2021 EJC Prelims Paper 1 Worked Solutions

2021 EJC JC2 Prelim Exam H2 Chemistry 9729 Paper 1 Worked Solution 1 A 1s2 2s2 2p3 → 1s2 2s2 (not octet) B✓ 1s2 2s2 2p6 3s2 3p1 → 1s2 2s2 2p6 (octet) C 1s2 2s2 2p6 3s2 3p6 → 1s2 2s2 2p6 3s2 3p3 (not octet) D [Ar] 3d2 4s1 (not ground state; ground state is [Ar] 3d1 4s2) → [Ar] (octet)  B 2 1 16 1 3 8HO + 2 16 1 3 8HO + 2 16 1 2 8HO 16 1 81OH − p 1 3 8 11 + = 1 3 8 11 + = 1 2 8 10 + = 81 9 + = e 1 3 8 1 10  + − = 1 3 8 1 10  + − = 1 2 8 10 + = 8 1 1 10 ++ = n 0 3 8 8 + = 1 3 8 11 + = 1 2 8 10 + = 80 8 + =  A 3 nR gains 1 electron to form stable n −R , which has (n+1) electrons  R is in Group 17. 2n+ S forms a stable ion which is isoelectronic with n −R . Hence 2n+ S must have lost 1 electron, to form 2n + + S , which has (n+2)–1 = (n+1) electrons  S is in Group 1 of the next period. A✓ Since n −R and 2n + + S are iso - electronic, 2n + + S with a higher nuclear charge will have a smaller ionic radius as the effective nuclear charge experienced by the valence e–s is higher. B 2 nn e− − −+→RR will be endothermic due to repulsion of the incoming e–; 22nn e+− ++ +→SS will be exothermic due to attraction of the incoming e–. C As S is an element in the next period, with one additional filled principal quantum shell, S has a larger atomic radius despite the higher nuclear charge. D 2 22nn e+ − + ++ −→SS will be more endothermic than nn e−−−→RR since the e– is being removed from positively charged 2n + + S .  A 4 A Polar single covalent bonds involves unequal sharing of bonding electrons. B Single covalent bonds are also formed in atoms without a complete octet, e.g. BF3 with 6 outer shell electrons around B, or with an expanded octet, e.g. PF5 with 10 outer shell electrons around P. C H–H bond (436 kJ mol –1) is stronger than H–Cl bond (431 kJ mol–1) D✓ Due to the larger s-character of sp 2 hybridised C (33%) compared to sp 3 hybridised C (25%), the bonding electrons are closer to the nucleus ( s orbitals are closer to nucleus than p orbitals). Hence, the 2spC –H bond is stronger than a 3spC –H bond.  D 5 1 There are no H bonded to F, O or N in cinnamaldehyde  no H bonds between cinnamaldehyde molecules. 2✓ The non-polar chain of cinnamaldehyde can form strong instantaneous dipole -induced dipole solvent-solute interactions with the non-polar hydrocarbon parts of organic solvents. 3✓ All carbon atoms in cinnamaldehyde are sp 2 hybridised, thus all the bond angles are the same at 120º.  B 6 A S2O32– to S4O62– is an oxidation as the oxidation number of S increases from +2 to +2.5. B✓ Br2 is unable to oxidise Cl–. Hence Br2(g) dissolves in water to give orange Br2(aq). C Although C l2 molecules possess lone pairs on C l. However, C l is not sufficiently electronegative (F, O or N) to form H bonds with the H in water. D Volatility is related to the ease to vaporisatio