Physical Chemistry

Physical chemistry carries the largest share of the Chemistry paper (our estimate: about 9 of 22 questions) and is the most numerical part of it. The eight syllabus topics reduce to a handful of calculation types: mole and stoichiometry, a gas-law problem, quantum numbers and configurations, oxidation numbers, pH and buffers, titration, Faraday and cell potential, and rate or equilibrium. The official booklet's own chemistry samples (a limiting-reagent count of HCl molecules, the mass of copper deposited by 2 faradays, the number of electrons in a p-orbital) are all of these types.

1.1 Chemical arithmetic

Dalton's atomic theory: atoms are indivisible (modified by subatomic particles and isotopes), atoms of one element are identical, compounds form in simple whole-number ratios. Laws of chemical combination: conservation of mass (Lavoisier); definite proportions (Proust); multiple proportions (Dalton: CO and CO₂ have O in 1 : 2 for a fixed C); reciprocal proportions (Richter); Gay-Lussac's law of combining volumes; Avogadro's hypothesis: equal volumes of gases at the same T and P hold equal numbers of molecules.

Relative atomic mass is on the carbon-12 scale (1 u =\frac{1}{12} of C-12 =1.66\times10^{-27} kg). Mole: 6.022\times10^{23} entities. n=\frac{m}{M}=\frac{N}{N_A}=\frac{V_{STP}}{22.4\ \text{L}}; molar volume 22.4 L at STP (0 °C, 1 atm; 22.7 L at 1 bar) and 24.6 L at 27 °C, 1 atm. Gas density \rho=\frac{PM}{RT}; vapour density =\frac{M}{2}.

Empirical formula: divide the percentages by atomic masses, divide by the smallest, clear fractions. 40 % C, 6.7 % H, 53.3 % O gives CH₂O; molecular formula =n\times empirical, n=\frac{M}{\text{empirical mass}}. Limiting reagent: compute the product from each reactant; the smaller yield decides. Percentage yield =\frac{\text{actual}}{\text{theoretical}}\times100.

Equivalent mass E=\frac{M}{n\text{-factor}}: element \frac{\text{atomic mass}}{\text{valency}}; acid \frac{M}{\text{basicity}} (H₂SO₄: 49); base \frac{M}{\text{acidity}}; oxidant/reductant \frac{M}{\text{electrons exchanged}} (KMnO₄ in acid: \frac{158}{5}=31.6). Normality N=\frac{w}{E\,V(L)}, N=M\times n-factor. Concentration: molarity, molality \frac{n}{\text{kg solvent}}, mole fraction, % w/w, ppm.

1.2 States of matter

Gas laws. Boyle P_1V_1=P_2V_2; Charles \frac{V_1}{T_1}=\frac{V_2}{T_2}; Gay-Lussac \frac{P}{T} constant; combined; ideal gas PV=nRT, R=0.0821 L atm K⁻¹ mol⁻¹ =8.314 J K⁻¹ mol⁻¹. Dalton: P_{total}=\sum P_i, P_i=x_iP (2 mol He + 3 mol Ne at 5 atm: P_{He}=2 atm). Graham: \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}} (H₂ effuses \sqrt{8}=2.83 times as fast as CH₄). Kinetic theory: \bar K=\frac{3}{2}kT; v_{rms}=\sqrt{\frac{3RT}{M}}; v_{mp}:v_{av}:v_{rms}=1:1.128:1.224.

Real gases: van der Waals \big(P+\frac{an^2}{V^2}\big)(V-nb)=nRT; a measures attraction, b the excluded volume; deviations are greatest at high P and low T; Z=\frac{PV}{nRT}; critical T_c=\frac{8a}{27Rb}; Joule–Thomson cooling. Liquids: vapour pressure rises with T; boiling at P_{vap}=P_{ext}; surface tension and viscosity fall with T; strength of intermolecular forces (hydrogen bonding in H₂O, HF, NH₃) raises boiling points. Solids: crystalline (ionic, covalent, metallic, molecular) and amorphous; unit cells: simple cubic (1 atom, 52 % packing), bcc (2 atoms, 68 %), fcc (4 atoms, 74 %); coordination 6, 8, 12; Bragg's n\lambda=2d\sin\theta; defects (Schottky, Frenkel).

