ASRJC 2021 J2Prelims H2Chem P1 soln

ASRJC JC2 PRELIMS 2021 9729/01/H2 [Turn over Anderson Serangoon Junior College 2021 Preliminary Examination H2 Chemistry Paper 1 Suggested Worked Solution 1 Two particles, A+ and B3+ are fired with equal velocity into an electric field. The information on two particles are given in the table below. particle number of electrons number of neutrons angle of deflection in an electric field A+ 10 12 2.0° B3+ ? 14 5.1° What is the number of electrons for B3+? A 10 B 11 C 12 D 13 Answer: A e p n (given) Mass (n+ p) Charge, q q (given) A+ 10 10+1 = 11 12 23 +1 2.0° B3+ ? ? 14 m +3 5.1° By proportion, for B3+ = ÷2 x 5.1 = 0.1109 \ = 0.1109 m = 27 Number of protons for B3+ = 27 -14 =13 Number of electrons for B3+ = 13-3 = 10 q m 1 23 3 m q m 1 23 3 m

2 ASRJC JC2 PRELIMS 2021 9729/01/H2 2 A to H are consecutive elements with atomic numbers less than 20. The graph below shows their second ionisation energies (2nd I.E.). Which of the following statements is correct? A The 2nd IE of G is lower than that of F due to the inter-electronic repulsion between its paired s electrons. B C exists as diatomic molecules at room temperature. C The compound formed between A and E has a low melting point. D Element B is from Group 17. Answer: D Since there is sharp decrease in 2nd IE between elements D & E, the electron removed from E+(g) is in an electron shell that is further from the nucleus than that from D+(g). D+ has noble gas configuration. D has 1 valence electron, hence it is from Group 1. Counting backwards, B is from Group 17. 0500100015002000250030003500400045005000 ABCDEFGH 2nd I.E. element

3 ASRJC JC2 PRELIMS 2021 9729/01/H2 [Turn over 3 The graph shows the logarithm of the first twelve ionisation energies (I.E.) for element J. What can be deduced about element J from the graph? 1 It can form a compound with oxygen with the formula J2O3. 2 J is likely to have a lower first I.E. than the element preceding it in the same period. 3 It is in the second period (Li to Ne) of the Periodic Table. 4 J has a half-filled p-subshell. A 1, 2 and 3 B 1, 2 and 4 C 1 and 2 only D 3 and 4 only Answer: C 1 Correct Element J belongs to Group 13. Largest energy difference between the 3rd and 4th I.E. This implies that the removal of the 4th electron is from an inner quantum shell which requires more energy. Hence, element J has 3 valence electrons and can form a compound with oxygen with the formula J2O3. 2 Correct Valence electronic configuration for J: ns2 np1 Valence electronic configuration for element preceding J: ns2 The first ionisation energy for element J requires removing 1 electron from p subshell which is further from the nucleus than s subshell and experience weaker electrostatic attraction. Hence, the first I.E of J is lower than the element preceding it in the same period. 3 Incorrect Period 2 elements have a maximum of only 10 ionisations energies but element J has at least twelve ionisation energies. 4 Incorrect Valence electronic configuration for J: ns2 np1 Element J does not have a half-filled p-subshell. log (I.E.) number of electrons removed

4 ASRJC JC2 PRELIMS 2021 9729/01/H2 4 Acrylonitrile, CH2=CHCN is a monomer used to made polyacrylonitrile. Which row correctly describes the bonding number of p bonds and hybridisation in a molecule of acrylonitrile? Number of π bonds Number of sp C atoms Number of sp2 C atoms A 2 2 1 B 3 1 2 C 1 1 2 D 1 2 1 Answer: B Double bond consists of 1 σ and 1 π. Triple bond consists of 1 σ and 2 π. Total no. of π bond = 1 + 2 = 3 The nitrile group has 1 sp C atom. The alkene has 2 sp2 C atoms. CC H H C H N

