IB Chemistry HL — All Flashcards

A pure substance made of **only one type of atom**; it cannot be broken down chemically.

**Two or more different atoms chemically bonded** together in a fixed ratio.

Two or more substances **physically combined but not chemically bonded**, in any ratio.

Only by **chemical** means (a reaction) — not by physical methods.

By **physical** means (e.g. filtration, distillation), because nothing is bonded.

Homogeneous = **uniform** (e.g. solution, air); heterogeneous = **not uniform**, parts visible (e.g. sand + iron).

Element = one type of atom; compound = different atoms **chemically bonded** in a fixed ratio.

A single element or compound — it has a **sharp, fixed** melting and boiling point.

A pure substance melts **sharply**; impurities **lower** it and spread it over a **range**.

A **mixture** (an alloy of copper and zinc) — the metals are not chemically bonded.

**Solid**, **liquid** and **gas** — they differ in how close the particles are and how freely they move.

**Packed** close in fixed positions; they only **vibrate**. A solid has a fixed shape and volume.

**Touching** but not fixed; they **slide** past each other. A liquid has fixed volume but takes the container's shape.

**Far apart**, moving **fast and randomly**. A gas fills its container and is easily compressed.

A model treating matter as **small particles in constant random motion**, with attractive forces between them that weaken as they spread apart.

The **average kinetic energy** of the particles — hotter means the particles move faster on average.

**Add 273.15** (≈ 273): T(K) = θ(°C) + 273.15.

**0 K** (about −273 °C) — the lowest possible temperature, where particle motion is at a minimum.

Gas particles are **far apart** with large gaps to close up; liquid particles are already **touching** with little space.

The added energy is used to **overcome the forces** between particles, not to speed them up, so the average kinetic energy (temperature) stays the same.

Its particles are **not held in fixed positions**, so they **slide** and flow to fit the container.

They **gain kinetic energy** and **vibrate more**, until they have enough energy to break free of their fixed positions and the solid melts.

Their components are **not chemically bonded**, so they keep their own properties and can be separated by **physical** methods.

An **insoluble solid** from a liquid — the solid is too large to pass through the **filter paper** (uses particle size).

A **dissolved (soluble) solid** from its solution — the **solvent boils off**, leaving the solid behind.

Liquids (or a liquid from a dissolved solid) by their difference in **boiling point**.

The dissolved components of a mixture by their difference in **solubility / attraction** to the paper.

R_{f} = distance moved by **spot** ÷ distance moved by **solvent** — a ratio with **no units**, between 0 and 1.

The component is **more soluble** in the solvent, so it was carried **further** up the paper.

Add water to dissolve the salt, then **filter** — the sand stays as the residue.

Use a **magnet** — iron is **magnetic**, sand and salt are not.

**Crystallisation** — dissolve in hot solvent, cool to form pure crystals, then filter them off.

**No** — the spot cannot move further than the solvent front, so 0 < R_{f} < 1.

Size → **filtration**; boiling point → **distillation**; solubility → **crystallisation / chromatography**.

Together in the tiny, dense central **nucleus** of the atom.

Moving around the nucleus in **shells** (energy levels); this region is mostly empty space.

Relative mass **1**, relative charge **+1**.

Relative mass **1**, relative charge **0** (neutral).

Relative mass ≈ **1/1836** (negligible), relative charge **−1**.

The number of **protons** in the nucleus; it defines the element.

The number of **protons + neutrons** (nucleons) in the nucleus.

**neutrons = A − Z** (mass number − atomic number).

**electrons = protons = Z** — the + and − charges balance.

Adjust electrons by the charge: **electrons = Z − charge** (lose e⁻ for +, gain e⁻ for −).

Top = **mass number A**, bottom = **atomic number Z**, X = element symbol.

Only the **electron** count; the protons and neutrons stay the same.

Atoms of the **same element** with the **same number of protons** but **different numbers of neutrons**.

The number of **protons** in an atom — it defines which element it is.

The total number of **protons + neutrons** in an atom.

The same atomic number Z, but a **different mass number A** (because they have different numbers of neutrons).

They have the **same number of electrons** and the **same electron arrangement**, and chemistry is controlled by the electrons.

**Physical** properties that depend on mass — e.g. **density** and rate of **diffusion** — because of the different number of neutrons.

Neutrons = **A − Z** (mass number minus atomic number).

37 − 17 = **20 neutrons**.

An isotope with an **unstable nucleus** that decays and gives out radiation; chemically it behaves like the stable isotope.

**Carbon-14** for dating, **cobalt-60** for radiotherapy/sterilising, or **iodine-131** for treating the thyroid.

No — neutrons have **no charge** and do not affect the electrons, so **bonding and reactions are unchanged**.

The **weighted average** mass of an element's isotopes, relative to one-twelfth of a carbon-12 atom. It has **no units**.

Because it averages **isotopes of different masses**, weighted by their **abundance** (e.g. Cl = 35.5).

It separates the atoms/ions of a sample by **mass**, producing a **mass spectrum**.

**m/z** (mass-to-charge ratio) on the x-axis; **relative abundance** on the y-axis.

For singly-charged ions, the **mass of that isotope**.

The **relative abundance** of that isotope — the taller the peak, the more common the isotope.

$A_{r} = \dfrac{\sum(\text{mass} \times \%\,\text{abundance})}{100}$ — weight each isotope mass by its abundance, sum, divide by 100.

**Three** peaks — one peak per isotope.

Divide the weighted sum by the **total of the abundances** instead of by 100.

It must lie **between** the lightest and heaviest isotope masses, closest to the **most abundant** one.

No — chlorine atoms are ³⁵Cl or ³⁷Cl; 35.5 is the **weighted average** (75% ³⁵Cl, 25% ³⁷Cl).

A tiny **packet of light energy**; its energy is given by E = hf (higher frequency → more energy).

A **fixed, allowed energy** an electron can have in an atom; energy levels are **discrete (quantised)**.

Continuous = an **unbroken rainbow** (all wavelengths). Line = a few **discrete bright lines** on black, from an excited element.

An excited electron **falls** from a higher to a lower energy level, emitting a photon of fixed energy (one line per allowed jump).

That the electron's energy levels are **discrete (quantised)** — fixed lines mean only fixed energy gaps are allowed.

The spectral lines get **closer together** toward **high frequency/energy**, because the energy levels bunch up at higher n.

The **biggest energy gap** — an electron falling **to n = 1** (the ground state).

$c = \lambda f$ — speed of light = wavelength × frequency, so **high f means short λ**.

$E = hf$ — photon energy = Planck's constant × frequency (higher f → higher E).

**Radio < red < violet** in frequency, so radio is lowest energy and violet is highest.

The lines merge; the electron gains just enough energy to **leave the atom** — this gives the **ionisation energy**.

The major 'shell' of an atom (n = 1, 2, 3, …); higher n means **higher energy** and **further** from the nucleus.

A subdivision of a main level, labelled **s, p, d, f**, differing slightly in energy (s < p < d < f).

A region around the nucleus that can hold up to **2 electrons**.

A **sphere** centred on the nucleus.

A **dumbbell** — two lobes pointing in opposite directions through the nucleus.

s = **1**, p = **3**, d = **5**, f = **7** orbitals.

s = **2**, p = **6**, d = **10**, f = **14** (2 electrons per orbital).

**2n²** — so n = 1 → 2, n = 2 → 8, n = 3 → 18, n = 4 → 32.

**4s** fills first — it is slightly lower in energy than 3d.

Electrons occupy orbitals of a sublevel **singly** (parallel spins) before any pairing up.

**s < p < d < f** (s is lowest, f is highest).

**3s, 3p and 3d** (max 18 electrons).

Electrons fill the **lowest-energy** sub-shell available first (build up: 1s, 2s, 2p, 3s, …).

Each orbital holds **at most 2 electrons**, and they must have **opposite spins**.

Within a sub-shell, electrons occupy orbitals **singly with parallel spins** before any pairing occurs.

1s, 2s, 2p, 3s, 3p, **4s, 3d**, 4p — note **4s fills before 3d**.

**s = 2**, **p = 6**, **d = 10** (each orbital holds 2).

1s² 2s² 2p⁶ 3s² 3p⁴.

Replace the inner electrons with the **previous noble gas** in [ ], then list the outer electrons — e.g. Ca = [Ar] 4s².

Start from the atom and **remove electrons from the highest main shell (largest n) first** — for transition metals, **4s before 3d**.

**[Ar] 3d⁶** — the two **4s** electrons are removed first, not the 3d.

**Add** the gained electrons to the next available sub-shell — e.g. O²⁻ = 1s² 2s² 2p⁶.

A **half-full** 3d⁵ sub-shell is extra stable, so one 4s electron promotes to 3d.

A **full** 3d¹⁰ sub-shell is extra stable, so one 4s electron promotes to 3d.

The energy to remove one mole of electrons from one mole of **gaseous +1 ions**: X⁺(g) → X²⁺(g) + e⁻.

The electron leaves an **increasingly positive ion** — the same protons pull on **fewer electrons**, so each remaining electron is held more strongly.

The next electron is removed from a **new, inner main energy level (shell)** that is closer to the nucleus and less shielded — evidence for **electron shells**.

A change of **sub-shell** (e.g. p → s) within the same main shell — finer evidence for **sub-shells**.

Count the electrons removed **before the first big jump** — that equals the number of **outer electrons = the group** (for a main-group element).

