Edexcel Chemistry — Book 1

Element = one type of atom; Compound = two+ elements chemically joined

Ion = charged atom/group; Molecule = atoms covalently bonded

State symbols: (s) solid (l) liquid (g) gas (aq) aqueous

n = m / M moles = mass(g) ÷ molar mass(g mol⁻¹)

n = N / Nₐ Nₐ = 6.022 × 10²³ mol⁻¹ (Avogadro)

Relative atomic mass (Aᵣ): weighted mean mass relative to ¹²C = 12

Molar mass (M) = Aᵣ or Mᵣ expressed in g mol⁻¹

Atom economy = M(desired) / ΣM(all products) × 100%

% yield = actual / theoretical × 100%

Theoretical yield: from mole ratios in the balanced equation

Empirical formula = simplest whole-number ratio of atoms

Molecular formula = n × empirical formula

To find n: n = Mᵣ(molecular) ÷ Mᵣ(empirical)

pV = nRT p (Pa), V (m³), T (K), R = 8.314 J mol⁻¹ K⁻¹

1 dm³ = 10⁻³ m³; 1 atm = 101 325 Pa

Molar volume at STP (0 °C, 1 atm): 22.4 dm³ mol⁻¹

Concentration: c = n / V (mol dm⁻³); ppm = mg dm⁻³ in solution

Memory Tip: Cover the symbol you want in the triangle → n = m/M, m = nM, M = m/n

n × Nₐ = number of particles; c = n/V for solution concentrations

Atom economy stays fixed for a given equation — % yield changes with conditions

For gases at STP: 1 mole = 22.4 dm³; at room T use pV = nRT

Common Mistake: Forgetting units in pV = nRT — p must be in Pa, V in m³, T in K

Atom economy ≠ % yield. Atom economy is about the equation; % yield is about experiment

Molar mass (g mol⁻¹) is numerically equal to Aᵣ but has different units — don't mix them

Proton: mass 1, charge +1 (nucleus); Neutron: mass 1, charge 0 (nucleus)

Electron: mass ≈0, charge −1 (shells around nucleus)

Mass number A = protons + neutrons; Atomic number Z = protons

Isotopes = same Z, different A → same chemistry, different mass

Mass spectrometry: ionise → accelerate → deflect → detect

Aᵣ = Σ(isotope mass × % abundance) ÷ 100

m/z ratio separates ions (z usually = 1, so m/z ≈ mass number)

Sub-shells: s (max 2e), p (max 6e), d (max 10e), f (max 14e)

Fill order: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ ...

Na: 1s² 2s² 2p⁶ 3s¹ or [Ne] 3s¹

Cr: [Ar] 3d⁵ 4s¹; Cu: [Ar] 3d¹⁰ 4s¹ (half-filled/filled stability)

1st IE: X(g) → X⁺(g) + e⁻ (always endothermic)

IE increases across period (more protons, same shielding)

IE decreases down group (outer e⁻ further, more shielding)

Exceptions: Al < Mg (3p vs 3s); S < P (paired 3p)

Periodic Table: Groups (columns) = same outer e⁻ config

Periods (rows) = same highest shell number

s-block: Groups 1 & 2; p-block: Groups 3–0; d-block: transition metals

Flame tests: Li=red, Na=yellow, K=lilac, Ca=orange-red, Cu=blue-green

Memory Tip: Aufbau = "building up" in German. Fill from lowest energy: 1s 2s 2p 3s 3p 4s 3d…

Electrons in same sub-shell: fill singly first (Hund's rule) then pair up

Large jumps in successive IE graph → shell boundary. Count electrons removed to find group

d⁵ and d¹⁰ are extra stable → Cr [Ar]3d⁵4s¹, Cu [Ar]3d¹⁰4s¹ break the pattern

Common Mistake: Writing Fe²⁺ as [Ar]3d⁴4s² — wrong! Remove 4s first: Fe²⁺ = [Ar]3d⁶

