CQ1. How many moles of CO₂ are present in 88 g of CO₂? (M = 44 g·mol⁻¹)
CQ2. What is the molarity of a solution containing 0.50 mol of NaCl dissolved in 250 mL of solution?
CQ3. The electronic configuration 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ describes which element?
CQ4. In methane (CH₄), the carbon atom is:
CQ5. For an ideal binary solution, if mole fraction of solvent x₁ = 0.80 and pure solvent vapour pressure P₁° = 100 mm Hg, the partial pressure of solvent is:
CQ6. A system absorbs 500 J of heat and does 200 J of work on the surroundings. Using ΔU = q − w (work by system positive), ΔU equals:
CQ7. The pH of 0.001 M HCl (complete dissociation) is:
CQ8. For a cell, E° = 1.10 V and n = 2. At 25 °C, if Q = 1, E_cell is approximately:
CQ9. Which oxide is most basic among the following?
CQ10. Down a group of p-block elements, atomic radius generally:
CQ11. In the laboratory preparation of ammonia from ammonium salts, a common reagent pair is:
CQ12. Among F₂, Cl₂, Br₂ and I₂, the strongest oxidising agent is:
CQ13. Many transition metal ions are coloured mainly due to:
CQ14. In [Cu(NH₃)₄]²⁺, the coordination number of copper is:
CQ15. The compound CH₃CH₂OH is named as:
CQ16. Propene treated with HBr (no peroxide) mainly gives:
CQ17. Which substrate is most likely to undergo SN1 rapidly?
CQ18. In the Lucas test, immediate turbidity is characteristic of:
CQ19. Tollen's reagent gives a silver mirror with:
CQ20. Hofmann bromamide reaction converts an amide RCONH₂ into an amine with:
3 marks · Apply concepts to new situations · Try first, then show model answer.
AQ1. A student needs 250 mL of 0.20 M NaOH. (a) How many moles of NaOH are required? (b) What mass of solid NaOH (M = 40 g·mol⁻¹) should be weighed? (c) Why is the solution diluted to the mark in a volumetric flask rather than adding 250 mL of water to solid?
Model Answer
(a) n = M × V = 0.20 × 0.250 = 0.050 mol. (b) m = nM = 0.050 × 40 = 2.0 g. (c) Volume of solution means final volume after dissolution; adding 250 mL water to solid would give total volume slightly greater than 250 mL and wrong molarity (L1 mole concept, L7 molarity).
AQ2. Explain why BeH₂ is linear while H₂O is bent, using hybridisation and lone pairs. Relate to electron-pair geometry.
Model Answer
Be in BeH₂ has two bonding pairs, no lone pairs → sp hybridisation → linear 180°. O in H₂O has two bonding pairs and two lone pairs → four electron pairs, tetrahedral electron geometry but bent molecular shape (~104.5°) due to lp–bp repulsion (L2 orbitals, L4 VSEPR/hybridisation).
AQ3. Dissolving a non-volatile solute in water lowers vapour pressure (Raoult) and also often involves heat of solution. How do colligative lowering of vapour pressure and enthalpy of solution differ in what they measure?
Model Answer
Colligative vapour-pressure lowering depends on mole fraction of solute particles (number), not chemical identity (ideal dilute). Enthalpy of solution (ΔH_sol) is energetic — heat absorbed/released when solute dissolves; can be endo or exo depending on lattice and solvation energies (L7 colligative, L9 thermochemistry).
AQ4. For the dissociation HA ⇌ H⁺ + A⁻, how does a large positive ΔG° relate to Ka and pH of a weak acid solution?
Model Answer
ΔG° = −RT ln K; large positive ΔG° means K (here Ka) ≪ 1 — weak acid. Equilibrium lies left; [H⁺] from partial dissociation is small → higher pH than a strong acid of same formal concentration (L9 free energy, L12 acid–base equilibrium).
AQ5. A buffer of CH₃COOH/CH₃COONa resists pH change. How does this idea connect to a hydrogen electrode used in electrochemical cells where [H⁺] affects potential?
