Compounds, Formulae and Ionic Charges

Spec Mapping — OCR H432 Module 2.1.3 — Amount of substance / formulae and equations, content statements covering the writing of formulae of ionic compounds from charges; prediction of ionic charges by periodic-table position; the names and formulae of common polyatomic ions; transition-metal nomenclature with Roman numerals; and IUPAC naming conventions distinguishing ionic from covalent compounds (refer to the official OCR H432 specification document for exact wording). This lesson is the bridge between Module 2.1.2 (relative masses) and the mole-stoichiometry lessons that follow — a wrong formula at this stage propagates into every subsequent quantitative answer.

This lesson establishes the formula-writing literacy without which no later quantitative chemistry calculation is possible. You must be fluent in three connected skills: predicting the charge of a monatomic ion from its periodic-table group; recognising and reproducing the formulae of the common polyatomic ions (memorisation, not derivation); and combining cations and anions into a neutral ionic formula using the cross-over rule, with bracket and simplification discipline. A formula error here is not a single-mark slip — it cascades into wrong $M_r$Mr​, wrong moles, wrong stoichiometry, and ultimately wrong answers across multi-part questions on Papers 1, 2 and 3. The historical thread is Mendeleev's periodic-table organisation (1869), which arranged elements by atomic mass and revealed the group patterns that explain why the alkali metals all form $+1$+1 ions; this organisation now sits at the conceptual centre of formula prediction. We also introduce transition-metal variable valency, the formal IUPAC use of Roman numerals to disambiguate Fe²⁺ from Fe³⁺, and the contrasting "mono/di/tri/tetra" prefix conventions of covalent nomenclature.

Key Rule: For any ionic formula to exist, total positive charge $=$= total negative charge. The compound is overall neutral.

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From Atoms to Compounds

Elements rarely exist as single atoms. Most combine chemically to form compounds — substances in which atoms of different elements are held together by chemical bonds. Compounds are of two main bonding types:

• Ionic — composed of cations (+) and anions (−) held together by electrostatic attraction in a three-dimensional lattice. Examples: NaCl, MgO, $\text{Al}_2(\text{SO}_4)_3$Al2​(SO4​)3​.
• Covalent (molecular) — composed of molecules held together by shared electron pairs. Examples: $\text{CO}_2$CO2​, $\text{H}_2\text{O}$H2​O, $\text{C}_6\text{H}_6$C6​H6​.

OCR's Module 2.1.3 focuses first on ionic formulae because this skill underpins all subsequent stoichiometry. A wrong formula propagates through every later mole calculation in the topic. Covalent nomenclature is treated more lightly here and re-examined in detail in Module 4 (organic chemistry).

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Predicting Ionic Charges from the Periodic Table

Main-group elements form ions by gaining or losing electrons to achieve a full outer shell (a noble-gas configuration). The group number predicts the charge:

Group	Typical ion	Electrons	Examples
1	$+1$+1	Lose 1	Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺
2	$+2$+2	Lose 2	Be²⁺, Mg²⁺, Ca²⁺, Sr²⁺, Ba²⁺
13	$+3$+3	Lose 3	Al³⁺ (B rarely forms ions)
14	Rarely ionic	—	C, Si predominantly covalent
15	$-3$−3	Gain 3	N³⁻, P³⁻ (mostly in binary nitrides / phosphides)
16	$-2$−2	Gain 2	O²⁻, S²⁻, Se²⁻
17	$-1$−1	Gain 1	F⁻, Cl⁻, Br⁻, I⁻
18	Generally none	Full outer shell	—

Pattern: Metals (left of periodic table) form cations; non-metals (right) form anions. The magnitude of charge equals the number of outer-shell electrons lost or gained to reach the nearest noble-gas configuration.

