1. Classification into Blocks
The Periodic Table is arranged in order of increasing atomic number. Elements are classified into blocks according to the subshell containing the highest energy valence electron:
- s-block elements: Groups 1 and 2 (valence electrons in \(s\) orbital).
- p-block elements: Groups 3 to 7 and Group 0 (Groups 13 to 18 in the newer numbering), with valence electrons in \(p\) orbitals.
- d-block elements: Transition elements between Groups 2 and 3.
- f-block elements: Lanthanoids and actinoids.
The block follows from the proton number. The proton number gives the number of electrons. This fixes the electron configuration, and so the sub-shell that the outer electrons occupy.
- Na (Z = 11): 1s² 2s² 2p⁶ 3s¹. Outer electron in 3s, so s block.
- Cl (Z = 17): 1s² 2s² 2p⁶ 3s² 3p⁵. Outer electrons in 3p, so p block.
- Fe (Z = 26): [Ar] 3d⁶ 4s². The 3d sub-shell is being filled, so d block.
2. Period 3 Element Trends (Na to Ar)
Atomic Radius Decrease
Why does atomic radius decrease across Period 3?
From sodium to argon, the number of protons in the nucleus increases by 1 each step (nuclear charge increases from +11 to +18). However, additional electrons enter the same third quantum shell (\(n = 3\)). Shielding from inner electron shells (\(1s^2 2s^2 2p^6\)) remains virtually constant. The increased nuclear attraction pulls the outer electron cloud closer to the nucleus, causing atomic radius to decrease.
Electronegativity Increase
Electronegativity increases across Period 3 because atomic radius decreases and nuclear charge increases. Bonding electrons in covalent bonds are attracted more strongly towards the smaller nucleus.
First Ionisation Energy Across Period 3
- General increase from Na to Ar: the nuclear charge increases, the shielding is similar and the atomic radius decreases. So the outer electron is held more strongly.
- Al is lower than Mg: the electron removed from Al is in a 3p orbital. This is higher in energy than 3s, and it is shielded by the 3s electrons.
- S is lower than P: the electron removed from S is paired in a 3p orbital. The repulsion between the two paired electrons makes one easier to remove.
The graph, data and the trend down Group 2 are on the ionisation energy notes.
3. Melting Point Variations Across Period 3
Melting point reflects the type of bonding and crystal structure possessed by the element:
Metallic Elements (Na, Mg, Al)
Melting point increases from Na to Al. Why?
- Ionic charge increases: \(\text{Na}^+ < \text{Mg}^{2+} < \text{Al}^{3+}\).
- Number of delocalised valence electrons contributed per atom increases from 1 to 3.
- Cation radius decreases, bringing delocalised electrons closer to the nucleus.
- Electrostatic attraction between cations and electrons strengthens.
Macromolecular Silicon (Si)
Silicon has the highest melting point in Period 3 (1687 K). Silicon forms a giant macromolecular covalent crystal lattice similar to diamond. Melting silicon requires breaking thousands of strong covalent bonds throughout the 3D network, which requires enormous thermal energy.
Simple Molecular Non-Metals (P₄,S₈,Cl₂) and Argon
These elements exist as discrete molecules or single atoms held together only by weak London dispersion forces:
- Sulfur (S₈): Contains 8 atoms per molecule (128 electrons). Its large electron cloud produces strong London dispersion forces, giving it a higher melting point than phosphorus.
- Phosphorus (P₄): Contains 4 atoms per molecule (60 electrons), with weaker London dispersion forces than S₈.
- Chlorine (Cl₂): Diatomic molecule (34 electrons), a gas at room temperature.
- Argon (Ar): Exists as monatomic atoms (18 electrons) with the weakest London dispersion forces, exhibiting the lowest melting point in Period 3.
Order of melting points: S₈ > P₄ > Cl₂ > Ar. This follows molecule size. Bigger molecules have more electrons, so the van der Waals forces between them are stronger.
- Na, Mg, Al (metallic): melting point rises, as the metallic bonding gets stronger.
- Si (macromolecular): the highest, because many strong covalent bonds must be broken.
- P₄, S₈, Cl₂, Ar (molecular): low, because only weak van der Waals forces between molecules are overcome.
4. Period 3 Properties Matrix
| Element | Symbol | Structure | Bonding | Melting Point (K) | Electrical Conductivity |
|---|---|---|---|---|---|
| Sodium | Na | Giant metallic | Metallic | 371 | Good (1 delocalised e⁻) |
| Magnesium | Mg | Giant metallic | Metallic | 923 | Good (2 delocalised e⁻) |
| Aluminium | Al | Giant metallic | Metallic | 933 | Good (3 delocalised e⁻) |
| Silicon | Si | Giant covalent | Covalent | 1687 | Semiconductor |
| Phosphorus | P₄ | Simple molecular | Covalent (London forces) | 317 | None |
| Sulfur | S₈ | Simple molecular | Covalent (London forces) | 392 | None |
| Chlorine | Cl₂ | Simple molecular | Covalent (London forces) | 172 | None |
| Argon | Ar | Monatomic | London forces only | 84 | None |
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
5. Practice Questions
Show Answer & Mark Scheme
Mark Scheme:
- Sulfur exists as S₈ molecules whereas phosphorus exists as P₄ molecules. (1 mark)
- S₈ molecules are larger and possess more electrons than P₄ molecules (128 electrons vs 60 electrons). (1 mark)
- Therefore, London dispersion forces between S₈ molecules are stronger and require more thermal energy to overcome. (1 mark)
Show Answer & Mark Scheme
Mark Scheme:
- Al: the electron is removed from a 3p orbital (Mg: from 3s). (1 mark)
- The 3p orbital is higher in energy and the electron is shielded by the 3s electrons. (1 mark)
- S: the electron is removed from a 3p orbital that holds a pair of electrons. (1 mark)
- Repulsion between the paired electrons makes it easier to remove. (1 mark)
Show Answer & Mark Scheme
Mark Scheme:
- Silicon is macromolecular, so many strong covalent bonds must be broken. (1 mark)
- Na to Al: the ionic charge increases (+1 to +3) and the ions get smaller. (1 mark)
- More electrons are delocalised, so the metallic bonding is stronger. (1 mark)
Show Answer & Mark Scheme
Mark Scheme:
- (a) The nuclear charge increases, but the shielding is similar (electrons are added to the same shell). (1 mark)
- So the outer electrons are pulled closer to the nucleus. (1 mark)
- (b) X is 1s² 2s² 2p⁶ 3s² 3p³, so its outer electrons are in a p sub-shell: p block. (1 mark)