1.3 Atomic structure and periodic classification

Subatomic particles: electron (Thomson), nucleus (Rutherford α-scattering), neutron (Chadwick). Bohr: E_n=-\frac{13.6}{n^2} eV, r_n=0.529\,n^2 Å. Quantum numbers: n (shell, 1, 2, …), l=0\ldots n-1 (subshell s, p, d, f), m_l=-l\ldots+l (2l+1 orbitals: s 1, p 3, d 5, f 7), m_s=\pm\frac{1}{2}. A shell holds 2n^2 electrons; a p-orbital holds at most two electrons of opposite spin. Radial nodes n-l-1, angular nodes l. de Broglie \lambda=\frac{h}{m}v, uncertainty principle.

Filling rules: Aufbau (n+l rule: 4s before 3d), Pauli (no two electrons with four equal quantum numbers), Hund (maximise unpaired spins in degenerate orbitals). Exceptions: Cr 3d⁵4s¹, Cu 3d¹⁰4s¹. Cations lose 4s before 3d: Fe²⁺ = [Ar]3d⁶, Fe³⁺ = 3d⁵. Magnetic moment \sqrt{n(n+2)} BM.

Periodic table: modern law (atomic number). s, p, d, f blocks. Trends: atomic radius falls across a period and rises down a group; ionisation energy rises across a period (exceptions Be > B, N > O) and falls down a group; electronegativity (Pauling) F 4.0 highest; electron affinity Cl > F; metallic character falls across. Isoelectronic ions: radius falls as Z rises (N³⁻ > O²⁻ > F⁻ > Na⁺ > Mg²⁺). Diagonal relationships: Li–Mg, Be–Al, B–Si.

1.4 Oxidation, reduction and equilibrium

Oxidation number rules: free element 0; H +1 (hydrides -1); O -2 (peroxides -1, OF₂ +2); sum equals the charge. Mn in KMnO₄ +7; Cr in K₂Cr₂O₇ +6; S in H₂SO₄ +6, Na₂S₂O₃ +2; N in HNO₃ +5, NH₄⁺ -3. Redox: oxidation is loss of electrons (increase in oxidation number); the oxidising agent is reduced. Disproportionation: Cl₂ + 2NaOH → NaCl + NaClO + H₂O. Balancing: ion-electron (half-reaction) or oxidation-number method. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.

Chemical equilibrium: K_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}; K_p=K_c(RT)^{\Delta n_g}; reaction quotient Q vs K. Le Chatelier: increasing pressure favours fewer gas moles (N₂ + 3H₂ ⇌ 2NH₃ shifts right); heating favours the endothermic direction; a catalyst changes the rate only. N₂O₄ ⇌ 2NO₂ at 20 % dissociation of 1 mol L⁻¹: K_c=\frac{0.4^2}{0.8}=0.2.

1.5 Volumetric analysis

Titration finds an unknown concentration: N_1V_1=N_2V_2 (or M_1V_1n_1=M_2V_2n_2 by mole ratio). Primary standards: oxalic acid, Na₂CO₃, K₂Cr₂O₇. Indicators: phenolphthalein (8.3–10, weak acid–strong base), methyl orange (3.1–4.4, strong acid–weak base), methyl red. KMnO₄ is a self-indicator (acidify with dilute H₂SO₄ only). Iodometry with starch (blue at the end point). Example: 25 mL of 0.1 M NaOH neutralised by 20 mL H₂SO₄ means M\times20\times2=2.5, M=0.0625 M. Titration curve of a strong acid with a strong base has a sharp jump at pH 7. EDTA titrations measure hardness.

1.6 Ionic equilibrium, acid, base and salt

Theories: Arrhenius; Brønsted–Lowry (proton donor/acceptor, conjugate pairs); Lewis (electron-pair acceptor/donor). K_w=[H^+][OH^-]=10^{-14} at 25 °C; pH =-\log[H^+]; pH + pOH =14. 0.01 M HCl: pH 2. Weak acid: K_a=\frac{[H^+][A^-]}{[HA]}, [H^+]=\sqrt{K_aC} (0.1 M CH₃COOH, K_a=1.8\times10^{-5}: pH 2.87); \alpha=\sqrt{\frac{K_a}{C}} (Ostwald). Mixtures: compute excess moles: 100 mL 0.1 M HCl + 100 mL 0.05 M NaOH leaves 0.005 mol in 0.2 L, pH 1.6.

