VJC 2022 H2 Chem Prelim P1 ans

1 Victoria Junior College 2022 H2 Chemistry Prelim Exam 9729/1 Suggested Answers 1 B 2 D 3 B 4 A 5 A 6 D 7 C 8 A 9 C 10 D 11 B 12 C 13 A 14 B 15 B 16 C 17 C 18 B 19 B 20 D 21 A 22 A 23 C 24 C 25 D 26 D 27 D 28 C 29 C 30 B 1 B element electronic configuration no. of unpaired e− A Al 1s22s22p63s23p1 1 O 1s22s22p4 2 B B 1s22s22p1 1 H 1s1 1 C Cu [Ar]3d104s1 1 I [Kr]4d105s25p5 1 D Fe [Ar]3d64s2 4 Cl 1s22s22p63s23p5 1 2 D Large dip in second I.E. from M to N → second electron removed from an outer shell in N → M is in Group 1 (Na) while N is in Group 2 (Mg). Option A: True L is a noble gas (Ne). It has a fully filled valence electron shell and does not form an oxide. Option B: True Q is silicon (Si) which has the highest melting point across period 3. Si has a giant molecular structure. Large amount of energy is required to break the strong covalent bonds. Option C: True The atomic radius of J (O) is larger than that of K (F). Both O and F are in the same period (similar shielding effect), but F has a greater number of protons (higher nuclear charge), hence higher effective nuclear charge, and valence electrons are pulled closer to the nucleus. So F has a smaller atomic radius than O. Option D: False The oxide of N (MgO) has a higher melting point than the oxide of R (P4O10). 3 B molecule no. of bp no. of lp shape bond angle A SO2 2 1 bent 118° OF2 2 2 bent 105° B OCS 2 0 linear 180° HCN 2 0 linear 180° C CCl4 4 0 tetrahe dral 109.5° XeF4 4 2 square planar 90° D CS2 2 0 linear 180° H2S 2 2 bent 105° 4 A Statement 1: Correct In all the three allotropes, each carbon atom is covalently bonded to 3 other carbon atoms. The remaining electron is delocalised and mobile. The p orbitals of the carbon atoms overlap to form a continuous π electron cloud allowing delocalisation of electrons and conduction of electricity. Statement 2: Incorrect Unlike graphite, graphene and carbon nanotube do not have layers held together by weak intermolecular forces of attraction . Hence they are not slippery to be used as lubricants. Statement 3: Incorrect Note that both carbon allotropes are non −polar. Energy released from the formation of dipole−induced dipole interactions between carbon allotropes and water is insufficient to overcome the strong C-C covalent bonds in the carbon nanotube and graph ene as well as the hydrogen bonds between water molecules. 5 A Since GA has a higher Mr than GB, and mass of the two gases is the same, amount of G A is lower than amount of GB. For an ideal gas, pV = nRT. Option A: Correct At constant p, 𝑉 = 𝑛𝑅 𝑝 𝑇 . Plotting V against T gives a straight line graph passing through the origin with constant gradient 𝑛𝑅 𝑝 . Option B: Incorrect At constant T, pV = nRT = constant. Plotting pV against V gives a horizontal line. Option C: Incorrect At constant T, pV = nRT = constant. Plotting p against pV gives a vertical line. Option D: Incorrect At constant V, 𝑝 = 𝑛𝑅 𝑉 𝑇 . Plotting Vp against T gives a straight line graph passing through the origin with constant gradient 𝑛𝑅 𝑉 . 6 D Option A: Correct Be in BeCl2 is electron deficient, hence it can accept lone pair of electrons to achieve stable octet configuration.