5 ASRJC JC2 PRELIMS 2021 9729/01/H2 [Turn over 5 The mechanism for a certain reaction is given below. Which of the statements are correct? 1 There is a decrease in the bond angle with respect to O atom in step 1. 2 The shape with respect to C atom in bold, changes from tetrahedral to trigonal planar and back to tetrahedral in the mechanism. 3 Dative bond is formed in step 1 and 3. A 1 and 3 B 1 and 2 only C 2 and 3 only D 2 only Answer: C Statement 1 is incorrect. O in reactant has 2 bond pairs and 2 lone pairs with bond angle 104.5o. It has increases to 3 bond pairs and 1 lone pair with bond angle 107o in the product. MCQ Tip J : Since option 1 is wrong, you can remove option A and B. Thus, all you have to do is to analyse option 3 to decide on the option to pick. Statement 2 is correct. Step reactant product 1 4 bp tetrahedral 4 bp tetrahedral 2 4 bp tetrahedral 3 bp only Trigonal planar 3 3 bp only Trigonal planar 4 bp tetrahedral Statement 3 is correct. Lone pair from O atom donates to empty orbital in H+. Lone pair from Cl- donates to empty orbital in positively charged C of carbocation. step 1 (CH3)3COH HCl+ fast (CH3)3COH2 Cl+ step 2 slow(CH3)3COH2 +(CH3)3C H2O step 3 (CH3)3C fastCl+ (CH3)3CCl (CH3)3COH+ H+ (CH3)3COH2step 1 +Cl-(CH3)3Cstep 3 (CH3)3CCl

6 ASRJC JC2 PRELIMS 2021 9729/01/H2 6 In which of the following does Statement II give a correct explanation for Statement I? Statement I Statement II 1 Magnesium has a higher melting point than sodium. Magnesium has more delocalised valence electrons which results in stronger metallic bonds. 2 Glycine, H2NCH2COOH, has a higher melting point than 2–hydroxyethanoic acid, HOCH2COOH. Glycine can form more extensive hydrogen bonds than 2–hydroxyethanoic acid. 3 Chloromethane undergoes nucleophilic substitution more easily than fluoromethane. The C–Cl bond in chloromethane is weaker than the C–F bond in fluoromethane. A 2 and 3 only B 1 and 2 only C 1 and 3 only D 1, 2, and 3 Answer: C Statement 2 is incorrect. Glycine forms zwitterions and has an ionic lattice structure with strong electrostatic forces of attraction between the zwitterions, hence has a higher melting point than 2–hydroxyethanoic acid

7 ASRJC JC2 PRELIMS 2021 9729/01/H2 [Turn over 7 The amount of carbon monoxide present in air can be determined by its reaction with iodine pentoxide, I2O5, to form carbon dioxide and iodine in the reaction below. 5CO + I2O5 ® I2 + 5CO2 The amount of iodine liberated is then determined by titration with a standard solution of sodium thiosulfate. I2 + 2S2O32– ® S4O62– + 2I– A 100 cm3 sample of polluted air is passed over solid iodine pentoxide and the iodine produced required 20.0 cm3 of 0.20 mol dm–3 of sodium thiosulfate for complete reaction. What is the concentration, in g dm–3, of carbon monoxide present in the sample of polluted air? A 0.100 B 1.12 C 2.80 D 11.2 Answer: C 5CO + I2O5 ® I2 + 5CO 2S2O32– + I2 ® S4O62– + 2I– 5CO º I2 º 2S2O32– nS2O32– = = 0.004 mol nCO = = 0.01 mol Mass of CO = 0.01 x (12 + 16) = 0.28 g Concentration of CO = = 2.80 g dm-3 20×0 . 21000 5× 0.0042 0.28 100 1000

8 ASRJC JC2 PRELIMS 2021 9729/01/H2 8 Two glass vessels M and N are connected by a closed valve. M contains helium at 25 oC at a pressure of 1 x 105 Pa. N has been evacuated, and has three times the volume of M. In an experiment, the valve is opened and the whole set-up placed in boiling water at 100 oC. What is the final pressure in the system? A 3.13 × 104 Pa B 3.76 × 104 Pa C 1.00 × 105 Pa D 1.33 × 105 Pa Answer: A Let the volume of M be v, hence volume of N is 3v. Total volume = 4v P!V!T!=P"V"T" 10#×v298= P"×4v373 P2 = 3.13 x 104 Pa N M

9 ASRJC JC2 PRELIMS 2021 9729/01/H2 [Turn over 9 Which of the following diagrams correctly describes the behavior of a fixed mass of an ideal gas at constant T? A B C D Answer: C The correct diagrams are as follow. Note that n and T are constant in all cases. A B pV = nRT è p = nRT()