Buffer (weak acid + its salt): \text{pH}=\text{p}K_a+\log\frac{[\text{salt}]}{[\text{acid}]} (Henderson); with ratio 10 and pK_a=4.76: 5.76. Best at pH \approx pK_a; blood buffer H₂CO₃/HCO₃⁻. Salt hydrolysis: strong acid + weak base gives acidic solution (NH₄Cl); weak acid + strong base basic (CH₃COONa); strong + strong neutral. Common-ion effect lowers solubility. Solubility product K_{sp}: AgCl s=\sqrt{K_{sp}}=1.34\times10^{-5} M for K_{sp}=1.8\times10^{-10}; precipitate forms when ionic product >K_{sp}; for A₂B s=\sqrt{[}3]{K_{sp}/4}.

1.7 Electrochemistry

Faraday's laws: m=ZIt=\frac{E}{96500}It; one faraday (96 485 C) deposits one equivalent (so 2 F deposits 63.5 g of Cu from Cu²⁺; 3 F per mole of Al). 2 A for 965 s (0.02 F) gives 0.635 g Cu. Electrolysis products: the species with the higher reduction potential is reduced at the cathode (copper over hydrogen from CuSO₄); in brine Cl₂ at the anode, H₂ at the cathode, NaOH in solution. Conductance: \kappa, \Lambda_m=\frac{\kappa\times1000}{M}, Kohlrausch's law for weak electrolytes.

Galvanic cell: anode (oxidation, negative), cathode (reduction, positive), salt bridge. E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}: Zn–Cu (Daniell) =0.34-(-0.76)=1.10 V. Nernst: E=E^\circ-\frac{0.0591}{n}\log Q at 25 °C. \Delta G^\circ=-nFE^\circ (for n=2, 1.10 V: -212 kJ); \log K=\frac{nE^\circ}{0.0591}. A more negative E^\circ means a stronger reducing agent (Li, K, Na, Zn, Fe, H₂, Cu, Ag, Au). Batteries: dry cell, lead-acid (PbSO₄), Li-ion; corrosion (rusting) and its prevention (galvanising, sacrificial anode); fuel cells.

1.8 Energetics, kinetics and bonding

First law, enthalpy: \Delta U=q+w; \Delta H=\Delta U+\Delta n_gRT; Hess's law (path independence); \Delta H^\circ_{rxn}=\sum\Delta H_f(\text{products})-\sum\Delta H_f(\text{reactants}); for CH₄ + 2O₂: -890=-393.5+2(-285.8)-\Delta H_f(\text{CH}_4), so \Delta H_f=-75 kJ mol⁻¹. Bond-energy method \Delta H=\sum D(\text{broken})-\sum D(\text{formed}). Entropy and free energy: \Delta G=\Delta H-T\Delta S; spontaneous if \Delta G<0; \Delta G^\circ=-RT\ln K.

Kinetics: rate =k[A]^m[B]^n, order =m+n (rate =k[A]^2[B] is third order); molecularity is a whole number for an elementary step. First order: [A]=[A]_0e^{-kt}, t_{1/2}=\frac{0.693}{k} (for k=10^{-3} s⁻¹: 693 s), independent of concentration; zero order t_{1/2}\propto[A]_0. Arrhenius: k=Ae^{-E_a/RT}, \ln\frac{k_2}{k_1}=\frac{E_a}{R}\big(\frac{1}{T_1}-\frac{1}{T_2}\big); a catalyst lowers E_a; the rate roughly doubles per 10 K.

Bonding: ionic (lattice energy, Born–Haber), covalent (Lewis, octet and its exceptions: BF₃, PCl₅, SF₆), coordinate, metallic, hydrogen bond. VSEPR / hybridisation:

Species	Electron pairs	Hybrid	Shape
BeCl₂, CO₂, C₂H₂	2	sp	linear
BF₃, C₂H₄, SO₃	3	sp²	trigonal planar
CH₄, NH₄⁺	4	sp³	tetrahedral
NH₃	3 + 1 lp	sp³	pyramidal (107^\circ)
H₂O	2 + 2 lp	sp³	bent (104.5^\circ)
PCl₅	5	sp³d	trigonal bipyramidal
SF₆	6	sp³d²	octahedral

Bond order from MO theory: O₂ has two unpaired electrons (paramagnetic), N₂ has the highest bond order (3). Ethyne has 3\sigma+2\pi bonds. Dipole moments: BF₃, CO₂, CH₄ are non-polar though bonds are polar; H₂O, NH₃ are polar.

