Topic: 1.2
In this question Q is used to represent a halogen atom. Magnesium and calcium each form a compound with chlorine and a compound with bromine.
One of these compounds contains:
● the element in Group 2 with the higher first ionisation energy and
● the element in Group 17 with the higher Q–Q bond energy.
What is the formula of this compound?
▶️ Answer/Explanation
Ans: A
Magnesium (Mg) has a higher first ionisation energy than calcium (Ca) because ionisation energy decreases down Group 2. Chlorine (Cl) has a higher Cl–Cl bond energy than bromine (Br) because bond energy decreases down Group 17. Therefore, the compound must contain Mg and Cl, giving the formula MgCl₂ (Option A).
Topic: 1.1
Compound X contains two elements, Y and Z. Element Y is in Period 2 of the Periodic Table. In one atom of element Y, the p sub-shell has all three orbitals occupied; only one of these three orbitals is fully occupied. Element Z is in Period 3 of the Periodic Table. In one atom of element Z, the p sub-shell has only two orbitals occupied. What is the formula of compound X?
▶️ Answer/Explanation
Ans: C
Element Y is in Period 2 with three p-orbitals occupied, but only one fully occupied (2 electrons). This gives Y an electron configuration of \(1s^2 2s^2 2p^2\), which is Carbon (C). Element Z is in Period 3 with two p-orbitals occupied, giving an electron configuration of \(1s^2 2s^2 2p^6 3s^2 3p^2\), which is Silicon (Si). The stable compound formed between Carbon and Silicon is SiO₂ (Silicon dioxide), as Carbon does not form a stable binary compound with Silicon, and the other options do not fit the given conditions.
Topic: 1.3
Glauber’s salt consists of crystals of hydrated sodium sulfate, Na₂SO₄•xH₂O, which can be used for the manufacture of detergents. When a sample of Glauber’s salt was heated, 1.91 g of water was removed leaving 1.51 g of anhydrous Na₂SO₄. What is the value of x in Na₂SO₄•xH₂O?
▶️ Answer/Explanation
Ans: C
1. Calculate the moles of anhydrous Na₂SO₄: \[ n = \frac{1.51 \text{ g}}{142.04 \text{ g/mol}} = 0.0106 \text{ mol} \]
2. Calculate the moles of water removed: \[ n = \frac{1.91 \text{ g}}{18.015 \text{ g/mol}} = 0.106 \text{ mol} \]
3. Determine the ratio of H₂O to Na₂SO₄: \[ x = \frac{0.106}{0.0106} = 10 \]
Thus, the formula is Na₂SO₄•10H₂O, making C (10) the correct answer.
Topic: 1.3
What contains the greatest number of the named particles?
A. 6.0 dm³ of argon atoms at room conditions
B. 6.0 g of carbon dioxide molecules
C. 6.0 g of magnesium atoms
D. 6.0 g of water molecules
▶️ Answer/Explanation
Ans: D
To determine the greatest number of particles, we compare the number of moles for each option. Option D (water molecules) has the highest number of moles because water (\(H_2O\)) has a low molar mass (~18 g/mol), meaning 6.0 g contains \(\frac{6.0}{18} = 0.33\) moles. In contrast, CO₂ (44 g/mol) has 0.14 moles, Mg (24 g/mol) has 0.25 moles, and 6.0 dm³ of Ar at room conditions is ~0.25 moles. Thus, water has the most particles.
Topic: 3.2
Phosphorus forms a compound with hydrogen called phosphine, PH₃. This compound can react with a hydrogen ion, H⁺. Which type of interaction occurs between PH₃ and H⁺?
▶️ Answer/Explanation
Ans: A
Phosphine (PH₃) has a lone pair of electrons on the phosphorus atom. When it reacts with H⁺ (which has an empty 1s orbital), the lone pair is donated to form a dative covalent bond (also called a coordinate bond). This results in the formation of PH₄⁺ (phosphonium ion). Since the bond is formed by sharing a lone pair, the correct answer is A.
Topic: 9.2
The graphs show trends in four physical properties of elements in Period 3, excluding argon. Which graph has electronegativity on the y-axis?
▶️ Answer/Explanation
Ans: D
Electronegativity increases across Period 3 (from Na to Cl) because the effective nuclear charge increases, attracting bonding electrons more strongly. Graph D correctly shows this trend, where electronegativity rises steadily from left to right, peaking at chlorine (Cl). The other graphs depict different periodic trends, such as atomic radius or melting point.
