Paper type: Essay Pages: 6 (1276 words)
A – Structural Questions: Question 1. (a)(i) The presence of isotopes 1M (ii) Let the abundance of 63X be a %. The % abundance of 65X. = ( 100 – a ) 1M Relative atomic mass = ( 62. 93 x a) + ( 64. 93 x ( 100 -a) ) 1M 100 63. 55 = 62. 93a + 6493 -64. 3a 100 6355 = -2a + 6500 a = 69. 0% 1M The % abundance of 65X = 100- 69. 0 = 31. 0 % Relative abundance 63X : 65X 1 : 2 1M (iii) Relative Abundance 63 64 65 Relative mass /m/e 2M SpeciesprotonsneutronsElectrons 20 Ne 10 10 10 10 16O2- 8 8 8 10 2 M The species have same number of electrons or isoelectronic. M —————- 10M 2. (a) (i) H2O2 + 2H+ + 2 I- > 2H2O + I21M (ii) Rate = k [H2O2] [I-]1M (iii) 0.
21M 0. 11M (iv) second order1M (b) (i) 121M (ii) 1s2 2s2 2p6 3s2. 1M (iii) +2 , X has two valence electrons2M (iv) X is a better electricity conductor. 1M —————- 10M 3. (a)
Atomic size increases, screening effect increases with more inner shells of electrons 1M effective nuclear charge decreases, ionisation energy lowered, valence electrons are more easily removed. 1M (b) i. Be2+ (aq) + 4H2O (l) > [ Be (H2O)4 ]2+ (aq)1M ii.
It is acidic, acting as a Bronsted-Lowry acid1M The Be2+ ion has a high charge density 1M and can strongly polarise large anions due to its smaller size. 1M The ions of other Group 2 elements have larger sizes and charge densities and weaker polarising power (d)i. platinum and rhodium1M ii. 4NH3(g) + 5O2(g) ? 4NO(g) + 6H2O (g)1M iii. low temperature1M low pressure1M ( Note : The reaction is exothermic reaction. According to le Chatelier principle, a low temperature will favour the formation of NO. For gaseous equilibrium, a decrease in pressure will favour the reaction which produces more gaseous molecules. Thus in the above equilibrium a low pressure will avour the formation of NO. ) ________ 10M 4. (a) i.
A is CH3CH2CH2COOH1M B is CH3CH2CH2COCl1M C is CH3CH2CH2COOCH2CH 1M ii. butanoyl chloride1M iii. Formation of ester: CH3CH2CH2COCl + CH3CH2OH > CH3CH2CH2COOCH2CH3 + HCl1M (b) i. H3N+CH2COOH + H2NCH(CH3)COOH > H2NCH2CONHCH(CH3)COOH + H2O1M Glycylalanine1M ( Note: Alanylglycine can also be formed ) ii. The amino group –NH2 which is basic group reacts with hydrochloric acid to form the ammonium chloride salt of alanine1M HOOCCH(CH3)NH2 + HCl > HOOCCH(CH3)NH3+Cl- 1M ___________ 10 M SECTION B – ESSAY 5. (a) (i) Orbitals with the same energy1M Example : 2p or 3d s orbitals1M ii) Nitrogen atom has 7 electrons 1M Fill 1s orbital with 2 electrons1M Fill 2s orbital with 2 electrons1M Fill 2px,2py and 2pz orbitals with 3 electrons1M / 6 1M (b) Fe 2+ 1s2 2s2 2p6 3s2 3p6 3d4 1M Fe 3+ 1s2 2s2 2p6 3s2 3p6 3d5 1M In terms of electronic configuration, Fe 3+ is more stable than Fe 2+ 1M Because it has half-filled 3d orbital which is more stable1M. / 4 (c ) The valence electronic configuration of the electrons for nitrogen atom is 2s2 2px1 2py1 2pz11M Nitrogen atom uses sp3 hybrid orbitals for forming covalent bonds between N and H atoms.
