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Chemistry Solutions (Multiple Choice), Papers of Chemistry

California Institute of Technology (Caltech)Chemistry

A-Level Chemistry teachers. Please find below a compilation of hard Chemistry Qs and Ans. Physical, Organic, Practical and MC. Pls RT if you find them useful

Typology: Papers

2024/2025

Available from 10/19/2024

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Chemistry Solutions (Multiple Choice)
Section A: Fast
1. Answer: 18 electrons, 19 protons, 22 neutrons
Working: Element K → proton number 19
Mass number 41 → neutron number 41 - 19 = 22
Ion +1 → 1 less electron → 18 electrons
2. Answer: [Ar] 3d9
Working: [Cu] = [Ar] 4s1 3d10 (copper and chromium are exceptions to Aufbau
principle)
Electrons are lost from 4s before 3d:
→ [Cu2+] = [Ar] 3d9
3. Answer: 1.8 x 1024
Working: Moles = 6/(14 + 14) = 0.2143
Molecules = 0.2143 * 6.02 * 1023 = 1.29 x 1023
Protons = 7 * 2 * 1.29 * 1023 = 1.806 x 1024 = 1.8 x 1024 (2 s.f.)
4. Answer: 7.76 x 10-2 mol dm-3
Working: Moles PbCl2 = 1.08/278.2 = 3.882 x 10-3
Mol Cl- = 2 * 3.882 x 10-3 = 7.764 x 10-3
[Cl-] = (7.764 x 10-3) / (100 / 1000) = 7.76 x 10-2 mol dm-3
5. Answer: butan-2-ol
Working: Butan-2-ol (hydrogen bonding) > butanal (induced dipole forces)
> 1-fluorobutane (induced dipole forces) > E-but-2-ene (VDW)
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Chemistry Solutions (Multiple Choice)

