This "cheat sheet" provides a guide to answering subjective questions in Paper 4 for A2-Level Chemistry. Format your answers according to what is written in this webpage.
1. Born-Haber cycle for formation of ionic compound
Draw the diagram in a way that includes:
(1) ΔH atomisation for metal
(2) ΔH ionisation for metal
(3) ΔH atomisation for non-metal*
(4) ΔH electron affinity for non-metal
(5) Lattice energy
(6) ΔH of formation
* ΔH atomisation for Cl₂, F₂ and O₂ are replaced with bond energy as they are gases in their standard state.
2. Born-Haber cycle for solution
Draw the diagram in a way that includes:
* Lattice energy
* ΔH hydration for metal
* ΔH hydration for non-metal
* ΔH of solution
3. Why lattice energy of XCl > YCl
* X⁻ has a smaller ionic radius and thus has a higher charge density.
* Ions approach each other more closely and forces of attraction are stronger.
4. Why entropy change is +ve / -ve
* -ve because no. of moles of gas/liquid in reactants is higher than in products.
* Reverse argument if +ve.
5. Why reaction is feasible
* Gibbs free energy is -ve, so reaction is feasible.
6. Why activation energy is high in a reaction
* +ve and -ve/charged reactants repel each other.
7. Standard conditions in half-cell
* 1 atm, 298 K
8. Why electrochemical reaction is feasible
* State half-cell eqn & E⁰ value
* E⁰ value is -ve/+ve, so reaction is not feasible/feasible.
Example:
Zn → Zn²⁺ + 2e⁻ -1.69 V
Sn⁴⁺ + 2e⁻ → Sn²⁺ +1.22 V
-1.69 + 1.22 = -0.47 V
E⁰ value is -ve, so reaction is not feasible.
9. Drawing electrochemical cell
* Draw a diagram of an electrochemical cell. The diagram must include:
* Electrode X in a solution of X²⁺ (1 moldm⁻³)
* Salt bridge
* H₂ gas bubbling into a solution of H⁺ (1 moldm⁻³) with a Pt electrode
* 1 atm, 298 K
* Voltmeter
Note: If one of the half-cells is fully ionic (etc. Sn⁴⁺(aq)/Sn²⁺(aq)), you must draw a beaker with a platinum electrode immersed in a solution of 1moldm⁻³ Sn⁴⁺ and 1moldm⁻³ Sn²⁺.
10. Effect of changing conc. of solution on E° of half-cell
\( E_{\text{cell}} = E^\circ - \frac{0.059}{n} \log_{10} \frac{[\text{oxidised species}]}{[\text{reduced species}]} \)
[conc. of oxidised species increases]
*Equilibrium shifts to the right
*e- are taken up at anode/cathode
*\(E_{\text{cell}}\) becomes more +ve
Reverse argument if conc. of oxidised ↓ or conc. of reduced reduced ↑
11. Transition metal characteristics
*Can form catalyst/complex ions, because it has partially empty d-orbitals which can accept lone pairs from reactants and form dative bonds.
*Can show colour
*Can form stable ions with partially full d subshells.
*Multiple oxidation states because 4s & 3d orbitals have a similar energy level.
*hard, ductile, shiny
12. How transition metals show colour
*When ligands form dative bonds with transition metal ion, energy of nearby d-orbitals are raised.
*Orbitals split into splitting occurs to form degenerate orbitals.
*e- absorbs λ of light energy corresponding to the energy gap and are promoted to a higher energy level.
*Remaining λ combine and the complementary colour is seen.
[If asked to explain why different metals have different colours]
*Different metals/ligands have different energy gap so diff. complementary colour is seen.
13. Why certain complex ions are polar/nonpolar
*Net dipole is present/dipoles cancel out so no net dipole.
*Cr/other ligands are on the same side.
14. Colour of transition metal complexes
$\begin{array}{|c|c|}\hline\textbf{Complex Ion} & \textbf{Color} \\\hline[\text{Cu}(\text{H}_2\text{O})_6]^{2+} & \text{blue solution} \\\hline[\text{Cu}(\text{OH})_2(\text{H}_2\text{O})_4] & \text{pale blue ppt} \\\hline[\text{Cu}(\text{H}_2\text{O})_2(\text{NH}_3)_4]^{2+} & \text{deep blue solution} \\\hline[\text{CuCl}_4]^{2-} & \text{yellow solution} \\\hline[\text{Co}(\text{H}_2\text{O})_6]^{2+} & \text{pink solution} \\\hline[\text{Co}(\text{OH})_2(\text{H}_2\text{O})_4] & \text{blue ppt} \\\hline[\text{Co}(\text{NH}_3)_6]^{3+} & \text{brown solution} \\\hline[\text{CoCl}_4]^{2-} & \text{blue solution} \\\hline \end{array}$
15. Concentration of Different Transition Metal Complexes / Stability of Different Complexes
* X has highest conc as it has highest Kstab so equilibrium is furthest to the right.
