Reactivity 3. What are the mechanisms of chemical change?

Pair species that differ by exactly one H⁺ to identify conjugate acid–base pairs. Charge alone does not decide the role: in NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, NH₄⁺ is the proton donor and water is the acceptor.

NH₄⁺/NH₃ and H₂CO₃/HCO₃⁻ are conjugate pairs because each pair differs by one H⁺. Removing H⁺ lowers charge by one; adding H⁺ raises it by one. Do not pair species merely because they occur on opposite sides of an equation—trace the specific proton transfer.

For HCO₃⁻, donation gives CO₃²⁻ whereas acceptance gives H₂CO₃. Showing both reactions is the evidence for amphiprotic behaviour; one acid–base equation alone is insufficient.

For [H⁺] = 2.0 × 10⁻³ mol dm⁻³, pH = 2.70; the leading 2 makes the answer non-integer. A colour indicator estimates a range, whereas a calibrated probe supports a numerical measurement.

At 25 °C, Kw = 1.0 × 10⁻¹⁴, so a neutral solution has [H⁺] = [OH⁻] = 1.0 × 10⁻⁷ mol dm⁻³. Neutrality always means equal ion concentrations; neutral pH is not necessarily 7 when temperature changes.

At a fixed temperature, Kw is constant, so [OH-] = Kw/[H+]: a higher [H+] means a lower [OH-]. For example, at pH 9.3 and 25 C, [OH-] = 2.0 x 10^-5 mol dm^-3. Classify a solution from the ion comparison; do not assume neutral pH is 7 at every temperature.

Represent a strong acid with essentially complete ionization and a weak acid with an equilibrium containing substantial undissociated acid. Strength is an equilibrium property, while concentration is an initial amount per volume; pH depends on both, so strength alone cannot rank arbitrary solutions.

Balance proton capacity as well as atoms: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O, while an acid–carbonate reaction also releases CO₂. To identify parents of Na₂SO₄, trace SO₄²⁻ to the acid and Na⁺ to the base rather than treating the salt name as a reaction equation.

For a strong acid titrated with strong base at 25 °C, calculate the initial pH from excess acid, locate equivalence from stoichiometric moles, and place the steep section around pH 7. Equivalence is a mole condition; it is not the same as equal solution volumes unless concentrations and stoichiometry make it so.

In Zn + Cu²⁺ → Zn²⁺ + Cu, Zn rises from 0 to +2 and is oxidized, so it is the reducing agent; Cu²⁺ falls from +2 to 0 and is reduced, so it is the oxidizing agent. Name agents from what happens to them, not from the process they cause in the other species.

For MnO₄⁻ → Mn²⁺ in acid, balance O with 4H₂O, H with 8H⁺ and charge with 5e⁻: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. After combining halves, cancel electrons and any identical H⁺ or H₂O, then recheck both atoms and net charge.

Zinc displaces Cu²⁺ because Zn more readily oxidizes: Zn + Cu²⁺ → Zn²⁺ + Cu. Chlorine displaces Br⁻ because Cl₂ more readily reduces. Keep the metal and halogen trends in their correct electron directions instead of using one vague 'more reactive' rule.

Balance the electron transfer behind the molecular equation: metal atoms are oxidized and 2H⁺ + 2e⁻ → H₂ is the reduction. Use the metal charge and acid anion to construct the salt rather than assuming every metal forms a 2+ ion.

Name electrodes from the half-reactions before assigning signs. Electrons leave the anode and reach the cathode through the external circuit; a power supply reverses the polarities in an electrolytic cell but never changes where oxidation and reduction occur.

In a Zn|Zn²⁺ || Cu²⁺|Cu cell, Zn is oxidized at the negative anode and electrons travel through the wire to the positive Cu cathode, where Cu²⁺ is reduced. Salt-bridge anions migrate toward the anode compartment and cations toward the cathode compartment to prevent charge buildup; electrons do not flow through the bridge.

Write the discharge half-equations first. Charging a secondary cell requires an external potential to drive their reverse, whereas a primary cell is not designed for safe efficient reversal and a fuel cell continues only while reactants are supplied. Rechargeability is a reaction-design property, not simply the presence of a power socket.

Molten MgCl₂ contains only Mg²⁺ and Cl⁻: Mg²⁺ + 2e⁻ → Mg at the cathode and 2Cl⁻ → Cl₂ + 2e⁻ at the anode. The melt conducts by ion migration; do not introduce H₂, O₂ or water-based competition into a molten-salt question.

In a primary-alcohol experiment, distil the aldehyde as it forms to limit further oxidation; heat under reflux when the carboxylic acid is required. Tertiary alcohols lack the required hydrogen on the carbon bearing –OH and are not oxidized in the same way.

Track the carbon functional group rather than only the reagent: an aldehyde gives a primary alcohol and a ketone gives a secondary alcohol. Hydride supplies an electron-rich H unit to the carbonyl carbon; named reducing agents and detailed mechanisms are outside this objective.

Count π bonds to determine hydrogen demand: one mole of H₂ saturates one C=C, while full conversion of one C≡C to C–C needs two moles of H₂. Keep the carbon skeleton unchanged when drawing the product.

The dot in Cl· or CH₃· represents one unpaired electron, not a positive or negative charge. Radicals react readily because pairing that electron can form a bond; track the dot through every equation so electron and atom accounting remain explicit.

Draw two single-barbed arrows from the breaking X–X bond, one toward each atom, to account for both electrons. The UV or heat step creates radicals and is initiation; a step that consumes one radical and forms another belongs to propagation.

For methane chlorination, initiation forms 2Cl· from Cl₂ under UV. Propagation uses Cl· + CH₄ → HCl + CH₃· and CH₃· + Cl₂ → CH₃Cl + Cl·; termination combines two radicals. Further substitution creates a mixture, so the mechanism does not guarantee only CH₃Cl.

Look for an available electron pair: OH⁻ and CN⁻ use a negative charge and lone pair, while NH₃ uses a lone pair without being an anion. A curly arrow must start at that pair and point toward the atom where the new bond forms.

For CH₃CH₂Br + OH⁻, the C–O bond forms as the C–Br bond breaks, giving CH₃CH₂OH + Br⁻. Account for charge and every atom in the product; the leaving group takes the bonding pair rather than departing as a neutral bromine atom.

Place the curly-arrow tail on the bond being broken and its head on the fragment receiving both electrons. Then assign charges from electron ownership: heterolysis creates ions, unlike homolysis, which gives one electron to each radical.

Identify the electron-poor atom, not merely a positive-looking formula. H⁺ and carbocations are electrophiles, and the δ⁺ carbon in a polar C–X bond can also accept a pair. The incoming curly arrow ends at this acceptor.

Treat the C=C as the reactive site and place the two added groups on its two carbon atoms. Bromine addition removes the double bond and forms a dibromoalkane; hydration forms an alcohol. At SL, deducing these products does not require a mechanism.

In BF₃ + NH₃ → F₃B←NH₃, NH₃ donates the pair and is the Lewis base; BF₃ accepts it and is the Lewis acid. Classify the roles from electron-pair movement rather than from whether H⁺ appears.

Show a coordination bond with an arrow from a ligand lone pair to the metal ion. The arrow records the origin of the shared pair; after formation the bond is not a different electrostatic species from other covalent bonds.

Read [Cu(NH₃)₄]²⁺ as one Cu centre with four NH₃ ligands and coordination number 4. Use ligand charges and the overall bracket charge to deduce the metal oxidation state; do not confuse coordination number with oxidation state.