Marking scheme

由 **Giraffe** 周日七月 26, 2009 11:04 am

Trial STPM 2008 CHEMISTRY PAPER 2

No.	Answer	Marks
1(a)	Electron deflected the most Because it is lightest	1m 1m
(b(i)	1s22s22p63s1	1m
(ii)	5	1m
(iii)	2s 2p	1m
(iv)	It is unstable	1m
(c )(i) (ii)	Correct of drawing. Correct of labeling. Correct of electronic configuration(represent by arrow)	1m 1m 1m
(iii)	Fluorine has very high ionization energy/high electron affinity	1m
2(a)(i)	Al3++3e→Al	1m
(ii)	1000x103x3x96500 27 =1.07x1010C	1m 1m
(iii)	Needed less energy/lower temperature	1m
(b)(i)	Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s)	1m
(ii)	E.m.f=E reduction-E oxidation =+0.80-(-0.76) =1.56V	1m
(iii)	Ecell=Eѳcell-0.059/2 log[Zn2+] /[Ag+2	1m
(iv)	Ecell=1.56+ 0.059/2log [Ag+] 2/[Zn2+] 1.21=1.56+ 0.059/2log[Ag+]2/1 [Ag+]=1.17x10-6moldm-6	1m 1m
(v)	Voltage will decrease Reason: Addition of hydrochloric acid decreases the conc. of Ag+ ions. *Ag+(aq)+Cl-→AgCl(s)	1m
3(a)(i)	2Mg(NO3)2→2MgO+4NO2+O2	1m
(ii)	1. Charge density decreases going down the group 2. Polarizing power decreases. Furthermore the lattice energy of the oxide formed becomes less exothermic.	1m 2m
(b)(i)	NH3+HCl→NH4Cl	1m
(ii)	Ammonia:sp3 hybridisation Ammonium ion: sp3 hybridisation	1m 1m
(iii)	Correct of drawing.	1m
(iv)	Make fertilizers	1m

由 **Giraffe** 周日七月 26, 2009 11:05 am

Marking scheme Q4-5

4(a)(i)	Nitrogen atom in ethanamine has a lone pair of electrons that can be donated /used to form a coordinate band with a proton	1m
(ii)	The lone pair electron of the nitrogen atom is delocalized in the benzene ring. They are less readily donated.	1m 1m
(b)(i)	Substitution	1m
(ii)	(CH3CH2)2NH (CH3CH2)2N (CH3CH2)2N+Br	3correct:2m 2correct:1m 1correct:0m
(iii)	CH3CN+2H2 (CH3CH2)2NH2 Catalyst: Ni Conditions:150°C Or CH3CN+4[H] (CH3CH2)2NH2 Catalyst: LiAl4 in dry ether Better because no other products will be formed.	1m @ 1m 1m
©	Add aqueous bromine/aqueous iron (III) chloride. Phenylamine forms a white precipitate /purple solution whereas ethanamine Give no visible change.	1m 1m
5(a)(i)	Hund’s rule states that in a given set of orbitals of equilvalent energy,electrons tend to occupy the orbitals singly first before pairing up. Pauli exclusion principle states that each orbital can be occupied by two electrons of opposite spin only . Aufbau principle states that electrons must occupy available orbitals of lower energy first before they fill orbitals of higher energy.	1m 1m 1m
(ii)	No. of electons in O2- ion=8+2=10 Step 1: Apply Pauli exclusion principle and Aufbau principle. Fill 1s orbital with 2 electrons. Step2: Fill 1s orbital with 2 electrons. Step 3: Apply Hund’s rule. Fill 2px, 2py, 2pz orbitals with 1 electron respectively. Step 4: Fill the remaining 3 electrons into 2px, 2py, 2pz orbitals respectively.	1m 1m 1m
(b)(i)	Equation for overall reaction 2O3⇌3O2 Consider step 1: Rate of forward reaction:k1[O3] Rate of reversed reaction:k-1[O2][O] At equilibrium, rate of forward reaction= rate of reversed reaction k1[O3]= k-1[O2][O] [O]= ……….(1) Consider step 2: Rate of reaction:k2[O3][o]………(2) Sub (1)into (2) Rate of reaction=k2x [O3] = [O3]2[O]-1	1m 1m 1m 1m 1m 1m
(ii)	Initial step CFCl3 ∙ CFCl2+∙Cl Propagation step ∙Cl+O3→ClO∙+O2 ClO∙+O→ O2+∙Cl (O3→ O2+∙O)	1m 1m 1m