IE increases across a period but has dips: Al < Mg (3p vs 3s) and S < P (paired 3p)

Isotopes have same number of protons (same Z) but different neutrons — same element

Ionic bond = electrostatic attraction between oppositely charged ions

Metal loses e⁻ → cation (+); Non-metal gains e⁻ → anion (−)

Ionic compounds: giant lattice, high mp, conduct when molten/dissolved

Covalent bond = shared pair of electrons (both atoms contribute 1 e⁻ each)

σ (sigma): end-on overlap — all single bonds, free rotation

π (pi): side-on overlap — in double/triple bonds, no rotation

Dative (coordinate): both electrons from one atom (e.g. NH₄⁺, H₃O⁺, Al₂Cl₆)

Electronegativity: tendency to attract shared electrons (Pauling scale)

Order: F > O > N ≈ Cl > Br > C > H

Polar bond: δ+ → δ− across bond (e.g. H–Cl, C–O, C–N)

Polar molecule: non-zero net dipole moment (asymmetric shape)

Metallic bond = positive ions in a sea of delocalised electrons

High mp/bp, good conductors, malleable, ductile

Strength increases with: more delocalised e⁻, smaller/higher-charge ion

Giant ionic (NaCl): high mp, conducts when molten/aq, brittle

Giant covalent: diamond (tetrahedral, very hard), graphite (layers, conductor)

Molecular (I₂, CO₂, S₈): low mp, non-conductor, weak intermolecular forces

Giant metallic: high mp, conducts solid, malleable

Memory Tip: Shapes: 0 lp = ideal angles; each lone pair reduces angle ~2.5°

2bp+0lp = linear 180°; 3+0 = trig planar 120°; 4+0 = tetrahedral 109.5°

3+1 = trig pyramidal 107°; 2+2 = bent 104.5° (water) — lone pairs push harder

Polar bond ≠ polar molecule — CO₂ has polar bonds but zero net dipole (linear)

Common Mistake: Confusing polar bond with polar molecule — shape determines net dipole

Dative bond is the same strength as a normal covalent bond once formed — just different origin

Dot-and-cross: show outer electrons only; use dots for one atom, crosses for the other

Metallic bonding: ALL electrons in the "sea" are delocalised — not just outer-shell ones

Homologous series: same functional group, differ by CH₂, same general formula

Displayed formula: shows ALL bonds; Skeletal: C at each joint/end

Structural formula: e.g. CH₃CH₂OH; Molecular: e.g. C₂H₅OH

Chain count: meth(1) eth(2) prop(3) but(4) pent(5) hex(6) hept(7) oct(8)

Suffix: -ane (alkane), -ene (alkene), -ol, -al (aldehyde), -one, -oic acid

Number from end nearest substituent/functional group

Substituents: methyl- (CH₃), ethyl- (C₂H₅), chloro-, bromo-, nitro-

Chain isomers: different C skeleton (e.g. butane / 2-methylpropane)

Position isomers: same group, different position (e.g. butan-1-ol / butan-2-ol)

Functional group isomers: different group (e.g. ethanol / methoxymethane)

Alkanes: CₙH₂ₙ₊₂ — saturated, unreactive except combustion & free-radical sub.

Complete combustion: CₓHᵧ + O₂ → CO₂ + H₂O

Incomplete: → CO (toxic) ± C (soot) — occurs when O₂ limited

SO₂, NOₓ from combustion → acid rain

Initiation: Cl₂ → 2 Cl• (UV light breaks Cl−Cl homolytically)

Propagation: CH₄ + Cl• → •CH₃ + HCl; •CH₃ + Cl₂ → CH₃Cl + Cl•

Termination: any two radicals combine (e.g. 2 Cl• → Cl₂)

Multiple substitution possible → mixture of products

Memory Tip: IUPAC naming: longest chain → number from end nearest substituent

Suffix: -ane (alkane), -ene (alkene), -ol (alcohol), -al (aldehyde), -one (ketone)