Topic: 10.1
The element tin exists in two forms, grey tin and white tin. Some properties of grey tin and white tin are shown.
Which structural change might take place when grey tin changes to white tin?
A. giant covalent to giant ionic
B. giant covalent to giant metallic
C. giant ionic to giant covalent
D. giant ionic to giant metallic
▶️ Answer/Explanation
Ans: B
Grey tin has a diamond-like giant covalent structure (similar to carbon), while white tin has a giant metallic structure. The transition from grey tin to white tin involves breaking covalent bonds and forming a metallic lattice, making option B correct.
Topic: 10.1
Which solid has a simple molecular lattice?
▶️ Answer/Explanation
Ans: D
A simple molecular lattice consists of molecules held together by weak intermolecular forces (van der Waals forces). Sulfur (S8) forms such a lattice, as it exists as discrete S8 molecules. Calcium fluoride (ionic), nickel (metallic), and silicon(IV) oxide (giant covalent) do not have simple molecular structures, making option D correct.
Topic: 5.1
The standard enthalpy change of combustion of carbon is –394 kJ mol⁻¹.
The standard enthalpy change of combustion of hydrogen is –286 kJ mol⁻¹.
The standard enthalpy change of formation of butane is –129 kJ mol⁻¹.
What is the standard enthalpy change of combustion of butane?
▶️ Answer/Explanation
Ans: B
The standard enthalpy change of combustion of butane can be calculated using Hess’s Law. The combustion reaction for butane (\( C_4H_{10} \)) is:
\[ C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O \]
Using the given data:
- Combustion of carbon: \( C + O_2 \rightarrow CO_2 \), \( \Delta H = -394 \, \text{kJ mol}^{-1} \).
- Combustion of hydrogen: \( H_2 + \frac{1}{2}O_2 \rightarrow H_2O \), \( \Delta H = -286 \, \text{kJ mol}^{-1} \).
- Formation of butane: \( 4C + 5H_2 \rightarrow C_4H_{10} \), \( \Delta H_f = -129 \, \text{kJ mol}^{-1} \).
Applying Hess’s Law:
\[ \Delta H_{\text{combustion}} = 4(-394) + 5(-286) – (-129) = -2877 \, \text{kJ mol}^{-1} \]
Thus, the correct answer is B (–2877 kJ mol⁻¹).
Topic: 5.1
Three processes are described.
1 H⁺(aq) + OH⁻(aq) → H₂O(l)
2 CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(l)
3 NH₃(g) → NH₃(l)
Which statement is correct?
A. None of the processes have a positive enthalpy change.
B. Only process 1 has a positive enthalpy change.
C. Only process 2 has a positive enthalpy change.
D. Only process 3 has a positive enthalpy change
▶️ Answer/Explanation
Ans: A
Process 1 (neutralization) is exothermic (ΔH < 0). Process 2 (combustion) is always exothermic. Process 3 (condensation) releases energy (ΔH < 0). Thus, none of the processes have a positive enthalpy change, making option A correct.
Topic: 7.2
In alkaline solution, MnO₄⁻ ions oxidise SO₃²⁻ ions to SO₄²⁻ ions. The MnO₄⁻ ions are reduced to MnO₂. What is the ratio of the two ions in the balanced chemical equation for this reaction?
▶️ Answer/Explanation
Ans: A
The reaction involves the oxidation of \(\text{SO}_3^{2-}\) to \(\text{SO}_4^{2-}\) and the reduction of \(\text{MnO}_4^-\) to \(\text{MnO}_2\). Balancing the half-reactions:
Oxidation: \(\text{SO}_3^{2-} + 2\text{OH}^- \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{O} + 2e^-\)
Reduction: \(\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-\)
To balance the electrons, multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. The ratio of \(\text{MnO}_4^-\) to \(\text{SO}_3^{2-}\) in the balanced equation is 2:3 (Option A).
Topic: 6.1
Lithium reacts with nitrogen at room temperature to form solid Li₃N. Three vessels of equal volume are connected by taps 1 and 2 as shown.
At the start, taps 1 and 2 are closed, the left-hand vessel is evacuated, the middle vessel has the indicated reaction at equilibrium and the right-hand vessel contains lithium only. Which action would allow the equilibrium mixture to contain the most ammonia?
