NJC 2022 H2 Chemistry Prelim P1 Ans

NJC/H2 Chem Preliminary Examination/01/2022 [Turn over NATIONAL JUNIOR COLLEGE SH2 PRELIMINARY EXAMINATION Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 1 Multiple Choice Additional Materials: Optical 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, highlighters, glue or correction fluid. Write your name, subject class and registration number on the Answer Sheet in the spaces provided unless this has been done for you. 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. Read the instructions on the Answer Sheet very carefully. 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. Instructions on how to fill in the Optical Mark Sheet Shade the index number in a 5 digit format on the optical mark sheet: 2nd digit and the last 4 digits of the Registration Number. Example: Student Examples of Registration No. Shade: 2105648 15648 This document consists of 15 printed pages and 1 blank page.

2 NJC/H2 Chem Preliminary Examination/01/2022 Suggestion solution for P1 (MCQ) 1 C 7 D 13 B 19 D 25 B 2 A 8 D 14 A 20 D 26 A 3 A 9 C 15 D 21 D 27 B 4 B 10 B 16 A 22 B 28 D 5 C 11 C 17 C 23 B 29 B 6 B 12 A 18 C 24 C 30 A

3 NJC/H2 Chem Preliminary Examination/01/2022 [Turn over 1 Technetium (Tc) is a second row transition element that does not occur naturally on Earth. One of its isotopes has 56 neutrons. What is the nucleon number of this isotope? A 43 B 56 C 99 D 112 Ans: C Refer to Periodic Table. Tc has atomic number 43, i.e. proton number = 43 nucleon number = proton + neutron = 43 + 56 = 99 2 Which row has the greater number of particles? A 1 mol of liquid sulfuric acid B 12000 cm3 of oxygen gas at s.t.p. C 32.5 g of zinc D 16 g of nitrogen gas Ans: A 1 mol of liquid sulfuric acid has 6.02  10−23 particles. There is no dissociation as the sulfuric acid is in liquid form. 12000 cm3 of oxygen gas at s.t.p has 12000 22700 = 0.5286 moles = 3.18  10−23 particles 32.5g of Zinc has 32.5 65.4 = 0.4969 moles = 2.99  10−23 particles 16 g of nitrogen gas has 16 28 = 0.5714 moles = 3.44  10−23 particles 3 Which species will have the smallest deflection when passed through the same electric field? A 23Na+ B 11B3+ C 19F− D 32S2− Ans: A Angle of deflection,   𝑐ℎ𝑎𝑟𝑔𝑒 𝑚𝑎𝑠𝑠 The species with the smallest charge to mass ratio will have the smallest deflection. 23Na+ : +1 23 11B3+ : +3 11 19F−: −1 19 32S2−: −2 32 = −1 16

4 NJC/H2 Chem Preliminary Examination/01/2022 4 0.0353 g of a sample of volatile organic compound was completely vaporized at 50 kPa and 300K. The vapour occupied a volume of 40 cm3. On complete combustion in excess oxygen, 80 cm 3 of carbon dioxide and 80 cm 3 of water vapour were formed. All gaseous volumes were measured under identical conditions. What would be the molecular formula of the compound? A C2H2O B C2H4O C C2H6O D C3H8 Ans: B Using Mr =mRT/ PV: Mr = 44.0 where m = 0.0353 g ,T = 300 K, p = 50  103 Pa, V = 40  10–6 m3, R = 8.31 J K–1 mol–1 From CO2(g) evolved: No. of moles of C = 80cm3 /40cm3 = 2 mol Mass of C = 12.0  2 = 24.0 g From H2O(g) produced No. of moles of H = (80cm3/40cm3)  2 = 4 mol Mass of H = 1.0  4 = 4.0 g Mass of oxygen = 44 – 24 – 4 = 16.0 g For A, Mr = 42.0 g For B, Mr = 44.0 g (2 carbon, 4 hydrogen and 1 oxygen atom) For C, Mr = 46.0 g For D, Mr = 44.0 g 5 Which species has the most number of unpaired electrons? A Co2+ B N C Fe D S Ans: C electronic configuration unpaired electrons box diagram Co2+ 1s22s22p63s23p63d7 3 3d      N 1s22s22p3 3 2p    Fe 1s22s22p63s23p63d64s2 4 3d      S 1s22s22p4 2 2p   