**Group 1** — one easily-removed outer electron, then the big jump into the inner shell.

**Group 13** — three outer electrons leave before the inner shell is reached.

Because the values span a **huge range** (hundreds to hundreds of thousands of kJ mol⁻¹); **log₁₀(IE)** fits them all on one axis as steps.

The number of **outer electrons**, and so the **group** of a main-group element.

Between the **1st and 2nd** (1 → 8) and between the **10th and 11th** (8 → 2) — the two jumps reveal three shells.

The **position** — after how many electrons the jump falls. The electrons removed before it are the outer electrons.

The **gaseous** atoms / ions — each step is X^{n+}(g) → X^{(n+1)+}(g) + e⁻.

The amount of substance containing **6.02 × 10²³** particles (Avogadro's constant, N_{A}).

N_{A} = **6.02 × 10²³ mol⁻¹** — the number of particles in one mole.

The mass of **one mole** of a substance, in **g mol⁻¹**; numerically equal to the relative atomic/formula mass.

$n = \dfrac{m}{M}$ — amount (mol) = mass (g) ÷ molar mass (g mol⁻¹).

$N = n\,N_{A}$ — number of particles = amount (mol) × Avogadro's constant.

Rearrange to $m = nM$ — multiply the amount (mol) by the molar mass.

$M = \dfrac{m}{n}$ — divide the mass by the amount in moles.

2 mol of O atoms = **1.20 × 10²⁴** atoms (each CO_{2} has 2 oxygens).

**g mol⁻¹** (grams per mole).

Forgetting to scale by the **number of that atom or ion in the formula** (e.g. 2 Cl⁻ per MgCl_{2}).

The **simplest whole-number ratio** of the atoms of each element in a compound.

The **actual number** of atoms of each element in one molecule of the compound.

The molecular formula is a **whole-number multiple** of the empirical formula (molecular = empirical × x).

**CH_{2}O** — divide every subscript by 6 to get the simplest ratio.

Treat % as g per 100 g → divide each by A_{r} (n = m/M) → divide by the **smallest** → round / scale to whole numbers.

**n(C) = n(CO_{2})** — every carbon atom ends up in one CO_{2}.

**n(H) = 2 × n(H_{2}O)** — each water molecule contains two H atoms.

By **difference**: subtract the masses of C and H from the sample mass, then divide the leftover by 16.00.

$x = \dfrac{M_{r}}{\text{empirical formula mass}}$, then multiply every subscript by x.

Multiply the **whole ratio** by 2 (for .5) or 3 (for .33) to clear it into whole numbers.

Atoms combine in whole-**number** ratios, which only show up once masses are turned into **moles** (÷ A_{r}).

An **empirical** formula — ionic compounds have no separate molecules, so the formula is the simplest ratio.

How much **solute** is dissolved in a given volume of **solution** — usually in **mol dm⁻³**.

$n = CV$ — amount (mol) = concentration (mol dm⁻³) × volume (**dm³**).

Rearrange to $C = \dfrac{n}{V}$ — divide the amount in moles by the volume in dm³.

**Divide by 1000** (1 dm³ = 1000 cm³). E.g. 250 cm³ = 0.250 dm³.

**Multiply by the molar mass M**: g dm⁻³ = mol dm⁻³ × M.

$C_{1}V_{1} = C_{2}V_{2}$ — the amount of solute is unchanged when you add solvent.

Diluting only adds solvent, so the **moles of solute (n = CV) stay constant**.

**1 mg dm⁻³** (1 part per million) — used for very dilute solutions.

A solution of **precisely known concentration**, made up in a **volumetric flask**.

Forgetting to convert the **volume from cm³ to dm³** (÷ 1000) before using n = CV.

The **total** final volume. Water added = V_{2} − V_{1}.

**Dissolve** the weighed solid → **transfer** to a volumetric flask (rinse beaker in) → **make up** to the mark → **invert** to mix.

At **constant temperature** (and amount), the pressure of a gas is **inversely proportional** to its volume: $P_{1}V_{1} = P_{2}V_{2}$.

Pressure is **directly proportional** to the **kelvin** temperature: $\dfrac{P_{1}}{T_{1}} = \dfrac{P_{2}}{T_{2}}$.

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$ — with T in **kelvin**. It is given in the data booklet.

**T/K = θ/°C + 273** — always do this before a gas-law calculation.

The particles have **no volume** of their own and there are **no forces** between them.

At **high temperature** and **low pressure** — particles are far apart and fast-moving.

At **low temperature** and **high pressure** — particle volume and intermolecular forces become significant.

The pressure **halves** (Boyle's law: P ∝ 1/V).

Only the **kelvin** scale starts at true zero (0 K), so only it gives the correct proportionality; °C would give wrong ratios.

Because P ∝ kelvin T — at 0 K the particles would stop and the pressure would be **zero**.

The **temperature** and the **amount** of gas; only P and V change.

Use the combined gas law: $P_{2} = P_{1}\times\dfrac{V_{1}}{V_{2}}\times\dfrac{T_{2}}{T_{1}}$ (T in kelvin).

$PV = nRT$ — links pressure, volume, amount and temperature of an ideal gas.

**Standard temperature and pressure**: 273 K (0 °C) and 100 kPa.

V_{m} = **22.7 dm³ mol⁻¹** — the volume of one mole of any ideal gas at STP.

$n = \dfrac{V}{V_{m}}$ — divide the volume (in dm³) by 22.7.

$V = n\,V_{m}$ — multiply the amount (mol) by 22.7 dm³ mol⁻¹.

**Pa** (pressure), **m³** (volume) and **K** (temperature), because R = 8.31 J K⁻¹ mol⁻¹ is in SI units.

R = **8.31 J K⁻¹ mol⁻¹** (given in the data booklet).

**Multiply by 1000** — e.g. 101 kPa = 1.01 × 10⁵ Pa.

**Divide by 1000** — e.g. 24.0 dm³ = 0.0240 m³.

**Add 273** — e.g. 25 °C = 298 K.

**At STP** use V_{m} = 22.7; at **any other conditions** use PV = nRT with SI units.

Find n from PV = nRT, then $M = \dfrac{m}{n}$ using the sample mass.

An atom (or group of atoms) with an overall **charge** because it has **lost or gained electrons**.

A **positive** ion, formed when an atom **loses** electrons (more protons than electrons).

A **negative** ion, formed when an atom **gains** electrons (more electrons than protons).

**Cations** — metals **lose** their outer electrons to form **positive** ions.

**Anions** — non-metals **gain** electrons to form **negative** ions.

To reach a **full outer shell** — the stable **noble-gas configuration** of a Group 18 atom.

**1+, 2+, 3+** — these metals lose 1, 2 or 3 outer electrons.

**3−, 2−, 1−** — these non-metals gain 3, 2 or 1 electrons.

The **electrostatic attraction between oppositely charged ions** (a cation and an anion).

Group 17, so it **gains 1** electron → a **1−** ion (reaching 2, 8, 8).

**Na⁺** (sodium loses 1 e⁻) and **Cl⁻** (chlorine gains 1 e⁻).

Ionic = **attraction between charged ions** (electrons transferred); covalent = a **shared pair** of electrons.

A **positively** charged ion (a metal, or NH_{4}⁺).

A **negatively** charged ion (a non-metal, or a polyatomic ion).

A charged group of **bonded atoms** that acts as a single unit (e.g. SO_{4}²⁻, NO_{3}⁻).

The ratio of ions is chosen so the **total positive and negative charges cancel** to zero.

Write each ion with its charge, **cross over** the charge sizes as subscripts, then **simplify** to the smallest whole-number ratio.

**MgCl_{2}** — Mg²⁺ needs two Cl⁻ to balance.

**Al_{2}O_{3}** — crossover of Al³⁺ and O²⁻ (6+ balances 6−).

When a **polyatomic ion appears two or more times**, e.g. Ca(NO_{3})_{2}, (NH_{4})_{2}SO_{4}.

**SO_{4}²⁻** — a 2− polyatomic ion.

**NH_{4}⁺** — a 1+ polyatomic cation.

Cation name first, then the non-metal anion ending in **-ide** (e.g. magnesium ox**ide**).

**Ca_{3}N_{2}** — Ca²⁺ and N³⁻ crossover (6+ balances 6−).

A regular, repeating **3-D array** of oppositely charged ions, with each ion surrounded by ions of the opposite charge.

**Strong electrostatic forces of attraction** between the oppositely charged ions (this is the ionic bond).

Many **strong electrostatic attractions** between the ions must be broken, which needs a **large amount of energy**.

**Higher ionic charge** and **smaller ionic radius** — both increase the electrostatic attraction.

When **molten** or **dissolved in water (aqueous)** — the ions are then **free to move**. Not as a solid.

The ions are held in **fixed positions** in the lattice, so no charged particles are free to move.

A force makes layers **shift**, bringing **like charges** next to each other; they **repel** and split the crystal.

Water is **polar**: its δ⁻ oxygen attracts cations and δ⁺ hydrogens attract anions, pulling ions out of the lattice (hydration).

Solid = ions **fixed**, does **not** conduct. Molten = lattice broken, ions **free to move**, **conducts**.

High melting point + does **not** conduct as a solid + **conducts when molten/aqueous** = ionic.

Mg^{2+} and O^{2−} carry **higher charges** than Na^{+} and Cl^{−}, so the electrostatic attraction is much stronger.

A **shared pair of electrons** between two (usually non-metal) atoms.

A **non-bonding** pair of electrons that stays on one atom (drawn as two dots).

A **bonding pair** (one shared pair of electrons).

Atoms tend to gain a full outer shell of **8 electrons** by sharing (or transferring) electrons.

