RVHS 2026 Prelim P2 Ans

2026 RVHS Preliminary Examination H2 Chemistry Paper 2 Suggested Answers 1 (a) (i) White ppt of AgCl dissolves in aqueous ammonia to give a colourless solution. AgCl(s) + 2NH3(aq) → [Ag(NH3)2]+(aq) + Cl(aq) (ii) The thermal stability of hydrogen halides decreases down the Group. As the size of the halogen atoms increases from Cl to I, their valence orbitals become more diffuse. This results in less effective overlap of the orbitals between the small H atom and the larger halogen atom and less energy is required to break the weaker HX bond. (b) In decreasing relative rate of hydrolysis: R > P > Q R is the most reactive as the acyl carbon atom is highly electron deficient as it is bonded to two electronegative atoms , O and C l, and hence is most susceptible to nucleophilic attack. P is less reactive than R because the electron deficient carbon is bonded to only one electronegative (Cl) atom. Q does not undergo hydrolysis. Due to sideway overlap of p -orbital of C l with the  electron cloud of benzene ring, the lone pair of electrons on C l atom delocalises into the benzene ring . This results in CCl bond having partial double bond character , which makes the bond stronger. (c) Since the OH group is highly activating, it increases electron density on the benzene ring as well as stabilises the arenium ion , making the ring more susceptible to electrophilic attack. (d)

2 (e) (i) Free radical substitution Limited Br2, UV light at room temperature (ii) relative ratio 3 × 1 = 3 4 × 7 = 28 6 × 1 = 6 2 (a) Circle N Tetrahedral (b) Energy released by ion-dipole interactions between (protonated dextromethorphan and Br−) ions and (polar) water molecules is sufficient to compensate for the energy required to overcome the electrostatic attraction between oppositely -charged ions/ ionic bonds and hydrogen bonds between water molecules. (c) (i) A B C D (ii) PCl5 (d) (i) Anion X is resonance−stabilised as the negative charge on C can delocalise into −CN and CO.

3 (ii) (iii) Carbonyl C atom in cyclohexanone is less electron −deficient as it is attached to 2 electron −donating alkyl groups while the carbonyl C atom in hexanal is attached to 1 electron−donating alkyl group. The 2 alkyl groups bonded to carbonyl C atom in cyclohexanone poses greater steric hindrance to the incoming nucleophile compared to 1 alkyl group bonded to carbonyl C atom in hexanal. Cyclohexanone is less susceptible to attack by nucleophiles and rate of reaction is slower. (iv) −CN is electron-withdrawing/ N atom in −CN is electronegative. Negative charge on −COO− is dispersed by the −CN in cyanoacetate anion, which makes it a more stable anion than CH3CH2CH2COO−. 3 (a) (i) Let x be the initial amount of N2. 3H2(g) + N2(g)  2NH3(g) Initial amount/ mol 3x x 0 Change in amount/ mol −0.9x −0.3x +0.6x Equilibrium amount/ mol 2.1x 0.7x 0.6x = 0.6x 0.3205 x = 2.667 mol Equilibrium concentration of H2 =  2.1 2.667 5 = 1.12 mol dm–3 Equilibrium concentration of N2 =  0.7 2.667 5 = 0.373 mol dm–3 (ii) Kc = 2 3 0.320 (1.12) (0.373) = 0.195 mol–2 dm6

4 (b) (i) Since there is no change in the concentration/ partial pressure of N 2, H 2 and NH3, the position of equilibrium remains unchanged. Therefore, the composition of the reaction mixture remains unchanged. (ii) Kc remains unchanged. (c) (i) The forward exothermic reaction is favoured by low temperature to absorb some heat. However, at low temperature, the reaction would be slow . Hence, an optimal moderately high temperature of 450 C is used so that a reasonable yield can be obtained at a fairly high rate. A high pressure would shift the position of equilibrium to the right increasing the yield of NH 3. It also increases the rate at which equilibrium is established. However, the cost of equipment to produce and maintain high pressure would be greater/costly to maintain high pressure . Hence, a relatively high pressure of 250 atm is used as a compromise between yield and cost/ optimal pressure. (ii) A large amount of energy is required to break the (very) strong NN bond. (d) (i) 26Fe: 1s2 2s2 2p6 3s2 3p6 3d6 4s2 (ii) Iron has energetically accessible/low-lying partially filled 3d subshells/ 3d and 4s electrons which allow the N2 and H2 molecules to adsorb onto the surface of the catalyst/ form weak temporary bonds with the surface of the catalyst. (iii) A catalyst provides an alternative reaction pathway with a lower activation energy . As a result, more reactant particles possess the energy required for an effective collision* and the frequency of effective collisions increases . The rate of reaction is thus increased and the rate constant increases. (e) NH3 is a stronger ligand than N2. In NH3, the N atom is sp3 hybridised, whereas in N2, the N atom is sp hybridised. The lone pair of electrons in N 2 is in a sp hybrid orbital which has greater s-character/ smaller in size/ closer to the nucleus (50%) compared to the sp3 hybrid orbital (25%) in NH3.

