MINISTRY OF EDUCATION
FIJI SEVENTH FORM EXAMINATION - 2005
EXAMINER’S REPORT
CHEMISTRY
A total of 1853 candidates appeared for the Chemistry Examination.
The 2005 paper closely followed the prescribed syllabus and tested a wide range of abilities of the candidates. The standard and format of the paper were similar to the previous year.
Candidates continued to make similar errors as in the past and it appears that very little attention has been given to the comments that have been made in the previous examiner’s report. The overall performance was below expectation.
As in the past, candidates often failed to read and understand the questions well and therefore not able to answer as expected.
Common weaknesses noted were :
• Poor concept of different types of bonds and strengths.
• Not showing working in questions involving calculations and their units.
• Inability to draw Lewis structures correctly. Shared electrons need to be drawn in pairs and atoms sharing drawn close together and not far apart.
• Failure to show “positive” or “negative” sign in calculations involving electrode potentials (Eo), oxidations numbers and enthalpy change (ΔH).
• Inability to write correct formulae, balanced equations and not using “reversible reaction” sign in dissociation and equilibrium reactions.
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PAPER-1 SECTION A [24 marks]
The responses to the multiple choice questions that were given in a random sample of 100 scripts are tabulated below. The correct response for each question is marked by an asterisk (*).
Question No. A B C D
1 62* 12 17 9
2 31 37 10 22*
3 15* 21 42 22
4 34 10 44* 12
5 24 50* 21 5
6 16 11 26 47*
7 72* 10 7 11
8 11 58* 20 11
9 10 28 33* 29
10 51* 13 21 15
11 14 38 19 29*
12 17 63* 13 7
13 62* 19 12 7
14 37 10 23 30*
15 14 46* 17 23
16 29 13 31* 27
17 8 7 39 46*
18 8 21* 64 7
19 23 29 38* 10
20 51* 7 14 28
21 15 10 20 54*
22 12 59* 7 22
23 51* 17 16 16
24 7 10 35* 48
Candidates did not score well in the multiple choice questions. The following questions proved very challenging and were poorly done : 2, 3, 9, 11, 14, 16, 18 and 24.
• It was surprising to note that candidates found questions 2 and 3 on electronegativity trend and ionisation energy difficult to answer.
• Response to questions 9 and 11 showed that chemistry of transition metals is still a problem area.
• The use of standard electrode potentials in predicting whether a reaction takes place in question 14 was not well understood by most candidates.
• Response to questions 16 and 18 showed that section on Aqueous Chemistry is still a weak area. In question 18 most of the candidates failed to realize that a mole of H2SO4 dissociates to give two moles
of H+ and hence the [H+] = 2 x 0.05 = 0.1 = 10-1, pH = 1.
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SECTION B [36 marks] QUESTION 1
Generally well done but candidates failed to score well due to the following common errors :
• Failure to convert temperatures to Kelvin.
• Using wrong formula PV = nRT.
• Not explaining why the balloon will burst.
1 1 T V = 2 2 T V 1
V = 975 cm3 V2 = ?
1
T = 5 + 273 = 278 K T2 = 25 + 273 = 298 K
∴ V2 = 1 2 1 T T V = 278 298 975x
= 1045 cm3
Balloon will burst, because it can hold only 1000 cm3 whereas the new volume is greater.
QUESTION 2
(a) Many candidates wrote + 2 1
for the spin quantum number, and hence could not gain any mark.
5, 0, 0, + 2 1
OR 5, 0 , 0, – 2 1
(b) Discussing and applying the Paulis Exclusion Principle was a problem for some students. Some candidates got confused with the Hund’s Rule.
Paulis Exclusion Principle states that no two electrons in the same atom can have the same four
quantum numbers.
For the above case the 3 quantum numbers are 5, 0, 0, but the spin has to be either + 2 1 or – 2 1 . QUESTION 3
(a) Lewis structure for NH+4 was poorly done. Usually electrons were not drawn in pairs, and brackets and charges were missing.
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(b) Candidates failed to provide concise explanations. Concept of a “lone pair” which involves two electrons was not well known by most candidates.
(i) NH3 has a lone pair of electrons which pushes down the bonded pairs to give pyramidal
shape.
BF3 has only three bonded pairs and no lone pair of electrons and hence equal repulsion
giving planner shape.