由 **Giraffe** 周日七月 26, 2009 11:05 am

Marking scheme Q6-8

6(a)	C(s)+ O2→C O2(g) Hf=-394kJmol-1 Heat released by burning 2.40g of carbon= x394 =78.8kJ Heat =heat capacity x increase in temperature 78.8kJ=10kJ°C-1x ѳ Ѳ=7.88°C	1m 1m 1m 1m 1m
(b)(i)	Correct labeling of y-axis and x-axis Correct labeling vapour and liquid Draw the curve correctly.	1m 1m 1m
(ii)	Show the path of fractional distillation shown in the diagram. The solutions boils at a temperature lower than azeotopic mixture. The vapour will contain more than 68% HBr. Distillate: pure HBr Residue: azeotopic mixture.	1m 1m 1m 1m 1m
(iii)	Total pressure of the mixture is lower than that predicted based on Raoult’s law Because intermolecular force in mixture is stronger than the pure liquids mixture (show negative deviation)	1m 1m
7(a)(i)	The d-orbitals are split into 2 groups of slightly different energies. Electrons from lower energy orbitals absorb visible light. Jump to a higher energy orbitals in a d-transition ,and reflect blue light	1m 1m 1m
(ii)	Blue colour of the solution is due to the presence of [Cu(H2O)4]2+complex. When aqueous ammonia is added, substitution of H2O molecules in the complex by NH3 molecules take place forming [Cu(NH3)4]2+complex that is blue. This show that NH3 is a stronger ligand than H2O	1m 1m 1m 1m
(b)(i)	Cobalt(II), because standard electrode potential has a positive value.	1m 1m
(ii)	Fe3++e⇌ Fe2+ E=o.77V 1/2 O2+2H+⇌ H2O E=1.23V 4 Fe2++ O2+4 H+→4 Fe3++2 H2O E=2.ooV Valence electronic configuration of Fe2+ is 3d6and Fe3+ is 3d5 Fe3+ is more stable because its d-orbitals are filly filled.	1m 1m 1m 1m 1m

由 **Giraffe** 周日七月 26, 2009 11:05 am

Marking scheme Q8-9

8(a)	Lead-has metallic structure. -consist of metallic bond between the lead atoms. - The metallic bonds are strong and needs high energy to be overcome thus high melting point PbCl3-ionic lattice -ions are held together by very strong ionic bond. PbCl4-simple molecular structure -molecules are held together by weak Van der waals forces.	1m 1m 1m 1m 1m 1m
(b)(i)	3 Cl2+6NaOH→5Na Cl+ Na ClO3 Disproportionation reaction take place. Cl2 is oxidized to ClO3- in sodium chlorate and reduce to Cl- in sodium chloride .Both products are soluble in water forming colourless solution.	1m 1m 1m
(ii)	Cl2(g)+2KBr(aq)→ 2K Cl (aq)+Br2(l) A displacement reaction has taken place. This is also a redox reaction where chlorine undergoes reduction (Cl2+2e→2 Cl-) and bromide ions have undergoes oxidation(2Br → Br2 +2e)	1m 1m 1m
(iii)	CH2= CH2+ Cl2→ CH2 Cl CH2 Cl An electrophilic addition reaction takes place/chlorination reaction take place. Cl2 act as an electrophile.	1m 1m 1m
9(a)(i)	Rate=k[(CH3)3C-Br] Mechanism: CH3)3C-Br→ CH3)3C++Br- (slow) CH3)3C++OH-→ CH3)3C- OH (fast){correct label slow@fast-1m} The slow step is the rate determining step. The reaction is a first order reaction.	1m 1m 1m 1m 1m
(ii)	Aqueous sodium hydroxide and ethanolic silver nitrate 1- chlorobutane-No visible change 2- 2-chlorobutane-White precipitate after several minutes 3- 2-chloromethylpropane- White precipitate formed immediately.	1m 1m 1m 1m
(b)(i)	Saponification /alkaline hydrolysis C17H35COO CH2 CH2OH I I C17H35COOCH +3KOH→ C17H35COOK+ CHOH I I C17H35COOCH2 CH2OH Importance of C17H35COOH :as soap As milder and softer soap.	1m 1m 1m
(ii)	C17H35 CH2OH CH2OH CHOHCH2OH	1m 1m

由 **Giraffe** 周日七月 26, 2009 11:05 am

Marking scheme Q10

10(a)(i)	2-amino-3-methylpentamide	1m
(ii)	It show optical isomerism Because the presence of a chiral carbon. Drawing of isomers.	1m 1m 1m
(iii)	CH3 O CH3 O I I I PCl5 I I I CH3CH2CHCHC – OH→ CH3CH2C-C-C – Cl I I NH2 NH2 ↓conc. ammonia CH3 O I I I CH3CH2CCHC – NH2 I NH2	PCl51m 1m 1m
(b)(i)	The linking of monomers containing c=c bond to form one product.	1m
(ii)	Cl H Cl H I I I I n C=C → - - I I I I n Cl H Cl H	1m 1m 1m
(iii)	Cl aton is electon withdrawing group. It stabilies the intermediate anion formed.	1m 1m

Marking scheme

Marking scheme

回复： Marking scheme

回复： Marking scheme

回复： Marking scheme

回复： Marking scheme