2023 VJC H2 Chem Prelim P2 (Ans)

1 2023 VJC H2 Chemistry Prelim Paper 2 Answers 1 (a) State three ways in which an ideal gas differs from a real gas. Molecules/particles of an ideal gas have zero/negligible volume. There are negligible (or NO) intermolecular forces of attraction between ideal gas molecules/particles Collisions between ideal gas molecules are perfectly elastic (i.e. no loss of energy during collision) [3] (b) (i) Draw a labelled diagram to show the significant force of attraction between two molecules of hydrogen fluoride. Include the name of the attraction in your diagram. Label as hydrogen bond Lone pair and dipole [2] (ii) The value of pV/RT is plotted against p for 1 mol of an ideal gas and 1 mol of fluorine gas at 300 K, where p is the pressure and V is the volume of the gas. On the same axes, sketch the variation of pV/RT against p for one mole of hydrogen fluoride, HF, at the same temperature of 300 K. Briefly explain your answer. Correct sketch HF deviates more from ideal behaviour because it experiences stronger intermolecular hydrogen bonding between the molecules compared to F2 which has only instantaneous dipole−induced dipole interaction between its molecules. [2] [Total: 7] pV/RT p 1.0 ideal gas fluorine HF

2 2 (a) The type of bonding present in a binary compound can be predicted from the electronegativities of the elements involved. This can be shown on a van Arkel - Ketelaar triangle in Figure 2.1, which plots the difference in electronegativity, ∆𝜒, on the y-axis against the average electronegativity of the two elements, 𝜒̅, on the x-axis. In this triangle, the three corners represent the extremes of metallic, ionic, and covalent bonding, with caesium (Cs), caesium fluoride (CsF) and fluorine (F2) at these corners. Figure 2.1 (answer for (a)(iii) Position in the triangle NaF any point below and to the right of CsF SiF4 any point below and to the right of NaF AND higher and to the left of F2. (i) Describe and explain the variation in electronegativity across the third period of the Periodic Table. Electronegativity increases across a period. Nuclear charge increases (OR number of protons increases ) while shielding effect is relatively constant (OR number of inner shell electrons is the same). Hence there is an increase in effective nuclear charge leading to stronger attraction between the nucleus and the electrons, and so ability of the atom to attract electrons to itself increases. [2] ionic metallic covalent × Cs F2 Difference in electronegativity ∆𝜒 Average electronegativity 𝜒̅ CsF × × × × NaF SiF4

3 (ii) State the type of bonding present in sodium fluoride (NaF), silicon tetrafluoride (SiF4) and the magnesium-aluminium alloy (Mg-Al). Type of bonding in NaF: ionic Type of bonding in SiF4: covalent Type of bonding in Mg-Al: metallic [2] (iii) Mark the approximate positions of NaF and SiF4 on Figure 2.1 See Figure 2.1 in front [2] (b) Lattice energies can be obtained from constructing a Born-Haber cycle with the aid of experimental data. They can also be calculated theoretically from knowledge of the distances between the cations and anions in the crystal structure and the charge on each ion. Table 2.1 shows the values of lattice energies for some compounds. These have been either determined from experimental data or theoretically calculated. Table 2.1 compound experimental value / kJ mol─1 theoretical value / kJ mol─1 NaCl −781 −766 NaBr −743 −730 NaI −699 −685 CaCl (non-existent compound) − −687 CaCl2 − − AgF −967 −824 AgI −889 −618 Calcium(I) chloride, CaCl, is a hypothetical compound that does not exist. (i) Define, with the aid of an equation, the lattice energy of CaCl2. The lattice energy of CaCl2 refers to the energy evolved when 1 mole of ionic solid CaCl2 is formed from its isolated gaseous ions according to the equation: Ca2+(g) + 2Cl─(g) → CaCl2(s). [2]

4 (ii) The given data show the l attice energies of the sodium halides becoming less exothermic from NaCl to NaI. By quoting relevant data from the Data Booklet, predict and explain whether you expect the lattice energy of CaC l2 to be more or less exothermic than that of NaCl. Ca2+ (r+ = 0.099 nm) is bigger than Na+ (r+ = 0.095 nm). But CaCl2 comprises doubly-charged Ca2+ whilst NaCl comprises singly- charged Na+. Since ionic charge is the predominant factor in affecting magnitude of lattice energy, the lattice energy of CaCl2 is more exothermic than that of NaCl. [2] (iii) A larger difference between the experimental and theoretical values of the lattice energy is observed for AgI as compared to that for AgF. Suggest why. A greater difference between lattice energy values is observed for AgI due to a more polarisable electron cloud of the anion / smaller difference in electronegativity between Ag and I, leading to greater covalent character. [2]

5 (iv) Using the data below as well as relevant data from Table 2.1 and the Data Booklet, construct a Born -Haber cycle in the grid provided and calculate the enthalpy change of formation of solid calcium(I) chloride, CaCl(s). Enthalpy change of atomisation of calcium = +178 kJ mol−1 first electron affinity of chlorine = −349 kJ mol−1 By Hess’s Law, ∆Hf = (+178) + (+½ x 244) + (+590) + (−349) + (−687) = −146 kJ mol─1 Equations are balanced, correct cycle and values Correct application of Hess’s Law Correct answer including sign and units [3] / kJ mol–1

6 (v) Calcium(I) chloride is not known to exist as it readily reacts as shown in the following equation. 2CaCl(s) → Ca(s) + CaCl2(s) State, with justification, the type of reaction that has occurred. Disproportionation The oxidation state of Ca changes from +1 to 0 and +2 which shows that Ca is both reduced and oxidised. [1] (c) One of the Group 2 chlorides, beryllium chloride, has many similar properties with aluminium chloride. In the vapour state at high temperatures, beryllium chloride, like aluminium chloride, exists as a dimer with the formula Be 2Cl4. Around each beryllium atom is a trigonal planar arrangement. (i) Explain how new bonds are formed two molecules of beryllium chloride during the process of dimerisation. Cl atom donates a lone pair of electrons to the vacant p orbital of beryllium, leading to the formation of a dative bond. [1] (ii) Draw the structure of the Be2Cl4 dimer. correct structure showing two dative bonds between Be and Cl. [1] [Total: 18]

7 3 Quinidine, Q, (Mr = 324.4) is a diacidic organic base commonly used as a drug for the control of heart rhythm disturbance. Due to the low solubility of quinidine, quinidine must be fully protonated to form QH22+ before its concentration can be measured by carrying out a titration. The pKa values of the acidic groups in fully protonated quinidine are given below. QH22+ ⇌ QH+ + H+ pKa1 = 4.00 QH+ ⇌ Q + H+ pKa2 = 8.60 A 10.0 cm3 sample containing 0.662 g of the fully protonated quinidine was titrated against NaOH. The first endpoint was seen after 20.00 cm3 of aqueous NaOH was added. (a) Calculate the concentration of the fully protonated quinidine in the 10.0 cm 3 sample and hence determine its pH (ignore the effects of the second acid dissociation on the pH). Concentration of fully protonated quinidine = 𝟎.𝟔𝟔𝟐 𝟑𝟐𝟒.𝟒 ÷ 10.0  1000 = 0.204 mol dm−3 Ka1 = x2 / (0.