(ii) CCl4 has symmetrical (tetrahedral) shape which cause the dipoles (polarity) to cancel and
hence the molecule is non-polar.
QUESTION 4
(a) Explanation of the term ligand was very vague and most students did not know the change of CN−ion. Ligand is an anion or molecule, example CN−, bonded to central metal ion (Fe3+).
(b) In most cases oxidation number was given as 3 and not +3.
[Fe (CN)6]-3
x + 6 (-1) = -3 x – 6 = -3 x = +3
(c) Majority of the candidates correctly gave the coordination number as 6.
(d) Naming of the complex ion proved too difficult to most of the candidates. Only a few candidates correctly named it as hexacyno ferrate (III) ion.
QUESTION 5
On the whole this question was not done well reflecting poor understanding of transition metal chemistry.
(a) (i) Most common response was d-orbitals having unpaired electrons.
Only few candidates knew that paramagnetic means having weak attraction to magnet or affected by magnetic field.
(ii) Many candidates wrote unshared electrons instead of unpaired electrons.
(b) Common incorrect responses were :
• [Ar] 4s1 3d10
• --- 3d10 4s1
• --- 4s2 3d9
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QUESTION 6
Candidates failed to score full marks as all bonds broken or formed were not identified and giving incorrect sign for ΔH.
(a)
Bond broken : Bonds formed :
2 C – C = 2 x 346 = 692 6 C = 0 = 6 x 743 = 4458
8 C – H = 8 x 412 = 3296 8 O – H = 8 x 463 = 3704
5 O = O 5 x 497 = 2485 8162
6473
ΔH = 6473 – 8162 = -1689 kJ
(b) Most candidates identified the reaction as exothermic but failed to give correct reasoning. ΔH being negative was not acceptable.
It should be emphasised that for exothermic reaction the heat content of products is less than that for the reactants and hence extra energy given out.
QUESTION 7
In most cases the two half equations were swapped but the overall equation was correct. It was evident that candidates had difficulty in identifying the oxidation and reduction reactions.
(a) Oxidation half equation :
Al + 4H2O → Al(OH)−4 + 4H + + 3e
-(b) Reduction half equation : MnO−
4 + 4H
+ + 3e+→ MnO
2 + 2H2O
(c) Overall equation :
Al + MnO−4 + 2H2O → Al(OH)−4 + MnO2
QUESTION 8
It revealed poor understanding of Aqueous Chemistry section. Candidates failed to calculate concentrations of Ag+ and Cl- ions and were not able to write correct ionic product expression.
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IP = [Ag+] [C
l-]
= (1 x 10-3) (1.5 x 10-4) = 1.5 x 10-7
(b) The concept was well understood. AgCl will form a precipitate as the ionic product (IP) is greater
than Ksp.
QUESTION 9
This question on Aqueous Chemistry was again not well attempted.
(a) Candidates failed to list all the species and a common omission was CH3COOH. Some candidates
wrote H+ instead of H 3O+.
The species present in a solution of ethanoic acid are : CH3COOH, CH3COO -, H3O+ and OH
(excluding H2O).
(b) Most common incorrect response was : NH4Cl + 2H2O → NH+4 + Cl
-Only a few candidates correctly gave the equation :
(c) Most candidates identified CH3COONa as the weak electrolyte and not CH3COOH.
(d) Candidates who mentioned that the HCl is a strong acid did not gain any mark.
It was expected that students will mention that HCl completely ionizes while CH3COOH only
slightly dissociates.
QUESTION 10
A very poorly done question. In many cases candidates failed to write both the products.
(a) Quite well done.
CH3CH2Cl + NH3→ CH3CH2NH2 + HCl
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(c) Many students wrote instead of CH3CH2OH as one of the products.
QUESTION 11
(a) Poorly done. Instead of hexagon, many drew octagon or pentagon and some failed to show double bonding.
(b) Candidates who correctly identified the structure of phenol were able to predict that phenol will have higher boiling point as it will be able to form H-bonds while benzene could not.
(c) The expected property of phenol were : it is acidic; soluble in water; more dense and less volatile as compared to benzene.
QUESTION 12
A reasonably well attempted question.
(a) Some candidates wrote 2-chloro-butanal instead of 3-chloro-butanal.
