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BedokFunland JC's A Level H2 Chemistry Qns (Part 2)

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    H2 Chemistry @ BedokFunland JC (near VJC & TJC)
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    • A BedokFunland JC Original H2 Chemistry Challenge

      Considering only the 1st proton dissociation (ie. regardless of proticity), ie. magnitude of Ka1 value, H2SO4 is a stronger acid compared to H2SO3, just as HNO3 is a stronger acid compared to HNO2, which is indeed expected and concordant to the general trend of acidic strength in any given series of oxoacids. However, bucking the trend are the oxoacids of phosphorus, in which H3PO3 is actually a stronger acid compared to H3PO4. There are actually 2 independent reasons for this unusual behavior, the combined effect of both, outweighs the expected general trend for oxoacid families (I assume you know how to explain the much easier within-H2-Chemistry-syllabus question of why H2SO4 is stronger than H2SO3, and why HNO3 is stronger than HNO2... do you?).

      My BedokFunland JC Challenge for you, of course, is : Explain why H3PO3 is a stronger acid compared to H3PO4.

      If you can correctly explain just 1 of the 2 reasons, you're already beyond A grade Distinction level (because even 99% of A graders can't solve this challenge question). If you can correctly explain both of the 2 reasons, you're BedokFunland JC level.

      As with all my BedokFunland JC challenge questions, I won't reveal the correct answers here. My BedokFunland JC students can ask me during tuition, and all other students can go ask your school teacher or private tutor.

    • Originally posted by MapPwner:

      Can I ask a question regarding the 2017 A level H2 Chemistry Specimen Paper 2 Question 6(i),which involves formation of FAME from reaction between a fatty acid,RCO2H,and diazomethane,CH2N2.

      Reaction occurs via two step mechanism:

      -The fatty acid reacts with diazomethane to form a carboxylate ion intermediate in the first step.

      -N2 is formed in the second step

      Suggest the mechanism for this reaction.(4 marks)

      Am i right to say that in the first step of the mechanism,CH2N2 deprotonates RCO2H to form RCOO- and CH3N2 respectively?And then subsequently the C-N bond in CH3N2 breaks and the bond pair is donated to positive charged N atom to form N2,before the RCOO- attacks the CH3 thingy to form RCO2CH3?


      Almost correct, but your described mechanism would be 3 steps, when Cambridge already specified 2 steps. The reason for this is the instability of the methyl carbocation, which is why the next step (after the proton transfer) is SN2, not SN1. Ideally, for the student to demonstrate a correct understanding of this reaction, both resonance contributors of diazomethane should be drawn, with the student showing to Cambridge that he/she knows which resonance contributor more appropriately functions as the Bronsted-Lowry base in the 1st step of the reaction mechanism.

    • imagehttps://mothership.sg/

      A BedokFunland JC H2 Chemistry Challenge Question for you :

      Amos Yee drinks Teh-O-Ice-Limau according to 1st order kinetics...


    • A BedokFunland JC H2 Chemistry Question

      The amount of CN−(aq) ions in an analyte solution can be determined by titration against Ag+(aq), during which the soluble dicyanoargentate(I) coordination complex is first generated, until sufficient Ag+(aq) is added from the burette to generate the insoluble white precipitate of the silver(I) dicyanoargentate(I) coordination compound. If x mol of Ag+(aq) had been added at first sighting of the white precipitate, what was the amount of CN−(aq) ions present in the analyte solution?

      A) 0.5 x
      B) 1.0 x
      C) 2.0 x
      D) None of the above

    • An Original BedokFunland JC H2 Chemistry Challenge



      Q1. Explain fully the relative basicities of the 3 N atoms in Histidine (ie. explain which N atom on the R group imidazole ring is more basic, and explain whether the alpha N atom or the R group imidazole ring is more basic).

      Q2. Given pKa1 = 1.60, pKa2 = 5.97 and pKa3 = 9.28, calculate the isoelectric point of Histidine.

