ASR 2020 J2Prelim H2Chem P1 Soln

ASRJC JC2 PRELIMS 2020 9729/01/H2 [Turn over Anderson Serangoon Junior College 2020 JC2 Preliminary Examination H2 Chemistry (9729) Paper 1 Worked Solutions. 1 Beams of charged particles are deflected by an electric field. In an experiment, protons are deflected by an angle of +15º. In another experiment, under identical conditions, particle A is deflected by an angle of –5º. What could be the composition of particle A? protons neutrons electrons A 1 2 2 B 3 3 2 C 3 3 4 D 4 5 1 Answer: A Particle A is deflected to the opposite direction as proton  it is negatively charged (i.e. electrons are more than protons)  options A and C are possible answer. Next, note that both are singly charged  the mass of particle A must be 3 times that of proton (i.e. sum of proton and neutron is 3) Alternatively, work out the details as follow. Angle of deflection, z m For a proton, 1 1H , 1 11 z m    For particle A,  51 115 3 z m     A : 3 1 A B : 6 3 A C : 6 3 A D : 93 4 A protons neutrons electrons Charge overall m e/m A 1 2 2 –1 3 –1/3 B 3 3 2 +1 6 +1/6 C 3 3 4 –1 6 –1/6 D 4 5 1 +3 9 +1/3

2 ASRJC JC2 PRELIMS 2020 9729/01/H2 2 50 cm 3 of a 0.10 mol dm –3 solution of a metallic salt was found to react exactly with 25.0 cm3 of 0.10 mol dm–3 aqueous sodium sulfite. In this reaction, the sulfite ion is oxidised as follows. SO32–(aq) + H2O(l)  SO42–(aq) + 2H+(aq) + 2e What is the new oxidation number of the metal in the salt if its original oxidation number was +3? A +1 B +2 C +4 D +5 Answer: B Amount of sulfite ions = 25 0.101000 = 0.0025 mol Amount of metallic salt = 50 0.101000 = 0.005 mol Method 1 Let x be the new oxidation no of metal in salt. [R]: M3+ + (3–x)e → Mx Since moles of electrons gained = moles of electrons lost in a redox reaction, 3 0.0025 2 0.005  x x = +2 Method 2 2 mol of metallic salt reacts with 1 mol of SO32– 2 mol of metallic salt gains 2 mol of e– (since 1 mol of SO32– loses 2 mol of e–) 1 mol of metallic salt gains 1 mol of e– Final O.N. = +3 + 1 = +2 3 Which particle would, on losing two electrons, have a half–filled p subshell? A Ga B Se C Te+ D As2+ Answer: B Ga: [Ar] 3d10 4s2 4p2; on losing two electrons, there will be 0 electron in p subshell. Se: [Ar] 3d10 4s2 4p5; on losing two electrons, there will be 3 electrons in p subshell (half - filled). Te+: [Kr] 4d10 5s2 5p3; on losing two electrons, there will be 1 electron in p subshell. As2+: [Ar] 3d10 4s2 4p1; on losing two electrons, there will be 0 electron in p subshell.

3 ASRJC JC2 PRELIMS 2020 9729/01/H2 [Turn over 4 The successive ionisation energies (I.E.) of two elements, B and C, are shown below: I.E. / kJ mol–1 1st 2nd 3rd 4th 5th 6th 7th 8th B 1000 2252 3357 4556 7004 8496 27107 31719 C 578 1817 2745 11577 14842 18379 23326 27465 What is the likely formula of the compound formed when B and C reacts together? A B2C3 B B3C2 C BC3 D B3C Answer: B For element B: biggest increase between 6th and 7th ionisation energy. (Largest difference in IE between 6th and 7th I.E.) 7th electron is removed from the inner quantum shell which is closer to the nucleus. Thus the element has 6 valence electrons. Element B belongs to Group 16. For element C: biggest increase between 3rd and 4th ionisation energy. (Largest difference in IE between 3rd and 4th I.E.) 4th electron is removed from the inner quantum shell which is closer to the nucleus. Thus the element has 3 valence electrons. Element C belongs to Group 13. So the likely formula of the compound formed is B3C2.