(b) A common incorrect structure was :
Some candidates failed to show H-atom attached to C-atoms. The accepted structure for 3-hydroxy butanoic acid is
(c) Many candidates wrote functional group as carbonyl instead of carboxyl.
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SECTION C [40 marks] QUESTION 1
Candidates did not score well in this question.
(a) Identification of the substance to match the descriptions provided was very poorly done. This reflected limited understanding properties and related structures. Candidates who wrote the name of the substance and failed to follow the instruction were enalized.
(i) D (iv) E
(ii) C (v) D
(iii) B (vi) E
(b) (i) Enthalpy of formation definition was very poorly explained. Very few candidates could explain it as enthalpy change for the formation of a mole of substance from its elements in their standard states under standard conditions.
(a) Calculation of enthalpy change was well done except for the need to be careful with signs (+ or -).
Reverse first given equation :
2NO2→ N2O4 ΔH = -57.93 kJ
2NO + O2→ 2 NO2 ΔH = -113.14 kJ
2NO + O2→ N2O4 ΔH = -171.07 kJ
(b) Poorly done. Most candidates re-stated the question in their own words.
Candidates failed to explain that combustion of one mole of sulphur gives one mole of SO2
which is same as formation of sulphur dioxide when one mole of S reacts with one mole of O2. Both reactions are same : S + O2→ SO2
(c) Explanations provided were very general. Most candidates mentioned that Al is not stable while Mg was more stable. Some wrote that Al has less nuclear attraction than Mg. Electron configuration of both elements using s, p notation would assist in answering the question.
Al 1s2 2s2 2p6 3s2 3p1 Mg 1s2 2s2 2p6 3s2
It is easier to remove an electron from incomplete 3p orbital of Al than an electron from fully filled 3s orbital.
QUESTION 2
This question was poorly attempted.
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(i) Candidates often failed to identify the RHE and LHE and omitted the negative sign.
Eo
cell = E o RHE – E
O LHE
= (-0.76) – (+0.34)
= -1.10V
(ii) The negative value indicates that actual cell reaction is reverse of the galvanic cell drawn.
ie. Zn + Cu2+→ Zn2+ + Cu
(iii) Very poorly done. Most candidates wrote : use salt bridge; use power supply; external power source or add more reactants.
Expected responses were :
increase concentration of solution (cations)
increase the temperature
scrape impurities/coatings from electrodes
(b) This question on match substances with bond types was unsatisfactorily done.
(i) SrO (iv) BCl−4
(ii) BCl3/CF4 (v) BCl3
(iii) CS2 (vi) CF4
(c) (i) The structural formula of primary, secondary and tertiary amines was very poorly done. Most candidates gave general formulae or gave structures in line with classification of alcohols into primary, secondary and tertiary alcohols.
I.
II. OR
III.
(iii) Generally well done. Many candidates wrote “fishy smell” instead of fish-like smell.
QUESTION 3
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(a) Candidates response to this question revealed poor understanding of transition metal chemistry.
(i) Variable oxidation states. (ii) Formation of complex ions.
(iii) Catalytic property.
(iv) Zinc (symbol Zn was not accepted).
(v) Complex formation is due to vacant orbitals of transition metal cations which accept the lone pair of electrons of ligards.
(b) Candidates showed poor understanding of Chemistry of Group IV elements. Similar concepts have been tested over the years and yet most students failed to score well.
(i) Chain compounds are common with C but not Si. This is because C-C bonds are
comparable in strength to C-H and C-O bonds whereas for Si, the Si-Si bonds are relatively weaker compared to Si-O bonds.
(ii) Multiple bonds are common with C because C=C and C-C bonds are relatively strong compared to Si-Si multiple bonds which are weak and chemically unimportant.
(iii) Hydrides of C (hydrocarbons) are more common and stable than those of Si. The C-H bonds are relatively strong compared to C=C bonds, whereas in Si the Si-H bonds are weak
relative to Si-O bonds.
(c) Well done but however candidates lost marks for not writing the correct equation and not showing the calculation steps.
Zn + S → ZnS
Mole ration 1 : 1
Moles of subst.
32 0 . 16 : 65 25 . 16
0.25 : 0.5
Zinc is the limiting reagent as all zinc (0.25 moles) will be used up leaving 0.25 moles of unused
(unreacted) sulphur.