Number of **shared pairs**: 1, 2, 3 — bond order 1, 2, 3. Higher order → shorter, stronger.

O=C=O — **two double bonds**, two lone pairs on each oxygen, none on carbon.

N≡N — a **triple bond** with **one lone pair on each** nitrogen.

**BF_{3}** (boron has 6 electrons) and **BeCl_{2}** (beryllium has 4) — electron-deficient.

Count valence electrons → least electronegative atom central → single bonds → complete outer octets → multiple bonds if the centre is short.

**One** (three bonding pairs to H, one lone pair).

**V**alence **S**hell **E**lectron **P**air **R**epulsion.

Any group of electrons around the central atom — a single/double/triple **bond (each = 1 domain)** or a **lone pair**.

**Linear**, 180° (e.g. CO_{2}, HCN).

**Trigonal planar**, 120° (e.g. BF_{3}).

**Tetrahedral**, 109.5° (e.g. CH_{4}).

**Trigonal pyramidal**, ~107° (e.g. NH_{3}).

**Bent**, ~104.5° (e.g. H_{2}O).

Lone pairs repel **more** than bonding pairs, so they **reduce** the bond angle.

Each double bond is **one** electron domain; 2 domains, 0 lone pairs → linear, 180°.

CH_{4} (109.5°) > NH_{3} (107°) > H_{2}O (104.5°) — angle falls as lone pairs increase.

A measure of how strongly an atom **attracts the shared (bonding) electrons** in a covalent bond.

A **difference in electronegativity** between the two atoms — the electrons are pulled towards the more electronegative atom.

The **more electronegative** atom (it gets a bigger share of the electrons); the less electronegative atom is **δ+**.

Δχ = 0 → **pure covalent**; small Δχ → **polar covalent** (δ+/δ−); large Δχ → **ionic**.

The small separation of charge (δ+ → δ−) along a polar bond; drawn as an **arrow** pointing to the δ− atom.

When the molecule is **symmetrical**, so the bond dipoles **cancel** (e.g. CO_{2}, CCl_{4}, BF_{3}).

It is **linear** — the two equal C=O dipoles point in opposite directions and **cancel**.

It is **bent** (lone pairs on O), so the two O–H dipoles **do not cancel** and give a net dipole.

Yes — it is **trigonal pyramidal** (a lone pair on N), so the N–H dipoles do not cancel; NH_{3} is polar.

(1) the bonds are **polar** (electronegativity difference) and (2) the **shape** means the dipoles **do not cancel**.

No — both atoms are identical, so Δχ = 0; the bond is **non-polar** and there is no dipole.

A continuous lattice of atoms joined by **covalent bonds** in every direction — there are **no separate small molecules**.

Melting requires breaking **many strong covalent bonds**, which needs a large amount of energy.

Different structural forms of the **same element** — e.g. diamond and graphite are both pure carbon.

To **four** other carbons in a rigid **3-D tetrahedral** network.

Its **rigid 3-D framework** of strong covalent bonds cannot be pushed out of shape.

All **four** outer electrons of each carbon are used in bonds, so there are **no delocalised electrons** to carry charge.

To **three** others in flat **layers**; the **fourth** electron is **delocalised**.

The **delocalised electrons** between the layers are free to move and carry charge.

**Weak forces** between the layers let the **layers slide** over each other (the covalent bonds within a layer stay strong).

**Diamond**, **graphite** (carbon allotropes), **silicon (Si)** and **silicon dioxide (SiO_{2})**.

Giant covalent → break **strong covalent bonds**; molecular → only overcome **weak intermolecular forces**.

Diamond uses all 4 electrons in bonds (**no** delocalised e⁻ → no conduction); graphite has **1 delocalised** e⁻ per carbon (conducts).

A force of attraction **between** separate molecules — much weaker than the covalent bonds **inside** a molecule.

The strength of its **intermolecular forces** — stronger IMFs need more energy, so a **higher** boiling point.

**London (dispersion) < dipole–dipole < hydrogen bonding.**

Forces from **temporary, instantaneous dipoles**; present between **all** molecules and the **only** force in non-polar ones.

**More electrons** (a larger, more polarisable molecule) — so they increase **down a group** and with molecular size.

When it is **polar** — it has a **permanent dipole** (δ+ and δ− ends) from an electronegativity difference.

The **strongest** IMF: a very δ+ H bonded to **N, O or F** is attracted to a lone pair on the N, O or F of a neighbour.

Hydrogen bonded directly to **N, O or F** ('H bonds to NOF').

NH_{3} has **hydrogen bonding** (H on N); PH_{3} has only weaker dipole–dipole/London forces.

Larger molecules have **more electrons → stronger London forces → higher boiling point**.

**No** — boiling only **separates the molecules** by overcoming intermolecular forces; the covalent bonds stay intact.

N, O and F are very electronegative, so the H is very δ+ and the attraction to a lone pair is especially strong.

A bookkeeping label comparing the electrons an atom 'owns' in a Lewis structure (lone pairs + half of each bond) with its normal valence count. It is **not** a real charge.

$\text{FC} = V - N - \tfrac{1}{2}B$ — valence electrons − non-bonding electrons − ½(bonding electrons).

The atom's **valence electrons as a free atom** — equal to its **group number** (C → 4, N → 5, O → 6).

Total electrons in the atom's bonds: **2 per single, 4 per double, 6 per triple**, summed over all its bonds.

V=5, N=0, B=8 → FC = 5 − 0 − 4 = **+1** (matches the ion's +1 charge).

The one with formal charges **closest to zero**, with any **negative** formal charge on the **most electronegative** atom.

When ≥2 valid Lewis structures can be drawn; the real species is a **hybrid** with electrons **delocalized** over the positions (drawn with a ↔ arrow).

Carbonate CO_{3}^{2-}, nitrate NO_{3}^{-} (three forms each) and ozone O_{3} (two forms).

**Equal bond lengths** — e.g. all three C–O bonds in CO_{3}^{2-} are identical, between a single and a double bond.

A molecule whose central atom has **fewer than 8** outer electrons, e.g. BeCl_{2} (4 on Be) and BF_{3} (6 on B).

A central atom holding **more than 8** outer electrons — only **period-3-or-below** atoms can, e.g. PCl_{5} (10) and SF_{6} (12).

Carbon is **period 2** — it has no available extra valence space, so it is limited to 8 outer electrons.

By **head-on (end-to-end)** overlap of orbitals, with electron density **along the bond axis**. Every single bond is a σ bond.

By **sideways** overlap of two parallel **p orbitals**, with electron density **above and below** the bond axis.

Single = **1 σ**; double = **1 σ + 1 π**; triple = **1 σ + 2 π**. Every multiple bond has exactly **one** σ.

The **mixing** of an atom's atomic orbitals (1 s + some p) into a set of equal-energy **hybrid orbitals**, one per electron domain.

**4** domains (1 s + 3 p), **tetrahedral**, 109.5°, **0** π bonds (all single bonds).

**3** domains (1 s + 2 p), **trigonal planar**, 120°, leaves **1** p orbital → **1 π** bond.

**2** domains (1 s + 1 p), **linear**, 180°, leaves **2** p orbitals → **2 π** bonds.

**Count its electron domains** (σ bonds + lone pairs): 4 → sp³, 3 → sp², 2 → sp.

**σ = total number of bonds** between atoms; **π = (number of doubles) + 2 × (number of triples)**.

The two p orbitals must stay **parallel** to overlap; twisting breaks the π bond, so a C=C cannot rotate → causes **cis–trans isomerism**.

Both are **sp²** (3 domains each), trigonal planar; the C=C is 1 σ + 1 π.

**sp** — it has 2 electron domains (the triple bond + one single bond), so it is linear.

The electrostatic attraction between a lattice of **positive metal cations** and a **sea of delocalised electrons**.

Electrons that are **not fixed to one atom** — free to move throughout the whole lattice.

The **delocalised electrons are free to move**, so they carry charge through the metal (solid or molten).

The bonding is **non-directional**, so layers of cations can **slide** over each other while the electron sea keeps them bonded.

A lot of energy is needed to overcome the **strong attraction** between the cations and the delocalised electron sea.

Sliding an ionic lattice brings **like charges** together → they repel and crack; a metal's non-directional sea has no like-charge layer, so it bends.

**Higher cation charge** and **smaller cation radius** (and more delocalised electrons).

Mg²⁺ has a **higher charge**, donates **two** electrons (denser sea) and is **smaller** than Na⁺.

It **weakens** — the cation radius **increases**, so the electron sea sits further from the nucleus.

The **delocalised electrons** (the cations stay fixed) — unlike a molten ionic compound, where the **ions** move.

The mobile **delocalised electrons** transfer kinetic energy quickly through the lattice.

That ionic, covalent and metallic bonding are the three **extremes** of one **continuum** — real compounds sit in between.

**Metallic** (bottom-left), **covalent** (bottom-right) and **ionic** (top).

How strongly an atom **attracts a shared pair of electrons**; values are in the data booklet.

Average the two electronegativities: $\chi_{avg} = \dfrac{\chi_A + \chi_B}{2}$ — it sets the **horizontal** position.

Take the difference: $\Delta\chi = |\chi_A - \chi_B|$ — it sets the **vertical** (ionic) position.

Electrons are essentially **transferred** → the bonding is **ionic** (high up the triangle).

Electrons are **shared** between similar non-metals → **covalent** (bottom-right corner).

A sea of delocalised electrons among metal atoms → **metallic** (bottom-left corner).

