NECO Chemistry (OBJ & Essay) Questions and Answers 2026
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Objective Answers (OBJ)
Essay & Theory Answers
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*CHEMISTRY OBJ* https://whatsapp.com/channel/0029Vb8Whka2P59oetvhwN3B 1-10: C, B, A, D, D, B, A, C, E, C 11-20: D, D, C, D, B, C, D, A, E, A 21-30: D, E, B, D, A, D, C, E, E, D 31-40: B, B, B, C, B, B, C, D, C, B 41-50: D, E, E, A, A, A, E, A, C, C 51-60: A, C, C, D, E, B, C, C, D, D *USE THIS IF URS IS RESHUFFLED ❤️🔥* 1. C – Magnesium ion (Mg²⁺) loses 2 electrons: 1s²2s²2p⁶. 2. B – Catenation = carbon atoms combining to form long chains. 3. A – Allotropy = existence of element in two or more forms in same physical state. 4. D – Graphite is a good lubricant due to its planar molecular structure (weak van der Waals forces between layers). 5. D – Flood water is heterogeneous (contains suspended particles). 6. B – CuCO₃ decomposes on heating to CuO and CO₂. 7. A – Galvanising prevents rusting (coating with zinc). 8. C – Dissolution of salt in water is a physical change (no new substance formed). 9. E – H₂O has bent (V-shaped) shape, not linear. 10. C – Carbon (C) has highest ionisation energy among B, Be, C, Li, N (increases across period). 11. D – Industrial acidic wastes can be treated with slaked lime (Ca(OH)₂). 12. D – Alkaline hydrolysis of oil = saponification (soap making). 13. C – Catalytic hydrogenation of vegetable oil produces margarine. 14. D – Incomplete combustion of carbon produces CO (carbon monoxide). 15. B – CCl₄ has tetrahedral shape. 16. C – Diamond (non-magnetic) and iron filings (magnetic) can be separated by magnetisation. 17. D – SO₂: O% = (2×16)/(32+32) × 100 = 32/64 × 100 = 50%. 18. A – Silver (Ag) does not react with oxygen. 19. E – Sublimation of silica is a physical change (not chemical). 20. A – Hydrogen bond involves covalent bond (intermolecular). 21. D – Redox = electron loss (oxidation) and gain (reduction). 22. E – Linear molecules have bond angle of 180°. 23. B – Al³⁺ has 10 electrons → same as Ne (not Ar). 24. D – Fluorine is the most electronegative element. 25. A – NH₃ (ammonia) is an alkaline gas. https://whatsapp.com/channel/0029VbCpC6c5Ejy61HyxKZ0v *To all POLAC RC 13 aspirants 👆👆.* https://t.me/necovipanswer Neco vip 👆👆 *REACT, SHARE AND FORWARD TO EVERYONE ❤️🔥* FOR NECO EXPO 24HRS BEFORE EXAM, MESSAGE ME ON TELEGRAM👇👇: https://t.me/Officialkingjay1 JOIN OUR VIP CHANNEL 👇🏻👇🏻 https://t.me/allexamexporunz 26. D – Ca(HCO₃)₂: 40 + 2(1+12+48) = 40 + 2(61) = 40 + 122 = 162. 27. C – Functional group determines chemical properties of homologous series. 28. E – Standard solution = solution with known concentration. 29. E – SO₂ is an acidic oxide. 30. D – NO (nitric oxide) is a neutral oxide. 31. B – 150°C + 273 = 423 K. 32. B – Carbohydrate = compound made of carbon, hydrogen and oxygen only. 33. B – H₂SO₄ is strong acid, [H⁺] = 0.01 × 2 = 0.02. pH = –log(0.02) = 1.7. 34. C – Alkanols react with sodium to evolve hydrogen gas. 35. B – Diamond used in abrasives due to its hardness. 36. B – Greenhouse effect = excess carbon (IV) oxide (CO₂). 37. C – Sodium atom undergoes ionisation by losing one electron (Na → Na⁺ + e⁻). 38. D – Metals are malleable because they can be beaten into sheets. 39. C – Energy given off when substance burnt completely in oxygen = heat of combustion. 40. B – Helium is a noble gas, not a transition element. 41. D – Coal tar is a product of coal, not a type of coal. Types: anthracite, bituminous, brown coal, peat. 42. E – Strong electronegativity is NOT a property of transition metals. 43. E – Sodium (Na) reacts vigorously with cold water. 44. A – Atomic radius increases down the group. 45. A – C₆H₁₂O₆ (glucose) is a non-electrolyte. 46. A – Equation: CO₂ + 2NaOH → Na₂CO₃ + H₂O. Z = 2. 47. E – Charles's law: V₂ = V₁ × T₂/T₁ = 250 × 300/273 = 274.7 cm³. 48. A – Carbon (II) oxide (CO) is NOT an acid anhydride. 