Chemistry Stpm Essay

A Structural Questions Question 1. (a)(i) The presence of isotopes 1M (ii) Let the teemingness of 63X be a %. The % abundance of 65X. = ( carbon a ) 1M Relative fragmentic 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 total 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 instal1M (b) (i) 121M (ii) 1s2 2s2 2p6 3s2. 1M (iii) +2 , X has two valence electrons2M (iv) X is a better electrical energy conductor. 1M - 10M 3. (a)Atomic size increases, screening effect increases with to a greater extent inner shells of electrons 1M rough-and-ready nuclear wake up decreases, ionisation energy lowered, valence electrons are to a greater extent easily removed. 1M (b) i. Be2+ (aq) + 4H2O (l) Be (H2O)4 2+ (aq)1M ii. It is social diseaseic, acting as a Bronsted-Lowry acid1M The Be2+ ion has a advanced charge density 1M and can strongly polarise large anions collectable to its smaller size. 1M The ions of opposite 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 reaction is heat-releasing reaction. According to le Chatelier principle, a low temperature willing favour the institution of NO. For gaseous equipoise, a decrease in pressure will favour the reaction which produces to a greater extent gaseous molecules. Thus in the supra correspondence 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 ii i. Formation of ester CH3CH2CH2COCl + CH3CH2OH CH3CH2CH2COOCH2CH3 + HCl1M (b) i. H3N+CH2COOH + H2NCH(CH3)COOH H2NCH2CONHCH(CH3)COOH + H2O1M Glycylalanine1M ( Note Alanylglycine can overly be formed ) ii. The amino gathering 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) newton section 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 fragment is 2s2 2px1 2py1 2pz11M Nitrogen atom uses sp3 intercros sed 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 solitary 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 plat 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 number 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) atomic number 13 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 aluminiferous structure, strong metallic bonf.Silicon giant 3 D covalent structure. Strong covalent bond between silicon atomes. higher melt down point Phosphorus and sulphur Both are simple molecules. sluttish 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 atomic number 35 E of chlorine is more positive than that of bromine. Cl2 + 2Br- - > 2Cl- + Br24M (b)iodine forms triodes complex in KI. I2 + I- - > I3- single d oes not form any complex ions in water. I2 + 2H2O > I- + HIO + H3O+4M (c)HCl is released in wintry acidNaCl + H2SO4 a NaHSO4 + HCl If heated more HCl released. NaHSO4 + NaCl -a Na2SO4 + HCl4M (d) Iodide is oxidized to iodine olympian 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 group is an electron donor by inductive effect. Y is more basic than W because the lone(prenominal) 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. front 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 appliance HNO3 + H2SO4 NO2+ + HSO4 + H2O NO2+ is an electrophile. H + NO2+ NO2 H NO2 + HSO4 NO2 + H2SO4 + HNO3 NO2 + H2O 6M Total 15 mark 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 functional group -OH isomers CH3CH2CH2OH and CH3CHCH3OH arm isomers separately with alkalic 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 silvern 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