RI 2023 Y5 Timed Practice Suggested Solutions

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2023 9729/J/23 1 2023 Y5 H2 Chemistry July Common Test – Suggested Solutions Section A Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer D A C B C B D A C A D B C D A MCQ worked solutions Q1 (D) Relative molecular mass is the ratio of the average mass of a molecule to 1 12 the mass of a carbon -12 atom. Q2 (A) Mass of SO2 molecules in 1 m3 of air = 0.570 x 1 = 0.570 mg = 0.000570 g Amount of SO2 molecules in 1 m3 of air = 0.000570 g ÷ (32.1 + 16  2) = 0.00000889 mol Number of SO2 molecules in 1 m3 of air = 0.00000889  6.02  1023 = 5.35 x 1018 Q3 (C) % abundance of 62Ni = 100 − 69.51 − 26.78 = 3.71% Relative atomic mass of Ni in sample = (69.51  58 + 26.78  60 + 3.71  62) ÷ 100 = 58.68 Relative formula mass of Ni35Cl2 = 58.68 + 35+ 35 = 128.68 Q4 (B) 1 is correct: Mg: [Ne] 2s2 P: [Ne] 2s2 2p3 Same number of electron pairs 2 is correct: As: [Ar] 3d10 4s2 4p3 Ge: [Ar] 3d10 4s2 4p2 Same number of electron pairs 3 is incorrect: Ba: [Xe] 6s² Pb2+: [Xe] 4f¹⁴ 5d¹⁰ 6s² Different number of electron pairs Q5 (C) Option A is incorrect as X has a higher nucleon number of 36 while Y has a smaller nucleon number of 32. Option B is incorrect as X has (33 − 14) = 19 neutrons while Y has (36 − 18) = 18 electrons. Option C is correct as it fulfils all 3 criteria. Option D is incorrect as X has 14 – 4 = 10 electrons while Y has 18 electrons. Q6 (B) │Angle of deflection│is proportional to│ q m│ . 2H+:│ q m│= 1 2 18O2–:│ q m│= 2 18 = 1 9 9Be2+:│ q m│= 2 9 Statement 1 is incorrect as 2H+ particles are deflected to the largest extent. Statement 2 is correct as the │ q m│ of 9Be2+ is larger than that of 18O2–, hence 9Be2+ particles are deflected to a larger extent than 18O2– particles. Statement 3 is incorrect as 9Be2+ particles are attracted and , hence, deflected towards the negatively charged plate. Statement 4 is correct as 18O2– particles are attracted and , hence, deflected towards the positively charged plate. Hence, option B (2 and 4 only) is correct. Q7 (D) Option A is incorrect as 3 dz2 orbital consists of a dumb -bell surrounded by a small doughnut -shaped ring at its waist and aligned along the z axis. Option B is incorrect as 3 dy orbital does not exist. Option C is incorrect as the lobes of 3dx2- y2 orbital are aligned along the x and y axes. Option D is correct as 3 dxz orbital has a 4 -lobed shape and its lobes are pointing between the x and z axes. Q8 (A) Real gases deviate from ideal gas behaviour as the intermolecular attractive forces between the gas particles are significant. Ammonia has the strongest intermolecular forces of attractions (hydrogen bonding) and, hence the greatest deviation from ideality, amongst the 4 gases. The other 3 gases are non -polar simple covalent molecules with smaller electron cloud size than ammonia, thus they have weak er instantaneous dipole-induced dipole interactions.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2023 9729/J/23 2 Q9 (C) For an ideal gas, pV = nRT. For a fixed mass of gas, n is constant. Option 1 is incorrect as V = nRT p , since T is constant, nRT is constant, thus V = k( 1 p). Graph of V against 1 p should be a straight line passing through the origin. Option 2 is correct as p = nRT V , since T is constant, nRT is constant, thus p = k( 1 V). Graph of p against 1 V will be a straight line passing through the origin. Option 3 is correct as V = nRT p , since p is constant, nR p is constant, thus V = k(T) = k ( 1 1 T ). Graph of V against 1 T will be of the same shape as y = k ( 1 x). Option 4 is incorrect. as p = nRT V , since V is constant, nR V is constant, thus p = k(T). Graph of p against T should be a straight line passing through the