49. C – Moles of C₆H₁₂ = 14.4/84 = 0.171. Moles of H₂O = 0.171 × 6 = 1.03 ≈ 1.20 (closest). 50. C – MnO₄⁻ is reduced → oxidizing agent; SO₃²⁻ is oxidized → oxidized species. 51. A – Temporary hardness removed by addition of slaked lime. 52. C – 0.11 g CO₂ = 0.11/44 = 0.0025 mol. Volume = 0.0025 × 22400 = 56 cm³. 53. C – C: 40/12 = 3.33, H: 6.67/1 = 6.67, O: 50.33/16 = 3.15. Ratio = 1:2:1 → CH₂O. 54. D – Solid product of destructive distillation of coal = coke. 55. E – Reaction between alkanes and halogens = substitution. 56. B – Sulphur hardens rubber by cross linkage. 57. C – Moles = 0.05 × 250/1000 = 0.0125. Mass = 0.0125 × 106 = 1.325 g ≈ 1.33. 58. C – Coating surface of one metal with another = electroplating. 59. D – PV = nRT → n = (2×10)/(0.082×283) = 20/23.206 = 0.862 mol. 60. D – Fluorine atom becomes fluoride ion by adding one electron to outer orbital. *REACT, SHARE, FORWARD ❤️ ✅* *NECO CHEMISTRY (PAPER II – ESSAY)* *COMPLETE ANSWERS – QUESTIONS 1 TO 6* https://whatsapp.com/channel/0029Vb8Whka2P59oetvhwN3B =========================================================== QUESTION 1 =========================================================== (a)(i) Classify the following into physical and chemical change (4 marks) I. Dissolution of salt in water – Physical change (no new substance is formed; salt can be recovered by evaporation) II. Burning of paper – Chemical change (new substances such as ash, carbon dioxide, and water vapour are formed) III. Magnetization of iron – Physical change (no new substance is formed; iron can be demagnetized) IV. Rusting of iron – Chemical change (new substance, iron(III) oxide/hydrated iron oxide, is formed) (a)(ii) Which state is mercury at room temperature? (1 mark) Mercury is in the liquid state at room temperature. (b)(i) Calculate the relative molecular mass of: (2 marks) I. Calcium hydroxide – Ca(OH)₂ Ca = 40, O = 16, H = 1 RMM = 40 + 2(16 + 1) = 40 + 34 = 74 g/mol II. Lead(II) trioxonitrate(V) – Pb(NO₃)₂ Pb = 207, N = 14, O = 16 RMM = 207 + 2(14 + 3×16) = 207 + 2(14 + 48) = 207 + 124 = 331 g/mol (b)(ii) Define isotopy (2 marks) Isotopy is the phenomenon in which atoms of the same element have the same atomic number (same number of protons) but different mass numbers due to different numbers of neutrons in their nuclei. Isotopes have the same chemical properties but different physical properties. (b)(iii) Consider the following pairs: 6A & 8B, 11C & 16D, 3E & 19F (2 marks) I. Which pair will form ionic compound? 11C & 16D (C has 11 electrons = Na, D has 16 electrons = S). Sodium (metal) and sulphur (non-metal) form an ionic compound. II. Name the compound formed. Sodium sulphide (Na₂S) (c)(i) State Charles' law (2 marks) Charles' law states that at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature. Mathematically, V/T = k (or V₁/T₁ = V₂/T₂). (c)(ii) Calculate the volume of a gas at s.t.p. if it occupies 15 cm³ at 7.50 × 10⁴ Nm⁻² and 50.0°C (2 marks) Using combined gas law: P₁V₁/T₁ = P₂V₂/T₂ P₁ = 7.50 × 10⁴ Nm⁻² = 0.75 × 10⁵ Nm⁻² = 0.75 atm V₁ = 15 cm³ T₁ = 50 + 273 = 323 K At s.t.p.: P₂ = 1.0 × 10⁵ Nm⁻² = 1.0 atm, T₂ = 273 K (0.75 × 15)/323 = (1 × V₂)/273 V₂ = (0.75 × 15 × 273)/323 = (3071.25)/323 = 9.51 cm³ Answer: 9.51 cm³ (d)(i) State two phenomena that support kinetic theory of gases (2 marks) 1. Brownian motion – random movement of particles in a fluid. 2. Diffusion – movement of gas particles from an area of high concentration to low concentration. (d)(ii) Define ionisation energy (2 marks) Ionisation energy is the minimum amount of energy required to remove the most loosely bound electron from a gaseous atom in its ground state to form a gaseous ion. (d)(iii) Complete Table 2.1 (4 marks) Periodic properties Down a group Across the period Atomic radius Increases Decreases Electron affinity Decreases Increases Electronegativity Decreases Increases =========================================================== https://whatsapp.com/channel/0029VbCpC6c5Ejy61HyxKZ0v *To all POLAC RC 13 aspirants 👆👆.