Section A: Fast

  1. Answer: 18 electrons, 19 protons, 22 neutrons Working: Element K → proton number 19 Mass number 41 → neutron number 41 - 19 = 22 Ion +1 → 1 less electron → 18 electrons
  2. Answer: [Ar] 3d^9 Working: [Cu] = [Ar] 4s^1 3d^10 (copper and chromium are exceptions to Aufbau principle) Electrons are lost from 4s before 3d: → [Cu2+] = [Ar] 3d^9
  3. Answer: 1.8 x 10^24 Working: Moles = 6/(14 + 14) = 0. Molecules = 0.2143 * 6.02 * 10^23 = 1.29 x 10^23 Protons = 7 * 2 * 1.29 * 10^23 = 1.806 x 10^24 = 1.8 x 10^24 (2 s.f.)
  4. Answer: 7.76 x 10-2^ mol dm- Working: Moles PbCl 2 = 1.08/278.2 = 3.882 x 10- Mol Cl-^ = 2 * 3.882 x 10-3^ = 7.764 x 10- [Cl-] = (7.764 x 10-3) / (100 / 1000) = 7.76 x 10-2^ mol dm-
  5. Answer: butan-2-ol Working: Butan-2-ol (hydrogen bonding) > butanal (induced dipole forces) > 1-fluorobutane (induced dipole forces) > E-but-2-ene (VDW)
  1. Answer: (CH 3 ) 2 CO Working: CCl 4 and CO 2 are symmetrical molecules and so their dipole moments cancel out. C 2 F 4 is also symmetrical on both sides of the double bond. Due to the CO bond in CH 3 COCH 3 , there is a permanent dipole.
  2. Answer: Li (s) + ½ F 2 (g) → LiF (s) Working: Enthalpy of formation: 1 mole of a compound formed in its standard state from its elements in their standard states.
  3. Answer: rate = k [ X ] [ Y ] Working: (1): rate = k [ W ]^2 [ X ]: false since would be x8. (2): rate = k [ W ]^2 [ Y ]: false since also would be x (4): rate = k [ X ] [ Z ]: false since cannot typically include product in rate equation. By process of elimination, third option is correct.
  4. Answer: mol-2^ dm^6 s- Working: Third order equation has the form: rate = k [reactants]^3 [k] = [rate] / [concentration]^3 = mol dm-3^ s-1^ / (mol dm-3)^3 [k] = mol-2^ dm^6 s-
  5. Answer: Low pressure and high temperature Working: Enthalpy change positive → endothermic → high temperature shifts equilibrium towards products. 3 gas molecules in products vs 2 in reactants → low pressure shifts equilibrium towards products
  1. Answer: methyl methanoate Working: Methanol: CH 3 OH = CH 4 O (no match) Methyl methanoate: HCOOCH 3 = C 2 H 4 O = 2 CH 2 O (match) Ethane-1,2-diol: CH 2 OHCH 2 OH = C 2 H 6 O 2 = 2 CH 3 O (no match) Butanal: CH 3 CH 2 CH 2 CHO = C 4 H 8 O (no match)
  2. Answer: 3 Working: 1: electron pair from nucleophile attacks partial positive C (correct) 2: elimination of bromide (correct) 3: N+-H electrons should move to N not H to resolve + charge 4: deprotonation by another ammonia molecule (correct)
  3. Answer: 2 Working: Ethanolic potassium hydroxide → elimination mechanism Elimination requires a hydrogen to be removed from an adjacent carbon from the halogen carbon: But the 1 and 2 are the same product → 2 alkenes.
  4. Answer: ethylamine > ammonia > phenylamine Working: Basicity of amines: 2o^ > 1o^ > NH 3 (inductive effect) and NH 3 > aromatic (delocalisation) so ethylamine (1o) > ammonia > phenylamine (aromatic)
  1. Answer: ClF 4 - Working: CH 4 , NH 4 +^ and AlCl 4 - : tetrahedral (bond angle 109.5o) ClF 4 - : 6 pairs (4 b.p., 2 l.p.) → square planar → 90o
  2. Answer: CH 3 OH Working: HCHO: planar carbonyl group CH 3 +: trigonal planar CH 3 OH: free rotation of O-H and C is tetrahedral → not planar C 2 H 4 : restricted rotation and trigonal planar
  3. Answer: Ammonium chloride Working: A + NaOH → NH 3 + NaCl + H 2 O A must contain nitrogen and chlorine as it is not on the other side. It also likely contains lots of H since there is only 1 on the left and 5 on the right → ammonium. Likely does not contain O since already has 1:1 reacting ratio. → ammonium chloride seems reasonable. Check: NH 4 Cl + NaOH → NH 3 + NaCl + H 2 O → simple 1:1 ratio balances. (This could also be deduced by knowledge of acid-base reactions.)
  4. Answer: boron Working: Reducing agent gets oxidised → loses e-^ / loses H / gains O Boron (B) gains O to B 2 O 3.
  5. Answer: 3-bromohexane Working: Secondary haloalkane: carbon bonded to halogen is bonded to 2 other carbons.
  1. Answer: propanal and copper(I) oxide Working: The reactions are: (By Markovnikov’s rule, primary carbocation is minor.) (Alkaline CuSO 4 is Fehling’s reagent.) The red precipitate indicates the presence of a transition metal (Cu) ion. Cu2+^ is blue and Cu2+^ must have been reduced (since it is an oxidising agent) so the only possible ion is Cu+. Cannot be copper(I) dichromate since all chromium was used up (it was limiting in the mild oxidation conditions) so must be oxide (Cu 2 O).