NaCl → **ionic** (top, large Δχ); Cl_{2} → **covalent** (bottom-right); Na → **metallic** (bottom-left).

It uses the **actual χ values**, so it correctly classifies polar-covalent metal compounds like BeCl_{2}.

Ionic = electrons **transferred** (large Δχ); covalent = electrons **shared** (small Δχ).

A **mixture** of a metal with one or more other elements (it is **not** a compound — no fixed ratio).

Its **different-sized atoms disrupt the regular layers**, so the layers **cannot slide** over each other as easily.

Yes — they keep **metallic bonding** (a sea of delocalised electrons); they are just **harder** than the pure metal.

**Brass** = copper + zinc; **steel** = iron + carbon (also bronze = copper + tin).

A **small molecule** that joins to many others to form a **polymer** (a giant molecule).

A long-chain molecule made by joining many **alkene monomers** (with **C=C**), with **no atoms lost**.

The **double bond opens up** — one bond becomes a single bond, the other joins to the next monomer.

The part of the polymer chain that **repeats**; get it by **opening the C=C** and drawing a bond out of each end.

**Monomer** has the **C=C double bond**; **repeat unit** has a **single** C–C bond with a bond out of each end.

Take **one repeating unit** and **put the C=C double bond back** between the two carbons.

**Ethene, CH_{2}=CH_{2}** — the repeat unit is –CH_{2}–CH_{2}–.

It is **chemically unreactive (inert)** and waterproof, so it resists corrosion — useful for packaging and containers.

By **increasing atomic number** (number of protons), not by relative atomic mass.

A horizontal **row**; the period number equals the highest occupied **main energy level (n)**.

A vertical **column**; elements in a group have the **same number of outer (valence) electrons**.

The **sublevel** that the outermost electrons are filling (s, p, d or f).

Groups **1 and 2** (plus H and He) — outer electrons fill the **s** sublevel.

Groups **13–18** — outer electrons fill the **p** sublevel.

The **centre** of the table (groups 3–12) — the **transition metals**, filling the d sublevel.

The **two detached rows** at the bottom — the **lanthanides and actinides**, filling the f sublevel.

Name the **sublevel the outermost electron enters** (e.g. …3p⁵ → p-block; …3d⁶ → d-block).

**Period** number = n of the outer shell; **group** number = number of outer electrons (group 17 → 7).

The **s-block** — its next electron would enter the **8s** sublevel (group 1, period 8).

**Nuclear charge** (proton pull) and **shielding/distance** (inner shells + extra shells).

The energy needed to remove one mole of electrons from one mole of **gaseous** atoms: X(g) → X⁺(g) + e⁻.

**Half** the distance between the nuclei of two bonded atoms — a measure of atom size.

How strongly an atom attracts a **bonding pair** of electrons (Pauling scale).

The energy change when one mole of gaseous atoms **gains** an electron: X(g) + e⁻ → X⁻(g).

**Decreases** — greater nuclear charge with similar shielding pulls the outer shell in.

**Increases** — each element has an extra electron shell.

**Increases** across a period (stronger pull); **decreases** down a group (further out, more shielded).

**Increases** across a period, **decreases** down a group (fluorine is the most electronegative).

A cation is **smaller** than its atom (it often loses a whole shell).

An anion is **larger** than its atom (extra electron–electron repulsion spreads the shell out).

Compare **nuclear charge**, compare **shielding/distance**, then state the **net effect** (held more/less tightly).

The same number of **outer (valence) electrons**, so they react in similar ways.

It **increases** — the outer electron is further out and more shielded, so it is **lost more easily**.

It **decreases** — the atom is bigger, so an incoming electron is **harder to gain**.

K is lower in group 1: **bigger atom + more shielding** → outer electron lost more easily.

F is smaller with less shielding, so it **gains** an electron more easily.

Able to act as **both an acid and a base** — reacts with acids **and** alkalis (e.g. Al_{2}O_{3}).

It **decreases** — elements change from **metallic** (Na) to **non-metallic** (Cl, Ar).

**Basic → amphoteric → acidic** left to right (Na_{2}O/MgO basic, Al_{2}O_{3} amphoteric, SO_{3} acidic).

**Basic** (e.g. Na_{2}O, MgO). Non-metal oxides are **acidic** (e.g. SO_{3}, P_{4}O_{10}).

**Caesium + fluorine** — lowest (most reactive) metal + top (most reactive) halogen.

Li < Na < K < Rb < Cs (increases down).

F > Cl > Br > I (decreases down).

A **d-block metal** that forms **at least one stable ion with a partially filled d sub-shell**.

Their only ions are **Sc³⁺ ([Ar] 3d⁰)** and **Zn²⁺ ([Ar] 3d¹⁰)** — empty/full d, never **partially filled**.

**4s fills first** (slightly lower energy when empty), so atoms end in **3d^{x} 4s²**.

**Remove 4s electrons before 3d.** e.g. Fe²⁺ = [Ar] 3d⁶ (the two 4s electrons go first).

**[Ar] 3d⁵ 4s¹** — an anomaly; a **half-full 3d⁵** is extra stable.

**[Ar] 3d¹⁰ 4s¹** — an anomaly; a **full 3d¹⁰** is extra stable.

The **4s and 3d sub-shells are close in energy**, so electrons can be removed in steps for similar energies → several stable states.

**+2 and +3** (Fe²⁺ = [Ar] 3d⁶; Fe³⁺ = [Ar] 3d⁵, a stable half-full sub-shell).

**+1 and +2** (Cu⁺ in Cu_{2}O, Cu²⁺ in CuSO_{4}).

**+7** — four O at −2 (−8) with an overall −1 charge forces Mn to +7.

The **partially filled d sub-shell** splits in a ligand field and **absorbs visible light**.

They can **change oxidation state** and use empty/part-full **d orbitals** to bind reactants (e.g. Fe in the Haber process).

Having **unpaired d electrons** — these are drawn into a magnetic field.

The ligands **split** the five d orbitals into two energy levels separated by a gap **Δ** (the splitting energy).

A molecule or ion (e.g. H_{2}O, NH_{3}, CN⁻) that bonds to a central metal ion by donating a **lone pair** of electrons.

The **splitting energy** — the energy gap between the two split d-orbital levels in a complex.

A d electron **absorbing a photon** of energy equal to Δ and jumping from the lower d-orbital level to the upper one.

Δ matches the energy of **visible light**, so the complex absorbs part of the visible spectrum in a d-d transition.

The **complement** of the colour **absorbed** — the leftover light. Absorbs red → looks green; absorbs blue → looks orange.

The **metal + its oxidation state**, the **ligand** (spectrochemical series), and the **number/geometry** of ligands.

Ligands ranked by the size of Δ they cause: I⁻ < Cl⁻ < H_{2}O < NH_{3} < CN⁻ (weak-field → strong-field, small Δ → large Δ).

**Weak-field** (I⁻, Cl⁻) → **small** Δ; **strong-field** (CN⁻, CO) → **large** Δ.

Larger Δ needs a **higher-energy** photon → a **shorter** wavelength of light is absorbed.

A different oxidation state changes Δ, so a **different wavelength** is absorbed and the complementary colour seen changes.

A **tetrahedral** complex has a **smaller Δ** than the equivalent **octahedral** one, so it absorbs different light and shows a different colour.

The chemistry of **carbon compounds**.

A family of organic compounds with the **same general formula** and **functional group**, each member differing by **CH_{2}**.

The reactive atom or group of atoms that gives a series its **characteristic chemistry** (e.g. C=C, –OH).

Same **general formula**; differ by **CH_{2}**; **gradual change** in physical properties; **similar chemical** properties.

**C_{n}H_{2n+2}** (saturated — only single C–C bonds).

**C_{n}H_{2n}** (unsaturated — one C=C double bond).

**C_{n}H_{2n+1}OH** (functional group –OH).

**Saturated** = only single C–C bonds (max H); **unsaturated** = at least one **C=C** double bond (fewer H).

Longer chains are bigger/heavier, so **intermolecular forces** are stronger → **higher boiling point**.

**Two** fewer hydrogens in the alkene — the C=C double bond replaces two C–H bonds.

**Ethene, C_{2}H_{4}** (alkenes start at n = 2).

The **simplest whole-number ratio** of the atoms in a compound (e.g. CH_{2}O for glucose).

The **actual number** of each type of atom in one molecule (e.g. C_{6}H_{12}O_{6} for glucose).

A diagram showing **every atom and every bond** in the molecule.

Atoms written **grouped in a line** with the bonds implied (e.g. CH_{3}CH_{2}OH).

Only the **carbon skeleton** drawn as lines; carbons are corners/ends and **H on carbon is implied**; functional groups are shown.

A **carbon** atom (each with enough H to make four bonds, not drawn).

Divide **every** subscript by their **highest common factor** (e.g. C_{6}H_{12}O_{6} ÷ 6 = CH_{2}O).

Molecules with the **same molecular formula** but a **different arrangement** of atoms (different connectivity).

Chain **branching**, **position** of a group, or different **functional group / class**.

Keep the **same molecular formula** but **change the connectivity** — never just rotate or flip the original.

No — CH_{2}O is an **empirical** formula; C_{2}H_{4}O_{2} (ethanoic acid) is one **molecular** formula with that ratio.

**No** — it is the same molecule; an isomer must have a genuinely different structure.

The **reactive atom or group of atoms** that defines an organic molecule's class and chemistry.

A family of compounds with the **same functional group** and the same general formula.

Saturated = only single C–C bonds (alkane); unsaturated = has a **C=C** (or C≡C) bond (alkene).