* https://t.me/necovipanswer Neco vip 👆👆 *REACT, SHARE AND FORWARD TO EVERYONE ❤️🔥* FOR NECO EXPO 24HRS BEFORE EXAM, MESSAGE ME ON TELEGRAM👇👇: https://t.me/Officialkingjay1 JOIN OUR VIP CHANNEL 👇🏻👇🏻 https://t.me/allexamexporunz QUESTION 2 =========================================================== (a)(i) Give two examples of iron ores (2 marks) 1. Haematite (Fe₂O₃) 2. Magnetite (Fe₃O₄) (a)(ii) State three adverse effects of chemical products (3 marks) 1. They can cause pollution of air, water, and soil. 2. They can be toxic to humans and animals, causing poisoning and diseases. 3. Some chemical products are carcinogenic and can cause cancer. (b)(i) Arrange Na, Li, K in decreasing order of their melting points (1 mark) Li > Na > K (b)(ii) Mention the three rules/principles of electron distribution in an atom (3 marks) 1. Aufbau principle – Electrons fill orbitals in order of increasing energy (lowest energy first). 2. Pauli exclusion principle – An orbital can hold a maximum of two electrons with opposite spins. 3. Hund's rule – Electrons occupy orbitals singly before pairing up. (c)(i) Differentiate between oxidizing agent and reducing agent in terms of electron transfer (2 marks) An oxidizing agent is a substance that accepts electrons (is reduced) during a redox reaction. A reducing agent is a substance that donates electrons (is oxidized) during a redox reaction. (c)(ii) Give the ions formed when the following compounds dissociate (3 marks) I. KOH → K⁺ + OH⁻ II. Pb(NO₃)₂ → Pb²⁺ + 2NO₃⁻ III. Na₂CO₃ → 2Na⁺ + CO₃²⁻ (d)(i) Define the following (2 marks) I. Anode – The positive electrode in electrolysis to which anions are attracted. II. Cathode – The negative electrode in electrolysis to which cations are attracted. (d)(ii) Calculate the volume of hydrogen produced at s.t.p. during electrolysis of dilute H₂SO₄ (4 marks) Time = 15 minutes = 15 × 60 = 900 seconds Charge = I × t = 1.0 × 900 = 900 C Moles of electrons = Q/F = 900/96500 = 0.009326 mol 2H⁺ + 2e⁻ → H₂ 2 moles of electrons produce 1 mole of H₂ Moles of H₂ = 0.009326/2 = 0.004663 mol Volume of H₂ = moles × 22.4 = 0.004663 × 22.4 = 0.1044 dm³ Answer: 0.104 dm³ =========================================================== QUESTION 3 =========================================================== (a) With the aid of an energy profile diagram, illustrate the effect of a catalyst on an exothermic reaction (5 marks) In your answer booklet, draw an energy profile diagram showing: · Reactants on the left at a certain energy level · Products on the right at a lower energy level (exothermic) · A curve showing the activation energy without catalyst (high peak) · A lower curve showing the activation energy with catalyst (lower peak) · Label: reactants, products, activation energy (Ea without catalyst), activation energy with catalyst, ΔH (negative), reaction pathway A catalyst provides an alternative reaction pathway with lower activation energy. The overall enthalpy change (ΔH) remains the same. (b)(i) Give one difference between endothermic and exothermic reactions (2 marks) Endothermic reactions absorb heat energy from the surroundings, while exothermic reactions release heat energy to the surroundings. (b)(ii) State Le Chatelier's principle (2 marks) Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium position will shift to counteract the effect of the change. (b)(iii) Mention three factors that affect the equilibrium position of a chemical reaction (3 marks) 1. Change in concentration of reactants or products. 2. Change in temperature. 