Mock Paper 1 — 40 MCQ

Answer: B — 20. Neutrons = mass number − atomic number = 39 − 19 = 20. This atom is potassium (K). The atomic number (19) = protons = electrons in a neutral atom. The remaining nuclear particles are neutrons.

Answer: D — Same mass number but different neutron numbers. Wait — isotopes have the SAME proton number (same element) but DIFFERENT mass numbers (different neutron numbers). Option D is partially wrong as stated — isotopes have DIFFERENT mass numbers. The correct definition: isotopes have same atomic number (protons) but different mass numbers (neutrons). Among the options, D is closest but imprecise. Best answer available: D (different neutron numbers) — the distinguishing feature. Chemical properties are identical (same electrons).

Answer: C — Group VI, Period 3. Group number = number of outer shell electrons = 6 → Group VI. Period number = number of occupied electron shells = 3 shells → Period 3. This element is sulfur (S), atomic number 16 (2+8+6=16). Common error: confusing group and period from the configuration.

Answer: B — Same number of outer shell electrons. Chemical properties depend on electron arrangement, specifically the outer (valence) shell electrons. Elements in the same group have the same number of outer electrons → same valency → similar reactions. Same number of electron shells = same period (not group).

Answer: C — 132. (NH₄)₂SO₄: 2N = 28, 8H = 8, 1S = 32, 4O = 64. Total = 28+8+32+64 = 132. Common error: forgetting the subscript 2 outside the bracket doubles the NH₄ unit — 2×(14+4) = 36, not 18.

Answer: B — 4Fe + 3O₂ → 2Fe₂O₃. Check: Left: 4Fe, 6O. Right: 4Fe (in 2Fe₂O₃ = 4Fe), 6O (3×2=6). Balanced ✓. Option A: 1Fe+2O left, 1Fe+1O right — no. Option C: 2Fe+2O left, 2Fe+3O right — no. The formula for iron(III) oxide is Fe₂O₃.

Answer: C — H⁺ + OH⁻ → H₂O. In strong acid + strong alkali neutralisation, all ions fully dissociate. Na⁺ and SO₄²⁻ are spectator ions (unchanged on both sides) and are cancelled. The net ionic equation for ALL strong acid + strong alkali reactions is H⁺ + OH⁻ → H₂O. Option A is the full molecular equation, not the ionic equation.

Answer: D — (aq). State symbols: (s) = solid, (l) = liquid (the pure substance in liquid form), (g) = gas, (aq) = aqueous (dissolved in water). Note: water itself is (l) — pure liquid. HCl(aq) means hydrochloric acid solution (HCl dissolved in water).

Answer: B — Loses one electron to form Na⁺. Sodium (Group I, config 2,8,1) loses its single outer electron to achieve a stable noble gas configuration (2,8). This forms Na⁺. Chlorine gains the electron to form Cl⁻. The ionic bond is the electrostatic attraction between Na⁺ and Cl⁻. Ionic bonding always involves electron TRANSFER (metal loses, non-metal gains).

Answer: C — 3. Nitrogen has electronic configuration 2,5. It needs 3 more electrons to complete its outer shell. It forms 3 single covalent bonds (one with each H atom in NH₃), sharing one pair of electrons with each H. Nitrogen also has one lone pair. The dot-cross diagram shows N with 3 bonding pairs + 1 lone pair.

Answer: B — Ions are fixed in the lattice and cannot move. NaCl has Na⁺ and Cl⁻ ions (charged particles exist). In the solid state, ions are held rigidly in the giant ionic lattice — they cannot move to carry charge. When melted or dissolved, ions become free to move → conducts. NaCl has no covalent bonds — option D is wrong.

Answer: B. Diamond has a giant covalent structure. Every C atom forms strong covalent bonds to 4 other C atoms in all directions throughout the lattice. To melt, ALL these covalent bonds must be broken — requiring enormous energy (~3550°C). Simple molecular covalent substances have low melting points because only weak intermolecular forces (not covalent bonds) need to be overcome.

Answer: B — Copper(II) nitrate. Acid + Metal oxide → Salt + Water. HNO₃ + CuO → Cu(NO₃)₂ + H₂O. The salt name: first part from the base (copper(II)), second from the acid. Nitric acid → nitrate. HCl → chloride; H₂SO₄ → sulfate; HNO₃ → nitrate. "Nitride" is a different compound (N³⁻ ion) — not formed by acid reactions.

Answer: C — It is a strongly acidic solution. pH 2 = acidic (pH < 7). pH 2 is very acidic (close to 1 = most acidic). Important: pH 2 tells us the concentration of H⁺ ions — it does NOT tell us whether the acid is strong or weak. A weak acid at high concentration can also give pH 2. Option B confuses pH with acid strength. Option D is wrong — all aqueous solutions contain both H⁺ and OH⁻; at pH 2, [OH⁻] is just very low.