Isomers: chain (skeleton differs), position (same group, different C), functional group

Combustion products: always H₂O; CO₂ if complete; CO/C if incomplete

Common Mistake: Mixing up initiation, propagation, termination — memorise which step uses UV

Initiation only needs UV light; propagation is a chain (no UV needed)

Homolytic fission = each atom gets 1 electron (produces radicals, not ions)

Numbering: always give substituents the lowest possible locant number

Alkenes: CₙH₂ₙ — unsaturated, contain C=C double bond

C=C = 1 σ bond + 1 π bond; π bond restricts rotation → stereoisomerism

π electron cloud is electron-rich → attracts electrophiles

E/Z isomerism: caused by restricted rotation around C=C

E (entgegen): higher-priority groups on opposite sides

Z (zusammen): higher-priority groups on the same side

Priority by atomic number (CIP rules)

Br₂(aq): orange → colourless — test for C=C (unsaturation)

H₂ + Ni catalyst, 150 °C: hydrogenation → alkane

HBr: Markovnikov — H to C with more H's (more stable carbocation)

H₂O(g) + H₃PO₄, 300 °C, 60 atm: → alcohol (hydration)

Addition polymerisation: monomers join through C=C to form long chain

Repeat unit = monomer structure with bonds instead of double bond

Poly(ethene) from ethene; PVC from chloroethene; polypropene from propene

Non-biodegradable → environmental concerns; use of biofuels/biodegradable polymers

Memory Tip: E/Z — E (entgegen = "opposite"), Z (zusammen = "together") — higher priority same side = Z

Priority by atomic number (CIP rules): highest atomic number = highest priority

Br₂(aq) decolourises: orange/brown → colourless = C=C confirmed (or other oxidisable group)

Markovnikov: H adds to C with more H's → more stable (more substituted) carbocation

Common Mistake: Calling it cis/trans when E/Z is required — use E/Z for systematic nomenclature

Addition reaction: C=C is consumed — product has NO double bond

Polymerisation: monomer must have C=C; repeat unit = monomer with double bond replaced by single bonds

Electrophilic addition ≠ substitution — no H is removed, an atom is added

Enthalpy change (ΔH) = heat exchanged at constant pressure

Exothermic: ΔH < 0 products at lower energy; heat released

Endothermic: ΔH > 0 products at higher energy; heat absorbed

Standard conditions: 298 K, 100 kPa, concentrations 1 mol dm⁻³

ΔcH° (combustion): 1 mol burns completely in O₂ — always negative

ΔfH° (formation): 1 mol compound from elements in standard states

ΔfH° of elements in standard state = 0 by definition

ΔnH° (neutralisation): −57 kJ mol⁻¹ for strong acid + strong base

Hess's Law: ΔH is path-independent (1st Law of Thermodynamics)

Route A = Route B — use known ΔfH° or ΔcH° to find unknown ΔH

ΔH_rxn = ΣΔfH°(products) − ΣΔfH°(reactants)

Bond enthalpy = energy to break 1 mol of bonds in gaseous molecules

ΔH = Σ(bonds broken) − Σ(bonds formed)

Breaking bonds: endothermic (+); Forming bonds: exothermic (−)

Mean bond enthalpies = averages (less precise than Hess cycles)

Memory Tip: Hess's Law: ΔH is path-independent (1st Law of Thermodynamics)

ΔfH° of all elements in their standard state = 0 by definition

Bond enthalpy: broken = endothermic (+); formed = exothermic (−). Net = ΔH

Standard conditions: 298 K, 100 kPa, 1 mol dm⁻³ concentrations

Common Mistake: Forgetting to multiply ΔfH° values by stoichiometric coefficients in the balanced equation

Bond enthalpies are mean (average) values — less accurate than Hess cycle using ΔfH°