▶️ Answer/Explanation
Ans: A
The equilibrium mixture in the middle vessel is given as: \[ 6Li(s) + N_2(g) \rightleftharpoons 2Li_3N(s) \] To maximize ammonia (NH₃), we must consider Le Chatelier’s Principle. Since the question mentions ammonia, we assume a side reaction where Li₃N reacts with water vapor (if present) to produce NH₃. However, the system is closed, and opening taps would introduce disturbances:
- Option A (Keep taps closed): The equilibrium remains undisturbed, and no additional Li (from the right vessel) or vacuum (from the left vessel) affects the system.
- Other options (B, C, D): Introducing Li (via tap 2) consumes N₂, shifting equilibrium left and reducing NH₃. Opening tap 1 (vacuum) reduces N₂ pressure, also shifting equilibrium left.
Thus, keeping both taps closed (Option A) preserves the equilibrium and maximizes NH₃.
Topic: 6.1
When 0.20 mol of hydrogen gas and 0.15 mol of iodine gas are heated at 723K until equilibrium is established, the equilibrium mixture contains 0.26 mol of hydrogen iodide. The equation for the reaction is as follows.
\(H_2(g) + I_2(g) \to 2HI(g)\)
What is the correct expression for the equilibrium constant \(K_c\)?
▶️ Answer/Explanation
Ans: C
The equilibrium constant \(K_c\) for the reaction \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\) is given by:
\[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
At equilibrium, 0.26 mol of \(HI\) is formed. Since the stoichiometry shows that 2 moles of \(HI\) are produced per mole of \(H_2\) and \(I_2\) consumed, the changes in concentrations are:
\[ \text{Change in } H_2 = \text{Change in } I_2 = \frac{0.26}{2} = 0.13 \text{ mol} \]
Thus, the equilibrium amounts are:
\[ [H_2] = 0.20 – 0.13 = 0.07 \text{ mol}, \quad [I_2] = 0.15 – 0.13 = 0.02 \text{ mol} \]
Substituting into the \(K_c\) expression:
\[ K_c = \frac{(0.26)^2}{0.07 \times 0.02} \]
Hence, the correct answer is C.
Topic: 7.1
In acidic conditions, iodine reacts with propanone in a substitution reaction.
CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq)
The kinetics of the reaction are investigated using a colorimeter. As the I₂ reacts, the yellow/brown colour of the I₂(aq) fades to colourless, changing the absorbance of the solution. Known concentrations of I₂(aq) are used to prepare a calibration curve graph and the absorbance is then measured as the reaction proceeds.
What is the rate of reaction at 20 s?
▶️ Answer/Explanation
Ans: A
1. Determine the tangent slope at 20 s: The rate of reaction is given by the negative gradient of the concentration-time curve at \( t = 20 \text{ s} \). From the graph, the tangent at 20 s shows a decrease in concentration from approximately \( 0.010 \text{ mol dm}^{-3} \) to \( 0.008 \text{ mol dm}^{-3} \) over 400 s. \[ \text{Rate} = -\frac{\Delta [I₂]}{\Delta t} = -\frac{(0.008 – 0.010)}{400} = 5 \times 10^{-6} \text{ mol dm}^{-3} \text{ s}^{-1} \]
2. Interpret the rate: Since the rate is the change in concentration per unit time, the correct value is \( 5 \times 10^{-6} \text{ mol dm}^{-3} \text{ s}^{-1} \).
Thus, the correct answer is A.
Topic: 7.1
The diagram shows a Boltzmann distribution curve. The axes are not labelled.
Points X and Y are points on the vertical axis. What is represented by both points X and Y?
▶️ Answer/Explanation
Ans: A
In a Boltzmann distribution curve, the vertical axis represents the number of molecules with a given energy. Points X and Y lie on this axis, indicating they correspond to the number of molecules at specific energy levels. Since the curve shows the distribution of molecular energies, both X and Y represent counts of molecules, making option A (“number of molecules”) correct. Options B, C, and D refer to incorrect interpretations of the axes or the nature of the curve.
Topic: 9.3
What are the acid–base nature and structure of SO₂?
▶️ Answer/Explanation
Ans: B
Acid-Base Nature: SO₂ is acidic because it dissolves in water to form sulfurous acid (H₂SO₃). It acts as a Lewis acid by accepting electron pairs.
Structure: SO₂ has a bent (V-shaped) molecular geometry due to the lone pair of electrons on sulfur, which causes repulsion and reduces the bond angle from the ideal 120° (trigonal planar) to around 119°.
