x-axis atomic number Na (11) to Ar (18). Points plotted at Na 496, Mg 738, Al 578, Si 786, P 1012, S 1000, Cl 1251, Ar 1521. Dips at Al and S explicitly annotated with the underlying cause (sub-shell energy and spin-pair repulsion respectively). Connecting line shows the upward trend with the two anomalies as local minima.
(b) Mg: [Ne]3s²; Al: [Ne]3s²3p¹. Two factors lower IE₁(Al) below IE₁(Mg) despite Z increasing by one. First, the 3p orbital lies at slightly higher energy than 3s — penetration of the 3p sub-shell into the inner shells is less effective, so the 3p electron experiences a smaller Z_eff. Second, the filled 3s² pair provides additional shielding for the 3p electron over and above the 1s²2s²2p⁶ core. The net effect is that the 3p electron is more easily removed, lowering IE₁(Al). The same pattern recurs at the s → p boundary in Period 2 (Be → B) and Period 4 (Ca → Ga) — a general periodic feature, not a Period-3 quirk.
(c) Si is a giant covalent solid with the diamond-cubic structure: every Si atom covalently bonded to four neighbours (bond length 235 pm, bond energy 226 kJ mol⁻¹). Melting requires partial disruption of this 3-D bond network, demanding very high temperatures (1410 °C). P exists as discrete P₄ tetrahedra in white phosphorus; strong P–P bonds (200 kJ mol⁻¹) operate within each tetrahedron, but only weak London (dispersion) forces operate between P₄ units. Melting white phosphorus (44 °C) overcomes only these weak intermolecular forces; the strong covalent bonds within each P₄ unit are preserved into the liquid and gas phases. The Si → P drop is therefore not an electronic anomaly but a direct consequence of the change in structure type — from giant covalent network to discrete molecular solid.
(d) Na₂O is fully ionic; the O²⁻ ion is a strong base and reacts with water to give NaOH (pH ≈ 13–14). MgO is also ionic but the Mg²⁺ ion has higher charge density and lower solubility; the saturated solution of Mg(OH)₂ reaches pH ≈ 9–10 only. The intermediate position of Al₂O₃ (amphoteric) reflects the increasing polarising power of the cation: Al³⁺ has sufficient charge density to give the oxide significant covalent character, allowing it to react with both acids and alkalis. P₄O₁₀ is molecular covalent; the electronegativity of P (2.19) is close enough to O (3.44) that the P–O bond is polar covalent rather than ionic, and the oxide reacts vigorously with water: P₄O₁₀ + 6 H₂O → 4 H₃PO₄ (pH ≈ 1–2). The basic-to-acidic trend across Period 3 oxides therefore tracks directly with electronegativity difference Δχ(E–O): large Δχ → ionic → basic; small Δχ → covalent → acidic.
Editorial commentary (Grade A):* Genuinely A*: invokes Z_eff and orbital penetration in (b); quotes bond lengths and energies in (c) and explicitly distinguishes intramolecular from intermolecular forces; in (d) connects the basic-to-acidic trend to electronegativity difference and the polarising power of cations, mentioning Al₂O₃ even though it was not strictly required. The pattern-recognition in (b) (same anomaly recurring at the s → p boundary in every period) demonstrates true synoptic insight.
Three undergraduate-adjacent extensions:
This lesson aligns with AQA A-Level Chemistry §3.2.1 (Periodicity) and provides the foundation for the Group 2 (L1), Group 7 (L2), and Period 3 oxides (L3) treatments that follow. Master the structural framework here — metallic, giant covalent, molecular, monatomic — and the rest of inorganic Period 3 will follow without further memorisation.