**–OH** (hydroxyl); name ends in **-ol** (e.g. propan-1-ol).

**–COOH** (carboxyl); name ends in **-oic acid** (e.g. propanoic acid).

Both have C=O. Aldehyde = carbonyl at the **end** (-al); ketone = carbonyl in the **middle** (-one).

An alkane with a **halogen** (–F, –Cl, –Br, –I) in place of an H; named with a prefix (chloro-, bromo-…).

**-ene**, because it contains a **C=C** double bond (e.g. propene).

**Stem** (number of carbons) + **suffix** (functional group) + **locant** (where the group is).

1 = meth-, 2 = eth-, 3 = prop-, 4 = but-.

The **number** in a name showing the position of the functional group on the chain (e.g. the 2 in but-2-ene).

Give the functional group the **lowest possible locant**.

Its m/z value is the **relative molecular mass (Mr)** — M⁺ is the peak at the **highest** m/z.

The **mass lost** (M⁺ − fragment) shows which **group broke off** (e.g. loss of 15 = CH_{3}).

Loss of a **CH_{3}** (methyl) group.

Loss of an **OH** group.

The **functional group**, from a characteristic absorption **wavenumber** (cm⁻¹) given in the data booklet.

An **O–H** group (an alcohol). A carboxylic acid O–H is even broader, ~2500–3000.

A **C=O** (carbonyl) — aldehyde, ketone, acid or ester.

The **number of peaks = number of different hydrogen environments** in the molecule.

**Three** — the CH_{3}, CH_{2} and OH hydrogens are three different environments.

By **symmetry** the two CH_{3} groups are equivalent, so all six H are in one environment.

**MS** (Mr + fragments), **IR** (functional group), **¹H NMR** (number of H environments) — used together.

In the **data booklet** — you read the wavenumbers off, you don't memorise them.

Different compounds with the **same molecular formula** but a different arrangement of atoms.

**Structural** isomers differ in **connectivity** (which atom bonds to which); **stereoisomers** have the same connectivity but a different **arrangement in space**.

**Chain** (different carbon skeleton), **position** (group in a different place), and **functional-group** (a different functional group/family).

Ethanol (CH_{3}CH_{2}OH, an alcohol) and methoxymethane (CH_{3}OCH_{3}, an ether) — both C_{2}H_{6}O.

A **C=C double bond** (restricted rotation) **and** two **different** groups on **each** doubly-bonded carbon.

**cis** = the two like groups on the **same** side of the C=C; **trans** = on **opposite** sides.

But-2-ene has a CH_{3} and an H on each C=C carbon (two different groups each); but-1-ene has two identical H atoms on one carbon, so no isomers.

**Priority** by Cahn–Ingold–Prelog: the atom of **higher atomic number** is higher priority. Z = higher-priority groups on the same side; E = on opposite sides.

A carbon bonded to **four different** atoms or groups (a stereocentre, often marked C*).

The **two non-superimposable mirror-image** forms of a molecule that has a chiral carbon (optical isomers).

They **rotate plane-polarised light** by the same angle in **opposite directions** (one +, one −); all their other physical properties are identical.

A **50:50 mixture** of the two enantiomers, which shows **no net rotation** of plane-polarised light (the effects cancel).

**sp²** — three σ bonds in a plane at **120°**, leaving one electron in a **p-orbital** perpendicular to the ring.

A **planar (flat) regular hexagon** of six carbons, each bonded to one hydrogen.

Two ring-shaped electron clouds **above and below** the ring, formed by sideways overlap of the six carbon p-orbitals.

**All six are equal** (~140 pm) — **intermediate** between a single (~154 pm) and a double bond (~134 pm).

**(1)** All six C–C bonds are the **same length**. **(2)** The **enthalpy of hydrogenation** is **less negative** than expected for three isolated C=C bonds.

Real benzene is **lower in energy (more stable)** than a Kekulé structure — the difference is the **delocalisation (resonance) stabilisation**.

**Kekulé** = alternating single/double bonds (two bond lengths). **Delocalised** = a circle in the hexagon (all bonds equal) — the **accepted** model.

Addition would **break the delocalised π system** and lose its stability, so benzene avoids it.

**Substitution** — a ring H is replaced (usually with a catalyst) and the **delocalised ring stays intact**.

Micro **6.4.3** (electrophilic substitution — the curly-arrow mechanism); 3.2.6 only sets it up qualitatively.

**(1)** add HX (e.g. HBr) → halogenoalkane; **(2)** warm with aqueous NaOH → alcohol (–X replaced by –OH).

**Orange → green** (Cr_{2}O_{7}^{2-} is reduced to Cr^{3+}).

The **heat energy** released or absorbed by a reaction at **constant pressure** (units: kJ mol⁻¹).

A reaction that **releases** energy to the surroundings, so they get **hotter**; **ΔH is negative**.

A reaction that **absorbs** energy from the surroundings, so they get **colder**; **ΔH is positive**.

Exothermic → **ΔH < 0** (negative); endothermic → **ΔH > 0** (positive).

**Endothermic** — energy must be **put in** to break a bond.

**Exothermic** — energy is **released** when a new bond forms.

When **making** the new bonds releases **more** energy than **breaking** the old bonds absorbed (net energy out).

The **minimum** energy reactants need to react — the reactant level up to the **peak** of the profile.

It is the energy gap between the **reactant** and **product** levels (down for exothermic, up for endothermic).

**Exothermic** products are **lower** in energy and so **more stable** than the reactants.

**Endothermic** — energy is absorbed from the surroundings, so **ΔH is positive**.

**Combustion** and **neutralisation** (also respiration) — they release energy.

Measuring the **temperature change** of a known mass of water (or solution) to find the heat transferred by a reaction.

The energy needed to raise **1 g** of a substance by **1 K** (1 °C). For water, **c = 4.18 J g⁻¹ K⁻¹**.

$Q = mc\Delta T$ — heat (J) = mass (g) × specific heat capacity × temperature change.

ΔT = **T_{final} − T_{initial}**. A change of 1 °C equals a change of 1 K, so the number is the same.

$\Delta H = -\dfrac{Q}{n}$ — divide Q (in kJ) by the amount that reacted, and add the sign.

**Exothermic** — heat released to the water — so **ΔH is negative**.

**Endothermic** — heat absorbed from the water — so **ΔH is positive**.

The mass of **water** (the substance heated), **not** the mass of fuel or reactant.

Q from $mc\Delta T$ is in **joules**; enthalpy changes are quoted in **kJ mol⁻¹**, so divide by 1000.

**Heat loss** to the surroundings and apparatus — so the measured ΔH is **less exothermic** than the true value.

All the heat goes to the **water**, and the **specific heat capacity** (and density) of the solution equals that of water.

ΔT → **Q = mcΔT** → ÷1000 for kJ → **÷ n** for per mole → add the **sign**.

The energy needed to **break one mole** of a particular bond in the **gaseous** state (always a positive value).

**Endothermic** — breaking a bond always **costs** (absorbs) energy.

**Exothermic** — forming a bond always **releases** energy.

$\Delta H = \Sigma(\text{bonds broken}) - \Sigma(\text{bonds made})$.

The reaction is **exothermic** — more energy was released making bonds than was used breaking them.

The reaction is **endothermic** — breaking bonds cost more energy than was released making them.

A bond (e.g. C–H) exists in many molecules with slightly different strengths, so the booklet gives an **average**; ΔH is therefore an **estimate**.

Only when **all species are gaseous**, because bond enthalpy is defined for the gaseous state.

Only the bonds that **break or form** — unchanged bonds (spectator bonds) cancel out.

**Higher** — a larger bond enthalpy means a stronger bond that needs more energy to break.

Because the bond enthalpies are **averages**, so the calculated ΔH is only an **estimate**.

The total enthalpy change of a reaction is the **same** whatever route is taken, because ΔH depends only on the initial and final states.

A property that depends only on the **current state** of the system, not on the path taken to reach it (enthalpy is one).

Because enthalpy is a **state function**, so ΔH is **path-independent** — you can add up the steps of an alternative route.

Its **sign is reversed** (the magnitude stays the same).

ΔH is **doubled** — multiply ΔH by the same factor as the equation.

$\Delta H^{\ominus} = \Sigma\,\Delta H_{f}^{\ominus}(\text{products}) - \Sigma\,\Delta H_{f}^{\ominus}(\text{reactants})$.

**Zero** by definition (e.g. O_{2}(g), C(s) graphite).

**With** an arrow → **add** its ΔH; **against** it (reverse) → **subtract** its ΔH (flip the sign).

Forgetting to **multiply** a step by the number of moles in the target equation.

To find a ΔH that **cannot be measured directly** (e.g. the reaction is too slow or has side reactions).

The ΔHf equation **is** a Hess cycle — going down to the elements (reverse ΔHf of reactants) and up to the products (ΔHf of products).

The enthalpy change when **1 mol** of a compound forms from its **elements in their standard states** (100 kPa, stated T).

The enthalpy change when **1 mol** of a substance is **completely burned in oxygen** under standard conditions; always **negative**.

**Zero** — e.g. O_{2}(g), N_{2}(g), C(graphite); there is nothing to form.

$\Delta H^{\ominus} = \sum \Delta H_{f}^{\ominus}(\text{products}) - \sum \Delta H_{f}^{\ominus}(\text{reactants})$.

$\Delta H^{\ominus} = \sum \Delta H_{c}^{\ominus}(\text{reactants}) - \sum \Delta H_{c}^{\ominus}(\text{products})$.

