WAEC Chemistry 2021
Questions & Answers

Distractor check: A student might select glucose, sucrose, or fatty acids by confusing protein building blocks with carbohydrates or lipids. Reasoning to the answer: Proteins are large polymers built from amino acid monomers linked together by peptide bonds. When proteins undergo hydrolysis in the presence of dilute mineral acids, these peptide bonds are cleaved to release free amino acids. Common mistake: Confusing the monomer units of proteins with those of carbohydrates or fats.

Distractor check: A student might pick carbon monoxide or hydrogen peroxide by mistakenly associating common pollutants or household chemicals with precipitation acidity. Reasoning to the answer: Nitrogen(IV) oxide is a major contributor to acid rain because it reacts directly with atmospheric water and oxygen to produce corrosive nitric acid according to the equation 4NO₂ + 2H₂O + O₂ → 4HNO₃. Common mistake: Confusing nitrogen dioxide's acid-rain-forming properties with other gaseous pollutants.

Distractor check: A student might pick C₂H₆, C₂H₄, or CH₂ by miscalculating the molecular mass scaling factor or confusing empirical and molecular formulas. Reasoning to the answer: A carbon-to-hydrogen ratio of 1:2 gives an empirical formula of CH₂ with a mass unit of 14. Dividing the given molecular mass of 56 by this empirical mass yields a multiplier of 4, resulting in the molecular formula C₄H₈. Common mistake: Stopping at the empirical formula instead of scaling it up using the molecular mass.

Distractor check: A student might pick 30, 44, or 74 by miscalculating individual atomic contributions or miscounting the constituent atoms of ethanoic acid. Reasoning to the answer: The relative molecular mass of ethanoic acid, CH₃COOH, is determined by summing the atomic masses of its component atoms: two carbon atoms at 12 each, four hydrogen atoms at 1 each, and two oxygen atoms at 16 each, giving 2(12) + 4(1) + 2(16) = 60. Common mistake: Incorrectly summing or omitting individual atomic weights during molecular mass calculation.

Distractor check: A student might select electrical conductivity or the nature of oxides formed by assuming corrosion prevention relies on surface coatings rather than electrochemical principles. Reasoning to the answer: Cathodic protection works by attaching the metal you want to protect to a more reactive sacrificial metal that acts as an anode. This strategy relies on the relative oxidation tendencies and standard electrode potentials of the metals involved. Common mistake: Confusing sacrificial cathodic protection with physical barrier coatings.

Distractor check: A student might choose solid NO₂, saturated NaCl, or acidified KMnO₄ by guessing based on general chemical reagents without matching the specific reduction reaction needed for chlorine gas. Reasoning to the answer: Discharged chlorine gas in humid air is a strong oxidizing agent that can be safely neutralized and removed by reduction when treated with acidified iron(II) sulfate, producing chloride ions and iron(III) ions. Common mistake: Choosing generic oxidizing reagents instead of a reducing agent like Fe²⁺ to neutralize chlorine gas.

Students might mistakenly choose B because they see multiple methyl groups, assuming any branching automatically makes an alcohol tertiary. To determine the correct classification, we must look at the specific carbon that bears the hydroxyl (-OH) group; in (CH₃)₃COH, this central carbon is directly linked to three distinct methyl groups rather than just two or one. Because tertiary alcohols are defined by having their -OH group bonded to a carbon that is itself connected to three other carbon atoms, option C is the right choice. Common mistake: Confusing a secondary alcohol (which has the carbon bonded to two carbons) with a tertiary alcohol due to the presence of branching.

A student might be tempted by option A or B, thinking that mixing random salt solutions will always yield a visible reaction. When we combine potassium sulfate and barium chloride, however, a double displacement reaction takes place where barium ions react with sulfate ions to produce barium sulfate, represented by the balanced equation K₂SO₄ + BaCl₂ → BaSO₄ + 2KCl. This specific compound, barium sulfate, forms an insoluble white precipitate in water, whereas the other pairs exclusively produce fully soluble products that remain dissolved. Common mistake: Assuming all ionic combinations will produce a precipitate without checking the solubility rules for the resulting products.

