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