      Q3. Calculate the pH at all significant points (ie. initial, 1st maximum buffer capacity, 1st equivalence point, 2nd maximum buffer capacity, 2nd equivalence point, 3rd maximum buffer capacity, 3rd equivalence point) in
      (i) the titration of the fully deprotonated form of Histidine against a strong monoprotic acid added in excess, as well as in
      (ii) the titration of the fully protonated form of Histidine against a strong monoprotic base added in excess.
      Use 0.05M as the molarities for Histidine and both titrants, and 25cm3 as the analyte volume of the Histidine solution for both titrations. Include a sketch of both titration curves, with all significant pH values labeled.

      Q4. By considering the enthalpy and entropy changes during decarboxylation, explain how temperature affects the thermodynamic feasibility and Kc value for the decarboxylation reaction.

      Q5. Suggest and draw out the curved-arrow electron-flow reaction mechanism for the (non-enzyme catalyzed) decarboxylation of Histidine.

      Edited by UltimaOnline 22 Oct `17, 4:20AM
    • Originally posted by Ellipsisss:

      On a side note, here is the link to my answers to the questions above. Can i trouble you to point out the mistakes? Thank you!!



      Ellipsisss of HCJC,

      Here are my comments on your attempted answers.

      For Q1, your order of basicity is wrong, and your explanation has multiple errors (including an internal inconsistency, prolly just ur careless typo), and a word of advice : for A levels, never use "it", but specify exactly which, otherwise Cambridge will think you're trying to smoke your way through and you'll kena penalized (even if your answer is otherwise correct). Hint : your answer would be somewhat less wrong (but there are still errors) for a typical alpha-amino acid, but Histidine happens to be anomalous amongst alpha-amino acids (this is afterall a BedokFunland JC Challenge, ie. difficult even for A grade students, and for some private tutors and school teachers as well), but still totally explainable, if you understand your chemistry well enough. You can 'cheat' a little and google out where the 3 pKa values belong, and thus 'reverse engineer' the basicity of the 3 N atoms (the least basic N atom doesn't even have a Kb value, if Cambridge asks why, do you know how to explain?).

      Q2. Correct, well done. More than half of all JC H2 students wouldn't know how to solve this.

      Q3. Ok, I was admittedly being unfairly sadistic in setting this question, as the pH values for the more extreme ends of the titration cannot be correctly obtained by using the A level acid-base equilibria within-syllabus methods, which are only approximations that fail when dealing with fairly strong acids and bases. In addition, do note that if the initial molarity of the acid or base is not sufficiently larger than the change in molarity (ie. as indicated by the magnitude of the relevant Ka or Kb value), it becomes inaccurate and indeed invalid to neglect the change in molarity, and to simply approximate equilibrium molarity back to initial (as taught by Singapore JCs, and as you had done in some of your calculations). But no worries, Cambridge probably wouldn't be as sadistic, and would also usually set initial molarities to be much larger than the change, so you usually wouldn't need to solve quadratic or cubic equations for H2 Chemistry (though all students should still bring 2 calculators as a backup, a non-graphing and a graphing). So as far as A levels are involved, very well done, your titration pH values are mostly correct.

      Q4. Correct, well done. Most Singapore JC students are confused by the effect of temperature change on the Gibbs free energy change value VERSUS the equilibrium constant Kc value, because they simplistically assume that the larger the Kc value, the more negative the Gibbs free energy change. This is only true for constant temperature. When temperature changes, you must consider both the enthalpy and entropy changes. The effect of temperature on Kc is controlled by the enthalpy change, while the effect of temperature on Gibbs free energy change is controlled by the entropy change. Interested A grade H2 students, H3 students and Olympiad students, can explore this aspect of thermodynamics deeper for themselves, including the Van 't Hoff equation.

      Q5. Almost correct, but not quite. Did it not occur to you that the decarboxylation (both with and without enzyme catalyst) is kinetically feasible only due to the zwitterionic nature of amino acids? An intermediate with a uninegative formal charge on a C atom that is not delocalized by resonance is prohibitively destabilizing making your mechanism kinetically unfeasible. On a related note : writing C- or H- for Grignard reagents and LiAlH4, is an oversimplification, and does not actually exist as an intermediate, but only as a transition state (so to speak). On another related note : investigate the mechanism for decarboxylation of beta-keto-carboxylic acids (which Cambridge used in a Singapore TYS Qn) followed by prototropic tautomerism, notice there is no C- intermediate there either.