ΔcH° is always negative (combustion releases heat) — if you get a positive value, check sign

q = mcΔT gives heat in joules — convert to kJ and divide by moles for ΔH in kJ mol⁻¹

London (dispersion) forces: temporary dipole → induced dipole

Present in ALL molecules; increase with more electrons / larger surface area

Explains increasing bp down homologous series

Permanent dipole-dipole: between polar molecules with net dipole

Stronger than London forces for similarly-sized molecules

Hydrogen bond: X−H ··· Y where X, Y ∈ {N, O, F}

Requires: H bonded to N/O/F AND lone pair on N/O/F of neighbour

Strongest IMF; explains high bp of H₂O, HF, NH₃

Water: 4 H-bonds per molecule → open hexagonal lattice in ice → ice floats

bp order for Group 6 hydrides: H₂O >> H₂Te > H₂Se > H₂S

H₂O anomalously high bp, low density as solid, high surface tension

Alcohols mix with water (H-bonding); longer alkyl chain → less soluble

Memory Tip: London forces exist in ALL molecules — they're just the weakest IMF

H-bonding needs: H bonded to N/O/F AND lone pair on N/O/F of a neighbour

Water: 2 O–H donors + 2 lone pairs on O = 4 H-bonds per molecule

Boiling point order for Group 6: H₂O >> H₂Te > H₂Se > H₂S (H₂O is anomalously high)

Common Mistake: Alkanes have NO hydrogen bonding — they only have London dispersion forces

London forces increase with molecular size/surface area, not just molecular formula

A hydrogen bond is NOT a covalent bond — it's an intermolecular attraction, much weaker

Dipole-dipole only exists in polar molecules — check shape before claiming dipole

Oxidation state rules: element = 0; O = −2 (except peroxides: −1); H = +1

H in metal hydrides = −1; sum of ox. states = overall charge

Fe(III)Cl₃: Fe = +3; Cr₂O₇²⁻: Cr = +6; MnO₄⁻: Mn = +7

OIL RIG: Oxidation Is Loss | Reduction Is Gain (of electrons)

Oxidising agent: gets reduced (gains e⁻); e.g. MnO₄⁻, Cr₂O₇²⁻, Cl₂

Reducing agent: gets oxidised (loses e⁻); e.g. Fe²⁺, I⁻, SO₂

Half-equation: balance atoms → add H₂O (for O), H⁺ (for H), then e⁻

Example: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (reduction)

Multiply half-eqs so e⁻ cancel → add → cancel spectators

Ionic equation: remove spectator ions from the full equation

Check ionic equation: atoms AND charges must balance on both sides

Disproportionation: one species is both oxidised AND reduced simultaneously

Memory Tip: OIL RIG = Oxidation Is Loss, Reduction Is Gain (of electrons)

Oxidising agent: gets reduced (gains e⁻). Reducing agent: gets oxidised (loses e⁻)

Half-equation steps: balance atoms → add H₂O for O → add H⁺ for H → balance charge with e⁻

Common ox. agents: MnO₄⁻ (Mn goes +7→+2), Cr₂O₇²⁻ (Cr +6→+3), Cl₂ (0→−1)

Common Mistake: Confusing the oxidising agent with what's being oxidised — they're opposites!

O is −2 in most compounds, but −1 in peroxides (e.g. H₂O₂, Na₂O₂) and +2 in OF₂

When combining half-equations, multiply so electrons cancel — don't just add as-is

H in metal hydrides (e.g. NaH) = −1, not +1 — this is the exception

Group 1 (Li → Cs): reactivity increases down group

2M + 2H₂O → 2MOH + H₂ (K floats & ignites; Cs explosive)

4Li + O₂ → 2Li₂O; 2Na + O₂ → Na₂O₂ (peroxide); K → KO₂ (superoxide)

Group 2 (Mg → Ba): reactivity increases down group

M + O₂ → MO; M + 2HCl → MCl₂ + H₂; Mg + H₂O(steam) → MgO + H₂

Thermal stability of carbonates: increases down group (BaCO₃ most stable)