3. Change in pressure (for gaseous reactions). (c)(i) Define the following (2 marks) I. Saturated solution – A solution that contains the maximum amount of solute that can dissolve at a given temperature. II. Molar solution – A solution that contains one mole of solute dissolved in one cubic decimetre (1 dm³) of solution. (c)(ii) Calculate mass of Na₂CO₃ needed to prepare 250 cm³ of 0.2 mol/dm³ solution (3 marks) Molar mass of Na₂CO₃ = 2(23) + 12 + 3(16) = 46 + 12 + 48 = 106 g/mol Moles needed = concentration × volume (dm³) = 0.2 × 0.250 = 0.05 mol Mass = moles × molar mass = 0.05 × 106 = 5.30 g Answer: 5.30 g (d)(i) I. Name the separation technique (1 mark) Chromatography (paper chromatography) II. How many components are resolved? (1 mark) Four components (from the diagram showing four spots) III. What material is used as the absorbent medium? (1 mark) Filter paper (or chromatography paper) IV. Which label indicates the point of application? (1 mark) II (where the sample was spotted at the origin) (d)(ii) Mention one separation technique suitable for: (2 marks) I. Insoluble solid from a liquid – Filtration II. Soluble solid from a solution – Evaporation (or crystallisation) =========================================================== QUESTION 4 =========================================================== (a)(i) What is a buffer solution? (2 marks) A buffer solution is a solution that resists changes in pH when small amounts of acid or alkali are added. It contains a weak acid and its salt (or a weak base and its salt). (a)(ii) Give two uses of buffer solution (2 marks) 1. Maintaining the pH of blood in the human body. 2. Used in the preservation of food and pharmaceutical products. (b)(i) Name the process used in the production of ammonia (1 mark) Haber process (or Haber-Bosch process) (b)(ii) Give two physical properties of nitrogen (2 marks) 1. It is a colourless, odourless, and tasteless gas. 2. It is slightly soluble in water. (b)(iii) List the two allotropes of carbon (2 marks) 1. Diamond 2. Graphite (b)(iv) Mention the structure of each allotrope (2 marks) 1. Diamond – Each carbon atom is bonded to four other carbon atoms in a tetrahedral structure (giant covalent structure). 2. Graphite – Each carbon atom is bonded to three other carbon atoms in hexagonal layers (layered structure with weak van der Waals forces between layers). (b)(v) Give one use of charcoal (1 mark) Charcoal is used as a fuel. (Also acceptable: Used as an adsorbent, in water purification, in making gunpowder) (c)(i) Give the general molecular formula for: (1 mark) I. Alkenes – CₙH₂ₙ II. Alkynes – CₙH₂ₙ₋₂ (c)(ii) Write out the possible structural isomers of C₄H₁₀ (2 marks) 1. Butane (CH₃CH₂CH₂CH₃) 2. 2-methylpropane (CH₃CH(CH₃)CH₃) (c)(iii) Give the IUPAC names of the isomers (2 marks) 1. Butane 2. 2-methylpropane (c)(iv) Calculate the molecular formula of a compound with 60% C, 13.3% H, 20.7% O, molar mass 60 (2 marks) Moles: C = 60/12 = 5 H = 13.3/1 = 13.3 O = 20.7/16 = 1.29 Divide by smallest (1.29): C = 5/1.29 = 3.87 ≈ 4 H = 13.3/1.29 = 10.3 ≈ 10 O = 1.29/1.29 = 1 Empirical formula = C₄H₁₀O Empirical mass = 4(12) + 10(1) + 16 = 48 + 10 + 16 = 74 Given molar mass = 60 → no, check: C₃H₈O would be 72? Let's recalculate: C: 60/12 = 5, H: 13.3/1 = 13.3, O: 20.7/16 = 1.29 Divide by 1.29: C = 3.87 ≈ 4, H = 10.3 ≈ 10, O = 1 C₄H₁₀O = 48 + 10 + 16 = 74 (not 60) Try C₃H₈O: 36 + 8 + 16 = 60 ✓ Empirical formula = C₃H₈O Answer: C₃H₈O =========================================================== QUESTION 5 =========================================================== (a)(i) Define octane rating (2 marks) Octane rating is a measure of the anti-knock properties of petrol (gasoline). It indicates the ability of a fuel to resist engine knocking (pre-ignition) during combustion. Higher octane numbers indicate better anti-knock performance. (a)(ii) State two differences between cracking and reforming (4 marks) 1. Cracking breaks down large hydrocarbon molecules into smaller ones, while reforming rearranges hydrocarbon molecules without changing their size. 2. Cracking