Why these charges? Consider sodium $[\text{Ne}]\,3s^1$[Ne]3s1. Losing one electron gives $[\text{Ne}]$[Ne] — the same configuration as the noble gas neon, a stable closed-shell arrangement. Gaining seven electrons to reach $[\text{Ar}]$[Ar] would require enormous energy and produce a highly unstable Na⁷⁻; losing one electron is overwhelmingly preferred. Symmetrically, chlorine $[\text{Ne}]\,3s^2\,3p^5$[Ne]3s23p5 gains one electron to reach $[\text{Ar}]$[Ar] — stable Cl⁻; losing seven electrons is energetically prohibitive.

Transition Metals — Variable Valency

Transition metals (Fe, Cu, Cr, Mn, Ni, Co, Zn, etc.) can form multiple stable ions because their 3d and 4s orbitals lie at similar energies. OCR will indicate the charge using Roman numerals in parentheses after the metal name:

Roman numeral	Cation	Example compound
Iron(II)	Fe²⁺	FeSO₄ (green)
Iron(III)	Fe³⁺	FeCl₃ (yellow-brown)
Copper(I)	Cu⁺	Cu₂O (red)
Copper(II)	Cu²⁺	CuSO₄·5H₂O (blue)
Chromium(III)	Cr³⁺	Cr₂(SO₄)₃ (green)
Chromium(VI)	found in CrO₄²⁻, Cr₂O₇²⁻	K₂Cr₂O₇ (orange)
Manganese(II)	Mn²⁺	MnSO₄ (very pale pink)
Manganese(IV)	in MnO₂	MnO₂ (black)
Manganese(VII)	in MnO₄⁻	KMnO₄ (deep purple)

Note that Zn forms only Zn²⁺ (full $3d^{10}$3d10) and is conventionally not counted as a "true" transition metal at A-Level for this reason.

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Common Polyatomic Ions — LEARN THESE

Polyatomic ions behave as a single charged unit in formulae. OCR does not provide these on the data sheet; you must memorise the table.

Name	Formula	Charge
Ammonium	NH₄⁺	$+1$+1
Hydroxide	OH⁻	$-1$−1
Nitrate	NO₃⁻	$-1$−1
Nitrite	NO₂⁻	$-1$−1
Hydrogencarbonate	HCO₃⁻	$-1$−1
Hydrogensulfate	HSO₄⁻	$-1$−1
Manganate(VII)	MnO₄⁻	$-1$−1
Cyanide	CN⁻	$-1$−1
Carbonate	CO₃²⁻	$-2$−2
Sulfate	SO₄²⁻	$-2$−2
Sulfite	SO₃²⁻	$-2$−2
Dichromate(VI)	Cr₂O₇²⁻	$-2$−2
Chromate(VI)	CrO₄²⁻	$-2$−2
Peroxide	O₂²⁻	$-2$−2
Phosphate	PO₄³⁻	$-3$−3

Memory trick: -ate ions usually have one more oxygen than -ite ions (nitrate NO₃⁻ vs nitrite NO₂⁻; sulfate SO₄²⁻ vs sulfite SO₃²⁻).

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Writing Ionic Formulae — The Cross-Over Rule

For an ionic compound to exist, total positive charge must equal total negative charge. The procedural shortcut:

1. Write the cation and anion with their charges.
2. Cross the magnitude of each charge to become the subscript of the other ion.
3. Simplify if both subscripts share a common factor.
4. Use brackets around polyatomic ions when the subscript is greater than 1.