Answer: C. CuSO₄ is soluble, so cannot be made by precipitation. Since CuO is insoluble and does not need exact quantities, the excess solid method is used: add excess CuO to warm dilute H₂SO₄ (ensures all acid is used up), filter off excess CuO, evaporate filtrate to obtain CuSO₄ crystals. Titration is used when both reactants are solutions (no excess solid method available).

Answer: B — Mixing BaCl₂(aq) with Na₂SO₄(aq). Insoluble salts are made by precipitation: mix two solutions containing the required ions. BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq). Filter off the white precipitate, wash with distilled water, dry. Do NOT add BaSO₄ source to H₂SO₄ — BaSO₄ formed would coat the surface and stop the reaction.

Answer: C — Hydrogen. Acid + Metal → Salt + Hydrogen. Zn + 2HCl → ZnCl₂ + H₂↑. Tested by: holding a lit splint near the test tube mouth — hydrogen burns with a squeaky pop. CO₂ is produced when acid reacts with a CARBONATE (not a metal). Chlorine is NOT released when metals react with HCl.

Answer: B — 0.5 mol. Mr of CO₂ = 12 + (2×16) = 12 + 32 = 44 g/mol. Moles = mass / Mr = 22 / 44 = 0.5 mol. Always calculate Mr first by adding up all Ar values. Common error: using Ar of C (12) instead of Mr of CO₂ (44).

Answer: B — 6 dm³. CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Mole ratio: 1 mol CaCO₃ produces 1 mol CO₂. So 0.25 mol CaCO₃ → 0.25 mol CO₂. Volume = 0.25 × 24 = 6 dm³. Key: use the mole ratio from the balanced equation before calculating volume.

Answer: C — 8 mol/dm³. c = n/V. V must be in dm³: 500 cm³ ÷ 1000 = 0.5 dm³. c = 4 / 0.5 = 8 mol/dm³. The most common error is not converting cm³ to dm³ — using 500 gives 4/500 = 0.008 mol/dm³ (wrong). Always convert cm³ to dm³ by dividing by 1000.

Answer: C — 75%. % yield = (actual yield / theoretical yield) × 100 = (6.0/8.0) × 100 = 75%. Percentage yield is always ≤ 100%. Reasons yield is less than 100%: reversible reactions don't go to completion; product lost during transfer/filtration; side reactions; incomplete reactions.

Answer: D. Exothermic: energy released to surroundings → temperature RISES. ΔH is negative. Products have LOWER energy than reactants (the difference is released). Examples: combustion, neutralisation, respiration, many oxidation reactions. Endothermic is the opposite: temperature falls, energy absorbed, products higher energy, ΔH positive.

Answer: B — Exothermic, ΔH = −170 kJ/mol. ΔH = energy absorbed (bonds broken) − energy released (bonds formed) = 450 − 620 = −170 kJ/mol. Negative ΔH = exothermic (more energy released forming bonds than absorbed breaking them). If bonds formed release more energy than broken → net energy released to surroundings → exothermic.

Answer: B — Decreasing the activation energy. A catalyst provides an alternative reaction pathway with LOWER activation energy. More particles therefore have sufficient energy to react at any given temperature → faster rate. The catalyst does NOT change ΔH (the overall energy change) or the energy of reactants/products. It is not consumed and can be reused.

Answer: B. Higher concentration → more particles per unit volume → more frequent collisions between reactant particles → more frequent successful collisions (those with energy ≥ activation energy) per second → faster rate. The particles do NOT move faster (that would require higher temperature). Activation energy is unchanged by concentration.

Answer: B. Smaller particles have a larger total surface area. More surface area means more CaCO₃ particles are exposed to HCl acid particles simultaneously → more frequent collisions per second → faster rate. The total mass (and moles) of CaCO₃ is the same — so the total amount of CO₂ produced is identical, just produced more quickly.

Answer: B. Temperature increases particle KE. TWO effects: (1) particles move faster → collide more frequently; (2) more particles have KE ≥ activation energy → greater proportion of collisions are successful. The activation energy itself does NOT decrease with temperature — it is a fixed energy barrier for that reaction. Both effects together cause a large increase in rate (roughly doubles per 10°C rise).

Answer: B. Fe is more reactive than Cu → Fe displaces Cu from its salt solution. Fe + CuSO₄ → FeSO₄ + Cu. Iron nail becomes coated in copper (reddish-brown deposit). The blue colour of Cu²⁺ fades as FeSO₄ (pale green) forms. Displacement: more reactive metal displaces less reactive metal from its salt solution.

Answer: C. Carbon can only reduce oxides of metals BELOW it in the reactivity series. Al is above C → Al is more reactive than C → C cannot displace Al from Al₂O₃. Therefore, the more powerful method of electrolysis of molten Al₂O₃ (with cryolite to lower melting point) is used. Metals below C (Fe, Cu, Pb, Zn) can be reduced by carbon (smelting).