produces lighter products such as gasoline, while reforming produces high-octane aromatic compounds. (b)(i) Draw a labelled diagram to illustrate the laboratory preparation of carbon(IV) oxide from marble (5 marks) In your answer booklet, draw: · A flask containing marble chips (calcium carbonate) and dilute hydrochloric acid · A delivery tube from the flask to a gas jar · A gas jar to collect the gas by downward delivery (upward displacement of air) · Label: marble chips, dilute HCl, delivery tube, gas jar, CO₂ gas (b)(ii) Write the equation for the reaction (2 marks) CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) (c)(i) Define vulcanization (2 marks) Vulcanization is a chemical process in which natural rubber is heated with sulphur (or other curatives) to improve its strength, elasticity, and resistance to heat and wear. (c)(ii) Give one example each of: (2 marks) I. Natural rubber – Latex from Hevea brasiliensis (rubber tree) II. Synthetic rubber – Neoprene (or Buna-S, Buna-N, Styrene-butadiene rubber) (c)(iii) Solubility of NaCl: 39.8 g at 100°C, 35.9 g at 15°C in 80 g water (6 marks) At 100°C: mass of NaCl dissolved in 80 g water = (39.8/100) × 80 = 31.84 g At 15°C: mass of NaCl dissolved in 80 g water = (35.9/100) × 80 = 28.72 g Mass precipitated = 31.84 - 28.72 = 3.12 g Answer: 3.12 g =========================================================== QUESTION 6 =========================================================== (a)(i) Define peroxide (2 marks) A peroxide is a compound that contains the peroxide ion (O₂²⁻) or the -O-O- group. It contains oxygen in an oxidation state of -1. (a)(ii) Give: (4 marks) I. One example of peroxides – Hydrogen peroxide (H₂O₂) (or sodium peroxide Na₂O₂, barium peroxide BaO₂) II. One example of monobasic acids – Hydrochloric acid (HCl) (or HNO₃, CH₃COOH) III. Two physical properties of acids (2 marks): · They have a sour taste. · They turn blue litmus paper red. (a)(iii) Define strong electrolyte (2 marks) A strong electrolyte is a substance that completely dissociates into ions when dissolved in water, conducting electricity very well. (a)(iv) Mention one example of strong electrolyte (1 mark) Sodium chloride (NaCl) (or HCl, NaOH, KNO₃) (b)(i) Define pH of a solution (2 marks) pH is a measure of the acidity or alkalinity of a solution. It is the negative logarithm (base 10) of the hydrogen ion concentration: pH = -log₁₀[H⁺]. (b)(ii) Solutions A(2), B(5), C(7), D(8), E(12) (3 marks) I. The most acidic – A (pH = 2) II. The most basic – E (pH = 12) III. Neutral – C (pH = 7) (b)(iii) State one difference between deliquescent and efflorescent compounds (2 marks) Deliquescent compounds absorb moisture from the atmosphere and dissolve in it to form a solution. Efflorescent compounds lose water of crystallisation to the atmosphere and become powdery. (b)(iv) Give one example of each (2 marks) I. Deliquescent compound – Calcium chloride (CaCl₂) (or NaOH, KOH) II. Efflorescent compound – Copper(II) sulphate pentahydrate (CuSO₄·5H₂O) (or sodium carbonate decahydrate Na₂CO₃·10H₂O) (c)(i) List two physical properties of water (2 marks) 1. Water is a colourless, odourless, and tasteless liquid at room temperature. 2. Water has a boiling point of 100°C and a freezing point of 0°C at standard atmospheric pressure. (c)(ii) Glass vessel filled with gas X weighed 28.92 g, empty vessel weighed 26.42 g at 25°C. Mass of hydrogen at 25°C = 0.31 g. Calculate relative molecular mass of gas X (5 marks) Mass of gas X = 28.92 - 26.42 = 2.50 g Mass of hydrogen at same conditions = 0.31 g Relative molecular mass of X = (Mass of X / Mass of H₂) × Molar mass of H₂ = (2.50/0.31) × 2 = 8.064 × 2 = 16.13 g/mol Answer: 16.1 g/mol =========================================================== *END OF ANSWERS – QUESTIONS 1 TO 6* ==============================================
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