2025 RI H2Chem Prelims P1 Answers

1 © Raffles Institution 2025 9729/01/S/25 2025 Y6 H2 Chemistry Preliminary Exams Paper 1 – Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer A D C C B D C B B A D B A A C MCQ worked solutions Q1 (Ans: A) The 37Cl+ ion has 37 nucleons which comprise 17 protons and 20 neutrons. It has one positive charge (having lost one electron) and hence it has a total of 16 electrons. Q2 (Ans: D) angle of deflection ∝ | q m| For 17O+, | q m| = 1 17 (= 0.0588) For [16O18O]2–, | q m| = 2 (16+18) = 1 17 (= 0.0588) Hence [16O18O]2– will be deflected to the same extent as 17O+. Option 1 is incorrect. Option 2 is correct since positively charged species will deflect in an opposite direction to negatively charged species. Each beam of charged particles will travel in a curved path, not a straight path. Option 3 is incorrect. Q3 (Ans: C) The cis-isomer (Q) has a higher boiling point as it has a net dipole moment and has permanent dipole-permanent-dipole interactions, whereas the trans-isomer (P) is non-polar. R has instantaneous dipole-induced dipole interactions, as well as the smallest electron cloud and hence has the weakest interactions amongst the three. Thus, it has the lowest boiling point. Correct order of increasing boiling point: R < P < Q. Q4 (Ans: C) BeF2 acts as a Lewis acid (as NH3 donates its lone pair to Be to form H3N→BeF2). A is incorrect. It is tetrahedral around the N atom in BeF2 • NH3, hence the molecule is not planar. B is incorrect. In BeF42–, two F– ions form one co-ordinate bond each with BeF2. C is correct. It is possible for the lone pair on F to form a hydrogen bond with the H in H2O. D is incorrect. Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer C B C B D B D B C A C D A D A

2 © Raffles Institution 2025 9729/01/S/25 Q5 (Ans: B) pV = nRT p = 1 V (nRT) Since n, R and T are constants, p = 1 V (k) Hence graph is y = kx (straight line that passes through origin) Option A pV = nRT Since n, R and T are constants, pV = constant at a particular T Option C pV = nRT p = mass MV RT = density( RT M ) 1 T =density( R Mp) Since R, M and p are constants, 1 T = density(k) Option D pV = nRT pV T = nR Since n and R are constants, pV T = constant at a particular n density 1 T pV T p p pV T pV 0 0 0

3 © Raffles Institution 2025 9729/01/S/25 Q6 (Ans: D) element identity X Si Y Na Z P ✘ A Y (Na) has a smaller atomic radius than Z (P) ✘ B Y (Na) has a lower electrical conductivity than X (Si) ✘ C X (Si) has a higher first ionisation energy than Z (P) ✔ D X (Si) has a higher electronegativity value than Y (Na)

4 © Raffles Institution 2025 9729/01/S/25 Q7 (Ans: C) A E value becomes more negative down the Group. M2+(aq) + 2e ⇌ M(s) E The tendency of backward react