Chemistry Stpm Essay

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 visual modality = ( 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) piss2 + 2H+ + 2 I- 2H2O + I21M (ii) govern = 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 than than inner shells of electrons 1M effective nuclear wake up decreases, ionisation energy lowered, valence ele ctrons 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 robustly 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) ? 4 zero(prenominal)g) + 6H2O (g)1M iii. low temperature1M low pressure1M ( Note The answer is exothermic reaction. According to le Chatelier principle, a low temperature will favour the composition 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 Al some(prenominal)lglycine can also be create ) ii. The aminic 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 loan-blend orbitals for forming covalent bonds between N and H atoms.Energy 2p sp3 hybrid p orbitals 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 - totality 15 M 6. (a) Dyanamic equilibrium . a reversible reaction , in a closed system frontward 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 -a t 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 take aims) light Resistant to corrosion Strong alloy2M (b)aluminium A giant metal(prenominal) 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 a cidNaCl + 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 inducive 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. strawman 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 attach 0. ( a ) ( i ) Terylene/Dacron O CH2 CH2 O C C O CH2 CH2 O C C 3M O O O O ( ii ) Condensation polymerisationTo make cloth/sleeping bags, etc 2M ( b ) ( i ) K useful group -OH isomers CH3CH2CH2OH and CH3CHCH3OH arm isomers one at a time 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 Tollens reagent. CH3CH2CHO gives a silver mirror but CH3COCH3 does not. CH3CH2CHO + 2Ag(NH3)22+ + OH CH3CH2COO + 2Ag + 2NH4+ + 2NH3 5M Note Can also accept other suitable chemical test. Total 15 marks