  2. Answer: 1 mole of each electrode metal Working: Standard conditions are: Temperature 298 K, 101 kPa for gases, 1 mol dm-3^ for solutions.
  3. Answer: A and B only Working: α-helices and β-pleated sheets are formed by hydrogen bonds in different geometries. Disulfide bridges are -S-S- bonds.
  1. Answer: CH 3 OCH 3 Working: PCl 5 , C 2 H 4 and CO 2 are symmetrical. In CH 3 OCH 3 , the C-O-C bond is non-linear (~104.5o) so there is a net dipole moment towards the O-side of the molecule.
  2. Answer: Calcium sulfide Working: Most exothermic → most stable (strongest ionic bonds). LiF: small, singly charged ions. LiCl: singly charged ions with covalent character → stronger Magnitude of charge has greater effect than size. CaO: doubly charged ions CaS: doubly charged ions with covalent character → strongest.
  3. Answer: 1 and 2 only Working: LiAlH 4 is a stronger reducing agent than NaBH 4. We know that LiAlH 4 and NaBH 4 both reduce aldehydes/ketones and only LiAlH 4 can reduce carboxylic acids. Without knowledge of reduction of esters, the answer can be deduced.
  4. Answer: phosphate Working: Carbonate: CO 3 2-. Phosphate: PO 4 3-. Sulfide: S2-. Hydrogencarbonate: HCO 3 -.
  1. Answer: fluoride Working: Enthalpy of hydration: X-^ (g) → X-^ (aq) Most exothermic → most stable in solution → forms strongest bonds with water molecules. Fluoride forms strongest ion-dipole forces with partial positive hydrogens.
  2. Answer: Li | Li+^ || Li+, CoO 2 | LiCoO 2 | Pt Working: Equations are: Li+^ + CoO 2 + e-^ → Li+[CoO 2 ]-^ (reduction) Li → Li+^ + e-^ (oxidation) Combining: Li + Li+^ + CoO 2 → Li+[CoO 2 ]-^ + Li+ Cancelling out Li+, Li + CoO 2 → LiCoO 2 Forming the cell, both solutions must contain Li+: Li | Li+^ || CoO 2 , Li+^ | LiCoO 2 | Pt (since aqueous phases)
  3. Answer: triethylammonium chloride Working: This is an ionic compound → quaternary ammonium salt.
  4. Answer: 2 n - 1 Working: When n = 1 (simple reaction), 1 molecule of water is lost. When one of each reactant is added to this product, waters are lost from both ends (total 2 per additional molecule) → 2n - 1 waters. (1, 3, 5 … etc)
  5. Answer: Cu2+^ ions are reduced to Cu+^ ions. Working: Cu oxidation state before: + After positive test result, colour change → oxidation state changed to +1 → Cu2+^ was reduced.
  1. Answer: 4-chlorobutanal, butanoyl chloride Working: Functional group isomers: same molecular formula, different functional groups. Cycloalkanes are isomers to alkenes but hexa-2,4-diene is a diene so no match. Propanoic anhydride and hexanedioic acid contain different numbers of oxygens → no match. 4-chlorobutanal and butanoyl chloride match: it is like a H has swapped place with the Cl. 2-methylpropan-2-ol and butan-2-ol are chain isomers not functional group isomers.
  2. Answer: gas chromatography Working: Volatile → easily vaporised → suitable to heat into gas phase and travel through gas chromatography apparatus.
  3. Answer: 1 only Working: SiO 2 is an acidic oxide so reacts with bases/basic oxides. SiO 2 + 2 NaOH → Na 2 SiO 3 + H 2 O SiO 2 + MgO → MgSiO 3
  4. Answer: oxidation/reduction Working: Acid/base neutralisation: yes since SO 3 is an acidic oxide and NH 3 is a base. Dative bond formation: yes since N lone pair enters empty S shell. Ionic: yes since it is an ionic salt product. Redox: no (S: +6, N: -3) all oxidation states remain constant.
  1. Answer: Br 2 O 5 Working: Br : O molar ratio is 1.6/80 : 0.8/16 = 0.02 : 0.05 = 2 : 5 → Br 2 O 5
  2. Answer: 1/ Working: Average oxidation/charge state of Fe ions in Fe 3 O 4 = +8/ Let x be mole fraction of Fe ions in 2+ state: 2x + 3(1 - x) = 8/3 → 6x + 9 - 9x = 8 → -3x = -1 → x = 1/
  3. Answer: The N-N σ-bond in N 2 H 4 is stronger than the P-P σ-bond in P 2 H 4 Working: Bi 2 O 5 is more ionic than N 2 O 5 (more different electronegativities) so is the more basic (alkaline) oxide. NF 3 is more covalent (less different electronegativities) than BiF 3. PH 3 has lower boiling point than NH 3 since NH 3 has strong hydrogen bonding but PH 3 does not (only dipole forces) N-N bond is weaker than P-P bond since the N atom is smaller than the P atom and so the lone pair-bonding pair repulsion is stronger
  4. Answer: 2 Working: The dot-and-cross structures are 2 lone pairs in BrF 2 +^ + 0 lone pairs in SNF 3 = 2 lone pairs
  1. Answer: 2 only Working: Exothermic reactions → temperature of reaction decreases and temperature of surroundings increases (heat is released) → entropy of reaction decreases and entropy of surroundings increases. Endothermic reactions → temperature of reaction increases and temperature of surroundings decreases (heat is absorbed) → entropy of reaction increases and entropy of surroundings decreases.
