RI 2018 Y5 Promo MCQ Suggested Solutions

17 2018 Year 5 Promotion Examinations Section A Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer B D B D B D A C C A C D B A C Q1(B) Oxidation half-equation 6OH– + S ⟶ SO32– + 3H2O + 4e– Reduction half-equation 4e– + 2S ⟶ 2S2– Overall equation 6OH– + 3S ⟶ 2S2– + 1SO32– + 3H2O Q2(D) Balanced equation: CxHySz(g) + (x + z + y/4)O2(g) ⟶ xCO2(g)+ zSO2(g) + y/2H2O(l) After cooling to room temperature, the gases which remain are CO2, SO2 and unreacted O2 i.e. 𝑉𝐶𝑂2 + 𝑉𝑆𝑂2 + 𝑉𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑂2 = 90𝑐𝑚3 The acidic gases, CO2 and SO2, react with the aqueous alkali. Therefore, 𝑉𝑢𝑛𝑟𝑒𝑎𝑐𝑡𝑒𝑑 𝑂2 = 70𝑐𝑚3. 𝑉𝑂2 𝑟𝑒𝑎𝑐𝑡𝑒𝑑 = 100 − 70 = 30𝑐𝑚3 𝑉𝐶𝑂2 + 𝑉𝑆𝑂2 + 70 = 90 ⇒ 𝑉𝐶𝑂2 + 𝑉𝑆𝑂2 = 20𝑐𝑚3 CxHySz(g) + (x+z+y/4)O2(g) ⟶ xCO2(g) + zSO2(g) + y/2H2O(l) V / cm3 10 30 20 -- mole ratio 1 3 2 -- x+z+y/4 = 3 and x +z = 2 ⇒ y = 4 x = z = 1. Therefore, compound is CH4S. Q3(B) A Incorrect. Using data from Data Booklet, radius of Ni2+ = 0.069nm and radius of Fe2+ = 0.061nm. B Correct. Using data from Data Booklet, 1st IE of Ni = 736 kJ mol–1 and 1st IE of Fe = 762 kJ mol–1. C Incorrect. Number of electrons in Fe 2+ = 24. Number of electrons in Ni atom = 28. They do not have the same no. of electrons i.e. they are not isoelectronic. D Incorrect. 2+ 2+ Angle of deflection 2(Ni ) 0.034158.7 2(Fe ) 0.035855.8 q m q m q m    Fe2+ will be deflected more than Ni2+. Q4(D) From question, presence of unpaired e – ⇒ paramagnetic. A Electronic configuration of Cl+ : 1s2 2s2 2p6 3s2 3p4. 2 unpaired electrons in 3p subshell ⇒ Cl+ is paramagnetic. B Electronic configuration of V 3+ : [Ar] 3d 2. 2 unpaired electrons in 3d subshell ⇒ V3+ is paramagnetic. C Electronic configuration of O – : 1s2 2s2 2p5. 1 unpaired electron in 2p subshell ⇒ O– is paramagnetic. D Electronic configuration of Cu + : [Ar] 3d 10. No unpaired electrons in Cu + ⇒ Cu+ is not paramagnetic. Q5(B) 1 Incorrect. Hence, the molecule is not planar. 2 Correct. From question: “…form the planar cyclopentadienyl anion…”. In order for the anion to be planar, every carbon atom in the anion is sp 2 hybridised. 3 Incorrect. Due to the resonance stabilisation in the anion, the pi electron density is evenly distributed over the entire anion. Hence, the carbon -carbon bond lengths in the anion are equal. Q6(D) The line is represented by an equation of the form y = mx where m is the gradient. If pV is the y -axis and T is the x -axis, the ideal gas equation can be manipulated to fit the form y = mx. This document is copyrighted, please do not reproduce it without permission

18 Q7(A) Refer to the defi nitions in Energetics I lecture notes to help you remember whether a process is endothermic or exothermic. It is also helpful to consider from a Chemical Bonding perspective whether some processes are endothermic or exot