VJC 2020 H2 Chem Prelim P1 MCQ [Detailed Ans] final

VICTORIA JUNIOR COLLEGE 2020 JC2 Preliminary Examinations H2 CHEMISTRY PAPER 1 ANSWERS 1 A 11 B 21 D 2 C 12 A 22 A 3 C 13 A 23 A 4 C 14 B 24 C 5 B 15 A 25 D 6 D 16 D 26 D 7 A 17 B 27 B 8 D 18 C 28 C 9 B 19 B 29 D 10 D 20 A 30 C 1 A Option A: Cr has 24 electrons . Electronic configuration is 1s22s22p63s23p63d54s1 Cr has 6 unpaired electrons. Option B: Cu has 29 electrons. Electronic configuration is 1s22s22p63s23p63d104s1 Cu2+ has 27 electrons. Electronic configuration is 1s22s22p63s23p63d9 Cu2+ has 1 unpaired electron. Option C: Mn has 25 electrons. Electronic configuration is 1s22s22p63s23p63d54s2 Mn2+ has 23 electrons. Electronic configuration is 1s2 2s22p63s23p63d5 Mn2+ has 5 unpaired electrons. Option D: Te has 52 electrons. Electronic configuration is 1s22s22p63s23p63d104s24p64d105s25p4 Te2– has 54 electrons. Electronic configuration is 1s22s22p63s23p63d104s24p64d105s25p6 Te2– has 0 unpaired electron. 2 C Option Ion Electron Neutrons Protons Nucleons A W– 20 18 19 37 B X2+ 15 17 17 34 C Y3+ 14 16 17 33 D Z3– 18 16 15 31 3 C 4 C XeF2 has 2 bond pairs and 3 lone pairs around Xe , geometry is linear. Option A: Cl2O has 2 bond pairs and 2 lone pairs around O, geometry is bent. Option B: NO2 has 2 bond pairs and 1 lone electron around N, geometry is bent. Option C: IBr2– has 2 bond pairs and 3 lone pair s around I, geometry is linear. Option D: SO32– has 3 bond pairs and 1 lone pair around S, geometry is trigonal pyramidal. 5 B Option 1: Correct Since volume and temperature are kept constant, using the gas equation 𝑝𝑣 = 𝑛𝑅𝑇 pressure is proportional to the number of moles gas. Option 2: Correct Manipulation of pV=nRT will give the equation 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝑝𝑀𝑟 𝑅𝑇 . Density is inversely proportional to the temperature as seen from the equation when the pressure is kept constant. Option 3: Incorrect As the units for the temperature in the ideal gas equation is in Kelvins, raising the temperature from 25 oC to 50 oC implies a n increase in temperature from 298 K to 323 K. The volume will increase by 323 298 = 1.08 times only.