Model Answer
Buffer keeps [H⁺] nearly constant via common-ion equilibrium (L12). Nernst equation for H⁺/H₂ depends on [H⁺]; stable [H⁺] gives stable electrode potential — important for reproducible measurements and for cells involving H⁺ (L13).
AQ6. During electrolysis of aqueous CuSO₄ with copper electrodes, Cu dissolves at anode and deposits at cathode. Why is copper suitable, and what oxidation states are typical?
Model Answer
Cu is a d-block metal that forms Cu²⁺(aq) readily. Anode: Cu → Cu²⁺ + 2e⁻; cathode: Cu²⁺ + 2e⁻ → Cu. Preferential deposition of Cu over H₂ when potential and overvoltage allow (L13). Transition metals show variable oxidation states; Cu commonly +1/+2 (L21).
AQ7. Compare the nature of Na₂O and SO₂ as oxides and predict the pH of their aqueous solutions qualitatively.
Model Answer
Na₂O is a basic oxide of an s-block metal: Na₂O + H₂O → 2NaOH (alkaline). SO₂ is an acidic oxide of a p-block non-metal: forms H₂SO₃ (acidic solution). Trend: metallic oxides basic, non-metallic oxides acidic (L17, L18).
AQ8. Chlorine water bleaches by oxidation; ammonia forms complexes and is basic. Using one reaction each, show oxidising behaviour of Cl₂ and basic behaviour of NH₃.
Model Answer
Cl₂ + H₂O ⇌ HCl + HOCl; HOCl oxidises dyes (bleaching) (L20). NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ (weak base); also NH₃ as ligand e.g. [Cu(NH₃)₄]²⁺ (L19, links L22).
AQ9. Why is [Ti(H₂O)₆]³⁺ coloured while [Sc(H₂O)₆]³⁺ is colourless? Use d-electron configuration.
Model Answer
Ti³⁺ is d¹ — can undergo d–d transition absorbing visible light → coloured. Sc³⁺ is d⁰ — no d–d transition in the visible → colourless (L21, L22 ligand field context).
AQ10. Write the IUPAC name of [Co(NH₃)₆]Cl₃ and state the oxidation number of cobalt. How does naming differ from simple binary salts?
Model Answer
Hexaamminecobalt(III) chloride; Co is +3 (six neutral NH₃, three Cl⁻). Coordination nomenclature lists ligands with metal + oxidation number in parentheses, then counter ions — not like NaCl-style binary naming (L22, L23 principles).
AQ11. Outline a two-step path from ethene to ethanol via a haloalkane intermediate, naming reagents.
Model Answer
Ethene + HBr → bromoethane (electrophilic addition, L24). Bromoethane + aq. KOH (or NaOH) → ethanol by nucleophilic substitution (L25). Alternative: direct hydration of ethene (acid) also gives ethanol.
AQ12. tert-Butyl bromide is converted to an alcohol with aqueous KOH, then treated with Lucas reagent. What is observed and why? Contrast with 1-bromopropane under the same sequence.
Model Answer
tert-Butyl bromide → 2-methylpropan-2-ol (3° alcohol). Lucas: immediate turbidity (3° ROH forms RCl fast). 1-Bromopropane → propan-1-ol (1°); Lucas: no turbidity in cold (slow) (L25 SN, L26 Lucas order 3° > 2° > 1°).
AQ13. How can you distinguish primary, secondary and tertiary alcohols using oxidation products and mild tests on the carbonyls formed?
Model Answer
1° ROH → aldehyde → acid (further oxidation); aldehyde gives Tollen's/Fehling. 2° → ketone (no Tollen's under mild conditions). 3° resistant to mild oxidation. Controlled oxidation of 1° can stop at aldehyde (L26, L27).
AQ14. Ethanal undergoes aldol condensation; methanal undergoes Cannizzaro. Explain the difference in terms of α-hydrogen, then state one test that distinguishes ethanal from ethanamine.
Model Answer
Ethanal has α-H → enolate → aldol (dil. base). Methanal has no α-H → Cannizzaro (conc. base, disproportionation) (L27). Ethanal: Tollen's silver mirror; ethanamine: carbylamine foul isocyanide with CHCl₃/KOH (L28).
