RI 2024 Y5 Timed Practice Suggested Solutions

2024 H2 Chemistry Y5 Timed Practice – Suggested Solutions Section A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C C B D B A B A B D D D C D C 1 Answer: C Amt of atoms in 0.2 g O2 = 0.2 / 32.0 x 2 = 0.0125 mol Amt of atoms in 0.225 g H 2O = 0.225 / 18.0 x 3 = 0.0375 mol Amt of ions in 1.275 g of Al2O3 = 1.275 / 102 x 5 = 0.0625 mol Amt of atoms in 0.300 dm3 of Ne at r.t.p. = 0.300 / 24 x 1 = 0.0125 mol Amt of atoms in 0.284 dm3 of N2 at s.t.p. = 0.284 / 22.7 x 2 = 0.0250 mol 2 Answer: C amt of e – transferred = 2 x amt of Sn2+ reacted = 2 x (31.20 / 1000 x 0.0400) = 2. 496 x 10–3 mol amt of IO3− reacted = 25.0 / 1000 x 0.0250 = 6.25 x 10–4 mol mol ratio of e – : IO3− = 2.496 x 10–3 : 6.25 x 10–4 = 3.99 : 1 ≈ 4 : 1 (i.e. 1 mol of IO3− gains 4 moles of e–.) oxidation number of I in product = (+5) – 4 = +1 3 Answer: B There is a large decrease in the 5th IE from R to S ⇒ R4+ has a noble gas configuration after losing 4 electrons. ⇒ R is in Group 14. ⇒ Q is in Group 13. The chloride is QCl3. 4 Answer: D The angle of deflection is proportional to charge to mass ratio. For 7Li+, charge / mass = +1/7 and angle = +xo (towards +ve terminal) Option A: charge / mass = –2/14 = –1/7, so angle = –x o (towards –ve terminal) Option B: charge / mass = –1/16, so angle < –xo (towards –ve terminal) Option C: charge / mass = +3/27 = +1/9, so angle < +xo (towards +ve terminal) Option D: charge / mass = +7/35 = +1/5, so angle > +xo (towards +ve terminal) This document is copyrighted, please do not reproduce it without permission

5 Answer: B BrF2+ is bent with a bond angle of about 105o. NO2 is bent with a bond angle greater than 120o. XeF2 is linear with a bond angle of 180o. Hence bond angle of BrF2+ < NO2 < XeF2. 6 Answer: A Statement 1 is correct as there is an intramolecular hydrogen bond between one of the F atoms and the H atom of the –OH group. Statement 2 is correct as there are 4 bond pairs (3 single bonds and 1 dative bond) around B atom and so the bonds around B are tetrahedrally arranged. Statement 3 is correct as there are 4 bond pairs around N atom but 2 bond pairs and 2 lone pairs around O atom, causing the H–N– O angle to be 109.5° and the H –O–N angle to be 105° . This is because the lone pair exerts greater repulsion than the bond pair. 7 Answer: B Option 1 Si, AlF 3, HCl are giant molecular, ionic, simple molecular respectively. Option 2 SiC l4, Al2O3, HBr are simple molecular, ionic, simple molecular respectively. Option 3 AlC l3, SiO2, BaI2 are simple molecular, giant molecular, ionic respectively. 8 Answer: A • and ○ represent Na+ and Cl− respectively. (Note: Na+ is smaller than Cl− in size.) Only in structure A, all Na+ and Cl− are bonded to oppositely charged ions. 9 Answer: B At constant V and T, p is directly proportional to n. After gas Z is introduced, the total pressure increases by P atm (since the final is 2P atm). ⇒ Z has the same number of moles as the combined amounts of X and Y. i.e. mol ratio of X : Y : Z = 1 : 2 : 3 new partial pressure of X = [1 / (1 + 2 + 3)] x 2P = P/3 atm 10 Answer: D Statement A is incorrect, increasing pressure causes boiling point to increase, not decrease. Statement B is incorrect, kinetic energy is dependent on temperature, not pressure. Statement C is incorrect, increasing pressure does not decrease the size of molecules. intramolecular hydrogen bond This document is copyrighted, please do not reproduce it without permission