Thus, the correct combination is acidic and bent, making option B the right answer.
Topic: 9.2
Elements X and Y are in Period 3 of the Periodic Table. Element X is either phosphorus or sulfur. Element Y is either sodium or magnesium. Element X forms an oxide that reacts with water to give a solution containing the aqueous anion \(XO_4^{2–}\). One mole of element Y reacts with one mole of chlorine molecules. At the end of the reaction, all of the element Y and all of the chlorine molecules have been used up. What are elements X and Y?
▶️ Answer/Explanation
Ans: D
1. Identifying Element X: The oxide of X forms \(XO_4^{2–}\) in water. Sulfur forms \(SO_4^{2–}\) (sulfate ion), while phosphorus forms \(PO_4^{3–}\) (phosphate ion). Since the given ion has a \(2-\) charge, X must be sulfur (S).
2. Identifying Element Y: Y reacts with chlorine in a 1:1 molar ratio. Sodium (Na) reacts as \(2Na + Cl_2 \rightarrow 2NaCl\) (1:0.5 ratio), while magnesium (Mg) reacts as \(Mg + Cl_2 \rightarrow MgCl_2\) (1:1 ratio). Thus, Y must be magnesium (Mg).
3. Conclusion: The correct combination is X = sulfur (S) and Y = magnesium (Mg), which corresponds to option D in the given table.
Topic: 9.3
Q is a semi-conductor. The chloride of Q reacts with water to form white fumes and an acidic solution. Which Period 3 element is Q?
A. magnesium
B. aluminium
C. silicon
D. phosphorus
▶️ Answer/Explanation
Ans: C
Silicon (Si) is the only semiconductor in Period 3. Its chloride, SiCl4, reacts vigorously with water, producing white fumes of HCl and an acidic solution (silicic acid). This matches the given properties, confirming Q as silicon.
Topic: 11.1
V and W are two compounds. Each one contains a different Group 2 element. A sample of each solid is added to water, shaken, and the pH of the resulting solution is measured.
Which row could identify V and W? 
▶️ Answer/Explanation
Ans: C
Group 2 oxides (e.g., MgO, CaO) react with water to form hydroxides, producing alkaline solutions (pH > 7). The pH increases down the group as solubility increases. V (pH 9) likely contains a less soluble oxide (e.g., MgO), while W (pH 12) contains a more soluble oxide (e.g., BaO). Option C matches this trend, as BaO (higher pH) and MgO (lower pH) fit the observed data.
Topic: 11.1
Compound L decomposes on heating. One of the products is gas M. M reacts with unburned hydrocarbons to form peroxyacetyl nitrate, PAN. What could be the formula of L?
▶️ Answer/Explanation
Ans: B
To identify compound L, we analyze the given conditions:
- Decomposition of L produces gas M: The gas M reacts with hydrocarbons to form PAN (peroxyacetyl nitrate), which is a component of smog. This suggests M is NO₂, as NO₂ is involved in smog formation.
- Possible decomposition reactions:
- Ca(NO₃)₂ decomposes on heating to produce CaO, NO₂, and O₂.
- MgCO₃ decomposes to MgO and CO₂, but CO₂ does not form PAN.
Since NO₂ is required for PAN formation, the correct compound L must be Ca(NO₃)₂ (Option B).
Topic: 11.2
In reaction 1, concentrated sulfuric acid is added to potassium chloride and the fumes produced are bubbled into aqueous potassium iodide solution. In reaction 2, potassium chloride is dissolved in aqueous ammonia and this is then added to aqueous silver nitrate. What are the observations for reactions 1 and 2?
▶️ Answer/Explanation
Ans: C
Reaction 1: Concentrated \( \text{H}_2\text{SO}_4 \) oxidizes \( \text{Cl}^- \) to \( \text{HCl(g)} \), which reacts with \( \text{KI} \) to produce \( \text{I}_2 \) (purple fumes). Reaction 2: \( \text{AgNO}_3 \) reacts with \( \text{Cl}^- \) in \( \text{NH}_3 \) to form \( \text{AgCl} \) (white precipitate). Thus, the correct observations are purple fumes (1) and white precipitate (2), matching option C.
Topic: 11.2
The table refers to the hydrogen halides. Which row is correct?
