RI Prelim P1 (Ans)

1 © Raffles Institution 2020 9729/01/S/20 2020 RI H2 Chemistry Prelim Paper 1 – Suggested Solutions Question 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Answer C C D A A B D A B C B A B C C MCQ worked solutions Q1 (Ans: C) 28.0 m 25.0 131 27.0 132 20.0 134 131.3128.0 25.0 27.0 20.0         m = 129 Q2 (Ans: C) 5 X− + XOn−  halogen-containing product (either X2 or XO−) Step 1: Balance X; 5 X− + XOn−  3 X2 OR 5 X− + XOn−  6 XO− Step 2: Check if amt of electrons gained = amt of electrons lost Option A 5 X− + XO−  3 X2 amt of electrons gained by 1 mol of XO− to form ½ mol X2 = 1 mol amt of electrons lost by 5 mol of X− to form 5 2 mol of X2 = 5 mol not balanced Option B 5 X− + XO2−  6 XO− amt of electrons gained by 1 mol of XO2− to form 1 mol of XO− = 2 mol amt of electrons lost by 5 mol of X− to form 5 mol of XO−= 10 mol not balanced Option C 5 X− + XO3−  3 X2 amt of electrons gained by 1 mol of XO3− to form ½ mol X2 = 5 mol amt of electrons lost by 5 mol of X− to form 5 2 mol of X2 = 5 mol balanced Option D 5 X− + XO4−  6 XO− amt of electrons gained by 1 mol of XO4− to form 1 mol of XO− = 6 mol amt of electrons lost by 5 mol of X− to form 5 mol of XO− = 10 mol not balanced Question 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Answer C D D B D A B A D B D C B A D

2 © Raffles Institution 2020 9729/01/S/20 Q3 (Ans: D) ntotal = nAr + nXe PtotalVtotal/RT = PArVAr/RT + PXeVXe/RT At constant T, PtotalVtotal = PArVAr + PXeVXe P  (5 + 10) = (150  5) + (350  10) P = 283 kPa At constant V and n, pressure  Temperature 600 T 283 298 T = 632 K Q4 (Ans: A) U92 235  Rb37 96 + Cs55 135 + 4 XA Z 235 = 96 + 135 + 4Z  Z = 1 92 = 37 + 55 + 4A  A = 0 X is n0 1  neutron Q5 (Ans: A) Horizontal axis: P4O10 (pH 3), SiO2 and Al2O3 (pH 7 as they do not dissolve in water), MgO (pH 9) Vertical axis: P4 (simple molecular) < Mg (giant metallic) < Al (giant metallic) < Si (giant covalent) Q6 (Ans: B) The graph shows increased solubility in acidic and alkaline solutions  compound is amphoteric, i.e. Al(OH)3 With acid: Al(OH)3(s) + 3HCl(aq)  AlCl3(aq) + 3H2O(l) With base: Al(OH)3(s) + NaOH(aq)  Na+[Al(OH)4]–(aq)

3 © Raffles Institution 2020 9729/01/S/20 Q7 (Ans: D) Graph shows a sharp decrease in second IE from W to X  second electron removed from W is from an inner electronic shell  W is in Group 1 Statement 1 Not correct. V is in Group 18. However, if V is helium then U is hydrogen, but hydrogen does not have a second IE. As such, U to Z must be fluorine to silicon (with atomic numbers less than 20). Label Group No. Identity U 17 F V 18 Ne W 1 Na X 2 Mg Y 3 Al Z 4 Si Statement 2 Correct. Group 1 metals (e.g. Na) are more reactive than their adjacent Group 2 metals (e.g. Mg) as Group 1 metals require less energy to form their ions. Statement 3 Correct. Aluminium is a better electrical conductor than silicon (which is a semi - conductor) Q8 (Ans: A) Option A Correct. Down the group, EM2+/M becomes more negative  tendency of oxidation of M increases  reducing power increases Option B Not correct. Electronegativity increases across a Period but decreases down a Group. Option C Not correct. Down the Group, the charge of the Group 2 ions remain the same but ionic radii increases. Hence, the magnitude of ∆Hhydration decreases. Option D Not correct. Down the Group, the charge of the Group 2 ions remain the same but ionic radii increases. Hence, the magnitude of lattice energy decreases. |LE|  qq rr  

