RI H2 CHEM P2 ANS Prelim

Raffles Institution 2015 Year 6 Preliminary Examination Suggested Answers to H2 Chemistry Paper 2 1 (a) H2O2 acts as an oxidizing agent to oxidise Co(II) to Co(III) and itself reduced to H2O. (b) Preparation of the 100 cm3 crude mixture 1. Weigh out accurately about 1.45 g of Co(NO3)2 using an electronic balance and transfer this to a 250 cm3 beaker. 2. Measure out 45 cm3 of concentrated ammonia using a 50 cm 3 measuring cylinder and add this to the beaker. Stir with a glass rod to mix. 3. Measure out 25 cm 3 of hydrogen peroxide using a separate 50 cm 3 measuring cylinder and using a dropper, add this to the mixture dropwise/slowly. 4. Cool the mixture in an ice bath or carry out step 5 in an ice bath. 5. Measure out 30 cm 3 of concentrated nitric aci d using a separate 50 cm 3 measuring cylinder and using a dropper, add this to the mixture dropwise/slowly with stirring. 6. Remove the mixture from the ice bath, allow the mixture to sit for 10 mins for complete reaction. Precipitate the product from the crude mixture 7. Using a 100 cm 3 measuring cylinder, transfer 100 cm 3 of ethanol to the 250 cm3 beaker containing the crude mixture. 8. Allow the mixture to sit for 10 mins for complete precipitation. Separate the precipitated product from the mixture 9. Filter the precipitated product using vacuum filtration/gravity filtration and wash the residue with ethanol to remove the impurities. Recrystallisation 10. Transfer the residue into a 50 cm3 (or 100 cm3) conical flask with 2 boiling chips. 11. Add about 5 cm 3 (small amount) of ammonium nitrate solution into the conical flask and place it on the heating plate. (or a hot ammonium nitrate solution can be prepared prior to addition by placing it on a heating plate) 12. When the mixture is boiling and the solid does no t dissolve, add ammonium nitrate slowly to the mixture until the solid just dissolves. 13. Remove from heating plate and allow it to cool to room temperature slowly before cooling in an ice bath. Crystals will grow as the mixture is cooled. 14. Filter the cooled mixture using vacuum filtration/gravity filtration and wash the residu e with cold ethanol. 15. Dry the residue under the infra-red lamp.

(c) The concentrated nitric acid (or concentrated ammonia) is corrosive. Wear gloves and goggles when handling the chemical / do the experiment in a fumehood. OR Inhaling the corrosive fumes of concentrated ammonia . Carry out the experiment in a fumehood. OR Ethanol is flammable. Any heating should be done using a water bath and not with a naked flame. 2 (a)(i) sp (ii) The  bond in C=O bond of R-N=C=O is made up of sp sp2 head-on overlap whereas the  bond in C=O of CH3COCH3 is made up of sp2sp2 head on overlap. The sp sp2  bond in C=O bond of R -N=C=O therefore has a higher s character, hence leading to more effective overlap of orbitals and shorter bond length. (iii) Geometric (cis-trans) isomerism (iv) 3 stereoisomers (b) [Ag(NCO)2] [Ag(OCN)2] 180 o 105 o

p Ideal gas 3 (a)(i) HBr and HCl readily dissolve in water to form acids containing H3O+, Br– and Cl–. (ii) HBr is a stronger reducing agent than HCl. HCl does not reduce concentrated H2SO4. HBr reduces concentrated H 2SO4 to SO2 in which the oxidation state of S decreases from +6 (in H 2SO4) to +4 (in SO 2). HBr is oxidised to Br 2 in which the oxidation state of Br increases from –1 (in HBr) to 0 (in Br2). 2HBr(g) + H2SO4(l)  Br2(l) + SO2(g) + 2H2O(l) (iii) pV T HF forms strong intermolecular hydrogen bonding while SO 2 forms slightly weaker permanent dipole-permanent dipole interaction. HF, with stronger intermolecular forces of attraction deviate from ideality to a larger extent , than SO2. (b)(i) N≡N bond is strong and the reaction has a high activation energy. (ii) As temperature increases, Kp increases. The position of equilibrium shifts to the right. An increase in temperature favours the endothermic forward reaction. HF SO2