Both reactants and products burn down to the **same products** (CO_{2} + H_{2}O), so the Hess cycle runs the other way → **reactants − products**.

A pressure of **100 kPa** and a stated temperature (usually **298 K**), with all substances in their standard states.

Enthalpy is a **state function** — ΔH depends only on the start and end states, so a 'paper' Hess route gives the same answer as experiment.

Forgetting to multiply each value by its **stoichiometric coefficient** (e.g. 2 H_{2}O) or forgetting an **element is zero**.

**Negative** (exothermic) — a quick check that you used the correct rule.

**kJ mol⁻¹** (kilojoules per mole).

The enthalpy change when **1 mol** of a **solid ionic compound** is converted into its **gaseous ions** (e.g. NaCl(s) → Na⁺(g) + Cl⁻(g)); **endothermic (+)**.

There is **no single experiment** that turns a solid into a gas of free ions, so it is found **indirectly** via a Born–Haber cycle and Hess's law.

A closed **enthalpy cycle** linking ΔH_{f}⊖ of an ionic solid to atomisation, bond dissociation, ionisation energy, electron affinity and lattice enthalpy — solved with **Hess's law**.

**Endothermic (+)** — energy is needed to make gaseous atoms from an element.

**Always endothermic (+)** — removing an electron works against the nuclear attraction.

**Usually exothermic (−)** for the first electron added to a gaseous atom.

**Endothermic (+)** — an electron is forced onto an already-negative ion.

$\Delta H_{lat}^{\ominus} = \Delta H_{atom}^{\ominus} + \Delta H_{IE}^{\ominus} + \Delta H_{EA}^{\ominus} - \Delta H_{f}^{\ominus}$.

**Bigger charge → larger** lattice enthalpy (stronger electrostatic attraction); the **dominant** factor. MgO ≫ NaCl.

**Smaller ions → larger** lattice enthalpy — the ions sit closer together, so attraction is stronger. NaF > NaI.

Forgetting that subtracting a **negative** ΔH_{f}⊖ **adds** its magnitude — write every value with its sign in brackets.

**kJ mol⁻¹** (kilojoules per mole).

A substance that releases useful **energy** when it is **burned** (combusted) in oxygen.

The reaction of a fuel with **oxygen** that releases energy as heat; it is always **exothermic** (ΔH < 0).

**Carbon dioxide (CO_{2}) and water (H_{2}O)** only — with maximum energy released.

**Carbon monoxide (CO) and/or carbon (soot)** plus water — less energy is released.

When there is a **limited supply of oxygen**, so the carbon is not fully oxidised.

CO is a **toxic** gas that binds to haemoglobin, stopping the blood from carrying oxygen.

Energy released **per unit mass** of fuel (e.g. **kJ g⁻¹**) — matters when weight is important.

Energy released **per unit volume** of fuel (e.g. **kJ cm⁻³**) — matters when storage space is important.

Fossil fuels (coal, oil, gas) are **non-renewable**; biofuels (e.g. ethanol, biodiesel) are **renewable**.

The crop **absorbs CO_{2}** as it grows, roughly balancing the CO_{2} released when the fuel is burned.

They release carbon that was **locked away for millions of years**, adding **new** CO_{2} to the atmosphere.

**Ethanol** (from fermented sugar cane/corn) or **biodiesel** (from plant oils).

A measure of how **dispersed** (spread out) the matter and energy of a system are — more ways to arrange the particles and energy means higher entropy.

When matter/energy become more dispersed: solid → liquid → gas, dissolving, and especially when the **moles of gas increase**.

When the system becomes more ordered: gas → liquid → solid, or when the **moles of gas decrease**.

The change in the **number of moles of gas** — gases have far higher entropy than liquids or solids.

The entropy of **one mole** of a substance under standard conditions; units **J K⁻¹ mol⁻¹**.

**J K⁻¹ mol⁻¹** — note joules, not kilojoules (unlike ΔH).

**gas > liquid > solid** — a gas has the most dispersed matter and energy.

**S = 0** — a single perfectly ordered arrangement with no thermal motion (the only zero-entropy state).

$\Delta S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants})$, each S° × its coefficient.

Ignoring the **stoichiometric coefficients**, and mixing **J** (entropy) with **kJ** (enthalpy).

**Always positive** for any substance above 0 K (entropy is a measure of dispersal, never negative).

Gas particles can occupy far more positions and share energy over more ways of moving, so there are many more arrangements → higher entropy.

$\Delta G = \Delta H - T\,\Delta S$ — it combines the enthalpy change and the entropy change into one test for spontaneity. (Given in the data booklet.)

When **ΔG < 0**. ΔG > 0 is non-spontaneous; ΔG = 0 is at equilibrium.

**No** — spontaneous means thermodynamically **feasible**. Speed is decided by **kinetics** (activation energy), not by ΔG.

ΔH is in **kJ** mol⁻¹ but ΔS is in **J** K⁻¹ mol⁻¹ — convert ΔS to kJ by **dividing by 1000** before combining.

**Kelvin** (K = °C + 273) — never degrees Celsius.

**Always spontaneous** (at every temperature) — ΔG is negative at all T.

**Never spontaneous** — ΔG is positive at all T.

Spontaneous only at **low** temperature (the +TΔS term wins once T is large).

Spontaneous only at **high** temperature (the −TΔS term wins once T is large).

Set **ΔG = 0**, so $T = \dfrac{\Delta H}{\Delta S}$ — use matching units (both in kJ) so T comes out in K.

**Same** signs on ΔH and ΔS ⇒ **temperature decides**. **Opposite** signs ⇒ same answer at every temperature.

A large increase in **disorder** (ΔS > 0) makes −TΔS very negative, so ΔG can be negative even when ΔH > 0 (e.g. dissolving ammonium nitrate).

An equation with the **same number of each kind of atom** on both sides — atoms are conserved.

The study of the **whole-number ratios** in which substances react and are formed, read from a balanced equation.

The ratio of the **coefficients** in a balanced equation — how many moles of one substance react with or form another.

Only the **coefficients** (the big numbers in front) — **never** a subscript inside a formula.

Changing a subscript changes the **substance** itself (e.g. H_{2}O → H_{2}O_{2}), so it no longer describes the same reaction.

**(s)** solid, **(l)** pure liquid, **(g)** gas, **(aq)** aqueous (dissolved in water).

**Aqueous** — the substance is **dissolved in water** (different from a pure liquid, (l)).

The ratio N_{2} : H_{2} : NH_{3} is **1 : 3 : 2** — 1 mol N_{2} reacts with 3 mol H_{2} to make 2 mol NH_{3}.

Balance **C first, then H, then O last** (oxygen appears in more than one product), then reduce to smallest whole numbers.

**CH_{4} + 2 O_{2} → CO_{2} + 2 H_{2}O** — smallest whole-number coefficients.

The C : CO_{2} ratio is 1 : 1, so **0.5 mol** of CO_{2}.

Changing a **formula** (subscript) instead of a coefficient, or forgetting the **state symbols** when asked.

The reactant that **runs out first** — it controls (limits) the amount of product that can form.

The reactant **left over** once the limiting reactant has been used up.

The **maximum** amount (or mass) of product, calculated from the **limiting** reactant.

Convert each reactant mass to **moles**, divide each by its **coefficient**, and the **smallest** result is limiting.

It compares the reactants fairly against the **mole ratio** in the balanced equation, so you can see which runs out first.

Always the **limiting** reactant — never the one in excess.

$n = \dfrac{m}{M}$ — convert every mass to moles before using the mole ratio.

Balanced equation → mass to **moles** (n = m/M) → scale by the **mole ratio** → moles back to **mass** (m = nM).

From the **coefficients** of the balanced equation (e.g. N_{2} + 3H_{2} → 2NH_{3} is 1 : 3 : 2).

Working out the product from the reactant in **excess**, or forgetting the **mole ratio** when coefficients are not 1 : 1.

**A** (0.3 mol runs out first); B is in excess by 0.2 mol.

$\%\text{ yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100$ — how much product you actually obtained versus the maximum predicted by the equation.

The amount of product predicted from the balanced equation if the **limiting reactant** reacted completely.

The amount of product you really obtain — always **less** than theoretical, due to side reactions, reversible reactions and losses.

Side reactions, reversible reactions not going to completion, and losses during separation/purification.

$\%\text{ AE} = \dfrac{M(\text{desired product})}{M(\text{all reactants})} \times 100$ — the fraction of reactant atoms ending up in the wanted product.

Yield = **how much product you made**; atom economy = **how little reactant mass you wasted** as by-products. They are independent.

**Addition** reactions — all reactants combine into a single product, so there are no by-products.

Sum the molar masses of **all** reactants, each multiplied by its **coefficient** in the balanced equation.

Fewer atoms wasted as by-products → less raw material used and less waste to treat → more **sustainable and economical**.

No — actual yield cannot beat the theoretical maximum. A value over 100% signals an error (e.g. impure/wet product).

$\text{actual} = \dfrac{\%\text{ yield}}{100} \times \text{theoretical}$.

Putting only **one** reactant (or forgetting coefficients) on the bottom — you must sum **every** reactant's molar mass.

At the **same temperature and pressure**, equal **volumes** of gases contain equal numbers of **moles** — so the volume ratio equals the coefficient ratio.

Because at fixed T and P volume is **proportional to amount**, so the balanced **coefficients** give the **volume ratio** — no moles needed.

**22.7 dm³ mol⁻¹** at STP (273 K, 100 kPa) — given in the data booklet.

$n = \dfrac{V}{V_{m}}$ with $V_{m} = 22.7$ dm³ mol⁻¹ (volume in dm³).