6 D Since the options in the formula does not have the gas constant value of 8.31 for R, assume, that 1 mol of vapour occupies 22.7 dm3 at s.t.p. to find a formula for R. 𝑃𝑉 = 𝑛𝑅𝑇 𝑃𝑉 = 𝑉 𝑉𝑚 𝑅𝑇 𝑅 = 𝑃𝑉𝑚 𝑇𝑠.𝑡.𝑝. – (1) Substitute (1) into the formula 𝑃𝑉 = 𝑛𝑅𝑇, 𝑃𝑉 = 𝑛 𝑃𝑉𝑚 𝑇𝑠.𝑡.𝑝. 𝑇 𝑃𝑉 = 𝑚𝑎𝑠𝑠 𝑀𝑟 𝑃𝑉𝑚 𝑇𝑠.𝑡.𝑝. 𝑇 𝑀𝑟 = 𝑚𝑎𝑠𝑠 𝑃𝑉 𝑃𝑉𝑚 𝑇𝑠.𝑡.𝑝. 𝑇 𝑀𝑟 = 0.1 × 22.7 × 373 0.025 × 273 7 A From the graph shown, the biggest drop in boiling point occur from element F to G. This implies that element F is silicon while element G is phosphorous. Option 1: Correct Since element E is aluminum. The chloride of E will be aluminum chloride. Aluminum chloride reacts with sodium hydroxide to give a white precipitate of aluminum hydroxide as seen in the equation. AlCl3(s) + 3NaOH(aq) → Al(OH)3(s) + 3NaCl(aq) The excess sodium hydroxide added will react with the aluminum hydroxide to form a soluble complex ion of [Na+Al(OH)4]–. Option 2: Correct The oxide of F is silicon dioxide. SiO 2 does not dissolve in water as large amount of energy are required to break the strong Si –O covalent bonds in the giant molecular structure. Therefore, the pH of the resultant mixture is 7. Option 3: Correct The oxide of K is potassium oxide. K 2O is a basic oxide which readily dissolves in water to give a strongly alkaline solution of potassium hydroxide. K2O(s) + H2O(l) → 2KOH(aq) 8 D From the graph shown, the biggest drop in the 2 nd ionisation energy implies electrons are removed from a different shell. Hence the element after option B is sodium and the element before option C is magnesium. Since the elements are arranged consecutively, Option A is oxygen, Option B is neon, Option C is aluminum and Option D is silicon. Option D is element P as silicon has poor electrical conductivity and silicon chloride is a simple molecule with a simple molecular structure. 9 B Option A: Incorrect Although the statement is true, as barium loses electrons more easily than magnesium to form an ion. It does not explain about the solubility of the two ionic compounds. Option B: Correct The statement is true. ∆Hsol = –∆HLE + ∑∆Hhyd ∆Hhyd ∝ | 𝑞+ 𝑟+| Since magnesium ion has a higher charge density than barium ions, the enthalpy change of hydration of magnesium sulfate will be much more exothermic. This will result in the ∆Hsol of magnesium sulfate to be more exothermic than barium sulfate. Option C: Incorrect A numerically larger lattice energy implies that more energy is needed to break the lattice hence solubility of magnesium sulfate should be lower. Option D: Incorrect The statement is not true. ∆Hhyd ∝ | 𝑞+ 𝑟+| The magnitude of the ∆H hyd is proportional to the charge density of the ions. Since barium ion has a larger ionic radius than magnesium ion, it has a lower charge density and therefore less exothermic enthalpy change of hydration than magnesium ion. 10 D Option A: Incorrect Astatine has more electron shells than bromine. The orbital overlap between hydrogen and astatine atoms will be less effective than the orbital overlap between hydrogen and bromine atoms. Therefore, the bond strength of H–At will be weaker than H –Br implying a lower decomposition temperature. Option B: Incorrect Going down group 17, the physical colo ur of the element changes from a pale -yellow chlorine to a black solid iodine. Since the position of astatine is below iodine in the periodic table, it should a dark coloured solid. Option C: Incorrect Going down group 17, the solubility product of silver halides decreases. Since the position of astatine is below iodine in the periodic ta ble, when NH 3(aq) is added, the ionic product of silver astatide will still exceed the solubility product, making it insoluble. Option D: Correct Going down group 17, the Eo value becomes less positive, this implies that iodine should have a more positive Eo value than astatine. Therefore, iodine will undergo reduction instead, making it a stronger oxidising agent than astatine. 11 B No. of atoms = Amount of atoms x 6.02 x 1023  no. of atoms  amount of atoms Option that gives the largest amount of atoms is the answer.

Option A: Each O2 molecule has 2 O atoms. Amount of atoms = 24 24 x 2 = 2 mol Option B: Each N2 molecule has 2 N atoms. Amount of atoms = 65 14.0 14.0 x 2 = 4.64 mol Option C: Each CH3CH2OH molecule has 9 atoms. Amount of atoms = 18.75 x 0.8 2(12.0) 6(1.0) 16.0 x 9 = 2.93 mol Option D: Amount of atoms = 254 63.5 = 4 mol 12 A CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) 0.99y 0.99y CH4(g) + 3 2 O2(g)  CO(g) + 2H2O(g) 0.01y 0.01y Volume of O2 used = 2(0.99y) + 3/2(0.01y) = 2y(1–0.01) + 1.5(0.01y) = 2y –2y(0.01) + 1.5y(0.01) = (2y – 0.01y 2 ) dm3 13 A End point is reached when volume of H2O2 = 70 cm3 Thus volume of KMnO4 reacted = 100 – 70 = 30 cm3 Amount of H2O2 = 5/2 x 30 x 10–3 x 0.50 = 3.75 x 10–2 mol [H2O2] = 2 3 3.75 x 10 70 x 10   = 0.536 mol dm–3 Heat evolved = mcT = 100 x 4.2 x 13.5 = 5670 J H = – 2 5670 3.75 x 10 x 10–3 = – 151 kJ mol–1 14 B Let the change in Pazomethane be y Pa. CH3N=NCH3(g) → CH3CH3(g) + N2(g) Initial P 16 0 0 Change –y +y +y Eqm P 16–y y y When total pressure = 30 Pa, 16 – y + y + y = 30 y = 14 , i.e. Pazomethane = 16 – 14 = 2 Pa. 16 8 4 2 When Pazomethane decreases from 16 Pa to 2 Pa, time taken = 3t1/2 = 3 x 20 = 60 min 15 A Option 1: Correct Kc is dependent on temperature only. Since temperature remains constant, Kc values