A test-taker might select option D, knowing that higher temperatures generally speed up chemical processes, but the question specifically asks what increases the rate through direct particle frequency in this reaction setup. According to collision theory, raising the concentration of hydrochloric acid increases the total count of reactive H⁺ ions present in the mixture, which directly leads to a greater frequency of effective collisions with the magnesium surface. Volume alone will not alter this rate if the overall concentration remains unchanged. Common mistake: Confusing the effects of concentration and total solution volume on the frequency of reactant collisions.

One might incorrectly select option A, assuming reactions are most vigorous and therefore shift pH fastest right at the start. During the titration of a strong acid with a strong base, the steepest pH change actually happens at the end point—also known as the equivalence point—because a sudden stoichiometric shift occurs as you transition from a slight excess of acid to a slight excess of base with minimal buffering capacity in the mixture, producing a nearly vertical jump on the titration curve. Common mistake: Assuming the rate of pH change is uniform throughout the entire titration curve or highest at the very beginning.

It is easy to mistakenly select option A, as students often generalize that heating a mixture always enhances solubility. When a salt dissolves via an exothermic process, heat is released as a product of the dissolution, meaning Le Chatelier's principle dictates that raising the ambient temperature will shift the equilibrium position back toward the reactants (undissolved salt). Consequently, this temperature increase suppresses the process, meaning the solubility of the salt actually decreases. Common mistake: Forgetting that Le Chatelier's principle applies in reverse to exothermic dissolution processes compared to endothermic ones.

A student might choose option A, mistakenly thinking that removing something implies the reverse of an addition process. The dehydration of ethanol proceeds via the chemical equation CH₃CH₂OH → C₂H₄ + H₂O, which involves the removal of a hydrogen atom and a hydroxyl group from adjacent carbon atoms in the molecular chain. Since atoms are taken away to form a double bond rather than added to one, this classifies the transformation as a classic elimination reaction. Common mistake: Misinterpreting the removal of elements as a substitution or addition step because water is a product.

One might select option B or C, forgetting which gases can actively participate in intermolecular bonding with water solvents. Ammonia (NH₃) dissolves readily at room temperature because it readily forms strong hydrogen bonds with water molecules and further reacts to generate ammonium and hydroxide ions (NH₄⁺ and OH⁻). The other listed gases lack this specific capability, resulting in significantly lower solubility levels. Common mistake: Overlooking the role of hydrogen bonding and assuming all gases share similar solubility properties in aqueous solutions.

A student might guess option A due to the familiar diatomic structure of standard elemental gases like nitrogen or oxygen. White phosphorus, however, adopts a unique structural arrangement consisting of four atoms bound together in a spatial configuration known as P₄ tetrahedral molecules, making it tetraatomic. This specific molecular geometry is exceptionally stable and directly accounts for its unique chemical properties. Common mistake: Assuming all non-metal elements naturally exist as diatomic molecules in their elemental state.

A test-taker could mistakenly choose option C, confusing the general preparation of soluble salts via acid-base neutralization with the specific synthesis of insoluble ones. Insoluble salts are most commonly produced through double decomposition—also known as precipitation or metathesis—where two soluble salt solutions are mixed so that their ions swap partners, yielding an insoluble solid precipitate alongside a soluble salt, as seen in AgNO₃ + NaCl → AgCl ↓ + NaNO₃. Common mistake: Confusing neutralization methods with double decomposition when dealing with insoluble reaction products.

One might select option B or C by making a calculation error or misplacing decimal points when handling the gas constant. To determine the correct value, we apply the ideal gas law equation rearranged for moles: n = PV / (RT). Substituting the given values gives n = (1 atm × 8.2 dm³) / (0.082 atm dm³ mol⁻¹ K⁻¹ × 320 K), which simplifies to 8.2 / 26.24, resulting in approximately 0.3125 mol, which rounds to 0.32 mol. Common mistake: Failing to correctly divide the product of pressure and volume by the product of the gas constant and temperature.

A student might mistakenly pick option A, confusing efflorescence with moisture removal when observing physical changes in salts exposed to air. Drying agents function properly because they are hygroscopic, meaning they possess the chemical property of actively pulling and absorbing moisture directly from the surrounding atmosphere to form stable hydrates, such as when calcium chloride (CaCl₂) dries gases. Common mistake: Mixing up hygroscopic absorption of water with efflorescent loss of water of crystallization.

A person might select option C by remembering that ammonia forms precipitates with various metal ions, but forgetting the distinct color associated with copper chemistry. When aqueous ammonia is introduced dropwise to copper(II) ions, a light blue precipitate of copper(II) hydroxide forms via Cu²⁺ + 2NH₃, but adding excess ammonia causes this precipitate to redissolve completely, yielding a deep blue complex solution containing [Cu(NH₃)₄]²⁺. Common mistake: Confusing the color of copper precipitates or forgetting that they can redissolve in excess reagent to form coordination complexes.

A student might choose option A, assuming that mixing two different salt solutions containing common metal and sulfate/nitrate ions will yield no observable reaction. Upon combining copper(II) sulfate and lead(II) nitrate, a precipitation reaction occurs according to the equation CuSO₄ + Pb(NO₃)₂ → PbSO₄ ↓ + Cu(NO₃)₂. Lead(II) sulfate is famously insoluble in water, meaning it immediately precipitates out as a distinct white solid. Common mistake: Assuming all double displacement reactions between clear salt solutions result in clear, unchanged mixtures.

One might select option A, overlooking the fact that it represents the exact same straight-chain molecule rather than a structural isomer. Butane has the molecular formula C₄H₁₀, and while option A is simply n-butane, option B displays 2-methylpropane (isobutane), which shares that identical atomic formula but features a branched carbon skeleton. Options C and D are incorrect because they represent propane and pentane, possessing different carbon chain lengths. Common mistake: Selecting the straight-chain version of a compound when asked to identify its structural isomer.

A student might choose option A, thinking that increasing both temperature and pressure simultaneously will consistently expand a flexible container. According to the combined gas law relationship where volume varies directly with temperature and inversely with pressure (V ∝ T/P), the volume of gas inside a balloon will only increase if the absolute temperature is raised while the external pressure is simultaneously reduced. Common mistake: Forgetting that pressure exerts an inverse effect on gas volume compared to temperature.

One might choose option D, recognizing that particle spacing is a core postulate of gas laws, but missing the specific mechanical definition of elasticity. Kinetic theory dictates that collisions among ideal gas particles must be perfectly elastic so that no overall kinetic energy is lost during impacts, which preserves the foundational thermodynamic assumptions regarding constant temperature and pressure. Common mistake: Confusing particle separation distances with the energy-conserving nature of elastic collisions.

A student might pick option A or B by misinterpreting how periodic trends dictate electron stability. High ionization energy means it takes a tremendous amount of energy to strip an electron away, a phenomenon that occurs because the valence electrons are held exceptionally tightly by a high effective nuclear charge (Z_eff) pulling inward from the nucleus. Common mistake: Believing that atoms with high ionization energy have weak nuclear attraction or lose electrons easily.

A test-taker might select option D, incorrectly assuming all non-metals share the same molecularity as noble gases or simple gases. For Group VII elements (the halogens), reactivity actually declines progressively down the group because atomic radius increases and electronegativity drops, making elements like fluorine vastly more reactive than iodine. Common mistake: Assuming chemical reactivity increases down a non-metal group the way it often does for heavy alkali metals.

A student might mistakenly select option B, confusing where reduction takes place with the site of electron release in an electrochemical cell. In any galvanic cell, the anode serves as the designated negative electrode where oxidation—defined as the loss of electrons—takes place, freeing those electrons to travel through the external circuit. Common mistake: Mixing up the anode and cathode regarding whether oxidation or reduction occurs there.