Worked Examples 1–9 — Cross-over practice

Example	Ions	Cross-over	Formula	Charge check
1. Aluminium oxide	Al³⁺ + O²⁻	Al₂O₃	Al₂O₃	$2(+3) + 3(-2) = 0$2(+3)+3(−2)=0
2. Calcium chloride	Ca²⁺ + Cl⁻	CaCl₂	CaCl₂	$(+2) + 2(-1) = 0$(+2)+2(−1)=0
3. Ammonium sulfate	NH₄⁺ + SO₄²⁻	(NH₄)₂SO₄	(NH₄)₂SO₄	$2(+1) + (-2) = 0$2(+1)+(−2)=0
4. Iron(III) hydroxide	Fe³⁺ + OH⁻	Fe(OH)₃	Fe(OH)₃	$(+3) + 3(-1) = 0$(+3)+3(−1)=0
5. Aluminium sulfate	Al³⁺ + SO₄²⁻	Al₂(SO₄)₃	Al₂(SO₄)₃	$2(+3) + 3(-2) = 0$2(+3)+3(−2)=0
6. Magnesium nitride	Mg²⁺ + N³⁻	Mg₃N₂	Mg₃N₂	$3(+2) + 2(-3) = 0$3(+2)+2(−3)=0
7. Sodium phosphate	Na⁺ + PO₄³⁻	Na₃PO₄	Na₃PO₄	$3(+1) + (-3) = 0$3(+1)+(−3)=0
8. Chromium(III) sulfate	Cr³⁺ + SO₄²⁻	Cr₂(SO₄)₃	Cr₂(SO₄)₃	$2(+3) + 3(-2) = 0$2(+3)+3(−2)=0
9. Ammonium dichromate(VI)	NH₄⁺ + Cr₂O₇²⁻	(NH₄)₂Cr₂O₇	(NH₄)₂Cr₂O₇	$2(+1) + (-2) = 0$2(+1)+(−2)=0

Worked Examples 10–11 — Simplification

Example	Cross-over	Simplify	Final formula
10. Tin(IV) oxide	Sn₂O₄	$\div 2$÷2	SnO₂
11. Lead(IV) sulfide	Pb₂S₄	$\div 2$÷2	PbS₂

Always simplify ionic formulae to the smallest whole-number ratio — this is the empirical formula of the ionic lattice (and the only formula type that exists for an ionic substance).

Worked Example 12 — Iron(II) phosphate (compound cross-over)

Iron(II) phosphate forms between Fe²⁺ and PO₄³⁻. Crossing magnitudes: Fe³(PO₄)₂ — three Fe²⁺ contributes $+6$+6 and two PO₄³⁻ contributes $-6$−6, balanced.

Final formula: Fe₃(PO₄)₂.

Note that the subscript 3 outside Fe is fine without brackets (Fe is monatomic), but PO₄ — being polyatomic — must be wrapped in brackets because its subscript 2 needs to multiply both P and O.

Worked Example 13 — Ammonium dichromate(VI) thermal decomposition

The formula (NH₄)₂Cr₂O₇ has been derived above. The compound is famous for the "volcano" demonstration:

$(\text{NH}_4)_2\text{Cr}_2\text{O}_7(\text{s}) \rightarrow \text{Cr}_2\text{O}_3(\text{s}) + 4\text{H}_2\text{O(g)} + \text{N}_2(\text{g})$(NH4​)2​Cr2​O7​(s)→Cr2​O3​(s)+4H2​O(g)+N2​(g)

Confirming the formula here is the input to balancing the equation; mis-writing it as NH₄Cr₂O₇ (forgetting the (NH₄)₂ multiplicity) would invalidate the entire stoichiometry. This is a worked example of why "formula error cascades" matter for downstream calculation.

Worked Example 14 — Tin(II) chloride vs tin(IV) chloride

Tin shows two common oxidation states.

• Tin(II) chloride: Sn²⁺ + Cl⁻ → SnCl₂ (white solid, mild reducing agent in aqueous solution)
• Tin(IV) chloride: Sn⁴⁺ + Cl⁻ → SnCl₄ (volatile liquid, fumes in moist air — used as a chemical-vapour-deposition precursor)

The Roman-numeral disambiguation matters: "tin chloride" alone is ambiguous and OCR would not award a mark.

Worked Example 15 — Chromium(III) sulfate and the alums

Chromium(III) sulfate is Cr³⁺ + SO₄²⁻ → Cr₂(SO₄)₃. The hydrated form is Cr₂(SO₄)₃·18H₂O — a complex coordination compound where 12 water molecules are bound to two Cr³⁺ centres and 6 are loosely held in the crystal lattice. Determining this water content is a Module 2 empirical-formula exercise (covered in the empirical-formulae lesson).