Energy 2p sp3 hybrid porbitals N(ground state) 1M In sp3 hybrid orbitals of nitrogen atom,one of the orbitals Is occupied by a lone pair of electrons and three sp3 orbitals are half filled 1M Each N-H atom is formed by the overlapping of the s orbital of hydrogen atom with one of the half filled sp3 orbitals to give the ammonia molecule 1M Diagram of the bond formation in NH3 molecule. M /. 5 ——————- Total : 15 M 6. (a) Dyanamic equilibrium …. a reversible reaction , in a closed system forward and backward reactions have the same rate of reaction. 2M (b) (i) N2O4 – 2NO2 Kc = [NO2] 2 = [0. 12] 2 [N2O4] [0. 04] = 0. 36 mol dm-3 5M (ii) Using PV =nRT where n = 0. 12 +0. 04 = 0. 16 mol P = 0. 16 (8. 31) (383) 10 -3 = 509. 24 kPa. 3M (c) N2(g) + 3H2(g) 2NH3(g) – at low temperatures, % NH3 is higher – forward reaction is exothermic equilibrium position shifts to the right at higher temperature -forward reaction is accompanied by a reduction in volume of gas -at higher pressures, equilibrium position shifts to the right -at high pressures, % NH3 is higher5M ————— Total : 15M No. 7 (a)(i) Aluminium metal is extracted by electrolysis The electrolyte is molten bauxite in sodium hexafluoroaluminate. The electroyte has aluminium ion and oxide ions. Anode : 2O2- — > O2 + 4e Cathode : Al3+ + 3e — > Al5M (ii) (Any 2 points) light Resistant to corrosion Strong alloy2M `(b)aluminium : A giant metallic structure, strong metallic bonf.
Silicon : giant 3 D covalent structure. Strong covalent bond between silicon atomes. higher melting point Phosphorus and sulphur – Both are simple molecules. Weak van der waals between molecules Sulphur has a stronger intermolecular forces – S8 larger than P48M No 8. (a) chlorine – strong oxidation agent Bromide is oxidized to bromine E° of chlorine is more positive than that of bromine. Cl2 + 2Br- —- > 2Cl- + Br24M (b)iodine forms triodes complex in KI. I2 + I- —- > I3- Iodine does not form any complex ions in water. I2 + 2H2O — > I-¬ + HIO + H3O+4M (c)HCl is released in cold acid
NaCl + H2SO4 a NaHSO4 + HCl If heated more HCl released. NaHSO4 + NaCl -a Na2SO4 + HCl4M (d) Iodide is oxidized to iodine Purple Iodine is released Pungent smell of H2S is detected3M ————– Total : 15M 9. ( a ) ( i ) order : W, Y, X W, Y, X act as Lewis bases. X is the strongest base because ethyl group is an electron donor by inductive effect. Y is more basic than W because the lone pair electron on the N atom is not delocalised. W is less basic than Y because the lone pair electron on the N atom is delocalised into the benzene ring. M ( ii ) pKb value > 9. 39 Z is a weaker base than W. Presence of Cl – an electron withdrawing group reduces the donating potential of lone pair electron on the N atom through inductive effect. 4M ( b ) Concentated H2SO4 and HNO3. , 550C Mechanism: HNO3 + H2SO4 NO2+ + HSO4– + H2O NO2+ is an electrophile. H + NO2+ NO2 H NO2 + HSO4– –NO2 + H2SO4 + HNO3 –NO2 + H2O 6M Total : 15 marks 0. ( a ) ( i ) Terylene/Dacron ~~~~O – CH2 – CH2 – O – C – –C – O – CH2 – CH2 – O – C – –C~~~~ 3M || || || || O O O O ( ii ) Condensation polymerisation
To make cloth/sleeping bags, etc 2M ( b ) ( i ) K: functional group : -OH isomers : CH3CH2CH2OH and CH3CHCH3OH arm isomers separately with alkaline iodine, CH3CHCH3OH gives a yellow precipitate but CH3CH2CH2OH does not. CH3CH2CH2OH + 4I2 + 6OH– CHI3 + 5I– + 5H2O + CH3COO– 5M (ii ) L : functional group : ? C = O | Isomers : CH3CH2CHO and CH3COCH3 warm isomers separately with Tollen’s reagent. CH3CH2CHO gives a silver mirror but CH3COCH3 does not. CH3CH2CHO + 2[Ag(NH3)2]2+ + OH– CH3CH2COO– + 2Ag + 2NH4+ + 2NH3 5M Note: Can also accept other suitable chemical test. Total : 15 marks
Cite this page
Chemistry Stpm. (2018, Oct 12). Retrieved from https://studymoose.com/chemistry-stpm-essay