Multiply the amount by the molar volume: $V = n\,V_{m} = n \times 22.7$ dm³.

**1000 cm³** — divide a cm³ value by 1000 before using the molar volume 22.7 dm³ mol⁻¹.

**2 volumes** of NH_{3} (the volume ratio matches the 1 : 3 : 2 coefficients).

Subtract the volume that **reacted** (from the coefficient ratio) from the volume **supplied**.

**No** — only **gases** contribute; liquids and solids (like condensed water) add zero volume.

Forgetting to **subtract the gas that reacted** when asked for the volume remaining, or counting **liquid** products as gas.

**273 K and 100 kPa** (standard temperature and pressure).

A precise technique to find an **unknown concentration** by reacting it with a **standard solution** to the **end point** (an indicator colour change).

A solution of **precisely known concentration**, made up in a **volumetric flask**.

Delivers a **fixed, exact** volume of the solution being analysed (e.g. 25.0 cm³).

Delivers the **variable** volume of titrant (the **titre**), read to ±0.05 cm³.

$n = CV$ — amount (mol) = concentration (mol dm⁻³) × volume (**dm³**). Given in the data booklet.

**(1)** n = CV on the known reagent → mol. **(2)** Cross by the **mole ratio**. **(3)** C = n/V (or M = m/n) on the unknown.

The volume must be in **dm³** — divide a cm³ titre by **1000** first.

Titres that **agree** (typically within 0.10 cm³). Only the concordant titres are **averaged** — a rough trial is ignored.

Add a **known excess** of a reagent, let it react, then titrate the **leftover** excess. Amount reacted = **added − leftover**.

When the reaction is **slow** or the sample is an **insoluble solid** (e.g. a carbonate), making a direct titration impractical.

**2 : 1** — sulfuric acid is diprotic, so it needs **two** moles of NaOH per mole of acid.

Forgetting the **mole ratio** from the balanced equation, or leaving a volume in **cm³** instead of dm³.

The **change in concentration** of a reactant or product **per unit time**.

**mol dm⁻³ s⁻¹** — a concentration (mol dm⁻³) divided by a time (s).

It is the **gradient** (steepness) of the curve — the tangent at a point gives the instantaneous rate.

The **reactant concentration is highest** at t = 0, so effective collisions are most frequent and the curve is **steepest**.

**Average** = total change ÷ total time (slope of the **chord**); **instantaneous** = slope of the **tangent** at one moment.

Particles must **collide** to react, but only **effective** collisions (enough energy + correct orientation) lead to a reaction.

Energy **≥ the activation energy Eₐ**, AND the particles collide in the **correct orientation**.

The **minimum energy** that colliding particles must have for a reaction to occur.

Reactants are **used up**, so their concentration falls and effective collisions become **less frequent**; rate drops to zero when reactants run out.

Measure the **volume of gas** collected vs time, or the **mass lost** vs time.

Use a **colorimeter** to measure the **light absorbed** as it changes with time.

The rate at t = 0 — the **slope of the tangent drawn at the start** of a concentration–time graph (the steepest point).

Energy **≥ the activation energy (E_{a})** AND the **correct orientation**.

The **minimum** energy a colliding pair of particles must have for a reaction to occur.

**Concentration, pressure, surface area, temperature** and a **catalyst**.

They put more particles in the reaction space, so collisions are **more frequent** (the energy per collision is unchanged).

Particles move faster (collisions **more frequent**) AND the distribution shifts right so a **greater fraction** have energy ≥ E_{a} — the second effect is the main one.

A substance that speeds up a reaction by providing an **alternative pathway of lower E_{a}**, and is **not used up** itself.

**No** — the reactant and product energy levels are unchanged, so ΔH is the same.

How the **kinetic energies** of particles are **spread out**; only those to the right of E_{a} can react.

**Lower and shifted to the right** (broader/flatter), but with the **same area** underneath.

The **fraction of particles** with enough energy to react (energy ≥ E_{a}).

The curve is **unchanged**; the **E_{a} line moves left**, so a larger fraction lies to the right of it.

The reaction goes **faster**, AND the solid is **recovered unchanged** (same mass/nature) at the end.

**rate = k[A]^{m}[B]^{n}** — the rate equals the rate constant k times the reactant concentrations, each raised to its order.

The **power** to which that reactant's concentration is raised in the rate equation (found by **experiment**).

The **sum** of the individual orders (m + n).

**Experimentally** — from how the initial rate responds to changing each concentration; **never** from the stoichiometric equation.

**Zero order** (× 1 = 2⁰) — that reactant is not in the rate equation.

**First order** (× 2 = 2¹).

**Second order** (× 4 = 2²).

The proportionality constant in the rate equation; **fixed at a given temperature** (it changes only with temperature).

**s⁻¹** — from k = rate ÷ [A] = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³).

**mol⁻¹ dm³ s⁻¹** — from k = rate ÷ [A]².

**mol⁻² dm⁶ s⁻¹** — from k = rate ÷ [A]³.

The **slowest** step in a multi-step mechanism; it sets the overall rate. The rate equation shows the species **up to and including** the RDS.

**k = A·e^{−E_{a}/RT}** — it gives the rate constant in terms of the frequency factor A, the activation energy E_{a}, the gas constant R and the absolute temperature T.

The **frequency of collisions** and whether they occur with the **correct orientation** (the steric factor). A has the **same units as k**.

The **activation energy** — the minimum collision energy needed to react. It sits in the **exponent**, so it has a large effect on k.

R = the **gas constant** = **8.31 J K⁻¹ mol⁻¹**; T = the **absolute temperature in kelvin** (°C + 273).

Because E_{a}/RT shrinks as T rises and the term is **exponential** — a small T increase makes a large k increase (often k roughly doubles per 10 K).

**ln k = ln A − (E_{a}/R)(1/T)** — the equation of a straight line for ln k against 1/T.

**y-axis = ln k**, **x-axis = 1/T** (T in kelvin).

**−E_{a}/R** — a negative value (because E_{a} > 0).

**ln A** — the value of ln k when 1/T = 0; so A = e^{intercept}.

**E_{a} = −gradient × R**, then **÷ 1000** to convert J mol⁻¹ → kJ mol⁻¹.

**ln(k₂/k₁) = −(E_{a}/R)(1/T₂ − 1/T₁)** — used to find E_{a} from rate constants at two temperatures.

**(1)** Forgetting the minus sign — E_{a} = −gradient × R. **(2)** Forgetting to ÷ 1000, since R is in **J** K⁻¹ mol⁻¹, so E_{a} comes out in J mol⁻¹.

A reaction that can go in **both directions** — reactants can form products and products can re-form reactants. Shown with the **⇌** symbol.

The state, in a **closed system**, where the **forward and reverse reactions occur at equal rates**, so the **concentrations of reactants and products stay constant**.

Because **both** the forward and reverse reactions are **still happening** — the reaction has **not** stopped; the opposite changes simply cancel out.

**No** — they are **constant** (unchanging), but generally **not equal** to one another.

The **rate of the forward reaction = the rate of the reverse reaction**.

So that **nothing is added or escapes** (no reactant/product/heat leaves); an open system could never settle to constant concentrations.

**Colour (absorbance)**, **pressure** (for gases) and **pH** — all remain constant because the concentrations are constant.

**How far** a reaction has gone — the **relative amounts** of reactants and products. To the **right** = mostly products; to the **left** = mostly reactants.

Measure a **macroscopic property** over time (e.g. colour intensity); when it **levels off to a constant value**, equilibrium has been reached.

That the reaction has **stopped** — in fact both reactions continue (it is **dynamic**); the concentrations are merely constant.

The **forward rate falls** and the **reverse rate rises** until they **meet (become equal)** — that point is equilibrium.

Both the reactant and product curves **level off** (become flat) and stay constant — at **different** values.

If a system at equilibrium is disturbed, the position shifts in the direction that **opposes** (partly cancels) the change.

…towards the **products** (right) — the system uses up the added reactant.

…towards the **products** (right) — the system replaces the lost product.

The position shifts to the side with **fewer moles of gas**, to reduce the pressure.

Changing the pressure causes **no shift** in the position.

The position shifts in the **endothermic** direction (it absorbs the added heat).

A change in **temperature** — concentration, pressure and a catalyst leave K_{c} unchanged.

**No shift** in position and **no change** in K_{c}; it just reaches equilibrium **sooner** (speeds up both directions equally).

K_{c} **decreases** (the position shifts towards the reactants).

**Count the moles of gas** on each side; the position shifts towards the side with **fewer** gas moles when pressure rises.

Write **heat** as a species (exo: products + heat; endo: reactants + heat), then treat adding heat like adding that species.

**Endothermic** — adding heat favours the heat-absorbing direction.

The **fixed ratio** of product to reactant concentrations at equilibrium, at a given temperature — it shows **how far** a reaction goes.

**Products over reactants**, each concentration **raised to the power of its balancing coefficient**; use [ ] for equilibrium concentration in mol dm⁻³.

$K_{c} = \dfrac{[\text{NH}_{3}]^{2}}{[\text{N}_{2}][\text{H}_{2}]^{3}}$ — the 2 and 3 become powers.

**Products are favoured** — the equilibrium lies to the **right** (mostly products).

**Reactants are favoured** — the equilibrium lies to the **left** (mostly reactants).

The **reciprocal**: K_{reverse} = **1 / K_{forward}**.

A change in **temperature** — concentration, pressure and a catalyst all leave K_{c} unchanged.