Answer: C — Both oxygen and water. The classic experiment with three test tubes proves this: iron in dry air (no rust), iron in boiled water with oil seal (no rust), iron in normal air and water (rusts). BOTH O₂ AND H₂O are required. Rust = hydrated iron(III) oxide, Fe₂O₃·xH₂O. Saltwater accelerates rusting (electrolyte) but is not required.

Answer: B. Zinc is MORE reactive than iron (above Fe in reactivity series). TWO forms of protection: (1) Physical barrier — zinc coating prevents O₂ and H₂O reaching iron; (2) Sacrificial/electrochemical protection — even if zinc layer is scratched, zinc corrodes preferentially (more reactive), protecting the exposed iron. Iron cannot corrode while zinc is present.

Answer: B — Cathode, by reduction. Cathode (−): cations (Pb²⁺) are attracted and REDUCED (gain electrons): Pb²⁺ + 2e⁻ → Pb. Lead metal is deposited. Anode (+): anions (Br⁻) are attracted and OXIDISED (lose electrons): 2Br⁻ → Br₂ + 2e⁻. Bromine gas is produced. OIL RIG: Oxidation Is Loss (at anode), Reduction Is Gain (at cathode).

Answer: B. Mg → Mg²⁺ + 2e⁻ (Mg LOSES electrons → OXIDISED). Cu²⁺ + 2e⁻ → Cu (Cu²⁺ GAINS electrons → REDUCED). OIL RIG: Oxidation Is Loss, Reduction Is Gain. Mg is the reducing agent (it gets oxidised and reduces Cu²⁺). Cu²⁺ is the oxidising agent (it gets reduced and oxidises Mg). In displacement reactions, the more reactive metal is always oxidised.

Answer: C — Hydrogen. Cathode (−): H⁺ ions (from H₂SO₄ and water) are reduced: 2H⁺ + 2e⁻ → H₂↑. Hydrogen gas forms at cathode. Anode (+): OH⁻ ions from water are oxidised: 4OH⁻ → 2H₂O + O₂ + 4e⁻. Oxygen forms at anode. Volume of H₂ : O₂ = 2:1. Test H₂ with a lit splint — squeaky pop. Test O₂ with a glowing splint — relights.

Answer: B. Alkenes have a C=C double bond. They undergo addition reactions — Br₂ adds across the double bond: CH₃CH=CH₂ + Br₂ → CH₃CHBrCH₂Br. The orange-brown bromine water is decolourised (becomes colourless) as Br₂ is consumed. Alkanes do NOT react with bromine water in the dark (only in UV light, by substitution). This is the standard test for an alkene (unsaturation).

Answer: B. Addition polymerisation: the C=C double bond in each ethene monomer opens up; monomers join together to form a long polymer chain. No atoms are lost — the formula of the repeat unit is (–CH₂–CH₂–)ₙ. No water is released (that would be condensation polymerisation, e.g. nylon, polyester). Option C describes condensation polymerisation.

Answer: B — Ethyl ethanoate and water. Esterification: alcohol + carboxylic acid → ester + water (reversible). C₂H₅OH + CH₃COOH ⇌ CH₃COOC₂H₅ + H₂O. Concentrated H₂SO₄ acts as acid catalyst and dehydrating agent. Ethyl ethanoate has a fruity/sweet smell. The ester name: alkyl group from alcohol (ethyl) + acid name (ethanoate).

Answer: D — Ammonia. Ammonia (NH₃) is the only common gas that turns damp red litmus BLUE (it is alkaline). It also has a distinctive sharp/pungent smell. Gas tests: H₂ = squeaky pop with lit splint; O₂ = relights glowing splint; CO₂ = turns limewater milky; Cl₂ = bleaches damp litmus (red then white). Chlorine turns red litmus red THEN bleaches it — different from ammonia.

Answer: C — Na⁺ (sodium). Flame test colours: Li = crimson/red, Na = YELLOW (intense, persistent), K = lilac/violet, Ca = brick red, Cu = green. Sodium gives the most distinctive yellow colour — so intense it can mask other colours. Note: yellow from Na is very different from the pale/lemon yellow that some students confuse with K (which is lilac).

Answer: C — Dilute HNO₃ then AgNO₃ → white precipitate (AgCl). Test for Cl⁻: acidify with dilute HNO₃ (not HCl — would introduce Cl⁻ ions!), then add AgNO₃ solution. White precipitate of AgCl confirms Cl⁻. Bromide gives cream precipitate (AgBr); iodide gives yellow (AgI). Option A (BaCl₂ after HCl) is the test for SO₄²⁻. Always add dilute acid first to remove CO₃²⁻ interference.