JAMB Chemistry 2019 Past Questions & Answers

The oxidation of alkanols depends on their class. A secondary alkanol has its hydroxyl (-OH) group attached to a carbon atom that is bonded to two other carbon atoms. When oxidized, a secondary alkanol loses hydrogen atoms to form an alkanone, which is also known as a ketone. Unlike primary alkanols, the oxidation stops at this stage.

Common mistake: Students often mistakenly choose alkanal or alkanoic acid because they forget that only primary alkanols are oxidized to form these compounds, while secondary alkanols stop at the alkanone stage.

The reaction with an ammoniacal solution of copper(I) chloride is a chemical test used to identify terminal alkynes. Terminal alkynes are hydrocarbons with a triple bond at the end of the carbon chain, which means they have an acidic hydrogen atom attached to a triply-bonded carbon (C≡C-H). But-1-yne is a terminal alkyne, and its acidic hydrogen reacts with the copper(I) ion to form an insoluble reddish-brown precipitate of copper(I) butynide. The other options do not have this specific structure and will not react.

Common mistake: Students often mistakenly choose But-2-yne because they remember that alkynes undergo this reaction, but they forget that only terminal alkynes with an acidic hydrogen will form a precipitate.

The compound CH₃CH₂COOCH₂CH₃ is an ester, which is formed from an alkanoic acid and an alkanol. To name an ester, we first name the alkyl group that came from the alcohol, which is the part attached to the single oxygen atom; in this case, it is the ethyl group (CH₂CH₃). Then, we name the part that came from the alkanoic acid, which has the C=O group, by changing the '-oic acid' ending to '-oate'. Here, CH₃CH₂COO- comes from propanoic acid (3 carbons), so it becomes propanoate, and putting it all together gives ethyl propanoate.

Common mistake: Students often miscount the number of carbon atoms in the alkanoate part, forgetting to include the carbon in the carbonyl group (C=O), which leads them to choose ethylethanoate instead of ethylpropanoate.

First, calculate the moles of carbon and hydrogen from the products. Moles of CO₂ = 89.6 dm³ / 22.4 dm³/mol = 4 moles, which gives 4 moles of carbon atoms. Moles of H₂O = 54 g / 18 g/mol = 3 moles. Since each water molecule (H₂O) has two hydrogen atoms, this gives 3 x 2 = 6 moles of hydrogen atoms. Therefore, the ratio of carbon to hydrogen in the hydrocarbon is 4:6, making the formula C₄H₆.

Common mistake: A common mistake is forgetting to multiply the moles of water by 2 to find the moles of hydrogen atoms, which would lead to an incorrect C:H ratio.

Isomers are compounds that have the same molecular formula but different structures. To find the isomer, we first determine the molecular formula of 2-methylprop-1-ene, which is C4H8. We then check the options for a compound with the same formula; but-2-ene (CH3-CH=CH-CH3) also has the molecular formula C4H8. Since both compounds have the same formula (C4H8) but different structures, they are isomers.

Common mistake: A common mistake is miscounting the total number of carbon atoms in each molecule, leading students to incorrectly select an option like 2-methyl but-1-ene which has five carbons (C5H10) instead of four.

Tollen's reagent is a chemical test used to distinguish terminal alkynes from other hydrocarbons. Terminal alkynes are hydrocarbons with a carbon-carbon triple bond at the end of the molecule, and they have the general formula CₙH₂ₙ₋₂. The hydrogen atom attached to the triple-bonded carbon is slightly acidic and reacts with the silver ions in Tollen's reagent to form a whitish precipitate of silver acetylide. Therefore, hydrocarbons that conform to the general formula CₙH₂ₙ₋₂ will react with Tollen's reagent.

Common mistake: Students may mistakenly associate Tollen's reagent only with aldehydes, which are related to alcohols (CₙH₂ₙ₊₁OH), but they forget that the question specifically asks for a class of hydrocarbons.

To name this compound using IUPAC rules, we first find the longest carbon chain, which has two carbons, so the parent name is 'ethane'. Next, we number the chain. When different halogens are attached, we number from the end that gives the substituent appearing first in the alphabet the lower number. Since 'bromo' comes before 'chloro' and 'fluoro' alphabetically, the carbon with the bromine atom is numbered as carbon-1. Therefore, the correct name is 1-bromo-1-chloro-2,2,2-trifluoroethane.

Common mistake: A common mistake is to number from the other end (the CF₃ side) to give the substituents the lowest possible set of numbers, which would incorrectly lead to the name 2-bromo-2-chloro-1,1,1-trifluoroethane.

Synthetic rubbers are polymers made from small repeating units called monomers, usually through addition polymerization. This process requires the monomers to have carbon-carbon double bonds. Isoprene and butadiene are common monomers for making rubber, while neoprene is a type of synthetic rubber itself. Hexane, however, is a saturated alkane with only single bonds, so it cannot undergo addition polymerization to form a rubber.