AQ15. Starting from aniline, how is chlorobenzene prepared via diazonium salt? Name the key reaction.
Model Answer
Aniline + NaNO₂/HCl at 0–5 °C → benzenediazonium chloride; then CuCl (Sandmeyer) → chlorobenzene + N₂ (L28). Links aromatic substitution and Cu(I) catalysis (L22 context).
AQ16. Glucose is an aldohexose that exists mainly as a cyclic hemiacetal. How does this relate to carbonyl chemistry of aldehydes, and why does open-chain form still give some aldehyde tests?
Model Answer
Cyclic form: OH attacks carbonyl C → hemiacetal (Nu addition to aldehyde, L27). Equilibrium with open-chain aldehyde allows Tollen's/Fehling positive despite mainly cyclic structure (L29).
AQ17. Proteins are polyamides of α-amino acids. Relate the peptide bond to organic amide chemistry and state why essential amino acids must be in the diet.
Model Answer
Peptide bond is –CO–NH– amide from COOH of one AA and NH₂ of another (L28 amides, L29 proteins). Essential amino acids cannot be synthesised by the body — must come from diet (~10 of ~20).
AQ18. Soaps are made by saponification of fats. Relate this to triglyceride structure of lipids and explain why soap fails in hard water.
Model Answer
Fats/oils are triglycerides (glycerol + 3 fatty acids, L29). Saponification: fat + NaOH → RCOONa (soap) + glycerol (L31). Hard water Ca²⁺/Mg²⁺ form insoluble carboxylates (scum) — no lather.
AQ19. Terylene is a polyester from ethylene glycol and terephthalic acid. Is this addition or condensation polymerisation? How does it differ from polyethene formation?
Model Answer
Condensation (step-growth): bifunctional monomers lose H₂O to form ester links (L31). Polyethene is addition (chain-growth) of ethene with no small molecule eliminated — unsaturated monomer + initiator (L31; ester link also relates to L27 carboxylic derivatives).
AQ20. Natural rubber is a polymer of isoprene. Why is vulcanisation with sulphur important, and how does the C=C character of polyisoprene link to alkene chemistry?
Model Answer
Isoprene (2-methylbuta-1,3-diene) polymerises to polyisoprene with residual double bonds (L24 dienes, L31). Vulcanisation: S cross-links chains — harder, more elastic, solvent-resistant, wider temperature range than raw sticky rubber.
5 marks · Compare, contrast and evaluate · Deep understanding of chemistry concepts.
ZQ1. Analyse how molarity, mole fraction and pH each express composition of an aqueous acetic acid solution. Which is intensive, which depends on ionisation, and how does dilution affect each?
Model Answer
Molarity = mol solute / L solution — intensive for given T if volume fixed; dilution decreases M. Mole fraction x = n_i / n_total — composition without volume; dilution with water decreases x_acid. pH = −log[H⁺] depends on equilibrium Ka and actual [H⁺] from dissociation — not simply equal to −log C for weak acids. Dilution of weak acid increases degree of dissociation but [H⁺] usually falls, pH rises. Colligative effects track particle mole fraction; acid–base behaviour tracks free [H⁺] (L1, L7, L12).
ZQ2. Compare the roles of s, p and d orbitals in (i) hybridisation of carbon in ethene, (ii) colour of transition-metal ions. Why does carbon not show the same variable oxidation states as iron?
Model Answer
Ethene carbon: sp² hybridisation — three sp² + unhybridised p for π bond (L2, L4). Transition ions: incomplete d shells enable d–d transitions — colour (L21). Carbon has only 2s/2p valence orbitals and tends to form four covalent bonds (octet); Fe has accessible 3d + 4s electrons and variable oxidation states (+2, +3 etc.) because d electrons can be removed stepwise without extreme energy cost.
ZQ3. Connect ΔG, equilibrium constant K and cell potential E for a redox reaction. When is a reaction spontaneous in the electrochemical sense?