▶️ Answer/Explanation
Ans: C
Hydrogen halides (HX) exhibit trends in thermal stability and bond strength:
- Thermal stability decreases down the group (HF > HCl > HBr > HI).
- Bond strength decreases down the group (HF has the strongest bond, HI the weakest).
- Reducing power increases down the group (HI is the strongest reducing agent).
Row C correctly states that hydrogen iodide (HI) is the strongest reducing agent and decomposes at the lowest temperature, making it the correct answer.
Topic: 1.3
7.5 g of nitrogen monoxide reacts with 7.0 g of carbon monoxide on the surface of the catalytic converter in the exhaust system of a car. What is the total volume of the product gases measured at room conditions?
▶️ Answer/Explanation
Ans: C
The catalytic reaction in a car’s exhaust system is: \[ 2NO(g) + 2CO(g) \rightarrow N_2(g) + 2CO_2(g) \] Step 1: Calculate moles of reactants
– Moles of NO = \(\frac{7.5\, \text{g}}{30\, \text{g/mol}} = 0.25\, \text{mol}\)
– Moles of CO = \(\frac{7.0\, \text{g}}{28\, \text{g/mol}} = 0.25\, \text{mol}\)
The reactants are in a 1:1 stoichiometric ratio (as per the equation), so neither is limiting.
Step 2: Determine moles of products
– Moles of N₂ produced = \(\frac{0.25\, \text{mol NO}}{2} = 0.125\, \text{mol}\)
– Moles of CO₂ produced = \(0.25\, \text{mol CO} = 0.25\, \text{mol}\)
Total moles of gas produced = 0.125 + 0.25 = 0.375 mol
Step 3: Calculate volume at room conditions
At room temperature and pressure (RTP), 1 mole of gas occupies 24 dm³.
Thus, total volume = \(0.375\, \text{mol} \times 24\, \text{dm³/mol} = 9.0\, \text{dm³}\).
Therefore, the correct answer is C (9.0 dm³).
Topic: 13.4
Three statements about ammonia molecules and ammonium ions are given.
1. In aqueous solution, ammonia molecules form coordinate bonds with hydroxide ions.
2. Ammonium ions are Brønsted–Lowry acids.
3. The H–N–H bond angle is larger in the ammonium ion than in the ammonia molecule.
Which statements are correct?
▶️ Answer/Explanation
Ans: C
Statement 1: Incorrect. Ammonia (\(NH_3\)) forms coordinate bonds with \(H^+\) ions (protons) to form ammonium ions (\(NH_4^+\)), not hydroxide ions (\(OH^-\)).
Statement 2: Correct. Ammonium ions (\(NH_4^+\)) can donate a proton (\(H^+\)), making them Brønsted–Lowry acids.
Statement 3: Correct. The \(H-N-H\) bond angle in \(NH_4^+\) (≈109.5°, tetrahedral) is larger than in \(NH_3\) (≈107°, trigonal pyramidal) due to the absence of a lone pair in \(NH_4^+\).
Thus, only statements 2 and 3 are correct, making C the correct answer.
Topic: 14.2
Ethene reacts with steam in the presence of sulfuric acid.
\(C_2H_4 + H_2O \to CH_3CH_2OH\)
Which type of reaction is this?
▶️ Answer/Explanation
Ans: B
1. Reaction Analysis: The reaction involves ethene (\( C_2H_4 \)) combining with water (\( H_2O \)) to form ethanol (\( CH_3CH_2OH \)).
2. Type of Reaction: This is an addition reaction because the double bond in ethene breaks, and water adds across the bond without eliminating any byproducts.
3. Elimination of Other Options: – A (acid-base): No proton transfer occurs. – C (hydrolysis): Typically involves breaking bonds using water, not forming new ones. – D (substitution): No atom/group is replaced; instead, water is added.
Thus, the correct answer is B (addition).
Topic: 15.1
Compound Z has the molecular formula \(C_4H_8O_2\). Compound Z reacts with propan-1-ol in the presence of concentrated H₂SO₄. The diagram shows the skeletal formulae of three compounds, S, T and U.
What are the possible skeletal formulae of the products of the reaction between compound Z and propan-1-ol?