      Overall, a very good attempt by Ellipsisss of HCJC, you should be able to get your A grade if you don't screw up your remaining P2, P3 and P1.

    • Originally posted by CKTR:

      Hi UltimaOnline, for qns 10 VJC 2014 P3 qns 1a iii, why is it that the ratio of [Mg2+]/[Cu2+] = Kc?


      Because the equation is Cu2+(aq) + Mg(s) --> Cu(s) + Mg2+(aq), and the thermodynamic activity of solids is taken to be 1 (mathematical derivation of this is done at University level).

    • Originally posted by CKTR:

      Oh yeah I forgot thanks.

      Btw I found ACJC the most challenging imo

      Excellent recommendation by CKTR!

      The ACJC 2017 prelim papers certainly belong to the upper difficulty tier of 2017 prelim papers, together with RJC / VJC / DHS / SRJC etc.

      Indeed in some ways, this might be just be my favorite set of 2017 prelim papers, due to it's multiple extensive questioning (ie. 5 questions across P2 & P3) of unfamiliar (ie. not taught within the H2 syllabus) curved-arrow electron-flow reaction mechanisms, which as some of you might know, is in line with BedokFunland JC style (ie. I would be utterly delighted if the actual Singapore A Level exam papers used the ACJC 2017 prelim papers).

      So all JC2 students serious about scoring a distinction grade, be sure you thoroughly go through this upper-tier set of ACJC / RJC / VJC / DHS / SRJC papers before next week's P2 & P3. Any questions post em here.

    • Originally posted by MapPwner:

      Can I just ask whether in the formation of a triiodide ion,I3^-1,from Iodine and an iodide ion,one of the I in I2 forms a dative bond with the I^-1 to form a linear molecular geometry I3- structure?

      Likewise for [I2Br]^-1,one of the I in I2 forms a dative bond to the more electronegative Br- to form [I2Br]^-1 with the same molecular geometry?

      Are we expected to know grp 12/17 reactions in the current new syallabus,for instance such as reaction of Br2 with concentrated H2SO4?


      To be precise, "accepts... from", not "forms... with". But bear in mind I- is an anion, and the adduct is an anion, so for dot-and-cross diagrams (if required), while the dative bond should still be shown with a dot-dot or cross-cross, a hollow-dot (ie. electron accepted from the species which reduced 1/2 I2 to the I- reactant, presumed to be the counter-cation present throughout, eg. K+) must be clearly shown as part of a lone pair in both the I- reactant and I3- product.

      Yes, of course (isn't it obvious why the ionic geometry of the adduct is linear?), Cambridge can certainly ask you to draw out the full mechanism (in which case no 'dative' arrowhead bonds should be shown anywhere in the product, just a curved arrow from the I- or Br- nucleophile to the I2 electrophile, then all straight lines (ie. normal bonds) in the product, any 'dative' arrowhead bonds shown will be penalized), as this mechanism is already much easier than the other ones asked in 2017 ACJC Prelim papers.

      The conceptual basics (eg. trend of redox potentials), yes. The technical details (eg. memorization of equations), no. But distinction A grade students are indeed expected to be exposed beyond the syllabus, and/or are expected to be intelligent enough to figure out the reaction, equation and products (eg. using Data Booklet for hints) even without prior exposure. So yes, Cambridge can indeed ask on these not-in-syllabus reactions, although Cambridge would probably be a little kinder, and give slightly more hints than Singapore JC prelim papers.

      Edited by UltimaOnline 04 Nov `17, 7:24PM
    • Ellipsisss, here's a chance for your 'redemption' (you were game enough to attempt my earlier BedokFunland JC Challenge Qn, but as I said, I'll only reveal the answers for such Challenge Qns to my own BedokFunland JC students, so I left you hanging back there regarding the imidazole ring of histidine, well, here's a simpler question that has a higher probability of being asked by Cambridge for the new H2 syllabus).

