Structure 3.1 Flashcards

s-block (Groups 1-2): s-orbital filling. P-block (Groups 13-18): p-orbitals. D-block (Groups 3-12): d-orbitals. F-block (lanthanides/actinides): f-orbitals.

An element’s position gives its highest occupied subshell. E.g. Period 3, Group 16 = 3p⁴ = Sulfur. Group number tells valence electrons; period number tells the highest shell.

The minimum energy required to remove one mole of electrons from one mole of GASEOUS atoms to form one mole of gaseous 1+ ions. The word ‘gaseous’ is essential - omitting it loses the mark.

(1) Effective nuclear charge (Z_eff) - increases across a period (more protons, same shielding). (2) Shielding - increases down a group (more electron shells). These explain radius, IE, EA, and EN.

Across a period: Atomic radius decreases. IE increases. EA becomes more exothermic. EN increases. All driven by increasing Z(eff) with no extra shielding.

That restates the observation, not the explanation. You must state the FORCES: increasing Z(eff) (more protons, same shielding) pulls the electron cloud tighter.

Na₂O, MgO = basic (giant ionic). Al₂O₃ = amphoteric. SiO₂ = weakly acidic (giant covalent). P₄O₁₀, SO₃, Cl₂O₇ = strongly acidic (simple molecular).

Amphoteric, NOT neutral. Amphoteric = reacts with both acids AND bases. Neutral = reacts with neither. Calling it neutral is a common error that loses marks.

Group 1 (metals): reactivity INCREASES down (lower IE = easier to lose valence e⁻). Group 17 (non-metals): reactivity DECREASES down (larger radius + more shielding = harder to gain e⁻). Same factor, opposite directions.

A more reactive halogen displaces a less reactive halide from solution. E.g. Cl₂(aq) + 2KBr(aq) → 2KCl(aq) + Br₂(aq). Solution turns orange/brown as Br₂ forms.

2M + 2H₂O → 2MOH + H₂. Li: steady fizz. Na: molten ball on surface. K: lilac flame + vigorous reaction. Products are always a metal hydroxide + hydrogen gas.

The enthalpy change when one mole of electrons is added to one mole of gaseous atoms to form one mole of gaseous 1- ions. Usually exothermic for most elements.

Fluorine, with a Pauling electronegativity value of 4.0. Electronegativity increases across a period and up a group.

The similarity in chemical properties between diagonally adjacent elements in Periods 2 and 3 (e.g. Li and Mg, Be and Al). Occurs because the effects of crossing a period and descending a group cancel out.

Melting points increase from Na to Al due to strengthening metallic bonding (higher charge, more delocalised electrons). Si is highest because it has a giant covalent network with strong covalent bonds throughout.

Melting points depend on London dispersion forces, which increase with molecular size. Order: S₈ > P₄ > Cl₂ > Ar. S₈ is largest so has the strongest LDFs.

By common exam definition: a d-block element forming at least one stable ion with an incomplete d-subshell (d1 to d9), excluding Zn (Zn2+ is 3d10) and Sc3+ (d0). Note: IUPAC defines it as having an incomplete d-subshell in the atom or any cation, which includes neutral Sc ([Ar] 3d1 4s2).

1) Variable oxidation states. 2) High melting points (d-electrons strengthen metallic bonding). 3) Coloured compounds (d-d transitions). 4) Catalytic activity (alternative pathways, lower Ea). 5) Complex ion formation.

Cr: [Ar] 4s¹ 3d⁵ (not 4s² 3d⁴) - half-filled d is extra stable. Cu: [Ar] 4s¹ 3d¹⁰ (not 4s² 3d⁹) - fully filled d is extra stable.

Light is ABSORBED when electrons are PROMOTED between split d-orbitals. We see the COMPLEMENTARY colour. Do not say colour is emitted when electrons fall down. d⁰ and d¹⁰ ions are colourless.

Be: 1s² 2s² (e⁻ from 2s). B: 1s² 2s² 2p¹ (e⁻ from 2p). 2p is higher energy, less penetrating, further from nucleus than 2s. B electron is easier to remove → lower IE despite higher Z(eff).

N: 2p³ (all unpaired). O: 2p⁴ (one orbital paired). The paired electron in O experiences spin-pairing repulsion, raising its energy. Easier to remove → lower IE than expected.

Yes! Mg → Al (same as Be → B: s → p subshell change) and P → S (same as N → O: spin-pairing repulsion). You must explain both using orbital diagrams in the exam.

Energy is required to overcome the electrostatic repulsion of forcing an electron into an already negatively charged ion.

Thermal stability increases down the group. Larger cations have lower charge density and less polarising power, so they distort the carbonate anion less, making it harder to decompose.

Na₂O(s) + H₂O(l) → 2NaOH(aq) (pH 13-14, strongly alkaline). P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq) (strongly acidic).