SAJC 2022 P1 ANS

1 [TURN OVER ST ANDREW’S JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATIONS HIGHER 2 CANDIDATE NAME STUDENT WORKED SOLUTIONS CLASS 2 1 S CHEMISTRY Paper 1 Multiple Choice Additional Materials: Multiple Choice Answer Sheet Data Booklet 9729/01 15 September 2022 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name and class on the Answer Sheet in the spaces provided. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. This document consists of 19 printed pages (including this cover page) and 1 blank page.

2 [TURN OVER 1 2 3 4 5 6 7 8 9 10 C D C A B A A C B C 11 12 13 14 15 16 17 18 19 20 C B D C C D D A D C 21 22 23 24 25 26 27 28 29 30 C B D B A B B C B D 1 Use of the Data Booklet is relevant to this question. Which statements about chromium and its compounds are correct? 1 The valence electronic configuration of Cr contains both an unpaired s electron and an unpaired p electron. 2 One of the 3d orbitals in chromium atom has only 2 lobes. 3 There are 58 electrons and 60 neutrons in the 52Cr16O42− ion. A 1, 2, and 3 B 1 and 2 C 2 and 3 D 1 only 1. The electronic configuration of Cr is [Ar]3d54s1. Cr contains both an unpaired s electron and an unpaired d electron (not p electron). 2. Only the 3dz2 d orbital has 2 lobes. 3. No of electrons = 24 + (4 x 8) + 2 = 58 No of neutrons = (52 – 24) + 4(16 – 8) = 60 Answer: C

3 [TURN OVER 2 The following graph shows the second ionisation energy of eight consecutive elements in Period 2 and 3. Which of the options A, B, C or D is silicon? B must be group 1 as the second electron is removed from the inner shell hence resulting in a large energy gap. Hence, D is group 14 and belongs to silicon. Answer: D 3 Which option is correct? Molecule Planar Polarity A Xenon difluoride Yes Polar B Chlorine(I) oxide No Non-polar C Methanal Yes Polar D Dichloromethane No Non-polar Molecule Planar Polarity Diagram  A Xenon difluoride Linear (Yes) Non-polar  B Chlorine(I) oxide Bent (Yes) Polar ✓ C Methanal Trigonal Planar (Yes) Polar 2nd IE Atomic number A B C D

4 [TURN OVER  D Dichloromethane Tetrahedral (No) polar Answer: C 4 Which statements about ethanoic acid are correct? 1 1 molecule of ethanoic acid has 4 lone pairs of electrons. 2 Gaseous ethanoic acid forms a dimer comprising of two hydrogen bonds within a ring of 8 atoms. 3 The CꟷC bond in ethanoic acid is formed from a sp3−sp2 orbital overlap. A 1, 2 and 3 B 1 and 2 C 2 and 3 D 1 only 1. Ethanoic acid has 4 lone pairs of electrons as shown above. 2. The dimer of ethanoic as shown above. There is illustration of 2 hydrogen bonds and 8 atoms are within the hydrogen bond structure. 3. C1-C2 forms a sp3-sp2 orbital overlap as shown above. Answer: A

5 [TURN OVER 5 Which description of Dalton’s law is correct? A Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. B The total pressure of a mixture of gases is equal to the sum of the partial pressures of those gases. C The partial pressure of a gas in mixture is given by the product of its mole fraction and the total pressure. D The partial pressure of a gas in a mixture is given by the product of its percent by mass and the total pressure. Answer: B Dalton's Law of Partial Pressure: At constant temperature, for gases which do not react chemically, the total pressure of a mixture of gases in a given volume is equal to the sum of the partial pressures of the constituent gases. pT = pA + pB + pC + ….

6 [TURN OVER 6 Which graph shows the correct trends when the physical property of each of the elements Na, Al, Si and S is plotted against its atomic radius? A B C D Atomic radius decreases across the period. Thus, Atomic radius increases in the following order: S < Si < Al < Na. Thus options A and B in terms of atomic radius is accurate. Melting point depends on the structure and bonding of the elements. Si > Al > Na > S. Si is a giant covalent molecule, Al and Na are giant metallic lattice but Al has more delocalised electrons hence a higher melting point. S is a simple covalent molecule hence the lowest boiling point. Electronegativity depends on the ability of the n ucleus to attract the bonding pair of electrons. The trend is S > Si > Al > Na. Answer: A

7 [TURN OVER 7 Which property describes the trend for the hydrogen halides as stated below? HCl > HBr > HI A thermal stability B boiling point C ease of oxidation D acidity Option A is correct as down the group, the bond length of H -X increases, bond strength decreases and bond energy decreases. Hence thermal stability decreases down the group. Option B is wrong as down the group, the size of electron cloud of H-X compound increases, it is more easily polarised, hence the strength of instantaneous dipole-induced dipole (id-id) interactions between molecules increases, more energy is needed to overcome the id-id interactions, resulting in higher boiling point. Option C is wrong as the ease of oxidation of the compounds, X– increase down the group. Down the group, Eө(X2/X–) becomes less positive, X– is more readily oxidized to X2. Reducing power of X– increases down the group. Option D is wrong as acid strength increases down the group, i.e. HCl < HBr < HI. Down the group, size of halogen atom increases , -X bond length increases , H-X bond strength decreases less energy required to break H-X bond. H-X dissociates to give H+ more easily. Answer A 8 Use of the Data Booklet is relevant to this question. Which contains the largest number of molecules? A 810 cm3 of sulfur dioxide gas (measured at r.t.p.) B 1.56 g of methyl methanoate C 5.25 cm3 of ethanol where the density of ethanol is 0.755 g cm–3 D 1.70 x 1022 molecules of hydrogen peroxide. Amount of SO2 = 810 / 24000 = 0.03375 mol Amount of methyl methanoate = 1.56 / 60 = 0.0260 mol Mass of ethanol = 5.25 x 0.755 = 3.964 g Amount of ethanol = 3.964 / 46 = 0.0862 mol

8 [TURN OVER Amount of hydrogen peroxide = 1.70 x 1022 / 6.02 x 1023 = 0.0282 mol Since option C has the largest amount, the number of molecules will also be the largest. Answer: C 9 Silane, SiH4, exists as a gas at standard temperature and pressure. Hess’ Law can be used to calculate the average Si–H bond energy in gaseous SiH4. Which information is needed to perform the calculation? A ∆Hformation(SiH4) only B ∆Hatomisation(Si), ∆Hatomisation(H), ∆Hformation(SiH4) C ∆Hcombustion(Si), ∆Hcombustion(H2), ∆Hcombustion(SiH4) D ∆Hcombustion(Si), ∆Hcombustion(H2), ∆Hformation(SiH4) Answer: B 10 The decomposition of Arsine, AsH3, is catalysed by Arsenic, As, and can be represented by the following equation. AsH3 → As + 3 2 H2 Which graph does not describe the above reaction? A B

9 [TURN OVER C D Options A and D accurately describes an autocatalytic graph where the reaction or Arsine and the production of hydrogen shows the following feature. The [Arsine] is as shown below. Option B shows the correct shape for an autocatalytic graph. The rate increase significantly when the product is formed as the product catalyses the reaction. The rate starts to fall when the reactants are used up.