NJC Prelims Paper 1 Answers Final

NJC/H2 Chem Preliminary Examination/02/2021 1 [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 2021 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: 1st digit and the last 4 digits of the Registration Number. Example: Student Examples of Registration No. Shade: 2005648 25648 This document consists of 12 printed pages.

NJC/H2 Chem Preliminary Examination/02/2021 2 Answer keys for 2021 SH2 H2 Chemistry Prelim Paper 1 1 C 11 A 21 B 2 A 12 A 22 C 3 C 13 A 23 D 4 C 14 A 24 D 5 A 15 C 25 C 6 D 16 C 26 B 7 C 17 D 27 D 8 D 18 C 28 A 9 C 19 C 29 B 10 C 20 C 30 B 1 A sample of element sulfur contains four isotopes of the following composition. relative isotopic mass relative abundance 32 95.02 33 0.76 34 4.20 36 0.02 What is the relative atomic mass of sulfur in this sample? A 32.07 B 32.08 C 32.09 D 32.10 Ans : C Ar = ( ଽହ.଴ଶ ଵ଴଴ × 32 ) + ( ଴.଻଺ ଵ଴଴ × 33 ) + ( ସ.ଶ ଵ଴଴ × 34 ) + ( ଴.଴ଶ ଵ଴଴ × 36 ) = 32.0924 | 32.09 2 When 15 cm3 of a gaseous organic compound were completely burnt in an excess of oxygen, 30 cm3 of carbon dioxide and 15 cm3 of nitrogen were formed, all volumes being measured at the same temperature and pressure. Which could be the formula of the organic compound? A C2H4N2 B C2H7NO C C3H7NO D C3H6N2 Ans : A According to the question, all the carbon atoms in the organic compound are converted into CO2 when burnt completely in excess O2, and the nitrogen atoms are converted into N2. Assuming that the formula of the organic compound is CxHyNz, when one mole of CxHyNz undergoes combustion, x moles of CO2 will be produced and z/2 moles of N2 will be produced. The volume ratio of organic compound : CO2 : N2 = 15 cm3 : 30 cm3 : 15 cm3 Mol ratio = 1 : 2 : 1 Hence x = 2 and z = 2 which fits the molecular formula C2H4N2.

NJC/H2 Chem Preliminary Examination/02/2021 3 [Turn over 3 Which ion has less electrons than neutrons and less neutrons than protons? ion neutrons nucleons A A− 18 37 B B2+ 17 34 C C3+ 16 33 D D3− 16 31 Ans : C Identity Neutrons Protons Electrons A Aꟷ 18 37 - 18 = 19 19 + 1 = 20 B B2+ 17 34 - 17 = 17 17 – 2 = 15 C C3+ 16 33 - 16 = 17 17 – 3 = 14 D D3ꟷ 16 31 – 16 = 15 15 + 3 = 18 4 Covalent bonds are formed by orbitals overlap. CC C CH3 H H CC H H H1 2 3 4 5 6 Which statement does not describe the molecule above? A The V bond between C1–C2 is formed by 2sp2–2sp2 overlap. B The V bond between C2–C3 is stronger than that between C5–C6. C The S bond between C4–C5 is formed by 2sp–2sp2 overlap. D The V bond between C6–H is formed by 2sp3–1s overlap. Ans : C Option A is correct. The V bond between C1–C2 is formed by 2sp2–2sp2 overlap. Option B is correct. The V bond between C2–C3 is formed by 2sp2–2sp2 overlap while that between C5–C6 is formed by 2sp2–2sp3 overlap. 2sp2–2sp2 overlap is more effective due to the hybridised orbitals having greater s character and results in a stronger bond. Option C is wrong. The S bonds between C3–C4 and C4–C5 should be formed by side–on overlap of unhybridised 2p orbitals, NOT the hybridised orbitals. Option D is correct. The V bond between C6–H is formed by head on overlap of 2sp3 of C and 1s of H.

NJC/H2 Chem Preliminary Examination/02/2021 4 5 Which options contain a polar and a non-polar molecule? 1 CO2, H2O 2 SO2, PCl5 3 CH2Cl2 , SiCl4 A 1, 2 and 3 B 1 and 2 only C 2 and 3 only D 1 only Ans : A Option 1: CO2 is non-polar while H2O is polar Option 2: SO2 is polar while PCl5 is non-polar Option 3: CH2Cl2 is polar while SiCl4 is non-polar 6 The value of pV is plotted against p for two gases, an ideal gas and a non-ideal gas, where p is the pressure and V is the volume of the gas. Which gas shows the greatest deviation from ideal gas behaviour at 200 qC? A CO2 B Cl2 C CH3OH D N2H4 Ans : D Deviation from ideal gas behaviour is greatest for molecules with strongest intermolecular forces of attraction. Both CO2 and Cl2 are non-polar molecules with weaker instantaneous dipole-induced dipole interactions (id-id) as compared to CH3OH and N2H4, which have stronger intermolecular hydrogen bonding. N2H4 has more extensive hydrogen bonds (an average of 2 H-bonds per molecule) than CH3OH (an average of 1 H-bond per molecule), hence N2H4 exhibit greatest deviation from ideal gas behaviour. Non-ideal gas pV p Ideal gas

NJC/H2 Chem Preliminary Examination/02/2021 5 [Turn over 7 Astatine is an element in Group 17. Which statements are correct? 1 Silver astatide is insoluble in aqueous ammonia. 2 Hydrogen astatide is less stable to heat than hydrogen iodide. 3 Astatine is more electronegative than iodine. A 1, 2 and 3 B 2 and 3 only C 1 and 2 only D 1 only Ans : C Astatine is below iodine in Group 17. Option 1 is correct, solubility of AgX ppt in NH3 decreases down the group. Option 2 is correct, thermal stability decreases down the group due to weaker H−X bond. Option 3 is wrong, Electronegativity decreases down the group.

NJC/H2 Chem Preliminary Examination/02/2021 6 8 The graph below shows the variation in the boiling point for 8 consecutive elements in the Periodic Table, all with atomic number ≤ 20. What can be deduced from the graph? A The ions of A and E are isoelectronic. B The chlorides become less acidic from A to C. C When the oxide of D is added to water, the resulting solution has a pH greater than 7 D The oxide of A reacts with excess aqueous sodium hydroxide to form a soluble complex. Ans : D Drastic drop in boiling point from B to C => change in structure from giant covalent to simple covalent. B is Silicon, C is Phosphorus. Option A is wrong. A is Aluminium (Group 13). Ions of A (Al3+) and E(Cl‒) are not isoelectronic (Cl− have 8 more electrons than Al3+) Option B is wrong. Chloride of A (AlCl3) is acidic (pH = 3) due to significant hydrolysis to give H+ ions, chlorides of B (SiCl4) and C (PCl5) are more acidic as they produce strong acid HCl in water. Option C is wrong. Oxide of D is SO3. SO3 + H2O o H2SO4 which is acidic (pH < 7). Option D is correct. Oxide of A (Al2O3) is amphoteric and can react with both acids & bases. It reacts with excess aq NaOH to form soluble complex Na[Al(OH)4].

NJC/H2 Chem Preliminary Examination/02/2021 7 [Turn over 9 Given the following data: Lattice energy of magnesium chloride –2526 kJ mol–1 Standard enthalpy change of hydration of chloride –384 kJ mol–1 Standard enthalpy change of hydration of magnesium –1890 kJ mol–1 What would be the change in temperature measured when 0.49 g of magnesium chloride was dissolved in 50 g of wate