Common mistake: A student might mistakenly choose isoprene because it is the monomer for natural rubber, but it is also used to produce a synthetic version called polyisoprene.

When magnesium ribbon is burnt in air, it reacts with oxygen to form white magnesium oxide powder. Because magnesium is a highly reactive metal, it also reacts with the abundant nitrogen gas in the air to form magnesium nitride. Therefore, both oxygen and nitrogen are removed from the air during this reaction.

Common mistake: Students often mistakenly choose option C because they only remember the common reaction of combustion with oxygen and forget that very reactive metals like magnesium can also react with nitrogen.

The way metal trioxonitrates(V) break down with heat depends on the metal's position in the reactivity series. Nitrates of moderately reactive metals like lead (Pb) decompose to give the metal oxide, nitrogen(IV) oxide, and oxygen. Nitrates of very reactive metals like sodium (Na) and calcium (Ca) only decompose to the metal nitrite and oxygen. The decomposition of lead(II) trioxonitrate(V) is a classic laboratory method for preparing oxygen, making it the best choice.

Common mistake: Students often mistakenly select Na, remembering that its nitrate produces oxygen, but they overlook that the decomposition of lead(II) nitrate is a more common laboratory example that yields oxygen as one of its products.

Sodium ions (Na+) are the major positive ions in the fluid outside of cells, including blood plasma. They play a crucial role in maintaining the body's fluid balance through osmosis, which directly regulates blood volume and blood pressure. This process is known as osmoregulation, and sodium is the most essential metal for it.

Common mistake: Students often mistakenly choose Iron because they know it is vital for blood, but its main role is in haemoglobin for oxygen transport, not in regulating fluid balance and blood pressure.

Alkaline earth metals are in Group 2 of the periodic table, and their reactivity with water increases as you go down the group. This is because the atoms get larger, and the outermost electrons become easier to lose in a reaction. Among the options given, Strontium (Sr) is the lowest in the group, making it the most reactive and therefore the one that reacts most spontaneously with water.

Common mistake: Students may mistakenly believe that reactivity decreases down the group and incorrectly choose Beryllium, which is the least reactive metal among the options.

When chlorine gas dissolves in water, it first forms a mixture of hydrochloric acid (HCl) and hypochlorous acid (HOCl). However, the hypochlorous acid is unstable and, in the presence of bright sunlight, it decomposes. This decomposition reaction breaks down the HOCl to produce more hydrochloric acid (HCl) and oxygen gas (O₂). Therefore, the final products are HCl and O₂.

Common mistake: Students often incorrectly choose option D because they remember that hypochlorous acid (HOCl) is an intermediate product, but they forget that it is unstable and fully decomposes in sunlight.

Chlorine is a more reactive halogen than iodine, so it displaces iodine from the potassium iodide salt present on the paper. In this process, chlorine acts as an oxidizing agent, taking electrons from the iodide ions (I⁻) and oxidizing them to form iodine molecules (I₂). This newly formed iodine then reacts with the starch on the paper to produce the characteristic blue-black colour.

Common mistake: Students might mistakenly choose option B, thinking the reaction between chlorine and iodide directly produces the colour, without realizing that this reaction first produces iodine which then reacts with the starch.

Oxides are classified based on their reaction with acids and bases. Neutral oxides are those that show neither acidic nor basic properties. The compounds in option A, which are dinitrogen monoxide (N₂O), water (H₂O), and carbon monoxide (CO), are all standard examples of neutral oxides. The other options are incorrect because they contain non-neutral oxides, such as the acidic oxide SO₂ (in option B), the basic oxide Na₂O (in option C), and the amphoteric oxide ZnO (in option C).

Common mistake: Students often mistakenly choose option B, forgetting that while N₂O and CO are neutral, sulphur(IV) oxide (SO₂) is a common acidic oxide.

In the industrial Contact process, the reaction between sulphur(VI) oxide (SO₃) and water is extremely exothermic, releasing a large amount of heat very quickly. This violent reaction creates a dense, corrosive mist of sulphuric acid that is difficult to collect and dangerous to handle. To avoid this, the SO₃ is first dissolved in concentrated H₂SO₄ to form a less reactive intermediate called oleum (H₂S₂O₇), which is then safely diluted with water to produce more sulphuric acid.

Common mistake: Students often confuse the specific reason for this industrial step with the general lab safety rule of adding acid to water (option B), but the core issue here is the extreme violence of the SO₃-water reaction that forms an uncontrollable acid fog.

Rutherford's nuclear theory, based on his gold foil experiment, proposed that an atom has a small, dense, and positively charged nucleus containing protons. However, the neutron was discovered much later in 1932 by James Chadwick. Therefore, Rutherford's original account of the atom did not include neutrons, making this statement the one that does not fit his theory.

Common mistake: Students may mistakenly believe that since the modern atomic nucleus contains protons and neutrons, this was always known, forgetting that the neutron was discovered by James Chadwick years after Rutherford proposed his model.