Chromium also forms an alum with the formula KCr(SO₄)₂·12H₂O — a double salt of K⁺ and Cr³⁺ with sulfate counter-ions and 12 waters of crystallisation. Alums are a classic introductory inorganic synthesis target.

Decision tree — formula construction

flowchart TD
    A[Cation + Anion names given] --> B[Identify charge of cation]
    B --> C{Transition metal?}
    C -->|Yes| D[Read Roman numeral as charge]
    C -->|No| E[Read charge from periodic group]
    A --> F[Identify charge of anion]
    F --> G{Polyatomic?}
    G -->|Yes| H[Recall formula from memorised table]
    G -->|No| I[Read charge from periodic group]
    D --> J[Cross magnitudes for subscripts]
    E --> J
    H --> J
    I --> J
    J --> K{Common factor in subscripts?}
    K -->|Yes| L[Simplify by dividing by GCD]
    K -->|No| M{Polyatomic subscript greater than 1?}
    L --> M
    M -->|Yes| N[Add brackets around polyatomic]
    M -->|No| O[Final formula]
    N --> O

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Naming Conventions

Binary ionic compounds (metal + non-metal)

Metal name first, non-metal name with -ide ending:

• NaCl → sodium chloride
• CaO → calcium oxide
• MgBr₂ → magnesium bromide
• Al₂S₃ → aluminium sulfide

With polyatomic anions

Use the anion name directly:

• CaSO₄ → calcium sulfate
• Na₃PO₄ → sodium phosphate
• KMnO₄ → potassium manganate(VII)
• NaHCO₃ → sodium hydrogencarbonate

Variable-valence metals

Include Roman numerals showing the cation charge:

• CuO → copper(II) oxide
• Cu₂O → copper(I) oxide
• FeCl₂ → iron(II) chloride
• FeCl₃ → iron(III) chloride
• SnO₂ → tin(IV) oxide

Why some metals are variable-valency

Transition metals show variable valency because the 3d and 4s sub-shells lie at very similar energies, and electrons can be lost from either with comparable energy cost. Iron has the electron configuration $[\text{Ar}]\,3d^6\,4s^2$[Ar]3d64s2; losing the two 4s electrons gives Fe²⁺ ($[\text{Ar}]\,3d^6$[Ar]3d6), while losing one further 3d electron gives Fe³⁺ ($[\text{Ar}]\,3d^5$[Ar]3d5 — a stabilised half-filled d-subshell). The energy difference between the +2 and +3 states is small in chemical terms, so both can form under different conditions:

• Fe²⁺ is the kinetically favoured product of mild acid attack on iron metal in the absence of an oxidising agent (e.g. dilute HCl);
• Fe³⁺ is the thermodynamically favoured product under oxidising conditions (e.g. with dilute HNO₃ or chlorine present).

This variable-valency behaviour is the conceptual basis of the rich colour chemistry of transition-metal compounds covered in Module 5 — different oxidation states have different d-d transition energies and so different absorption colours.

The s-block elements (Groups 1 and 2) do not show variable valency because the energy gap between the outer s-electrons and the inner full noble-gas core is enormous. After losing one or two electrons to reach the noble-gas configuration, removing a third electron costs prohibitive amounts of energy. The p-block elements show variable valency only sporadically: Sn(II)/Sn(IV) and Pb(II)/Pb(IV) are the classic examples in Group 14, with Pb(II) becoming the thermodynamically favoured state at the bottom of the group due to the inert pair effect (a relativistic phenomenon affecting the 6s electrons of heavy elements).

Covalent compounds (brief overview)

For covalent compounds between two non-metals, Greek prefixes are used: mono-, di-, tri-, tetra-, penta-, hexa-: