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Question 1

Topic: 9.2

The elements phosphorus, sulfur and chlorine are in Period 3 of the Periodic Table. Table 1.1 shows some properties of the elements P to Cl . The first ionisation energy of S is not shown.

(a) (i) Complete Table 1.1 to show the number of electrons in the 3p subshell and the total number of unpaired electrons in an atom of P, S and Cl .
(ii) Construct an equation to represent the first ionisation energy of P.
(iii) Three possible values for the first ionisation energy of S are given.
1000 kJ/mol, 1160 kJ/mol, 1320 kJ/mol
Circle the correct value.
Explain your choice by comparing your chosen value to those of P and Cl .

(b) \(P^{3–}\), \(S^{2–}\) and \(Cl^–\) have the same number of electrons.
(i) Give the full electronic configuration of P3–.

(ii) State the trend in ionic radius shown by \(P^{3–}\), \(S^{2–}\) and \(Cl^–\). Explain your answer.
(c) A student does three tests on separate samples of NaCl (aq). Complete Table 1.2 with the observations the student makes in each test.

(d) POCl₃ shows similar chemical properties to PCl₅. POCl₃ has a melting point of 1°C and a boiling point of 106°C. POCl₃ reacts vigorously with water, forming misty fumes and an acidic solution.
(i) Explain how the information in (d) suggests the structure and bonding of POCl₃ is simple covalent.
(ii) Construct an equation for the reaction of POCl₃ with water.
POCl₃ + …………………. \(to\) ……………………..
(iii)POCl₃ contains a double covalent bond between P and O. Complete the dot-and-cross diagram, in Fig. 1.1, to show the bonding in POCl₃. Show outer shell electrons only.

(e) POCl₃(g) forms when PCl₃(g) reacts with O₂(g).
\(2PCl_3(g) + O_2(g) \to 2POCl_3(g)\)
Table 1.3 gives some relevant data

(i) Define enthalpy change of formation, \(\Delta H_f\).
(ii) Calculate the bond energy of P=O in POCl₃ using the data in Table 1.3. Show your working

▶️ Answer/Explanation

Solution

(a)(i)

Explanation: Phosphorus (P) has 3 electrons in the 3p subshell with 3 unpaired electrons. Sulfur (S) has 4 electrons (2 unpaired). Chlorine (Cl) has 5 electrons (1 unpaired).

(a)(ii) \(P(g) \to P^+(g) + e^{(–)}\)

Explanation: The first ionisation energy equation represents the removal of one electron from a gaseous phosphorus atom.

(a)(iii) 1000 (kJmol⁻¹)

Explanation: Sulfur’s ionisation energy is lower than phosphorus due to electron-electron repulsion in the paired 3p electrons. Chlorine has a higher ionisation energy due to greater nuclear charge.

(b)(i) \(1s^2 2s^2 2p^6 3s^2 3p^6\)

Explanation: \(P^{3-}\) gains 3 electrons, filling the 3p subshell to achieve a noble gas configuration.

(b)(ii) Ionic radius decreases from \(P^{3-}\) to \(Cl^-\).

Explanation: As nuclear charge increases across the series, the attraction for electrons increases, reducing the ionic radius despite the same number of electrons.

(c)

Explanation: Br₂ shows no reaction (colorless to orange), conc. \(H_2SO_4\) remains colorless, and dil. AgNO₃ forms a white precipitate (AgCl).

(d)(i) Low melting/boiling points indicate weak intermolecular forces (simple covalent). Vigorous hydrolysis suggests covalent bonding.

Explanation: Simple covalent molecules have weak van der Waals forces and react readily with water.

(d)(ii) POCl₃ + 3H₂O → H₃PO₄ + 3HCl

Explanation: POCl₃ hydrolyzes to form phosphoric acid and hydrochloric acid.

(d)(iii)

Explanation: The dot-and-cross diagram shows a double bond between P and O, with single bonds to Cl atoms and lone pairs on O and Cl.

(e)(i) Enthalpy change when 1 mole of a compound forms from its elements in standard states.

Explanation: \(\Delta H_f\) is defined for formation under standard conditions.

(e)(ii) P=O bond energy = +551 kJ/mol

Explanation: Using Hess’s Law: \(2(-592) = 2(-289) + 496 – 2(P=O)\). Solving gives P=O = 551 kJ/mol.

Question 2

Topic: 11.2

Barium hydroxide, Ba(OH)₂, is a strong base used in inorganic and organic reactions. Fig. 2.1 shows a reaction scheme involving Ba(OH)₂.

(a) (i) State the variation in solubilities of group 2 hydroxides.
(ii) State what is observed in reaction 1.
(iii) Suggest a reactant for reaction 2.
(iv) Identify A.
(v) Ba(OH)₂ is made by the reaction of Ba with water. Write an equation for this reaction.

(b) The mineral barytocalcite contains both BaCO₃ and CaCO₃. Both compounds decompose on heating.
(i) State which compound decomposes first when barytocalcite is heated. Explain your answer.
(ii) Construct an equation for the complete thermal decomposition of barytocalcite. The formula of barytocalcite is BaCa(CO₃)₂.

(c) Ba(OH)₂ is used to hydrolyse organic compounds. Fig. 2.2 shows the reaction of B with Ba(OH)₂, followed by acidification. Draw the structures of the organic products of the process shown in Fig. 2.2

▶️ Answer/Explanation

Solution

(a)(i) The solubility of Group 2 hydroxides increases down the group.

Explanation: As we move from Mg(OH)₂ to Ba(OH)₂, the lattice energy decreases faster than hydration energy, making the hydroxides more soluble.

(a)(ii) The white solid disappears.

Explanation: Ba(OH)₂ dissolves in water, forming a clear solution as it is a strong base.

(a)(iii) CO₂ / carbon dioxide.

Explanation: Ba(OH)₂ reacts with CO₂ to form BaCO₃, a white precipitate.

(a)(iv) Ba(CH₃COO)₂ / barium ethanoate.

Explanation: The reaction between Ba(OH)₂ and CH₃COOH produces barium ethanoate and water.

(a)(v) Ba + 2H₂O → Ba(OH)₂ + H₂

Explanation: Barium reacts vigorously with water to form barium hydroxide and hydrogen gas.

(b)(i) CaCO₃ decomposes first.

Explanation: Thermal stability of carbonates increases down Group 2, so CaCO₃ (less stable) decomposes before BaCO₃.

(b)(ii) BaCa(CO₃)₂ → BaO + CaO + 2CO₂

Explanation: Heating barytocalcite produces barium oxide, calcium oxide, and carbon dioxide gas.

(c)

Explanation: The hydrolysis of the ester (B) with Ba(OH)₂ followed by acidification produces a carboxylic acid and an alcohol, as shown in the image.

Question 3

Topic: 7.1

Potassium chlorate, KClO₃, is widely used as an oxidising agent and to make O₂(g).
(a) Define oxidising agent.
(b) KClO₃(s) decomposes when heated. MnO₂(s) catalyses the exothermic decomposition reaction. Complete and label the diagram in Fig. 3.1 to show the effect of MnO₂(s) on the decomposition of KClO₃(s).

(c) When KClO₃ is heated without a catalyst, KClO₄ and KCl form.
4KClO₃ \(\to\) 3KClO₄ + KCl
Explain why this reaction is described as a disproportionation reaction.

(d) Molten KClO₃ reacts with glucose, C₆H₁₂O₆.
4KClO₃ + C₆H₁₂O₆ → 6CO₂ + 6H₂O + 4KCl
KClO₃ melts at 630K. At this temperature, both CO₂ and H₂O are gases.
(i) Use the ideal gas equation to calculate the volume, in m³, of one mole of gas at 630 K and \(1.00 × 10^5 Pa\). Show your working. Give your answer to 3 significant figures

(ii) 5.00g of C₆H₁₂O₆ reacts completely with molten KClO₃. Use your answer to (d)(i) to calculate the total volume of gas released at 630K and \(1.00 × 10^5 Pa\) in this reaction. (If you were unable to answer (d)(i), use 0.0463 m³ in this question. This is not the correct answer to (d)(i).)

(e) The structure of glucose, C₆H₁₂O₆, is shown in Fig. 3.2.

(i) Complete Table 3.1 to identify the number of primary, secondary and tertiary alcohol groups present in the structure shown in Fig. 3.2.

(ii) Separate samples of aqueous glucose are tested with the reagents shown in Table 3.2. Complete Table 3.2 with the observation for each reaction. Write “no reaction” if applicable.

(iii) There are many structural isomers of \(C_6H_{12}O_6\). Define structural isomers.

▶️ Answer/Explanation

Solution

(a) Oxidising agent: A compound/molecule/substance that oxidises another (i.e., it is reduced).

Explanation: An oxidising agent gains electrons and is reduced in a redox reaction, facilitating the oxidation of another species.

(b)

Explanation: The diagram shows the effect of MnO₂ as a catalyst, lowering the activation energy for the decomposition of KClO₃, making the reaction faster.

(c) Disproportionation reaction: Chlorine is both oxidised and reduced.

Explanation: In the reaction, the oxidation state of chlorine changes from +5 in KClO₃ to +7 in KClO₄ (oxidation) and -1 in KCl (reduction).

(d)(i) Volume calculation: Using the ideal gas equation \( pV = nRT \), we have \( V = \frac{nRT}{p} \). Substituting \( n = 1 \), \( R = 8.31 \), \( T = 630 \), and \( p = 1.00 \times 10^5 \), we get \( V = \frac{8.31 \times 630}{1.00 \times 10^5} = 0.0524 \, \text{m}^3 \).

(d)(ii) Total volume of gas: Moles of C₆H₁₂O₆ = \( \frac{5.00}{180} = 0.0278 \). From the reaction, 1 mole of glucose produces 12 moles of gas. Thus, total volume = \( 0.0278 \times 12 \times 0.0524 = 0.0175 \, \text{m}^3 \).

(e)(i) Alcohol groups in glucose: 1 primary, 4 secondary, 0 tertiary.

Explanation: The structure of glucose has one primary alcohol group (-CH₂OH) and four secondary alcohol groups (-CHOH-).

(e)(ii) Observations:

KMnO₄: Solution turns from purple to colourless.
Fehling’s solution: Orange/red precipitate forms.
Bromine water: No reaction.

(e)(iii) Structural isomers: Molecules with the same molecular formula but different structural formulae.

Explanation: Structural isomers differ in the arrangement of atoms, leading to different properties.

Question 4

Topic: 14.2

Compounds C and D are alkenes with the same molecular formula, C₅H₁₀.

(a) (i) Give the systematic name of D.
(ii) Explain why C and D do not show geometrical (cis/trans) isomerism.
(iii) Draw the structure of a molecule that is a positional isomer of C and D

(iv) Give the structural formula of the compound formed when D reacts with H₂(g) in the presence of a Pt catalyst.
(v) C can form an addition polymer. Draw the structure of one repeat unit of this addition polymer.

(b) The mass spectrum of C shows a molecular ion peak at m/e = 70. This peak has a relative intensity of 48.7. The relative intensity of the [M+1] peak is 2.7. Show that this information is consistent with the molecular formula of C.
(c) C and D both react with HBr.
(i) C reacts with HBr to form E. Complete the diagram in Fig. 4.2 to show the mechanism for this reaction. Draw the structure of the organic intermediate. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

(ii) D reacts with HBr to produce F, a chiral bromoalkane. Draw the structure of F.

(iii) Explain why the reaction of HBr with C and D produces different major products.

(d) C can be used to form H

One possible synthesis of H is shown in Fig. 4.5. Different portions of C are used in reactions 1 and 3. Some of the products are then combined to produce H. Fig. 4.5 does not show any of the inorganic products of the reactions.

Complete Table 4.1 with the reagents and conditions required for each of the reactions shown in Fig. 4.5.

▶️ Answer/Explanation

Solution

(a)(i) 3-methylbut-1-ene

Explanation: The systematic name follows IUPAC rules, where the longest carbon chain with the double bond is butene, and the methyl group is at the 3rd position.

(a)(ii) One end of the C=C bond has the same groups attached.

Explanation: For cis/trans isomerism, each carbon in the double bond must have two different groups. Here, one carbon has identical groups (H and H), preventing geometric isomerism.

(a)(iii)

Explanation: A positional isomer shifts the double bond location, e.g., pent-2-ene (CH₃CH=CHCH₂CH₃).

(b) \(\frac{2.7}{48.7} \times \frac{100}{1.1}=5(.04)\)

\(M_r = 5 \times 12 + 10 \times 1 = 70\)

Explanation: The [M+1] peak intensity ratio matches the expected carbon count (5), confirming the molecular formula C₅H₁₀ (Mᵣ = 70).

(c)(i)-(ii)

Explanation: The mechanism involves electrophilic addition, forming a carbocation intermediate. For D, the product (F) is chiral due to asymmetric carbon.

(c)(iii) C forms a more stable tertiary carbocation, while D forms a less stable secondary carbocation. The alkyl groups stabilize the intermediate in C, leading to different products.

(d) (reaction 2) NaOH(aq)
(reaction 3) hot acidified concentrated KMnO₄
(reaction 4) conc H₂SO₄ catalyst

Explanation: NaOH hydrolyzes the ester, KMnO₄ oxidizes the alkene to carboxylic acid, and H₂SO₄ catalyzes esterification.

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