Element X has the electronic configuration 1s²2s²2p⁴, meaning it has 6 valence electrons and needs to gain 2 electrons to be stable, giving it a valency of 2 (X²⁻). Element Y has the configuration 1s²2s²2p⁶3s²3p¹, with 3 valence electrons which it will lose to be stable, giving it a valency of 3 (Y³⁺). To form a neutral compound, the valencies are exchanged using the 'cross-over' method, where the valency of Y (3) becomes the subscript for X, and the valency of X (2) becomes the subscript for Y, resulting in the formula Y₂X₃.

Common mistake: Students often make the mistake of incorrectly assigning the subscripts after the cross-over, leading them to choose X₂Y₃ by swapping the valencies the wrong way around.

This question is solved using the combined gas law, which is (P₁V₁)/T₁ = (P₂V₂)/T₂. First, convert the initial temperature to Kelvin: T₁ = 27°C + 273 = 300 K. Let the initial pressure P₁ be 'P', so the final pressure P₂ is '2P', and we are given V₁=200cm³ and V₂=150cm³. Plugging in the values gives (P × 200) / 300 = (2P × 150) / T₂, which simplifies to T₂ = 450 K. The question asks for the increase in temperature, which is the final temperature minus the initial temperature (450 K - 300 K = 150 K), and a change of 150 K is equal to a change of 150°C.

Common mistake: Students often mistakenly select the final temperature (177°C) as the answer, forgetting that the question asks for the amount the temperature must be increased by.

From the balanced equation, 2 volumes of CO react with 1 volume of O₂ to form 2 volumes of CO₂. This means 50 cm³ of CO will completely react with only 25 cm³ of O₂ and produce 50 cm³ of CO₂. Since we started with 100 cm³ of oxygen, the unreacted oxygen is 100 cm³ - 25 cm³ = 75 cm³. The final mixture contains the product (50 cm³ of CO₂) and the leftover reactant (75 cm³ of O₂), so the total volume is 50 + 75 = 125 cm³.

Common mistake: Students may correctly calculate the volume of unreacted oxygen as 75 cm³ (Option A) but forget that the final mixture also contains the 50 cm³ of CO₂ that was produced.

In an ammonia (NH₃) molecule, the central nitrogen atom has a lone pair of electrons after forming three covalent bonds with hydrogen. A hydrogen ion (H⁺) is a proton with an empty orbital that needs two electrons to be stable. The nitrogen atom donates its entire lone pair of electrons to the hydrogen ion to form a new N-H bond. A covalent bond where one atom provides both of the shared electrons is called a dative covalent bond.

Common mistake: Students might mistakenly choose 'ionic bond' because the reaction involves an ion (H⁺), but the bond is formed by sharing electrons from one atom, not by the complete transfer of electrons between a metal and a non-metal.

Ammonium chloride (NH₄Cl) is made up of two parts: the ammonium ion (NH₄⁺) and the chloride ion (Cl⁻). The strong force of attraction between the positively charged ammonium ion and the negatively charged chloride ion is an ionic bond. Within the ammonium ion itself, the nitrogen atom is joined to four hydrogen atoms by covalent bonds, which involve the sharing of electrons. Therefore, the compound contains both ionic and covalent bonding.

Common mistake: Students may mistakenly choose 'ionic only' because they recognize ammonium chloride as a salt and focus solely on the attraction between the ammonium cation (NH₄⁺) and the chloride anion (Cl⁻).

According to Le Chatelier's principle, if you decrease the pressure on a system at equilibrium, the system will shift to the side with more moles of gas to oppose the change. In this reaction, the reactant side has 2 moles of gas (from 2B) while the product side has only 1 mole of gas (from D). Therefore, the equilibrium will shift to the left (the reactant side), which means the rate of the backward reaction will increase to produce more reactants.

Common mistake: A common mistake is to miscount the moles of gas on each side or forget that the shift is towards the side with more moles for a pressure decrease, leading to the incorrect conclusion that the forward reaction is favoured.

The equilibrium constant expression, Kc, is defined as the ratio of the concentration of products to the concentration of reactants for a reaction at equilibrium. Each concentration is raised to the power of its stoichiometric coefficient from the balanced chemical equation. Therefore, for the reaction xP + yQ ⇌ mR + nS, the expression is the concentration of products ([R]ᵐ[S]ⁿ) divided by the concentration of reactants ([P]ˣ[Q]ʸ).

Common mistake: Students often mistakenly choose option C because they incorrectly place the concentration of reactants in the numerator and the products in the denominator, which is the inverse of the correct expression.

Oxidation is a chemical process that involves the loss of electrons or an increase in the oxidation number of an element. In option B, the oxidation number of oxygen increases from -2 in the oxide ion (O²⁻) to 0 in the oxygen molecule (O₂). This increase in oxidation number signifies that an oxidation process has occurred. All other options show a decrease in oxidation number, which is reduction.

Common mistake: A common mistake is to only look for the loss of electrons, but the equation in option B is written incorrectly; students should instead check for an increase in oxidation number to correctly identify the oxidation process.