4 ASRJC JC2 PRELIMS 2020 9729/01/H2 5 Molecular dimerisation can be described as the process in which two identical molecules combine to give a single product. Examples of dimers are: Al2Cl6, N2O4 and (CH3CO2H)2. Which of the following descriptions about the above dimers is incorrect? A Hydrogen bonds hold the CH3CO2H molecules together in the dimer. B Each aluminium atom is surrounded by four chlorine atoms in Al2Cl6. C All the nitrogen–oxygen bonds in N2O4 are of equal length. D Al2Cl6 is a planar molecule. Answer: D Statement A is correct. H3C C O O H C CH3 OH O   Statement B is correct. Al Al Cl Cl Cl Cl Cl Cl Statement C is correct. N N O O O O N N O OO O N N OO O O Statement D is incorrect. as Al2Cl6 is not a planar structure. It is tetrahedral about each Al.

5 ASRJC JC2 PRELIMS 2020 9729/01/H2 [Turn over 6 Which of the following observations can be explained by intermolecular hydrogen bonding? 1. Ammonia (NH3) has a higher boiling point than methane (CH4). 2. Water has a lower density at 0 °C than at 25 °C. 3. Formation of H3O+ from water. A 1,2 and 3 B 1 and 2 only C 2 and 3 only D 1 only Answer: B Option 1: Ammonia has intermolecular hydrogen bonding while methane only has intermolecular id−id. More energy is required to overcome the stronger intermolecular hydrogen bonding and thus, ammonia has a higher boiling point than methane. Option 2: In ice, each H2O molecule is hydrogen−bonded to four other H2O molecules in a tetrahedral arrangement, giving rise to an open structure. Hence having a lower density. Option 3: H3O+ is formed via dative bonding between water and H+. 7 Melphalan is a drug used in chemotherapy. When dissolved in blood, the decrease in its concentration has a constant half–life of 90 minutes. A 100 mg melphalan tablet is dissolved in 4.0 dm3 blood. What is the concentration of melphalan in the blood six hours later? A 1.56 mg dm−3 B 3.13 mg dm−3 C 12.5 mg dm−3 D 25.0 mg dm−3 Answer: A [melphalan]initial = 100 / 4 = 25 mg dm–3 No of half–lives = (6×60) / 90 = 4 Using C/Co = (1/2)n [melphalan]6 hours later / [melphalan]initial = (1/2)4 [melphalan]6 hours later / 25 = (1/2)4 [melphalan]6 hours later = 1.56 mg dm−3

6 ASRJC JC2 PRELIMS 2020 9729/01/H2 8 Hydrogen peroxide slowly decomposes at room temperature. 2H2O2 2H2O + O2 Two experiments were performed to study the effects of adding lead( IV) oxide on the decomposition of 1.5 mol dm–3 hydrogen peroxide at constant temperature. Experiment 1: 20 cm3 of hydrogen peroxide. Experiment 2: 20 cm3 of hydrogen peroxide and 1.0 g of lead(IV) oxide. At the end of the experiment 2 the mixture was filtered and 1.0 g of lead( IV) oxide was recovered. Which row is correct? value of rate constant activation energy A equal in experiment 1 and 2 higher in experiment 1 B equal in experiment 1 and 2 higher in experiment 2 C higher in experiment 2 higher in experiment 1 D higher in experiment 2 higher in experiment 2 Answer: C Since rate of production of O 2 is faster in experiment 2 and lead(IV) oxide is recovered at the end of the experiment, lead(IV) oxide is a catalyst that speeds up the rate of the reaction. A catalyst increases the reaction rate by providing a different reaction pathway which has a lower activation energy (Ea). Since k = A 𝑒 −𝐸𝑎 𝑅𝑇⁄ , a lower Ea