  2. Answer: [He] 2s^2 2p^5 Working: [He] 2s^1 is a metal [He] 2s^2 2p^5 and [Ne] 3s^2 3p^5 are more reactive than [Ne] 3s^2 3p^2 since needs only 1 more electron to complete valence shell compared to 3 more [He] 2s^2 2p^5 is more reactive than [Ne] 3s^2 3p^5 since reactivity increases up group 7 (fluorine > chlorine)
  3. Answer: NO 3 - , S 2 O 3 2-, Cr 2 O 7 2- Working: Correct names of similar ions are: Nitronium: NO 2 +, Nitrate: NO 3 -^ , Nitride: N3- Sulfate: SO 4 2-, Thiosulfate: S 2 O 3 2-, Per(oxydi)sulfate: S 2 O 8 2- Chromate: CrO 4 2-, Dichromate: Cr 2 O 7 2-, Chromium(III): Cr3+
  4. Answer:^18 F Working:^16 O: 8 protons, 8 neutrons → no spin (^18) F: 9 protons, 9 neutrons → spin (^20) Ne: 10 protons, 10 neutrons → no spin (^24) Mg: 12 protons, 12 neutrons → no spin (Nuclear spin is the property similar to electron spin, which determines whether a nucleus can be used in NMR i.e. 1 H, 13 C.)
  1. Answer: 10% Working: Let x be abundance (from 0 = 0% to 1 = 100%) of 86 Sr = 87 Sr. → 87.7 = 86 x + 87 x + 88(1 - 2 x ) [since all abundances add to 1] → 87.7 = 88 - 3 x → x = 0.1 = 10%
  2. Answer: AlCl 4 -^ and CH 3 CH 2 CH 2 O-^ are formed as intermediates Working: Reactions listed are: (Intermediates are AlCl 4 -^ and CH 3 CH 2 C+O).
  1. Answer: 1 and 3 Working: Two products form (ignoring optical isomers) since either the NH 2 on alanine could react with the COOH on cysteine or the other way around. Each group undergoes condensation, eliminating H 2 O and forming an amide bond: In their usual state, the amino acids (and hence the dipeptide) form zwitterions so convert all NH 2 to NH 3 +^ and all COOH to COO-: → CH 3 CH(NH 3 +)CONHCH(CH 2 SH)COO- → HSCH 2 CH(NH 3 +)CONHCH(CH 3 )COO- (the disulfide 4 is not correct since it does not include alanine.)
  2. Answer: 13 Working: Adenine and thymine pair with two hydrogen bonds. Guanine and cytosine pair with three hydrogen bonds. → 2 + 3 + 3 + 2 + 3 = 13
  3. Answer: 0 Working: Possible reduction routes from nitriles are: R-CN → R-NH 2 (notice this requires only one N-C bond) Or Ph-CN → Ph-CH 2 NH 2 (notice the extra CH 2 ) (with H 2 / Pd or LiAlH 4 ). The 1st, 2nd, 4th and 6th have multiple C-C bonds attached so cannot be formed from a nitrile. The 3rd one could not form a ring from a nitrile alone. The 5th one is reduced from nitrobenzene (not a nitrile).
  1. Answer: 5 o Working: Aluminium trichloride dimer is Al 2 Cl 6 , with dative covalent (coordinate) bonds between a Cl and each Al: Cl-Al-Cl has Al as central atom: 4 electron pairs, 0 lone pairs → tetrahedral around Al → bond angle = 109.5o Al-Cl-Al has Cl as central atom: 4 electron pairs, 2 lone pairs → tetrahedral with distortion → 109.5 - 2(2.5) = 104.5o → difference = 5o.
  2. Answer: 2 and 3 only Working: Biodiesel is made by reacting vegetable oils (triglyceride esters) with small primary alcohols such as methanol. Motor fuels are made by catalytically cracking long-chain alkanes with zeolite. Soaps are long-chain sodium/potassium salts, formed from base hydrolysis (NaOH / KOH) of long-chain esters such as those in vegetable oils/animal fats.
  3. Answer: 1 , 2 and 3 Working: Ethanoyl chloride is less safe (it is corrosive), less easily stored (it is easily hydrolysed to give ethanoic acid and toxic HCl fumes). Ethanoic anhydride on the other hand produces only ethanoic acid (weak acid) and does so with a slower reaction (it is an equilibrium).

Section B: Standard

  1. Answer: It displays geometrical isomerism. Working: methyl 2-methylpropenoate: Geometric (E/Z) isomer: no (double bond has H and H on one side) Addition polymer: yes (almost all alkenes can) Reduction: yes (hydrogenation of the double bond and LiAlH 4 on the carbonyl are both possible reduction reactions) Decolourises bromine: yes (double bond reacts by electrophilic addition to form dibromoalkane)
  2. Answer: Peaks at m / z = 70, 72 and 74 in the ratio 9:6: Working: Possible combinations: (^35) Cl- (^35) Cl : m/z = 70 : probability = ¾ * ¾ = 9/ (^35) Cl- (^37) Cl : m/z = 72 : probability = ¾ * ¼ = 3/ (^37) Cl- (^35) Cl : m/z = 72 : probability = ¼ * ¾ = 3/ (^37) Cl- (^37) Cl : m/z = 74 : probability = ¼ * ¼ = 1/ But 35 Cl-^37 Cl and 37 Cl-^35 Cl are the same compound, so 6/16. Ratios are then 9/16 : 6/16 : 1/16 = 9 : 6 : 1. (Could also use binomial distribution.)