Statement D is correct, increasing the pressure pushes the molecules closer together and hence the intermolecular forces of attraction become significant and the gas condenses from gaseous to liquid state. 11 Answer: D q = mc∆T = (76 + 70)(1)(4.18)(70 – 25) = 27460 J ∆H = –q n nCaCl2 = –q ∆H = –27460 –83000 = 0.3309 mol mass of CaCl2 = (0.3309) × (40.1 + 35.5 × 2) = 36.8 g 12 Answer: D Lattice energy is the enthalpy change when one mole of ionic compound is formed from its constituent gaseous ions under standard conditions. 13 Answer: C Bonds broken Bonds formed 1 C=C 1 C–C 1 O–H 1 C–O 1 C–H ∆Hr = BE(C=C) + BE(O–H) – BE(C–C) – BE(C–O) – BE(C–H) 14 Answer: D ∆G = ∆H – T∆S To determine the sign of ∆S When ∆G is more negative at a higher temperature, ∆S must be positive. OR 1 mol of gas + 1 mol of solid → 2 mol of gases ⇒ ∆S > 0 because a gas has greater entropy than a solid To determine the sign of ∆H At a lower temperature, when “ –T∆S” is a small negative value, ∆G is positive. This implies that ∆H is positive. OR +78000 = ∆H – 378∆S Since ∆S > 0, – 378∆S < 0. Thus, ∆H must be positive. This document is copyrighted, please do not reproduce it without permission

15 Answer: C 1. ∆S < 0 Entropy decreases as the decrease in temperature causes the narrowing of the Maxwell -Boltzmann energy distribution of the particles. Thus there are fewer ways of arranging “energy quanta” in the cooler system. 2. ∆S > 0 When a solid melts into a liquid, the order in the solid is destroyed. Particles in a liquid are more randomly arranged and more disordered than those in the solid, resulting in an increase in entropy. 3. ∆S > 0 The entropy of a system increases as the number of particles in the system increases. With more particles, there are more ways to arrange the particles and more ways to distribute the energy in the system, and hence creating greater disorder in the system. This document is copyrighted, please do not reproduce it without permission

B1(a) The successive ionisation energy increases as the number of protons remains the same and hence nuclear charge remains the same. The number of electrons decreases and shielding experienced by the remaining outermost electrons decreases. Hence the electrostatic attraction between the nucleus and the remaining electrons increases, resulting in more energy required to remove each subsequent electron. Examiner comments • Students are required to explain the trend in successive ionisation energies in terms of nuclear charge and shielding, before concluding that the attraction between nucleus and remaining electrons increases. B1(b) The 6th electron in element A is located in an inner electron shell that is nearer to the nucleus, experiencing less shielding and is attracted more strongly by the nucleus. Comparing to element B, the 6th electron of element A experiences very little shielding due to the lack of inner shell electrons whereas the 6 th electron of element B experiences shielding by one inner electron shell. Examiner comments • From the large jump in Fig 1.1, students are required to identify that the 6th electron removed in element A is from the inner electron shell and hence, closer to nucleus. • Then, students are required to explain why this large jump is more much significant than in element B by stating that the 6 th electron in element A experiences almost no shielding (e.g. very close to the nucleus, no inner shell at all). • Some students misunderstood the question and considered the jump in element B to be a small jump which corresponds to the removal of the 6th electron to be from a different subshell instead. The question states that there is a big jump for both elements, hence any answer involving the removal of the 6th electron from different sub-shells (e.g. s vs p) is not applicable here. B1(c) 1s2 2s2 2p3 Examiner comments • Students are reminded to indicate the number of electrons in each subshell as the superscript (e.g. 2p3). B