Uses: Ca(OH)₂ neutralise soil; Ba(SO₄) barium meal X-ray; Mg(OH)₂ antacid

Group 7: F₂ > Cl₂ > Br₂ > I₂ (reactivity/oxidising power decreases)

Halide displacement: Cl₂ + 2KBr → 2KCl + Br₂ (Cl displaces Br and I)

Cl₂ + 2NaOH(cold, dilute) → NaCl + NaOCl + H₂O (bleach)

Cl₂ + H₂O ⇌ HCl + HClO (Cl disproportionates: 0 → −1 and +1)

Conc. H₂SO₄ + NaCl → HCl (steamy fumes only — no further reduction)

Conc. H₂SO₄ + NaBr → HBr + Br₂ + SO₂ (orange fumes + pungent gas)

Conc. H₂SO₄ + NaI → HI + I₂ + SO₂ + S + H₂S (purple vapour)

Reducing power: I⁻ > Br⁻ > Cl⁻ (I⁻ can reduce H₂SO₄ to H₂S)

Titrations: standard solution = accurately known concentration

n(substance) = c × V (V in dm³); n = mass / M

Accuracy: use volumetric flask, pipette, burette (not measuring cylinder)

Percentage error: (instrument precision / reading) × 100%

Memory Tip: Group 7 reactivity decreases down group — more shells, more shielding, weaker attraction for e⁻

Group 1/2: reactivity increases down group — outer e⁻ further away, easier to remove

Cl₂ + 2NaOH(cold dilute) → NaCl + NaOCl + H₂O (bleach = NaOCl)

Halide tests: add dilute HNO₃ first → then AgNO₃(aq) → Cl⁻ white, Br⁻ cream, I⁻ yellow

Common Mistake: Cl₂ cannot displace Cl⁻ — a halogen only displaces a LESS reactive halide below it

Cl₂ disproportionates in water (0 → +1 and −1) — it's both oxidised and reduced

BaCO₃ thermal stability: increases down Group 2 — NOT decreases. BaCO₃ most stable

NaBr + conc H₂SO₄: produces orange Br₂ fumes (HBr reduces H₂SO₄ to SO₂)

Rate = change in concentration / time (mol dm⁻³ s⁻¹)

Concentration ↑ → more collisions per second → rate ↑

Temperature ↑ → particles faster AND more exceed Eₐ → rate ↑↑

Surface area ↑ → more particles exposed → rate ↑; Catalyst → lower Eₐ

Collision theory: reaction needs correct energy (≥ Eₐ) + correct orientation

Activation energy Eₐ = minimum energy needed to break bonds in reactants

Not all collisions lead to reaction — orientation matters

Catalyst: provides alternative pathway with lower Eₐ

Homogeneous: same phase (e.g. H⁺ in ester hydrolysis)

Heterogeneous: different phase (e.g. Fe in Haber, Pt in catalytic converter)

Catalyst does NOT change ΔH, Kc, or equilibrium position

Memory Tip: Temperature ↑ by 10 K roughly doubles the rate (rule of thumb for A-level)

Maxwell-Boltzmann: higher T = curve broader + lower peak + more molecules exceed Eₐ

Catalyst provides alternative pathway with lower Eₐ — more molecules can react

Collision theory: need correct energy (≥ Eₐ) AND correct orientation — not every collision reacts

Common Mistake: Saying "particles have more energy" at higher T — more importantly, more particles EXCEED Eₐ

Catalyst does NOT shift equilibrium position or change Kc — only reaches it faster

Surface area only matters for heterogeneous reactions — has no effect on homogeneous reactions

Rate increases with concentration because more collisions per second, not higher energy