One might choose option C, picking a half-reaction where a species loses electrons, which actually describes an oxidation process. The cathode is universally defined as the site where reduction occurs, meaning it involves the gain of electrons by a positive ion to form a neutral solid, perfectly represented by option A where A³⁺(aq) + 3e⁻ yields A(s). Common mistake: Selecting an equation representing loss of electrons (oxidation) when asked for a cathode reaction.

A student might mistakenly select option D, thinking a strong bond implies high reactivity rather than stability. Fluorine's extreme reactivity is driven primarily by its exceptionally high electronegativity value of 4.0, which allows it to exert a massive attractive force on electrons and forcefully pull them away from other atoms during chemical bonding. Common mistake: Confusing bond strength with electronegativity when explaining why fluorine reacts so vigorously.

One might mistakenly choose option A or B by miscalculating the stoichiometric Faraday equivalence or misplacing decimal places during conversion. To find the required electricity, we first determine the moles of silver deposited by dividing the mass by its atomic weight: 1.08 g / 108 g/mol = 0.01 mol. Since silver ions require 1 Faraday (96500 C) per mole of atoms, multiplying 0.01 mol by 96500 C gives exactly 965 C. Common mistake: Forgetting to adjust the Faraday constant by the number of moles calculated from the mass.

A student might choose option C or D, focusing on subatomic particles discovered later rather than the specific structural breakthrough of this particular model. The Bohr model revolutionized atomic theory by proposing that electrons do not wander randomly, but instead travel in fixed, discrete circular orbits or electron shells around the central nucleus, which successfully explained observed atomic emission spectra. Common mistake: Attributing the discovery of neutrons or nucleons specifically to the Bohr atomic model.

A test-taker might select option B or D by miscalculating the volume conversion or molar mass ratios. First, calculate the moles of potassium carbonate obtained by dividing the recovered mass by its molar mass: 14 g / 138 g/mol ≈ 0.101 mol. Since this solute was extracted from a 100 cm³ volume (which equals 0.1 dm³), dividing the moles by this volume in cubic decimeters yields a solubility of 0.101 mol / 0.1 dm³, resulting in 1.01 mol dm⁻³ in option C. Common mistake: Forgetting to convert the solution volume from cubic centimeters to cubic decimeters when calculating molarity.

A student might choose option B, assuming that changing indicators has no mathematical impact on experimental titration volumes. Because sodium carbonate undergoes a two-step neutralization with hydrochloric acid, phenolphthalein changes color at the first equivalence point (pH 8.3, corresponding to a 1:1 mole ratio where Na₂CO₃ becomes NaHCO₃), whereas methyl orange changes at the second equivalence point (pH 4.4, requiring a 2:1 mole ratio for complete neutralization). Consequently, Student X's titre value using methyl orange must be exactly double the titre value obtained by Student Y. Common mistake: Overlooking the fact that indicators with different pH transition ranges measure different stages of a polyprotic base neutralization.

1) Distractor check: A student might mistakenly select A by simply dividing the mass by the molar mass without considering the volume in cubic decimetres, or choose C or D through incorrect decimal placements during calculation. 2) Reasoning to the answer: To find the concentration, start by converting the given volume into cubic decimetres by dividing 100 cm³ by 1000, giving 0.1 dm³. Next, calculate the number of moles of potassium hydroxide by dividing the mass of 0.28 g by its molar mass of 56 g/mol, which yields 0.005 mol. Finally, divide these moles by the volume in dm³ (0.005 / 0.1) to arrive at the final concentration of 0.05 mol dm⁻³. 3) Common mistake: Forgetting to convert cm³ into dm³ before calculating concentration.

1) Distractor check: Options A, B, and D are plausible if a student improperly simplifies the given mole ratio or mistakes the subscript numbers for the number of moles directly without establishing the simplest whole number relationship. 2) Reasoning to the answer: Determine the simplest whole number ratio of the elements by comparing the given moles of carbon and hydrogen, which are 0.160 and 0.640 respectively. Dividing both values by the smaller mole quantity (0.160) yields a carbon to hydrogen ratio of 1 to 4. Consequently, applying these subscripts results in the empirical formula CH₄. 3) Common mistake: Failing to divide by the smallest mole value when establishing the ratio.