QUESTION 4
A poorly attempted question in particular parts (a) and (b) on Aqueous Chemistry.
(a) This question on titration curve proved very challenging.
(i) Very few candidates correctly identified the titration curve as strong acid and strong base.
(ii) pKa = 7
(iii) Bromothymol blue.
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(b) This was the most difficult part of the question. While many candidates knew the function of Buffer solutions they could not apply the knowledge to other parts of the question.
(i) It resists small changes in pH of solution when small amounts of either an acid or a base is added.
(ii) H3O+ + HCO3− → CO2 + 2H2O
Extra acid (H3O+) reacts with hydrogen carbonate ion producing CO2 gas which escapes
(breathed out).
(iii) Since extra CO2 is produced, the patient will breath faster to remove CO2 or compensate
for it.
(iv) Common response was that water is neutral. Very few candidates knew that extra water alters the concentration of both the acid and its conjugate base equally.
Hence Ka and the ratio
] [ ] [ base conj acid
does not change and pH remains constant.
(c) Generally well attempted.
(i) 3
(ii) d-orbital
(iii) Transition metal/d-block
(iv) Catalyst
QUESTION 5
Response to this question was well below expectation.
(a) Poorly done. Candidates failed to give specific systematic names in part (i). In part (ii) candidates did not mention that reagent P must be acidified and often wrote formula instead of name.
(i) A. butyl butanoate
B. butan-l-ol
C. butanoic acid
D. 1-butene / but-l-ene
E. 1-bromo-butane
F. 1-amino butane
(ii) P – acidified potassium dichromate (or permanganate) Q – hydrogen bromide
(iii) I - Hydrolysis
II – Dehydration
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(iv) Sweet fruity smell.
(b) This part was very poorly done. Often candidates described two separate tests one each for alcohol and ketone. When students gave correct test, they failed to specify the special conditions like heating or acidifying the solution.
Some common possible tests were :
• To both samples add acidified K2Cr2O7 solution and warm the mixture.
→ No colour change ⇒ Ketone
→ Orange to green colour change ⇒ secondary alcohol
• Reacting both samples with Brady’s reagent. → No reaction ⇒ secondary alcohol
→ Orange precipitate ⇒ ketone
• Reacting both samples with Lucas reagent. → No reaction ⇒ ketone
→ Cloudiness after sometime ⇒ secondary alcohol
PAPER 2
The overall performance in Paper 2 was disappointing. It is evident that not much emphasis is placed on practical work and candidates lack hands on experience which will be fatal for those furthering their career in science. Although similar concepts have been tested in past papers, the student performance has not improved.
QUESTION 1
Candidates performance in this question was below expectation. Recognition of equipment and their uses will be perfected if students were given the opportunity to handle them and also perform the various practicals prescribed for them.
A. (i) Calorimeter.
(ii) Ensure no heat is lost to surrounding.
B. (i) Distillation.
(ii) Separate pure liquid from a mixture based on different boiling point.
C. (i) Separating funnel.
(ii) Separate two or more immiscible liquids from a mixture.
D. (i) Reflux.
(ii) Prevent loss of volatile liquids during heating.
E. (i) Pipette.
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QUESTION 2
(a) Correct observation was made but explanation were not very precise.
Ice is less dense than water or it floats on water because ice has open extended cage structure occupying more volume which cause density to decrease.
(b) (i) 45 mL was correctly recorded by most candidates.
(ii) Often candidates wrote that methylated spirit evaporated. Only some candidates wrote that water molecules form H-bonds with alcohol and become more closely coordinated with it. Hence there is reduction in the volume.
(c) This part was poorly explained. Most did not mention and could not distinguish between molecular and intra-molecular bonding. Often the explanations were incomplete and hence
students failed to score well.
Parachlorophenol forms intermolecular H-bonds with adjacent molecules forming chain links.
Ortho-chlorophenol forms intramolecular H-bonds which restricts the possibility intermolecular bonding and chain links. Hence it has lower boiling point.
QUESTION 3
(a) Generally well done except for part (iii) where students could not apply the inverse square law to determine how attractions and repulsions were affected.
(i) Sodium chloride – B Caesium chloride – A
(ii) Cs+ has longer ionic radius than Na+ and hence can pack more C
l- ions around it.