5 NJC/H2 Chem Preliminary Examination/01/2022 [Turn over 6 Rubidium and bromine form ions that are isoelectronic. 1 Ionic radius of rubidium is smaller than its atomic radius. 2 Ionic radius of bromine is larger than its atomic radius. 3 Ionic radius of bromine is smaller than that of rubidium. Which statements are correct? A 1, 2 and 3 B 1 and 2 only C 1 and 3 only D 2 and 3 only Ans: B Proton number Ion Rb 37 Rb+ For Rb vs Rb +, since the same nuclear charge is attracting less electrons for the ion, the outermost electrons are more tightly held in Rb+ Ionic radius is smaller. Option 1 is true. Br 35 Br− For Br vs Br −, since the same nuclear charge is attracting more electrons for the ion, the outermost electrons are less tightly held in Br− Ionic radius is larger. Option 2 is true. For Br − vs Rb +, there are less nuclear charge (less protons) attracting the same number of electrons in Br − than in Rb +, the outermost electrons are less tightly held by nucleus in Br−. Ionic radius of Br− is larger. Option 3 is not true. 7 Which row correctly shows the bond angles for α, β, γ in the molecule below? O CH3 NH2a b g α β γ A 109.5o 120o 109.5o B 120o 109.5o 107o C 120o 107o 105o D 120o 109.5o 105o Ans: D α: 3 Bond pairs ( 1 C=C, 1 C-C and 1 C-H) – Bond Angle of 120o β: 4 bond pairs ( 3 C-C and 1 C−H) – Bond Angle of 109.5o γ: 2 lone pairs & 2 bond pairs ( 2 C−O) – Bond Angle of 105o

6 NJC/H2 Chem Preliminary Examination/01/2022 8 In which pair is the melting point of the second species higher than that of the first species? A C and Si B K and Rb C NaCl and NaBr D SiCl4 and SiO2 Ans: D C and Si are both giant covalent structure and atoms held by covalent bonds. C atoms have a smaller orbital size than Si and hence greater degree of orbital overlap leader to stronger covalent bonds in C than Si. More energy is needed to break the stronger covalent bonds in C than Si. Hence, C has a higher melting point than Si. K and Rb are both metallic lattices. Both K and Rb contribute the same number of electrons to the sea of mobile electrons. K + has a larger charge to size ratio than Rb + due to having a smaller ionic radius. More energy is needed to break the stronger metallic bonds in K than in Rb. Hence, K has a higher melting point than Si. NaCl and NaBr are both giant ionic lattices. The product of the charges are the same for both NaCl and NaBr , however, the interionic distance for NaCl is smaller than NaBr. Since L.E ∝ | 𝑞+ × 𝑞– 𝑟+ + 𝑟– |, more energy is needed to break the stronger ionic bonds in NaCl than NaBr. Hence, NaCl has a higher melting point than NaBr. SiCl4 is simple molecular structure and SiO2 is giant covalent lattice structure. More energy is needed to overcome the covalent bonds in SiO2 than the weak id-id in SiCl4. Hence, SiO2 has a higher melting point than SiCl4. 9 When liquid N2F4 is heated, it decomposes into a single product, B. Which statements are correct? 1 N–F bonds are broken during this decomposition. 2 The enthalpy change when N2F4 decomposes into B is approximately +160 kJ mol–1. 3 Molecules of B are non-linear. A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only Ans: C N2F4 → 2NF2 From the equation, can see that only N-N bonds are broken. Hence statement 1 is incorrect. Based on Data Booklet, N–N bond is +160 kJ mol–1, so statement 2 is correct because that is the only bond broken in the reaction. Statement 3 is correct because there are lone pair(s) on N of NF2 molecule, so it has to be non-linear.