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 vaporisation, which has to do with overcoming intermolecular forces of attraction, and not breaking of covalent bonds.  B 7 A✓ NH3 + H2O  NH4+ + OH– (Brønsted- Lowry acid-base reaction) B✓ NH3 + BH 3 → H3N→BH3 (Lewis acid- base reaction) C✓ AlCl3 + Cl2  [Cl→AlCl3]– + Cl+ (Lewis acid-base reaction) D Both A lCl3 and BH 3 are electron - deficient species, capable of acting as Lewis acids only.  D 8 A 2 3 O 31 24 dm 1 mol24 dm moln −== AAno. of O atoms 1 mol 2 2 NN=   = B ( ) ( ) 2 3 C 1.5 101325 1 10 8.31 273 0.0670 mol n −  =  = l 2A A no. of C molecules 0.670 mol 0.670 N N = = l C✓ 4CaSO 68.05 40.1 32.1 16.0 4 0.500 mol n = + +  = CaSO4 → Ca2+ + SO42– AAno. of ions 0.500 mol 2 1.00NN=   = D 24 3 CH 31 22.7 dm 1 mol22.7 dm moln −== AAno. of C atoms 1 mol 2 2 NN=   =  C 9 pH of the period 3 chloride in water: no hydrolysis – NaCl : 7 slightly hydrolysis – MgCl2 : 6.5 extensive hydrolysis – AlCl3 : 3 complete hydrolysis – SiCl4, PCl5 : 2  C 10 A 6 P–P bonds are broken. So ( )( ) ( )r 136 B.E. P–P B.E. P–P42H =  = B✓ Phosphorus exists as P 4(s) under standard condition. C Phosphorus exists as P 4(s) under standard condition. D Enthalpy change of vapourisation is the energy required to transform one mole of a liquid substance into its gases state: P4(l) → P4(g)  B 11 1 3.33.08 298 1000 2.10 kJ mol H G T S − − =  +  = + +  =+ 1✓ 2 Graphite is more stable than diamond. C–C bonds in graphite are stronger than in diamond. 3 Since S is negative, –TS is always positive. Hence G increases (less spontaneous) with temperature.  A 12 Since    A B ,    rate A , where Bk k k==   1 2 ln2 ln2 Bt kk==  When [B] is doubled, 1 2 t will be halved.  B 13 1✓ Enzymes provide an alternative pathway with a lower activation energy. 2✓ When [substrate] is low, there are available active sites for additional substrate molecules to bind to. Hence, rate  [substrate]. 3 When [substrate] is high, all active sites are occupied. No available active sites for additional substrate molecules to bind to. Hence rate is constant.  B 14 Given       2 c 2 C AB K == x For the new equilibrium,                 1 2 1 2 1 2 c 2 2 A B 1 1 1 CC C AB AB K = = = =   x  C 15 1✓ ing the volume es the concentration of the reactants and products, leading to  in rates of both the forward and backward reactions. 2 adding a noble gas at constant pressure will  the partial pressure of all components  eqm shifts to the side with more gaseous particles. 3 Equilibrium constants are only affected by changes in temperature.  D

16 NaOH 20 0.500 0.0100 mol1000n =  = 32CH CO H 20 1.00 0.0200 mol1000n =  = CH3CO2H + NaOH → CH3CO2Na + H2O   3 32 0.0200 0.0100CH CO H 0.250 mol dm20 20 1000 −−== +   3 32 0.0100CH CO Na 0.250 mol dm20 20 1000 −== +     ( )3 2 5 a 32 CH CO NapH p lg lg 1.8 10 4.74CH CO HK −= + = −  =  C 17 Given solubility of AgCl and AgBr are x and y mol dm–3 respectively, ( )   ( )   2 2 6 sp 2 2 6 sp AgC Ag C mol dm AgBr Ag Br mol dm Kx Ky + − − + − − == == ll In a solution saturated with both AgCl and AgBr, let the solubility of AgCl and AgBr be x and y mol dm–3 respectively. Ag xy+  =+ ,  C x− =l ,  Br y− = ( )   ( )( ) ( )   ( )( ) 2 sp 2 sp AgC Ag C AgBr Ag Br K x y x x K x y y y +− +−   = = + =   = = + = ll 1 ( )( ) ( )( ) ( ) 22 2 22 x y x x y y x y x y x y      + + + = + + = + 22Ag x y x y x y+  = + = +  + 2✓   22 22 Br yyyy xy xy − = = = + + 3✓    Ag C Brxy+ − −  = + = + l  D 18 A There are no chiral centres in 2-methyl- N-phenylpropan-1-imine. B  bonds are formed by side-on overlap of unhybridised p orbitals. The C–N  bond is formed by head -on overlap between sp2 hybrid orbitals of C and N. C N forms 2  bonds and possesses 1 lone pair. Hence it is sp2 hybridised. D✓ The lone pair on N in 2-methyl-N- phenylpropan-1-imine is in a sp2 hybrid orbital, while that in NH 3 is in a sp 3 hybrid orbital . Due to the larger s- character of sp 2 hybrid orbital (33%) compared to sp 3 hybrid orbital (25%), the lone pair is closer to the nucleus (s orbitals are closer to nucleus than p orbitals) and less available for donation.  D 19 A✓ B C✓ . D✓  B 20 Dehydration upon heating with conc. H2SO4,  A 21 A The alkyl group is an activating group. Hence alkylbenzene will react faster. B✓ Anhydrous AlX3 and FeX3 can be used in Friedel-Crafts alkylation to generate the carbocation. C✓ Due to the bulky alkyl group, the 2 - position is sterically hindered and hence the CH 3CO– group will preferentially go to the 4-position. D✓ It is an electrophilic substitution: Ar–H + RCOCl → Ar–COR + HCl  A 22 and are resistant to nucleophilic substitutio