(iii) F α 12
d F α 42
1
⇒ F α
16 1
If d decreases by 4 then F increases by 16 times.
(b) Well done though some candidates wrote A, A, B, B respectively and failed to get any mark.
(i) atomic
(ii) atomic
(iii) bulk (iv) bulk
QUESTION 4
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(i) Most candidates correctly worked out the mass of vapour.
(43.86 – 43.68) = 0.18 g
(ii) It was evident that students who had done this practical could provide correct explanation.
The 100 mL flask is filled with water to the brim and than the liquid poured into a measuring cylinder to get the volume.
(iii) Generally well done. Candidates often failed to convert the temperature to Kelvin and volume into litres. They also lost marks for not giving the correct units for the molar mass.
PV = nRT T = 77 + 273 = 350 K
PV = mRT V = 118 mL = 0.118 L
M ie. M = mRT
PV
= 0.18 x 8.314 x 350 101.3 x 0.118 = 43.82 g/mol
Alternately students could have calculate the moles (n) and divided the mass (0.18) by n, (4.11 x 10-3), to obtain the molar mass.
(iv) The expected value of molar mass will be low because high pressure reduces mass of air in the flask to give a weight value too low.
(v) Cannot use water bath (temperature will be 100°C) but use other methods to get high temperature
like :
- paraffin bath
- electric oven
- electric heater
QUESTION 5
Generally well attempted question except for part (vi) which proved too difficult for most candidates.
(i) n = CV = 0.1 x 0.02 = 0.002 moles
(ii) From equation mole ratio : MnO−
4 : Fe 2+
1 : 5
ie. 0.002 : 5 (0.002)
= 0.01 moles
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(iv) ⎟
⎠ ⎞ ⎜ ⎝ ⎛ 100 65 . 0 56 . 0
x % = 86.15 %
(v) Prevent oxidation of Fe2+ (to stabilize Fe2+ ions).
(vi) Difficult to determine the end-point in alkaline solution as the product is green.
MnO−
4 Mn
2+ colourless
(purple) MnO2−
4 green
QUESTION 6
A well attempted question.
(i) The structural formula for 2 chloro 2 methylpropane was correctly given.
(ii) Candidates lost mark for not mentioning that the acid must be concentrated. Reagents : 2 methyl propan-2-ol and conc. hydrochloric acid.
(iii) (CH3)3− COH + HCl→ (CH3)3C – Cl + H2O
(iv) I. Candidates did not get mark for writing to neutralise (remove) acid, unless they mentioned to neutralise excess or unreacted acid.
II. Absorb water from the product (drying agent). No mark was awarded if candidates wrote it acts as dehydrating agent.
QUESTION 7
The whole question was very poorly done. It was again evident, as in Paper – 1, that the Chemistry of transition metal was not well understood.
(a) (i) I. VO−
3 + 4H
+ + e-→ VO2+ + 2H 2O
II. V3+ + H2O → VO2+ + 2H+ + e
-(ii) Relationship between the colour of compounds and the oxidation state was not very well known. Candidates failed to score any mark as the formula of ion, its colour and the oxidation state was not given.
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Possible answers :
CrO−4 yellow +7 MnO−4 purple +7 Fe2+ greenish +2
Cr3+ green +3 Mn2+ colourless +2 Fe3+ yellowish +3
CrO2−
7 orange +6 MnO24−green +6
QUESTION 8
Candidates response to this question was below expectation.
(i) Very few candidates scored full mark in this part. Fe → Fe2+ + 2e
-2Ag+ + 2e-→ 2Ag Fe + 2Ag+→ Fe2+ + 2Ag
(ii) Often candidates failed to put positive sign in the emf calculated. Eo
Cell = E o RHE - E
o LHE
= (0.80) – (-0.44)
= +1.24V
(iii) Some of the possible factors are :
• Temperature not 25°C.
• Solution not 1.0 mol L-1 .
• Impurities in solution or on electrodes.
• Electrical resistance in wire.
• Current drawn from voltmeter.
• Insufficient transfer of electrons from ions to electrodes and vice versa.
(iv) The diagram work was disappointing. The electrolytes were not named and even the electrodes were just labeled as anode and cathode instead of the names of metals used. Some even included power supply in the diagram with incorrect direction of electron flaw. A few candidates drew
electrolytic cell.