Formula and fact sheet

Topic	Result
mole	n=\frac{m}{M}=\frac{N}{N_A}=\frac{V}{22.4}
ideal gas	PV=nRT
Graham	\frac{r_1}{r_2}=\sqrt{M_2/M_1}
orbitals	s,p,d,f=1,3,5,7
pH	-\log[H^+]; buffer pK_a+\log\frac{\text{salt}}{\text{acid}}
Faraday	m=\frac{E\,I\,t}{96500}
cell	E^\circ=E^\circ_c-E^\circ_a; \Delta G=-nFE
first-order	t_{1/2}=0.693/k
Gibbs	\Delta G=\Delta H-T\Delta S

Chapter summary

• Convert everything to moles first; the limiting reagent decides the yield.
• Quantum numbers fix orbitals: 2l+1 per subshell; cations lose 4s before 3d.
• Count oxidation-number change to find n-factor and equivalent mass.
• For acid–base mixtures find the excess; weak acids use \sqrt{K_aC}; buffers use Henderson.
• One faraday = one equivalent; E^\circ_{cell} is cathode minus anode.
• Order is read from the rate law, not the equation; first-order half-life is constant.
• Shape comes from electron pairs including lone pairs.

Practice questions

Set A — Recall (16 questions)

1. The number of moles in 36 g of water isa2b1c4d0.5
2. The number of molecules in 0.5 mol of a substance isa6.02\times10^{23}b3.01\times10^{23}c1.5\times10^{23}d12.04\times10^{23}
3. The volume occupied by one mole of an ideal gas at STP isa11.2 Lb24.5 Lc22.4 Ld44.8 L
4. A compound has 40 % C, 6.7 % H and 53.3 % O. Its empirical formula isaCHObC₂H₄O₂cCH₄OdCH₂O
5. The equivalent mass of H₂SO₄ (M = 98) as an acid isa49b98c32.7d24.5
6. A gas occupies 4 L at 1 atm. At constant temperature and 2 atm it occupiesa8 Lb2 Lc4 Ld0.5 L
7. The number of orbitals in a d subshell isa3b7c5d1
8. The electronic configuration of Fe²⁺ (Z = 26) isa[Ar]3d⁵4s¹b[Ar]3d⁴4s²c[Ar]3d⁸d[Ar]3d⁶
9. The pH of 0.01 M HCl isa2b1c12d0.01
10. The oxidation number of Mn in KMnO₄ isa+2b+7c+4d+6
11. The charge required to deposit one mole of aluminium from Al³⁺ isa1 Fb2 Fc3 Fd6 F
12. For the rate law rate =k[A]^2[B], the overall order isa1b2c0d3
13. The carbon atoms in ethene areasp² hybridisedbsp hybridisedcsp³ hybridiseddsp³d hybridised
14. The shape of the NH₃ molecule isaplanarbpyramidalctetrahedraldlinear
15. For an exothermic reaction \Delta H isapositivebzerocnegativedinfinite
16. Increasing the pressure on N₂ + 3H₂ ⇌ 2NH₃ shifts the equilibrium to thealeftbno changecneither sidedright

Answers and explanations — Set A

1. (a) — \frac{36}{18}=2.

2. (b) — 0.5\times6.02\times10^{23}.

3. (c) — 22.4 L (24.5 L is at 25 °C, 1 atm).

4. (d) — C : H : O =\frac{40}{12}:\frac{6.7}{1}:\frac{53.3}{16}=3.33:6.7:3.33=1:2:1.

5. (a) — \frac{98}{2} (basicity 2).

6. (b) — P_1V_1=P_2V_2: V_2=\frac{4\times1}{2}.

7. (c) — 2l+1=5.

8. (d) — Fe is [Ar]3d⁶4s²; the 4s electrons go first.

9. (a) — -\log10^{-2}.

10. (b) — +1+x-8=0.

11. (c) — \text{Al}^{3+}+3e^-\to\text{Al}.

12. (d) — 2+1.

13. (a) — trigonal planar, three σ bonds around each carbon.

14. (b) — three bond pairs and one lone pair (trigonal pyramidal).

15. (c) — heat is released.

16. (d) — 4 mol of gas become 2 mol; Le Chatelier.