Dynamic equilibrium: forward rate = reverse rate in a closed system

Concentrations constant (not equal); symbol ⇌

Only reached in closed system at constant temperature

Le Chatelier: system opposes the imposed change

Add reactant → shifts right; remove product → shifts right

Increase P → shifts to side with fewer moles of gas

Increase T → shifts in endothermic direction

Exothermic (ΔH < 0): ↑T shifts equilibrium left → less product; Kc decreases

Endothermic (ΔH > 0): ↑T shifts equilibrium right → more product; Kc increases

Kc only changes with temperature — not concentration or pressure

Catalyst: equilibrium reached FASTER but same position

Increases rate of both forward and reverse equally → Kc unchanged

Industrial compromise: high T (kinetics) vs low T (yield) — optimal balance

Memory Tip: Le Chatelier: system opposes any imposed change to restore equilibrium

Pressure ↑ → shifts to side with fewer moles of gas; no gas = no pressure effect

Temperature ↑ → favours endothermic direction (absorbs added heat)

Kc only changes with temperature — concentration, pressure, catalyst all leave Kc unchanged

Common Mistake: Catalyst does NOT shift equilibrium — it increases rate of both directions equally

Kc expression: products over reactants, each raised to stoichiometric power

Adding more catalyst or changing concentration does NOT change Kc

Exothermic reaction: ↑T gives LESS product at equilibrium — Kc decreases

Primary (1°): −X on C bonded to 1 alkyl group; General: RX

Secondary (2°): −X on C bonded to 2 alkyl groups

Tertiary (3°): −X on C bonded to 3 alkyl groups

Reactivity: RI > RBr > RCl > RF (C−F bond strongest — hardest to break)

Nucleophile: electron-pair donor attacks δ+ carbon (C−X is polar)

NaOH(aq)/reflux: R−X + OH⁻ → R−OH + X⁻ (alcohol)

NaCN/ethanol: R−X + CN⁻ → R−CN + X⁻ (nitrile — adds 1 C)

NH₃(excess)/ethanol: R−X → R−NH₂ + HX (primary amine)

Elimination: hot KOH in ethanol → alkene (H and X removed)

Competes with substitution: hot/ethanol → elimination; aq/warm → substitution

SN2 (1°): one-step, inversion of configuration (backside attack)

SN1 (3°): two-step via carbocation intermediate — racemisation

Alcohols: −OH group; 1° (1 alkyl), 2° (2 alkyl), 3° (3 alkyl) attached to C−OH

Oxidation with K₂Cr₂O₇/H₂SO₄ (Cr₂O₇²⁻ orange → Cr³⁺ green):

1° alcohol → aldehyde (distil) → carboxylic acid (reflux)

2° alcohol → ketone (reflux); 3° alcohol → no reaction

Mass spectrometry: M⁺ (molecular ion) gives Mᵣ

Fragmentation: bonds break → characteristic m/z peaks

Base peak = tallest (most abundant) fragment

IR spectroscopy: different bonds absorb IR at characteristic wavenumbers (cm⁻¹)

O−H (alcohol): broad 2500–3300 cm⁻¹

O−H (carboxylic acid): broad 2500–3300; C=O at 1700–1750 cm⁻¹

C−H bonds: 2850–3100 cm⁻¹; N−H: 3300–3500 cm⁻¹

Fingerprint region 400–1500 cm⁻¹: unique to each compound

Memory Tip: IR peaks: O–H broad 2500–3300; C–H 2850–3100; C=O sharp 1700–1750

Nucleophiles attack the δ+ carbon of C–X (X is electronegative → C is δ+)

SN2 (primary): one step, backside attack → configuration inverted

SN1 (tertiary): two steps via carbocation → racemisation (equal +/− enantiomers)

Common Mistake: Distilling gives aldehyde; refluxing gives carboxylic acid — not the other way around

3° alcohol + oxidising agent = no reaction — do not try to oxidise tertiary alcohols

NaOH(aq)/warm → substitution (alcohol); KOH/ethanol/hot → elimination (alkene)

C–F bond is strongest in halogens → RF least reactive in nucleophilic substitution