* X > Y > Z
16. Trends in Thermal Stability of Nitrates / Carbonates
* Going down Group 2, thermal stability increases.
* Charge density decreases because ionic radius increases.
* NO₃⁻ / CO₃²⁻ anion is less polarized.
17. Trends in Solubility of SO₄²⁻ / OH⁻
* Going down Group 2, solubility of sulfates decreases.
* ΔH(sol) becomes more endothermic and ΔH(hyd) becomes more exothermic.
* Sulfate decreases at a lower rate than ΔH(hyd).
* Reverse argument for OH⁻.
18. Half-life of Reaction
* Reaction is first order → constant half-life.
19. Factors Affecting Rate of Reaction
* Temperature → frequency of effective collisions → higher rate of reaction → value of k increases.
* Pressure → particles closer to each other.
* Concentration → equilibrium shifts to the left/right so rate of the forward/backward reaction increases.
20. Comparison of Strengths as an acid
[Benzene]
* Benzoic group is e⁻ withdrawing.
* Lone pair on O atom is delocalized onto ring.
* Conjugate base more stable, stronger acid.
[Halogens]
* Halogen atom is electronegative so it draws e⁻ towards itself.
* Negative charge drawn away from O atom / -COO group.
* Conjugate base more stable, stronger acid.
[Alkyl Group]
* Allyl group is electron releasing.
* Electrons released towards O atom.
* Attracts H⁺ more strongly, weaker acid.
Note: No. of alkyl groups is more important than length of group.
[Carboxylic Acid vs. Alcohol]
* Carboxylic acid stronger acid than alcohol.
* p-orbitals of C and O atoms overlap to form a delocalized system.
* Conjugate base more stable, stronger acid.
21. Comparison of Strengths as a base
Bases are usually amines and amides.
[Benzene]
* Benzene is e⁻ withdrawing
* Low pair of e⁻ delocalized onto ring
* Weaker base because Nitrogen is less -ve / less able to donate low pair to H⁺
[Alkyl group]
* e⁻ releasing
* e⁻ charge is released towards N atom
* Stronger base because more able to accept H⁺
[Halogen]
* e⁻ withdrawing
* Electrons are drawn away from N atom
* Weaker base because less able to accept H⁺
[Amide vs amine]
* e⁻ are delocalised because p-orbitals of C, N and O overlap
* Electron density on N atom decreases
* Weaker base as less able to accept H⁺ ions.
21. Amino acid electrophoresis
[example image]
* Isoelectric point of X = 3.0
* Isoelectric point of Y = 2.0
Example solution:
* X is +vely charged at pH 2 as it accepts H⁺ ions to form NH₃⁺ group. Thus, it moves towards the -ve terminal.
* X moves further than X-Y as it has a lower Mr.
* Y does not move as it exists as a zwitterion at pH 2.0.
22. How bond angles come about in benzene?
* Bond angles are 120°
* Carbon atoms are sp² hybridized.
* Form 2 σ bonds with neighboring carbons and a σ bond with an H atom.
* p-orbitals overlap to form delocalized aromatic system.
23. Naming reactions/mechanism
[Reactions]
* Hydrogenation
* Oxidation
* Friedel-Crafts alkylation/acylation
* Halogenation
* Free radical substitution
* Reduction
* Hydrolysis
* Dehydration
* Esterification
[Mechanisms (If a question asks for the reaction name, these can also be used)]
* Nucleophilic addition/substitution/elimination
* Electrophilic addition/elimination/addition/elimination
* Ligand substitution
24. Differentiating btw HCOOH / (COOH)₂ and other carboxylic acids
* Test with Tollen's / Fehling's reagent
* Silver mirror / brick-red are formed for HCOOH and (COOH)₂. However, other acids will not show any chemical reactions.
* HCOOH and (COOH)₂ can be oxidized into CO₂ and H₂O