K_{c} **increases** (the position shifts towards products).

K_{c} **decreases** (the position shifts towards reactants).

Convert each **amount (mol)** to a **concentration (mol dm⁻³)** using **c = n/V**, then substitute.

Pure **solids** and pure **liquids** — only gases and dissolved (aqueous) species appear.

**No** — K_{c} describes the **extent** (how far), not the **rate** (how fast).

Products over reactants, each concentration raised to the power of its **stoichiometric coefficient**: $K_{c}=\dfrac{[\text{C}]^{c}[\text{D}]^{d}}{[\text{A}]^{a}[\text{B}]^{b}}$.

**Pure solids (s) and pure liquids (l)** — including the solvent in dilute solution — because their concentration is effectively constant.

K_{c} ≫ 1 → equilibrium lies to the **right**: mostly **products** at equilibrium.

K_{c} ≪ 1 → equilibrium lies to the **left**: mostly **reactants** at equilibrium.

**Temperature** only. Changing concentration, pressure or adding a catalyst does **not** change K_{c}.

**I**nitial, **C**hange and **E**quilibrium amounts (or concentrations) for each species.

They are in the **ratio of the stoichiometric coefficients** — reactants decrease (−), products increase (+).

The **same expression as K_{c}**, but evaluated with the concentrations at **any moment**, not only at equilibrium.

**Forward** (→), making more product, until Q rises to K_{c}.

**Backward** (←), making more reactant, until Q falls to K_{c}.

$\Delta G^{\circ}=-RT\ln K$ (given in the data booklet); R = 8.31 J K⁻¹ mol⁻¹, T in kelvin.

ln K > 0, so ΔG° = −RT ln K is **negative** — the reaction is **spontaneous** and favours products.

A **proton (H⁺) donor**.

A **proton (H⁺) acceptor**.

A **hydrogen ion, H⁺** — a hydrogen atom that has lost its electron.

Two species that differ by **exactly one H⁺** (an acid and the base left after it donates).

**Remove** one H⁺ from the acid (e.g. HCl → Cl⁻).

**Add** one H⁺ to the base (e.g. NH_{3} → NH_{4}^{+}).

A species that can **both donate and accept** a proton (e.g. H_{2}O, HCO_{3}⁻).

**HSO_{4}⁻** (remove one H⁺ — not SO_{4}^{2-}, which is two H⁺ away).

**H_{3}O^{+}** (the oxonium / hydronium ion).

**H_{2}O** and **HCO_{3}⁻** — both can donate or accept a proton.

**HCl** — it donates the proton; water is the base.

An acid can only **donate** H⁺ if a base is there to **accept** it — every proton transfer has both.

A measure of acidity based on hydrogen-ion concentration: $\text{pH} = -\log_{10}[\text{H}^{+}]$.

$\text{pH} = -\log_{10}[\text{H}^{+}]$ — given in the data booklet.

$[\text{H}^{+}] = 10^{-\text{pH}}$ — the rearranged given equation.

The ionic product of water, $K_{w} = [\text{H}^{+}][\text{OH}^{-}] = 1.0\times10^{-14}$ at 25 °C.

pH < 7 acidic · pH = 7 neutral · pH > 7 basic (alkaline), at 25 °C.

[H_{+}] changes by a factor of **10** (pH is a log scale).

Strong = **fully** dissociated into ions; weak = only **partially** dissociated.

No — strength is the **degree of dissociation**; concentration is the amount dissolved.

Single arrow, e.g. $\text{HCl} \rightarrow \text{H}^{+} + \text{Cl}^{-}$ (full dissociation).

Equilibrium arrows, e.g. $\text{CH}_{3}\text{COOH} \rightleftharpoons \text{H}^{+} + \text{CH}_{3}\text{COO}^{-}$ (partial).

Strong acid has a **lower pH**, **higher conductivity** and a **faster** reaction (more H_{+} ions).

It is fully dissociated, so it gives a **higher [H_{+}]**, and a higher [H_{+}] means a lower pH.

The reaction of an **acid with a base** to give a **salt and water**; the H⁺ and OH⁻ cancel out.

The ionic compound formed when the **H⁺** of an acid is replaced by a **metal ion** (or NH_{4}⁺).

**salt + hydrogen** (e.g. Mg + 2HCl → MgCl_{2} + H_{2}).

**salt + water** (neutralisation; the base is a metal oxide or hydroxide).

**salt + water + carbon dioxide** (e.g. 2HCl + CaCO_{3} → CaCl_{2} + H_{2}O + CO_{2}).

A **chloride** (e.g. NaCl, MgCl_{2}).

A **sulfate** (e.g. Na_{2}SO_{4}, MgSO_{4}).

A **nitrate** (e.g. NaNO_{3}, Ca(NO_{3})_{2}).

**Hydrogen** gives a squeaky **'pop'** with a lit splint.

**Carbon dioxide** turns **limewater milky** (cloudy).

It is **diprotic** — it provides **two H⁺**, so it neutralises two 1+ bases: 2NaOH + H_{2}SO_{4} → Na_{2}SO_{4} + 2H_{2}O.

Balance the **ionic charges** (e.g. Mg²⁺ with Cl⁻ → MgCl_{2}), then balance the whole equation.

The **acid-dissociation constant** — the equilibrium constant for HA ⇌ H⁺ + A⁻: $K_a = \dfrac{[H^+][A^-]}{[HA]}$.

A **stronger** weak acid — the dissociation equilibrium lies further to the right (more H⁺).

The **base-dissociation constant** for B + H_{2}O ⇌ BH⁺ + OH⁻; a larger K_{b} = a stronger weak base.

$\text{p}K_a = -\log_{10} K_a$ — the negative logarithm of the acid-dissociation constant.

A **stronger** acid — the minus sign flips the scale, so the lowest pK_{a} is the strongest acid.

Line up their **pK_{a}** values; the **lowest pK_{a}** is the strongest (largest K_{a}).

For a conjugate pair, $K_a \times K_b = K_w$ (= 1.0 × 10⁻¹⁴ at 298 K).

For a conjugate pair, $\text{p}K_a + \text{p}K_b = 14$ at 298 K.

$K_b = \dfrac{K_w}{K_a}$ — divide K_{w} by the acid's K_{a}.

$[H^+] = \sqrt{K_a\,c}$, then $\text{pH} = -\log_{10}[H^+]$.

**1.** [H⁺] = [A⁻] (acid is the only source of H⁺). **2.** [HA] ≈ c (dissociation is negligible).

$[OH^-] = \sqrt{K_b\,c}$ → pOH = −log[OH⁻], then **pH = 14 − pOH** (298 K).

**No** — like K_{c}, you write the expressions yourself; only K_{w} = 1.0 × 10⁻¹⁴ is a booklet constant.

A solution that **resists pH change** when a small amount of acid or alkali is added.

A **weak acid + its conjugate base** (e.g. ethanoic acid + sodium ethanoate), both present in appreciable amounts.

A **weak base + its conjugate acid** (e.g. ammonia + ammonium chloride).

The **conjugate base** reacts with it: A⁻ + H⁺ → HA (so the ratio [A⁻]/[HA] barely changes).

The **weak acid** reacts with it: HA + OH⁻ → A⁻ + H_{2}O.

$\text{pH} = \text{p}K_a + \log_{10}\dfrac{[A^-]}{[HA]}$ — buffer pH from pK_{a} and the base/acid ratio.

When **[A⁻] = [HA]** (equal amounts), so the log term is zero — the **half-equivalence point**.

**Exactly 7** — the salt formed is neutral.

**Greater than 7** — the salt of a weak acid hydrolyses to a basic solution.

**Less than 7** — the salt of a weak base hydrolyses to an acidic solution.

A **weak acid**, HIn ⇌ H⁺ + In⁻, whose acid form and conjugate base are **different colours**.

Pick one whose **colour-change range (≈ pK_{a}(In)) lies inside the vertical jump** at the equivalence point.

**Phenolphthalein** (range 8.3–10.0) — it changes within the basic vertical jump.

**Methyl orange** (range 3.1–4.4) — it changes within the acidic vertical jump.

A number tracking how many electrons an atom has **gained or lost** relative to the free element — the charge it would have if all bonds were ionic.

Always **0** (e.g. Na, O_{2}, Cl_{2}, S_{8}).

**Equal to its charge** (e.g. Mg²⁺ is +2, Cl⁻ is −1).

Oxygen is **−2**; hydrogen is **+1** — except peroxides (O is −1) and metal hydrides (H is −1).

They add up to the **total charge**: 0 for a neutral compound, the ion charge for a polyatomic ion.

An **increase** in oxidation state — the atom has **lost** electrons (OIL).

A **decrease** in oxidation state — the atom has **gained** electrons (RIG).

**O**xidation **I**s **L**oss, **R**eduction **I**s **G**ain (of electrons).

The species that **takes** electrons and is itself **reduced** (its oxidation state goes down), e.g. O_{2}, Cl_{2}.

The species that **gives** electrons and is itself **oxidised** (its oxidation state goes up), e.g. a reactive metal.

An atom's oxidation state **changes** during the reaction — so electrons have been transferred.

**+6** — four O at −2 (= −8) plus S must equal the ion charge −2, so S = +6.

An equation showing **just the oxidation or just the reduction** part of a redox reaction, with the electrons (e⁻) included.

On the **right** (product) side — oxidation is **loss** of electrons.

On the **left** (reactant) side — reduction is **gain** of electrons.

**O**xidation **I**s **L**oss, **R**eduction **I**s **G**ain (of electrons).