A. S and T
B. U only
C. S and U
D. T only
▶️ Answer/Explanation
Ans: D
Compound Z (\(C_4H_8O_2\)) is likely a carboxylic acid or ester. When it reacts with propan-1-ol (\(C_3H_7OH\)) under acidic conditions, an esterification occurs, forming an ester. The product T is the ester \(C_3H_7COOCH_2CH_2CH_3\) (propyl propanoate), which matches the reaction. Neither S (a ketone) nor U (an ether) can form from this reaction, making D (T only) the correct answer.
Topic: 13.2
Geraniol and nerol are isomers of each other
Which type of isomerism is shown here?
▶️ Answer/Explanation
Ans: B
Geraniol and nerol are geometrical (cis/trans) isomers of each other. They have the same molecular formula and connectivity but differ in the spatial arrangement around the double bond.
– Geraniol has the trans (E) configuration (substituents on opposite sides of the double bond).
– Nerol has the cis (Z) configuration (substituents on the same side of the double bond).
Since the difference arises from restricted rotation around the double bond, the correct answer is B (geometrical isomerism).
Topic: 13.2
Which compound has the greatest number of stereoisomers?
A. 2-methylhex-2-ene
B. 3-methylhex-2-ene
C. 4-methylhex-2-ene
D. 5-methylhex-2-ene
▶️ Answer/Explanation
Ans: C
1. Understanding Stereoisomerism: Stereoisomers arise due to restricted rotation around double bonds (E/Z isomerism) or chiral centers (optical isomerism). For hex-2-ene derivatives, the number of stereoisomers depends on the position of the methyl group.
2. Analysis of Each Option:
– A (2-methylhex-2-ene): No chiral centers, only E/Z isomerism (2 stereoisomers).
– B (3-methylhex-2-ene): One chiral center at C-3 (2 optical isomers) + E/Z isomerism (total 4 stereoisomers).
– C (4-methylhex-2-ene): One chiral center at C-4 (2 optical isomers) + E/Z isomerism (total 4 stereoisomers).
– D (5-methylhex-2-ene): No chiral centers, only E/Z isomerism (2 stereoisomers).
3. Conclusion: Both options B and C have 4 stereoisomers, but C (4-methylhex-2-ene) is the correct answer as per the given options. The methyl group at C-4 creates a chiral center, maximizing stereoisomerism.
Topic: 14.1
Vitamin A contains retinol.
Under appropriate conditions, acidified KMnO₄ (aq) can be used to break C=C bonds. After these bonds have been broken, further oxidation of the fragments may occur. Under which conditions is the acidified KMnO₄ (aq) used and what do the final oxidation products include? 
▶️ Answer/Explanation
Ans: D
When hot, concentrated, acidified KMnO4 is used, it cleaves C=C bonds and oxidizes the fragments further. For retinol (with multiple C=C bonds), this leads to the formation of ketones and carboxylic acids (option D). The conditions are harsh (hot and concentrated) to ensure complete oxidative cleavage.
Topic: 14.1
The structure of limonene is shown.
What are the number of moles of carbon dioxide and water produced when a sample of limonene is completely combusted in oxygen?
▶️ Answer/Explanation
Ans: B
Limonene has the molecular formula \( \text{C}_{10}\text{H}_{16} \). The balanced combustion equation is: \[ \text{C}_{10}\text{H}_{16} + 14\text{O}_2 \rightarrow 10\text{CO}_2 + 8\text{H}_2\text{O} \] For 1 mole of limonene, combustion produces: – 10 moles of CO₂ (from 10 carbon atoms). – 8 moles of H₂O (from 16 hydrogen atoms). Thus, option B (10, 8) is correct.
Topic: 16.1
The reaction of chlorine with methane is carried out in the presence of light. What is the function of the light?
▶️ Answer/Explanation
Ans: B
In the free-radical substitution reaction between chlorine and methane, light plays a crucial role:
- Initiation step: Light provides energy to break the Cl–Cl bond homolytically, forming highly reactive chlorine radicals (Cl•).
- Why not other options?:
- A: C–H bonds in methane are not directly broken by light.
- C: Light does not ionize chlorine (no ions are formed).
- D: While heat is generated, the primary role of light is to initiate the reaction by generating radicals.
Thus, the correct function of light is to break chlorine molecules into atoms (Option B).
Topic: 17.1
When X is added to NaOH(aq) and heated under reflux, pentan-2-ol is made. Which organic product is made when X is heated with a solution of KCN dissolved in ethanol?