5 Consequently, the lone pair of electrons in N 2 experience greater nuclear attraction, making it less available to form a coordinate/ dative bond. 4 (a) (i) 222 86Rn (Mass must balance: 226 − 4 = 222; atomic number must balance: 88 − 2 = 86. Element 86 corresponds to Rn on the periodic table). (ii) The beam will deflect towards the negatively charged plate because -particles are helium nuclei ( 4 2+ 2He ) which carry a positive charge. (b) The successive half-lives (t1/2) can be calculated directly from Table 4.1: - Time to drop from 5.12 × 1020 to 2.56 × 1020 atoms (100% to 50%): 1600 − 0 = 1600 years - Time to drop from 2.56 × 1020 to 1.28 × 1020 atoms (50% to 25%): 3200 − 1600 = 1600 years Since the half-life is constant at 1600 years, the order of reaction is first-order. (c) (i) Helium atoms occupy negligible volume compared to volume of the container. There is negligible IMF between the helium atoms. (ii) For an ideal gas, the product of pV must remain constant at a constant temperature (pV = constant). Experiment 1: pV = (2.026 × 105) × (3.274 × 10−9) = 6.6 × 10−4 J Experiment 2: pV = (4.052 × 105) × (1.623 × 10−9) = 6.6 × 10−4 J (d) (i) 59 7pV (2.026 10 )(3.274 10 )n = = = 2.924 10 molRT (8.31)(273) − −  17 23 1 A 7 1.88 10 = = 6.43 10 mol2.924 10N −−  (ii) 23 23 23 6.43 10 6.02 10% error = 100% = 6.81%6.02 10 −   (iii) Each decaying Radium atom produces four -particles. Total number of 226Ra atoms decayed = 171.88 10 4  = 4.70 x1016 atoms Initial number of 226Ra atoms = 23192 /1000 6.02 10226  = 5.114 × 1020 atoms

6 Percentage converted = 16 20 4.70 10 100%5.114 10  = 0.00919% 5 (a) (i) [CO2] = 0.01  2.10  10−3 = 2.10  10−5 mol dm−3 [CO32⁻] = 0.04  2.10  10−3 = 8.40  10−5 mol dm−3 [HCO3⁻] = 0.95  2.10  10−3 = 2.00  10−3 mol dm−3 (ii) The small amount of acid added is neutralised by the large amount of CO32− in seawater. Hence, [H+] does not increase appreciably, and the pH is kept approximately constant (b) (i) When [HCO3−] = [CO32⁻], pH = 9.5 Ka2 = 10−9.5 = 3.16  10−10 mol dm−3 (ii)     2 22 a1 a2KK −− − −            =       = ++ 33 2 3 + 3 2 HCO H CO H CO HCO CO H CO (iii) [H+] = 10–8.10 = 7.94  10−9 mol dm−3   22 52 69 ) + 3 2 8.10 CO H CO (8.40 10 )(10 (1.42 10 ) (1.08 10 ) − −− −−     =  =  a1 a2KK = 3.46  10−6 mol dm−3 (c) (i) 2 6 4 430 10 1 4.30 10 2CO totalp = p CO total n n − −  = = 1.00 atm [CO2] = 0.0340  4.30  10−4 = 1.46  10−5 mol dm−3 (ii) The concentration of dissolved CO2 in seawater sample, S1, will be lower. Seawater has higher salinity compared to pure water. Non-polar CO2 gas is less solu