(iii) In this reaction, 2 moles of gaseous molecules react to form 2 moles of gaseous products, n=0. S is very small. For the reaction to be feasible at high temperature, G < 0. Thus, G = H – TS < 0. Since H is positive, S must be slightly positive for G to be negative. (iv) O N O (v) NO and NO 2 are free radicals/have an unpaired electron. They will propagate a chain reaction to cause rapid and extensive ozone depletion. (vi) Step 1: C . OH .C C C Step 2: . . CC C C (c)(i) SCl2(g) + 2CH2=CH2(g) C4H8SCl2(g) Initial / mol 0.258 0.592 0 Change/ mol –0.0350 –0.070 +0.0350 Eqm / mol 0.223 0.522 0.0350 Total no. of moles of gas at equilibrium = 0.223 + 0.522 + 0.0350 = 0.780 mol (ii) pV = nRT p (5 x 10–3) = 0.780 x 8.31 x (20 + 273) p = 3.798 x 105 Pa = 3.80 x 105 Pa (3sf) x x x x x x : : xx xx xx .

(iii) PSCl2 = 0.223 0.780 x 3.80 x 105 = 1.086 x 105 Pa PCH2=CH2 = 0.522 0.780 x 3.80 x 105 = 2.543 x 105 Pa PC4H8SCl2 = 0.035 0.780 x 3.80 x 105 = 1.705 x 104 Pa Kp = 4 5 5 2 1.705 10 1.086 10 (2.543 10 ) x x x x = 2.43 x 10–12 Pa–2 (iv) With a larger flask, the volume of the system is greater, and the concentrations (or pressures) of all the gases are lower. The position of equilibrium will shift to the left with more gaseous molecules to increase the concentration (or pressure) of the gases. OR When volume increases, partial pressures of all the gases decreases Qp = 4 8 2 2 2 2 4 C H SCl SCl C H P P x P > Kp Hence, the position of equilibrium will shift to the left to decrease Qp. AND As the temperature of the system remains the same, and Kp is only affected by temperature, the Kp value is unchanged. (v) Addition (vi) Mustard gas: ClCH2CH2SCH2CH2Cl 4 (a)(i) Sn, conc HCl, heat then NaOH(aq)

(ii) (iii) AlBr3 (iv) Al has a energetically low-lying empty p-orbital which can accept a pair of electrons. (v) The –NH2 group activates the benzene ring due to the partial delocalisation of the lone pair of electrons on N into the pi electron cloud of the benzene ring. Hence the benzene ring is more susceptible to electrophilic attack in phenylamine and only mild conditions are required for reaction. (b)(i) Mechanism: Electrophilic substitution Generation of electrophile: 2H2SO4 + HNO3 NO2 + + 2HSO4 – + H3O+ Step 1: Electrophilic attack by NO2 + Step 2: Loss of a proton from carbocation a carbocationelectrophilebenzene + H NO2 NO2++ slow .. + H2SO4 NO2 fast S O OHO O + H NO2 + nitrobenzene (regenerated)

(ii) enthalpy (c)(i) The rate determining step (slow) in the nitration of benzene does not involve the breaking of the CH bond. Hence the deuterium isotopic effect is not present. (ii) Since 𝑘𝐻 𝑘𝐷 = 6, the rate-determining step involves breaking of a CH bond. Hence the rds is step 2. (iii) Rate = k’[CH3C(OH)CH3 +] = k[CH3COCH3][H+] (iv) The rate -determining step does not involve the breaking of an I-I bond, so the rate of reaction does not depend on the concentration of iodine. 5 (a) 1s2 2s2 2p6 3s2 3p6 3d9 (b) The presence of H2O ligands causes the splitting of the five 3d orbitals in Cu2+ ion into two sets of slightly different energy levels. Since these 3d orbitals are partially filled , electrons from the lower -energy d orbitals can absorb energy corresponding to certain wavelengths (OR red/orange/yellow) from the visible spectrum and get promoted to the higher-energy d orbitals The colour observed is the complement of the colour absorbed.

(c)(i) [Cu(NH3)4]2+ or [Cu(NH3)4(H2O)2]2+ (ii) Fe(OH)3 (iii) [Cu(NH3)4]2+(aq) + 2H2O(l)