4 © Raffles Institution 2020 9729/01/S/20 Q9 (Ans: B) Option A Not correct. C1 and C3 are sp 2 hybridised with a trigonal planar shape (three regions of electron density – 3 bond pairs, 0 lone pairs). Option B Correct. C2–C3 bond is formed from a sp2-sp2 overlap while C4–C5 bond is formed from a sp2-sp3 overlap. Since sp2 orbital has greater s -character than sp3 orbital, the sp2-sp2 overlap is more effective. Hence the C2–C3 bond is a stronger bond and has a shorter bond length. Option C Not correct. There are 12  bonds in a molecule of penta-1,3-diene. Note: The C-H bonds are also  bonds. Option D Not correct. The  electrons are only able to move freely between C1 and C4. This is because C1 to C4 are all sp 2 hybridised and each contain an unhybridised p orbital which allows for continuous p orbital overlap. This allows for the  electrons to move freely (delocalised) within the overlapping p orbitals. C5 is sp3 hybridised and does not have an unhybridised p orbital. Q10 (Ans: C) Option A Not correct  m.p. of SiCl4 < SiO2 SiCl4 has a simple molecular structure with weak intermolecular forces of attraction whereas SiO2 has a giant molecular structure with strong covalent bonds. Option B Not correct  m.p. of NaCl < Na2O Since O2− has a larger charge and smaller radius than C l−, Na2O has stronger ionic bonds (more exothermic LE) than NaCl. |LE|  qq rr   Option C Correct  m.p. of NaCl > AlCl3 NaCl is an ionic compound with strong ionic bonds whereas AlCl3 is a covalent compound with simple molecular structure and weak intermolecular forces of attraction. (Note: AlCl3 is a covalent compound due to the high polarising power of Al3+ and the high polarisability of Cl−.) Option D Not correct  m.p. of cis-but-2-ene < trans-but-2-ene The molecules of cis-isomer pack poorly in the solid lattice because the two bulky methyl groups are located on the same side of the C=C bond. The poor packing results in larger distances between molecules, which lead to weaker intermolecular forces and hence a lower melting point for the cis-isomer.

5 © Raffles Institution 2020 9729/01/S/20 Q11 (Ans: B) Option A Not correct. For step Q, 2Al(g)  2Al3+(g) + 6e− H = 2  sum of 1st, 2nd and 3rd IE of Al = 2(577 + 1820 + 2740) = +10274 kJ mol–1 Option B Correct. For step R, 3O(g) + 6e−  3O2−(g) H = 3  sum of 1st and 2nd electron affinity of oxygen = +2106 kJ mol–1 Sum of the first and second electron affinity of oxygen = +2106 3 = +702 kJ mol−1 Option C Not correct. The enthalpy change for step P should be: sum of 2Hatom(Al) and 3Hatom(O2) (Recall: Standard enthalpy change of atomisation, Hatom, of an element is the energy absorbed to form one mole of gaseous atoms from the element at 298 K and 1 bar.) Option D Not correct. Lattice energy = H (step S) = H (step T) – H (sum of steps P + Q + R) It is wrong to ignore the signs of the enthalpy changes by saying that lattice energy of aluminium oxide is the sum of the enthalpy changes in steps P, Q, R and T. A correct statement would be “The magnitude of lattice energy of aluminium oxide (step S) is the sum of the magnitude of the enthalpy changes in steps P, Q, R and T”. Q12 (Ans: A) Statement 1 Correct. (1) C2H6(g) + 7 2 O2(g)  2CO2(g) + 3H2O(l) H = –1542 kJ mol−1 (given) (2) C2H6(g) + 7 2 O2(g)  2CO2(g) + 3H2O(g) H = –1434 kJ mol−1 (3) 3H2O(l)  3H2O(g) H = 3y Since eqn (3) = eqn (2) – eqn (1), 3y = –1434 –(–1542) = +108 kJ mol−1 y = +108 3 = +36 kJ mol−1 Statement 2 Correct. Reaction 2 involves a change of physical state from H2O(l)  H2O(g), hence G = H – TS = 0 at boiling point, 373 K (phase transition temperature). S = H T = y 373 kJ mol−1 K−1a Statement 3 Not correct. S < 0 as t he number of gaseous particles de creases as reaction 1 proceeds. Since G = H – TS and H > 0, reaction 1 is not spontaneous at all temperatures.

6 © Raffles Institution 2020 9729/01/S/20 Q13 (Ans: B) Option A Correct. NO2 is a free radical as it con