1) Distractor check: Students often choose B, C, or D because they contain heavier elements or higher atomic numbers, assuming that more protons inherently mean a larger ionic radius. 2) Reasoning to the answer: All the listed species are isoelectronic, possessing 18 electrons in total, meaning their size is determined by the effective nuclear charge pulling the electron cloud inward. As nuclear charge increases from sulfur with 16 protons up to calcium with 20 protons, the electrostatic attraction strengthens and shrinks the radius. Therefore, S²⁻ has the fewest protons and the weakest inward pull, making its ionic radius the largest among them. 3) Common mistake: Assuming ionic radius always increases with atomic number regardless of electron count.

1) Distractor check: A student might select A or B because variations across periods and the formation of ions are commonly discussed alongside ionization energy trends in periodic table lessons. 2) Reasoning to the answer: Ionization energy refers to the energy required to remove an electron from a gaseous atom. Moving down a group, the atomic radius increases and the inner shell shielding effect becomes more pronounced, which weakens the attraction between the nucleus and the outermost valence electrons, making them progressively easier to remove and thus causing the ionization energy to decrease. 3) Common mistake: Confusing trends down a group with trends across a period.

1) Distractor check: Options like distillation, evaporation, or filtration might be chosen if a student confuses the separation techniques used for soluble salts versus insoluble mixtures or liquids. 2) Reasoning to the answer: Potassium trioxonitrate(V) is a soluble salt dissolved in an aqueous medium. To obtain pure crystals from this solution, one employs crystallization, which relies on cooling or evaporating a saturated solution to allow the solute to form solid, pure crystals. 3) Common mistake: Selecting evaporation to dryness, which can decompose heat-sensitive salts or trap impurities, instead of crystallization.

1) Distractor check: A student might pick A, C, or D by simply taking the mass of one isotope, averaging the two mass numbers directly without percentages, or confusing the percentage with the final mass. 2) Reasoning to the answer: The relative atomic mass is calculated by finding the weighted average of the isotopes' masses based on their abundance. Multiply the mass of the first isotope by its fractional abundance (63 × 0.69 = 43.47) and add it to the product of the second isotope's mass and its abundance (65 × 0.31 = 20.15). Summing these values gives 63.62, which rounds to 63.6, matching the value for copper isotopes. 3) Common mistake: Averaging the mass numbers directly (63 + 65 / 2) instead of using the percentage abundances.

1) Distractor check: Options B, C, or D might be selected if a student miscounts the electrons or confuses which ion has a different noble gas electron configuration compared to the rest. 2) Reasoning to the answer: Magnesium ion Mg²⁺ has 10 electrons, oxygen ion O²⁻ has 10 electrons, and aluminum ion Al³⁺ also has 10 electrons, meaning they all share the neon electron configuration of 1s²2s²2p⁶. In contrast, the chloride ion Cl⁻ has 17 original electrons plus one gained, totaling 18 electrons, which corresponds to the argon configuration. 3) Common mistake: Forgetting to adjust the electron count for the ionic charge when determining configuration.

1) Distractor check: Terms like spectra, quanta, or field might be chosen due to general familiarity with atomic physics terminology and energy levels. 2) Reasoning to the answer: According to quantum mechanics, electrons do not follow strict circular paths but rather occupy a three-dimensional region surrounding the atomic nucleus where the probability of finding an electron is highest. This specific region of high electron probability is defined as an orbital. 3) Common mistake: Confusing an atomic orbital with a fixed planetary orbit or an energy shell.

1) Distractor check: Students may choose B, C, or D because they describe measurable properties of subatomic particles rather than the historical definition of why they were termed fundamental. 2) Reasoning to the answer: In early atomic theory, before quarks and other subatomic discoveries, protons and electrons were regarded as the basic, indivisible building blocks that made up all matter, unlike composite particles. 3) Common mistake: Confusing modern subatomic particle classifications with historical atomic models.