origin since T is measured in K. Q10 (A) pV = nRT = m MRT m = 28.0  (p  105)  (V  10-6) (273 + T)  R Since 1 mol of an ideal gas occupies 22.7 dm3 at s.t.p (1 bar = 105 Pa, 0 C = 273 K), R can be derived from pV = nRT. Substituting R = 105  22.7  10-3 1  273 , = 28.0  (p  105)  (V  10-6)  273 (273 + T)  105  22.7  10-3 = 28.0  pV  273 (273 + T)  22700 Q11 (D) Center atom No. of bond pair electrons No. of lone pair electrons Electron pair geometry Shape Bond angle N 3 1 Tetrahedral Trigonal pyramidal 107o C (of CH2) 4 0 Tetrahedral Tetrahedral 109.5o C (of C=O) 3 0 Trigonal planar Trigonal planar 120o O 2 2 Tetrahedral Bent 105o Order of increasing angles: d<a<b<c. Q12 (B) Statement 1 is correct. More energy is required to overcome the strong C−C covalent bonds in diamond than the weak intermolecular forces between the layers in graphite. Statement 2 is correct. The freedom of the layers of carbon in graphite to slide results in greater entropy. Statement 3 is incorrect. Delocalised electrons from continuously overlapping p orbitals are required to conduct electricity. Graphite can conduct electricity while diamond cannot. Q13 (C) The C-H single bonds each comprise of a  bond. The C=C double bonds each comprise a  bond and a  bond. Thus there is a t otal of 6  bonds and 2  bonds. Q14 (D) Given that the solution turns cold, this is an endothermic reaction and ∆H is positive. Also, ∆G is negative since dissolution is spontaneous. Since ∆G = ∆H − T ∆S, ∆S needs to be positive in order for ∆G to be negative. Q15 (A) Option A is correct as the reaction results in an increase in the number of moles of gaseous particles (from 0 mol to 1 mol ) in the system , causing entropy to increase. Option B is incorrect as the reaction results in an decrease in the number of moles of gaseous particles (from 1 mol to 0 mol) in the system , causing entropy to decrease. Option C is incorrect as there is no change in the number of moles of gaseous particles in the system, so the entropy change is less than other options. Option D is incorrect as the reaction results in an decrease in the number of moles of gaseous particles (from ¼ mol to 0 mol) in the system , causing entropy to decrease.

This document is copyrighted, please do not reproduce it without permission © Raffles Institution 2023 9729/J/23 3 Section B B1(a)(i) Comments: • Students should follow the requirement of the question and label both graphs. • As the question only require d a sketch of the graphs, students should not waste time drawing the graphs to scale. • Note that the atomic and ionic radii of the Period 3 elements are given in the Data Booklet. B1(a)(ii) P: 1s2 2s2 2p6 3s2 3p3 P3−: 1s2 2s2 2p6 3s2 3p6 Both P 3− and P have the same nuclear charge . However, P 3− has more electrons than P leading to greater electron–electron repulsion and hence the electrostatic attraction between the nucleus and the outermost electron is weaker. Comments: • Students should note that both P and P3− have the same number of electronic shells. B1(a)(iii) Na+: 1s2 2s2 2p6 Mg2+: 1s2 2s2 2p6 Na+ and Mg2+ have the same number of electrons/ are isoelectronic and hence their outermost electrons experience the same shielding effect. However, Mg2+ has a higher nuclear charge than Na+. Consequently, the outermost electrons in Mg 2+ are more strongly attracted by the nucleus. Comments: • Several students erroneously compared the charges of the two ions, instead of their nuclear charge / proton number. Note that nuclear charge ≠ charge on the ion (e.g. K+ has a higher nuclear charge than Na+, K+ has a higher nuclear charge than Mg2+). • Students should also note that effective nuclear charge ≠ nuclear charge. Effective nuclear charge is the resultant positive charge experienced