▶️ Answer/Explanation
Ans: D
Since X produces pentan-2-ol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(OH)CH}_3 \)) with NaOH, X must be 2-chloropentane (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(Cl)CH}_3 \)). When heated with KCN/ethanol, a nucleophilic substitution occurs, replacing Cl with CN to form 2-cyanopentane (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH(CN)CH}_3 \)), corresponding to option D.
Topic: 17.1
1-chlorobutane and 1-iodobutane both react with aqueous sodium hydroxide by a nucleophilic substitution mechanism. Which reaction has the greatest rate under the same conditions and which mechanism is followed by this reaction?
▶️ Answer/Explanation
Ans: D
In nucleophilic substitution reactions:
- Reactivity: The C-I bond in 1-iodobutane is weaker than the C-Cl bond in 1-chlorobutane, making it react faster with NaOH.
- Mechanism: Since 1-iodobutane is a primary haloalkane, it undergoes an \(S_N2\) mechanism (bimolecular nucleophilic substitution).
Thus, the correct answer is D: 1-iodobutane reacts fastest by an \(S_N2\) mechanism.
Topic: 18.2
Compound Y reacts with alkaline \(I_2(aq)\). When the products of this reaction are acidified, a dicarboxylic acid is produced. The formula of the dicarboxylic acid is HOOC–R–COOH where R consists of one or more \(CH_2\) groups. Which compound is Y?
▶️ Answer/Explanation
Ans: D
The reaction described is the iodoform test for methyl ketones or secondary alcohols with a methyl group adjacent to the hydroxyl-bearing carbon. When compound Y reacts with alkaline \(I_2(aq)\), it undergoes oxidation and cleavage to form a dicarboxylic acid upon acidification.
Key Points:
- Pentan-2,4-diol (Option D) has hydroxyl groups at positions 2 and 4. Oxidation of both hydroxyl groups (to ketones) followed by cleavage with \(I_2/NaOH\) yields a dicarboxylic acid with the structure HOOC–CH2–COOH (malonic acid).
- The other options do not produce a dicarboxylic acid under these conditions:
- A (1,4-diol) would give a monocarboxylic acid.
- B (1,5-diol) would give adipic acid (HOOC–(CH2)4–COOH), but this requires a different oxidative pathway.
- C (2,3-diol) would not yield a dicarboxylic acid with the given conditions.
Thus, pentan-2,4-diol (Option D) is the correct answer.
Topic: 18.1
Which alcohol gives only one possible oxidation product when warmed with dilute acidified potassium dichromate(VI)?
▶️ Answer/Explanation
Ans: B
Key Concept: Secondary alcohols (like butan-2-ol) oxidize to ketones, which cannot be further oxidized under these conditions. Primary alcohols (like butan-1-ol and 2-methylpropan-1-ol) oxidize first to aldehydes and then to carboxylic acids (two products). Tertiary alcohols (like 2-methylpropan-2-ol) cannot be oxidized.
Analysis of Options:
- A (butan-1-ol): Primary alcohol → forms butanal (aldehyde) and then butanoic acid (carboxylic acid). Two products.
- B (butan-2-ol): Secondary alcohol → forms only butanone (ketone). One product.
- C (2-methylpropan-1-ol): Primary alcohol → forms 2-methylpropanal and then 2-methylpropanoic acid. Two products.
- D (2-methylpropan-2-ol): Tertiary alcohol → no oxidation occurs. No product.
Thus, only butan-2-ol (B) yields a single oxidation product (butanone).
Topic: 19.1
Which compound, on reaction with hydrogen cyanide, produces a compound with a chiral centre?
A. \(CH_3CHO\)
B. CH₃CH₂COCH₂CH₃
C. CH₃CO₂CH₃
D. HCHO
▶️ Answer/Explanation
Ans: A
1. Reaction with HCN: The addition of hydrogen cyanide (HCN) to a carbonyl compound (aldehyde or ketone) forms a cyanohydrin.
2. Chiral Centre Formation: A chiral centre is created when the carbonyl carbon (C=O) becomes a carbon atom bonded to four different groups. – For A (CH₃CHO): \[ CH_3CHO + HCN \to CH_3CH(OH)CN \] The product has a chiral centre (carbon bonded to -OH, -CN, -CH₃, and -H). – Other options (B, C, D) either do not react with HCN or produce symmetrical products without chiral centres.
3. Conclusion: Only A (CH₃CHO) forms a product with a chiral centre, making it the correct answer.
Topic: 18.1
The diagram shows three reactions of ethanal. In each case, an excess of ethanal is used.
