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 4