Set B — Understanding (10 questions)

1. 21 g of MgCO₃ is treated with 50 mL of 12 M HCl. The number of molecules of unreacted HCl isa6.02\times10^{23}b6.02\times10^{22}c6.02\times10^{21}d6.02\times10^{19}
2. 2 g of hydrogen and 16 g of oxygen react to form water. The mass of water formed isa9 gb20 gc36 gd18 g
3. The number of radial nodes in a 3p orbital isa0b2c1d3
4. 100 mL of 0.1 M HCl is mixed with 100 mL of 0.05 M NaOH. The pH of the mixture is nearlya1.6b1.0c2.0d7
5. An acetic acid–acetate buffer (pK_a=4.76) has salt : acid =10:1. Its pH isa3.76b5.76c4.76d6.76
6. For AgCl, K_{sp}=1.8\times10^{-10}. Its solubility in pure water is abouta1.8\times10^{-10} Mb9\times10^{-6} Mc1.8\times10^{-5} Md1.34\times10^{-5} M
7. A first-order reaction has k=1\times10^{-3} s⁻¹. Its half-life isa69.3 sb6930 sc693 sd0.693 s
8. Given E^\circ(\text{Zn}^{2+}/\text{Zn})=-0.76 V and E^\circ(\text{Cu}^{2+}/\text{Cu})=+0.34 V, the standard Daniell cell potential isa1.10 Vb0.42 Vc-1.10 Vd0.76 V
9. The volume of one mole of an ideal gas at 27 °C and 1 atm isa22.4 Lb24.6 Lc27.3 Ld30 L
10. The number of σ and π bonds in ethyne (C₂H₂) area3σ and 2πb5σ and 0πc2σ and 3πd4σ and 1π

Answers and explanations — Set B

1. (b) — 0.25 mol MgCO₃ uses 0.5 mol HCl of 0.6 mol: 0.1 mol =6.02\times10^{22} left.

2. (d) — 1 mol H₂ and 0.5 mol O₂ are in exact ratio 2 : 1: 1 mol H₂O = 18 g.

3. (c) — n-l-1=3-1-1.

4. (a) — excess H⁺ =0.010-0.005=0.005 mol in 0.2 L =0.025 M; pH =1.60.

5. (b) — 4.76+\log10.

6. (d) — s=\sqrt{1.8\times10^{-10}}=1.34\times10^{-5}.

7. (c) — \frac{0.693}{10^{-3}}.

8. (a) — 0.34-(-0.76).

9. (b) — \frac{nRT}{P}=\frac{0.0821\times300}{1}=24.6.

10. (a) — H–C≡C–H: two C–H σ, one C–C σ, two π.

Set C — Application (8 questions)

1. Given \Delta H_c(\text{CH}_4)=-890, \Delta H_f(\text{CO}_2)=-393.5 and \Delta H_f(\text{H}_2\text{O}_l)=-285.8 kJ mol⁻¹, the enthalpy of formation of CH₄ isa+75 kJ mol⁻¹b-890 kJ mol⁻¹c-75 kJ mol⁻¹d-965 kJ mol⁻¹
2. For a cell with n=2 and E^\circ=1.10 V, \Delta G^\circ is nearlya-106 kJb-212 kJc-424 kJd-96.5 kJ
3. 1 mol of N₂O₄ in 1 L is 20 % dissociated into NO₂ at equilibrium. K_c isa0.05b0.4c0.8d0.2
4. 25 mL of 0.1 M NaOH is neutralised by 20 mL of H₂SO₄. The molarity of the acid isa0.125 Mb0.0625 Mc0.25 Md0.05 M
5. A current of 2 A is passed for 965 s through a copper(II) sulphate solution. The mass of copper deposited isa0.635 gb0.3175 gc1.27 gd6.35 g
6. The pH of 0.1 M acetic acid (K_a=1.8\times10^{-5}) is abouta1.0b4.74c3.37d2.87
7. The ratio of the rates of diffusion of H₂ to CH₄ under the same conditions isa2b4c2.83d8
8. A mixture has 2 mol He and 3 mol Ne at a total pressure of 5 atm. The partial pressure of He isa2 atmb3 atmc2.5 atmd5 atm

Answers and explanations — Set C

1. (c) — -890=-393.5+2(-285.8)-\Delta H_f: \Delta H_f=-75.1.

2. (b) — -2\times96485\times1.10=-212 kJ.

3. (d) — [\text{N}_2\text{O}_4]=0.8, [\text{NO}_2]=0.4: \frac{0.16}{0.8}.

4. (b) — 2.5 mmol OH⁻ =2\times M\times20.

5. (a) — Q=1930 C =0.02 F \to0.01 mol Cu =0.635 g.

6. (d) — \sqrt{1.8\times10^{-6}}=1.34\times10^{-3}, pH 2.87.

7. (c) — \sqrt{\frac{16}{2}}.

8. (a) — \frac{2}{5}\times5.