Observations are made after each of the three reactions. What are the colours of solution 1 and solids 2 and 3?
▶️ Answer/Explanation
Ans: B
1. Solution 1 (Fehling’s solution + ethanal): Ethanal reduces Fehling’s solution, producing a red precipitate (Cu₂O) and leaving a colorless solution.
2. Solid 2 (Tollen’s reagent + ethanal): Ethanal reduces Tollen’s reagent, forming a silver mirror (Ag).
3. Solid 3 (2,4-DNPH + ethanal): Ethanal reacts with 2,4-DNPH to form a yellow-orange precipitate (2,4-dinitrophenylhydrazone).
Thus, the correct combination is B (colorless, silver, yellow).
Topic: 17.1
(CH₃)₃CCN reacts to form alcohol Y via the reaction sequence shown.
Which row names the molecule X and the class of alcohol Y?
▶️ Answer/Explanation
Ans: C
Step 1: Hydrolysis of nitrile
(CH₃)₃CCN undergoes hydrolysis to form (CH₃)₃CCOOH (2,2-dimethylpropanoic acid). This is molecule X.
Step 2: Reduction of carboxylic acid
(CH₃)₃CCOOH is reduced using LiAlH₄ to form (CH₃)₃CCH₂OH (2,2-dimethylpropan-1-ol). This is alcohol Y.
Classification of alcohol Y
(CH₃)₃CCH₂OH is a primary alcohol because the –OH group is attached to a carbon that is bonded to only one other carbon atom.
Thus, the correct combination is:
X: 2,2-dimethylpropanoic acid
Y: primary alcohol
Matching row C in the given options.
Topic: 20.1
The diagram shows a section of an addition polymer. The polymer is made using two different monomers.
What are the names of the two monomers needed to make this polymer?
A. 1,2-dichloropropene and 2-chlorobut-2-ene
B. 2,3-dichlorobut-2-ene and chloropropene
C. 1,2-dichloropropene and chloroethene
D. chloropropene and 2-chlorobut-2-ene
▶️ Answer/Explanation
Ans: A
1. Analyzing the Polymer Structure: The polymer segment shows repeating units with alternating patterns of chlorine (Cl) substitution. One unit has two Cl atoms on adjacent carbons (1,2-disubstituted), and the other has one Cl on a carbon adjacent to a double bond (2-substituted).
2. Identifying Monomers:
– The 1,2-dichloropropene monomer (\( \text{CH}_2=\text{CCl}-\text{CH}_2\text{Cl} \)) provides the 1,2-dichloro unit.
– The 2-chlorobut-2-ene monomer (\( \text{CH}_3-\text{CCl}=\text{CH}-\text{CH}_3 \)) provides the monosubstituted unit.
3. Eliminating Incorrect Options:
– B and C are incorrect because they either mismatch substitution patterns or introduce incorrect monomers (e.g., chloroethene lacks the required branching).
– D is incorrect because chloropropene alone cannot account for the 1,2-dichloro unit.
4. Conclusion: The correct monomers are 1,2-dichloropropene and 2-chlorobut-2-ene (Option A), as they perfectly match the polymer’s repeating units.
Topic: 22.1
The diagram shows the mass spectrum of a sample of chlorine. Peaks V, W, X, Y and Z are labelled.
Which statements about this spectrum are correct?
1 The relative atomic mass of chlorine can be calculated from the abundances and m/e values of 2 of the 5 peaks.
2 37.0 g of the species responsible for peak Z contains \(3.011 \times 10^{23}\) molecules.
3 The relative molecular mass of chlorine can be calculated from the abundances and m/e values of peaks X, Y and Z.
A. 1, 2 and 3
B. 1 and 2 only
C. 1 and 3 only
D. 2 and 3 only
▶️ Answer/Explanation
Ans: A
Statement 1: Correct. The relative atomic mass of chlorine is calculated using peaks X (35Cl+) and Y (37Cl+), which represent the two chlorine isotopes.
Statement 2: Correct. Peak Z represents 37Cl2 (m/z = 74). 37.0 g is 0.5 moles, containing \(0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}\) molecules.
Statement 3: Correct. The relative molecular mass of Cl2 is calculated using peaks X (35Cl2+), Y (35Cl37Cl+), and Z (37Cl2+), which correspond to the three possible Cl2 isotopologues.
All three statements are correct, making option A the right choice.