2 Option B: Correct Ba has a lower melting point than Sr. Less energy is required to overcome the weaker metallic bonds between the larger Ba2+ ions and electrons than the stronger metallic bonds between the smaller Sr 2+ ions and electrons. Option C: Correct Ba has more electron shells than Ca. Its valence electrons are further away from the nucleus, hence less attracted to the nucleus, and so need less energy to be removed. Option D: Incorrect The charge density and hence polarising power of Mg2+ is higher than that of Ca 2+ since Mg 2+ is smaller than Ca 2+. Hence there is a greater distortion of the anionic charge cloud and so MgCO3 is less stable to heat and decomposes at a lower temperature than CaCO3. 7 C Option A: Incorrect N2H4 is the Bronsted−Lowry base in Reaction 1 as it accepts a proton to form N2H5+. Option B: Incorrect N2H5+ and N 2H4, as well as NH 3 and NH 4+, are conjugate acid−base pairs. Option C: Correct Since the POE lies to the right for both reactions, HClO is a stronger acid than N2H5+ from Reaction 1 as it prefers to donate a proton. Likewise for Reaction 2 where N 2H5+ is a stronger acid than NH4+. Option D: Incorrect N2H4 is a weaker base than NH 3 because in Reaction 2, the POE lies to the right. 8 A Reactivity and oxidising power decreases down Group 17. Cl2 + 2Br− → 2Cl− + Br2 (reddish−brown gas) Br2 + 2I− → 2Br− + I2 (purple gas, dissolves in water to form brown solution) 9 C CxHy + (x+ y 4)O2 → xCO2 + y 2H2O Initial volume 10 90 0 Change in volume −10 −10(x+ y 4) +30 End volume 0 40 30 x = 3 90 − 10(x+ y 4) = 40 → y = 8 Hence formula is C3H8. 10 D Amount of TeO2 = 1.01 159.6 = 0.00633 mol Amount of K2Cr2O7 = 0.070 x 30 1000 = 0.0021 mol Since solution changes from orange (Cr 2O72−) to green (Cr3+), Cr2O72− + 14H+ + 6e− → 2Cr3+ + 7H2O, amount of electrons = 6 x 0.0021 = 0.0126 mol amount of e- amount of TeO2 = 0.0126 0.00633 = 2 Oxidation state of Te in TeO2 = +4 Since 1 mol of TeO2 loses 2 mol of e− in the reaction, oxidation state of Te in product = +6 11 B Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g); 2Hf(H2O) Hf(Ca(OH)2 Ca(s) + 2H2(g) + O2(g) By Hess Law, Hf(Ca(OH)2) = 2Hf(H2O) + Hr = 2Hc(H2) + Hr Hence, other than Hr, enthalpy change of combustion of hydrogen is needed. 12 C Option A (Wrong) S > 0 and H > 0 G= H – TS < 0 for large T only. Option B (Wrong) S < 0 and H < 0 G = H – TS < 0 for small T only. Option C (Correct) S > 0 and H < 0 G =H – TS < 0 for all T. Option D (Wrong) S < 0 and H > 0 G = H – TS > 0 for all T. 13 A Rate law is determined by slow step in proposed mechanism, i.e. Rate = k’[O3][O]. This rate law cannot be compared directly with the experimental rate equation because it contains the concentration of an intermediate, O. Thus we need to express rate law in a way that removes the intermediate O. From Step 1, K = [O2][O]/[O3]  [O] = K[O3]/[O2] Assuming that Step 1 equilibrium is established quickly before O is reacted with O3 in Step 2, Rate = k’[O3][O] = k’K[O3]2/[O2] = k[O3]2/[O2] where k = k’K ∆Hr

3 14 B = 23H S NHpp (NH4HS(𝑠) ⇌ H2𝑆(𝑔) + NH3(𝑔)) Thus, 23 1 H S NH 2 66.4 33.2 kPapp = =  = 1 ( ) ( )=  =  23 22 p H S NH 33.2 1100 kPaK p p 2 Removal of H 2S will cause more NH 4HS to dissociate forming not only H 2S but also NH 3. Hence total pressure will increase. 3 Addition or removal of a solid has no effect on the position of equilibrium so long as there is still solid present. 15 B A HA/A− buffer with maximum buffer capacity will have its pH = pKa of HA. So, pKa of HA = 6.38 => pKb of A− = 14 – pKa = 7.62 Hence, H2CO3 / HCO3− can be used to prepare a buffer of pH 6.28 with maximum buffer capacity. 16 C Stage I : oxidation of 2º alcohol to ketone Stage II : nucleophilic :OH– substitutes for the –CI3 group Stage III : basic :CI3– abstracts a proton from the acid RCO2H 17 C While all 3 compounds have 2 chiral centres , the compound in option one has a plane of symmetry and is hence not optically active. 18 B 19 B Step 1: Alkylation of benzene to form methylbenzene: Step 2: Nitration of methylbenzene to give 2–nitromethylbenzene and 4–nitromethylbenzene since –CH3 group is 2,4–director. For the formation of 4- nitroomethylbenzene: Step 3: Bromination of 4 – nitromethylbe