Model Answer
ΔG = −RT ln K = −nFE. Large K (products favoured) means large positive E (for the cell as written) and negative ΔG — spontaneous. At equilibrium E = 0, ΔG = 0, Q = K. Nernst equation links E to Q; pH and concentrations shift E (L9, L12, L13). Spontaneous cell: E > 0 under conditions of interest.
ZQ4. Trace the change from basic to acidic oxides across period 3 (Na₂O to SO₃/Cl oxides). How does this relate to metal/non-metal character and industrial acids?
Model Answer
Left: Na₂O, MgO basic; middle Al₂O₃ amphoteric; right SiO₂ weakly acidic/network, P₄O₁₀, SO₃, Cl oxides acidic (L17–L20). Metallic character decreases across period — oxides become more acidic. SO₃/H₂SO₄ and nitrogen oxides/HNO₃ are industrially important strong acids — p-block non-metal chemistry.
ZQ5. Analyse why coordination compounds of Cu²⁺ and Fe³⁺ are central both to colour chemistry and to redox/electrochemical behaviour.
Model Answer
Variable oxidation states and partially filled d orbitals enable redox (Cu²⁺/Cu, Fe³⁺/Fe²⁺) and coloured complexes (d–d or charge transfer) (L21, L22). Electrode potentials depend on complexation — ligands stabilise oxidation states differently, shifting E (Nernst/complex formation, L13). Example: [Cu(NH₃)₄]²⁺ deep blue — ligand field and complex stability.
ZQ6. Map a synthetic sequence: propene to propan-2-ol to propanone. Identify reaction types and justify regiochemistry where relevant.
Model Answer
Propene + H₂O/H⁺ (or oxymercuration) → propan-2-ol (Markovnikov; secondary carbocation preferred) (L24, L26). Or HBr then aq KOH. Propan-2-ol oxidation (acidic K₂Cr₂O₇ or PCC) → propanone (2° alcohol → ketone) (L26, L27). Avoid 1-propanol path if propanone is target — need 2° alcohol.
ZQ7. Compare strategies to introduce nitrogen: reduction of nitro compounds vs Hofmann bromamide vs nucleophilic substitution on alkyl halides. When is each preferred?
Model Answer
ArNO₂ + Sn/HCl → ArNH₂: best for aromatic amines (L28; nitration first). Hofmann: RCONH₂ → RNH₂ with one C less — chain shortening from amides (L28). RX + NH₃: aliphatic amines but polyalkylation risk — excess NH₃ (L25, L28). Choice depends on aromatic vs aliphatic and carbon-count goals.
ZQ8. Aldehydes/ketones appear in sugars; esters in fats and polyesters. Analyse how the same C=O chemistry underpins biomolecules and synthetic polymers.
Model Answer
Sugars: polyhydroxy aldehydes/ketones; hemiacetals, glycosides (L27, L29). Fats: glycerol triesters of fatty acids — esterification equilibrium (L27, L29). Terylene: polyester condensation of diol + diacid — repeating ester links (L31). Hydrolysis of esters (acid/base) links soap-making (saponification) to lipid chemistry.
ZQ9. Nitrogen appears in amines, amino acids, nucleic acid bases and ligands like NH₃. Analyse structural roles of N lone pairs across these contexts.
Model Answer
Amines: lone pair — basicity, H-bonding, nucleophilicity (L28). Amino acids: zwitterions, peptide bonds (L29). DNA bases: H-bonding A–T, G–C uses N and O lone pairs (L29). NH₃ ligands donate lone pair to metal — coordinate bonds, coordination number (L22). Same lone pair chemistry, different environments.
ZQ10. Evaluate detergents vs soaps for cleaning in hard water and environmental impact. How do micelle formation and polymer non-biodegradability both raise systems-level issues?
Model Answer
Micelles lower surface tension and emulsify grease (L31) — effective cleaning. Detergents work in hard water; soaps form scum. Branched ABS poor biodegradability — pollution; linear ABS better. Non-biodegradable plastics (addition polymers with inert C–C chains) accumulate waste — contrast PHBV ester links cleaved by enzymes. Manufacturing costs energy (L9); sustainability balances function with degradation pathways (L7 hard-water ions + L31).