VJC 2021 Prelim H2P1 Ans [FInal]

1 Victoria Junior College 2021 H2 Chemistry Prelim Exam 9729/1 Suggested Answers 1 A 2 B 3 A 4 A 5 B 6 D 7 C 8 D 9 C 10 B 11 B 12 D 13 C 14 D 15 A 16 C 17 B 18 D 19 A 20 D 21 B 22 D 23 B 24 C 25 D 26 C 27 C 28 A 29 A 30 C 1 A (1, 2 and 3) Option 1: Incorrect X2+ has 53 electrons, X atom has 55 electrons (protons), X is Cs while Xe has 54 electrons (or protons). Hence, the atom of X is not isoelectronic with the atom of Xe. Option 2: Incorrect Both XF and RbF have giant ionic structure. Since LE α |q+q–/(r+ + r–)|, as size of X+ is bigger than Rb+, hence XF has a less exothermic LE and ionic bond strength in XF is weaker. Hence, XF has a lower melting point than RbF. Option 3: Incorrect Since Fr is below Cs, Fr should have a lower first IE as first IE decreases down the group. Option 4: Correct Since angle of deflection α (charge / mass) ion Na+ X2+ charge / mass 1/23 = 0.0435 2/133 = 0.0150 Hence, X2+ has a smaller angle of deflection. 2 B Option A: Correct Ethanol, C2H5OH, which has a highly electron–deficient H atom (d+) that is covalently bonded to a highly electronegative O, can be attracted to the lone pair of a highly electronegative atom (d–), which is O in caffeine through hydrogen bonding. Option B: Incorrect Caffeine does not have a planar structure since there are 3 bond pair of electrons and 1 lone pair around each N atom having three single bonds around it, it is trigonal pyramidal about each of these N atoms. Option C: Correct Each double bond is made up of 1 sigma and 1 pi bonds while each single bond is made up of only 1 sigma bond. Option D: Correct There are 5 carbon atoms having 3 bond pair of electrons around each of them, implying that they are sp2 hybridised. 3 A Option A: Correct CH3CO2–NH4+ has a giant ionic structure. More energy is required to break the stronger electrostatic forces of attraction between oppositely charged ions. CH3CH2NH2 has a simple molecular structure with weaker hydrogen bonds between molecules. Hence, CH3CO2–NH4+ has a higher bp. Option B: Incorrect SiO2 has a giant molecular structure while SiCl4 a simple molecular structure. More energy is required to break the stronger and extensive covalent bonds between atoms in SiO2 than the id-id interactions between non-polar SiCl4 molecules. Hence SiO2 has a higher bp. Option C: Incorrect Both PH3 and SiCl3H exist as simple molecules. SiCl3H has a larger electron number, hence its electron cloud is more polarisable resulting in stronger id-id interactions between SiCl3H molecules. Thus SiCl3H has a higher bp. Option D: Incorrect Both Ni(OH)2 and NiSO4 exist as giant ionic structures. Since SO42– has a higher charge than OH– and LE α |q+q–/(r++r–)|, LE of NiSO4 is more exothermic and more energy is required to break the stronger ionic bonds in NiSO4. Hence, NiSO4 has a higher bp. 4 A When the vessel is dented, pV = nRT n = pV/RT = (52.8 x 103 x 38.5 x 10–6) / [8.31 x (27 +273)] = 8.15 x 10–4 mol Amount of gas remained unchanged when the shape of the vessel is restored. Hence, when the shape of the vessel is restored to 40.0 cm3 at 60 oC, p = nRT/V = [8.15 x 10–4 x 8.31 x (60 + 273)] / (40.0 x 10–6) = 56.4 kPa 5 B Al2O3 is an amphoteric oxide while SiO2 is an acidic oxide. Since both solutions prepared can dissolve Al2O3, but only one can dissolve SiO2, one solution must be basic while the other solution must be acidic. Y (chloride of Y) Z (oxide of Z) x A silicon (SiCl4: acidic) phosphorus (P4O10: acidic) ü B phosphorus (PCl5 : acidic) sodium (Na2O: basic) x C magnesium (MgCl2: acidic) phosphorus (P4O10: acidic) x D sodium (NaCl: neutral) sulfur (SO3: acidic)

2 Hence, Y is P while Z is Na. Chloride of Y: PCl5 + 4H2O ® H3PO4 + 5HCl Oxide of Z: Na2O + H2O ® 2NaOH For reaction with Al2O3: 2H3PO4 + Al2O3 ® 2AlPO4 + 3H2O 2NaOH + Al2O3 + 3H2O ® 2NaAl(OH)4 For reaction with SiO2: H3PO4 + SiO2 ® no reaction 2NaOH(conc.) + SiO2 ® Na2SiO3 + H2O 6 D Based on the melting point data, the four elements are Si (Group 14) (high melting point due to giant molecular structure), P4 (Group 15), S8 (Group 16) and Cl2 (Group 17). Property X is electronegativity since electronegativity increases across the period. Property Y is 3rd IE since there is an anomaly at Group 15 element. Si2+ ® Si3+ + e 3s2 3s1 P2+ ® P3+ + e 3s2 3p1 3s2 For P2+, an electron is removed from the 3p subshell which is further away from the nucleus. Thus, the 3p subshell is of a higher energy and its electron is less strongly attracted to the nucleus and hence less energy is required to remove it. 7 C Option A: Incorrect Ba2+ has a lower charge density than Mg2+ due to its smaller ionic radius, hence, BaCl2 undergoes smaller extent of hydrolysis than MgCl2, which results in a higher pH. Option B: Incorrect MCO3 ® MO + CO2 1 mol of MCO3 will decompose to form 1 mol of CO2. As Mr of MCO3 increases down the group, amount of 1 g of MCO3 will decrease down the group. Hence, smaller amount and volume of CO2 gas will be produced. Option C: Correct Mg reacts less vigorously with cold water than strontium as the Eo(Sr2+|Sr) is less positive / more negative than that of Eo(Mg2+|Mg). Hence, Sr tends to be oxidised to Sr(OH)2 more readily than Mg. Option D: Incorrect Group 2 sulfates become less soluble down the group since |DHhyd| decreases as cations become larger in size, while |DHLE| remains relatively constant due to large size of sulfate. Indeed, barium sulfate has a weaker ionic bond strength. It has less exothermic lattice energy since ionic radius of Ba2+ is larger than that of Mg2+. 8 D From expt 1, Y– is oxidised to Y2 by X2. With Y2 appearing as violet in organic layer, Y2 = I2 and oxidising strength: X2 > Y2. From expt 2, X– is oxidised to X2 by Z2. With X2 appearing as orange-red in organic layer, X2 = Br2 and oxidising strength: Z2 > X2. From expt 3, Y– is oxidised to Y2 by Z2. With Y2 appearing as violet in organic layer, Y2 = I2 and oxidising strength: Z2 > Y2. Hence, oxidising strength: Z2 > X2 > Y2 Þ Z2 = Cl2, X2 = Br2, Y2 = I2 Option A: Incorrect HY (HI) has the highest boing point among three hydrogen halides due to its strongest instantaneous dipole−induced dipole interactions between molecules since HY (HI) has the largest number of electrons. Option B: Incorrect Ksp of AgY (AgI) is so low that even at high concentration of NH3, the ionic product of AgY is still greater than its Ksp value. Hence, AgY (AgI) does not dissolve in both dilute and concentrated NH3. Option C: Incorrect HY (HI) is a stronger acid than HZ (HCl), hence HY has a larger Ka value. Since atomic radius of Y is larger than than of Z, H−Y bond is weaker than H−Z bond and less energy is needed to overcome H−Y bond to dissociate HY into H+ and Y–. Option D: Correct For expt 4, since Y2 is a weaker oxidising agent than X2, Y2 will not be able to oxidise X– to X2. Hence, no redox reaction occurs and Y2 (I2) will appear as violet in organic layer, which is the same as expt 3. 9 C CO32– in sodium percarbonate reacts with sulfuric acid to form CO2 gas: CO32– + 2H+ ® CO2 + H2O n(CO32–) = n(CO2) = 48 / 24000 = 2.00 × 10–3 mol H2O2 is oxidised by MnO4–: MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O H2O2 ® O2 + 2H+ + 2e– 2 MnO4– : 5 H2O2 n(H2O2) = 52 × n(MnO4–) = 52 × 241000 × 0.0500 = 3.00 × 10–3 yx = n(H2O2) / n(CO32–) = (3.00 × 10–3) / (2.00 × 10–3) = 32

3 10 B CSz + (1 + z)O2 ® CO2 + zSO2 Initial / cm3 20 80 0 0 Change / cm3 –20 –20(1 + z) +20 +20z End / cm3 0 80 – 20(1 + z) 20 20z Gas mixture after combustion contains unreacted O2, CO2 and SO2. Upon treatment with NaOH(aq), the acidic gases, CO2 and SO2, are removed. Hence, volume after treatment with NaOH(aq) = volume of unreacted O2 = 20 80 – 20(1 + z) = 20 60 – 20z = 20 z = 2 11 B (1 and 4 only) DHro = 790 = Σ(DHf)products – Σ(DHf)reactants = 3xDHf(CO) – DHf(Cr2O3) DHf(Cr2O3) = 3xDHf(CO) – 790 Hence, to