      Since the new syllabus specimen paper asked for the hybridization of N atoms (where in the old syllabus, questions were restricted to C atoms), almost all the 2017 Singapore JC Prelim Papers started following suit. So here's a BedokFunland JC question for all of you.

      State *and* explain the hybridization of the N atom in the amide functional group.

      Note : At Uni levels, the actual hybridization will be x% this, y% that, z% something else, but at A levels, you're expected to identify the single 'major' hybridization that contributes most to the actual hybridization.

      I'll leave it to you guys (feel free to attempt and post what you would write as your answer if this question comes out in this coming week's P2 or P3) to discuss this very easy question, I'm sure you guys don't need me to spoonfeed you the answer, but try to resist googling out the answer before attempting it yourself, that's just lazy.

    • Thanks for making the effort and taking the trouble giving a detailed look at the entire P2, MapPwner, appreciate it :)

      Originally posted by MapPwner:


       Incompletely Compiled chemistry suggested answers so far(Answers may not be fully correct so take it with a pinch of salt):

      Qn 1(Ionisation energies of element A(chlorine) and hydrogen properties)

      Increase in I.E as successive electrons are removed from element A(1m),as it is increasingly difficult to remove a negatively charged electron from a positively charged element A ion(1m).Large increase from 7th to 8th i.e as the 8th electron is located in an inner electron shell,which is closer to the nucleus and thus more energy required to overcome the strong electrostatic forces of attraction,resulting in large rise in I.E.(2m)


      Element A is chlorine(1m).1s^22s^22^p63^s23p^5.(electronic configuration-1m).


      Hydrogen can be classified as a grp 1 element as it has 1 valence electron in its outermost shell(1m).

      Cannot be classified:

      Grp 1 elements-metal lattice structure vs simple covalent molecule structure for H+describe(1m)

      Contrasting properites=grp 1 higher melting/boiling point than H/H2+accounting for the reason why(1m)


      Qn 2(Acids and buffers of HCO2- and blood)


      part i)ph=2.60+showing calculation(3m)

      (ii)forgot the answer

      (iii)ph will fall if no buffer system in a human while exercising

       Qn 3,4,5(C2H4O/cinammealdehyde and transition elements)



      (i)showing molecular formula is c2h4o(relatively straightforward)

      (ii) structure is a triangular molecule with c-o-c bond and c-c bond between both c for c2h4o,with remaining h bonded to the 2 c ensuring octet rule fufilled.

      Reaction with Br2(aq)=orange br2 decolourises rapidly to colourless

      Na2co3:No vigorous effervescence,production of co2 observed.

      Kc is small which thus show most of the product is the (LHS one-I forgot the name) and thus position of equilibrium lies more to the left(1m).Kc shows the extent of favourability of product formed in the equilibrium and can be used to determine the amount of product formed based on Kc value.''If Kc small=poe left while if Kc large=poe right.''(Stating something like this is 1m I think)


      (iv) memorise balanced eqn of triiodomethane with ch3cho

      cant really remember much for qn 3...



      for cinammealdehyde,1 structure still has c=c and aldehyde just that it is arranged at different positions(swapped)=positional isomerism,while the other only has an aldehyde group and no alkene group=structural isomerism(Or show cis-trans version for one of them).

      After that is Hcn(aq),trace amount of nacn(aq),in the cold,10 to 20 degree.

      when the nitrile compound is reduced as shown in question by h2,alkene and nitrile is reduced,but for lialh4 alkene is not reduced.

      behind the mechanism is nucleophilic addition(relatively straightforward)

      Generation of Hydride ion:BH4- ->(reversible arrow) H^- +BH3.

      Then the slow and fast step relatively standard.


      Why lialh4 a stronger red. agent than nabh4=orbital overlap of Al and H in AL-H bond or different polarising power of Li+ and Na+ of anion electron cloud to release hydride ions.


      the naoh thingy just dilute the thing(1m) and state how to deduce order+brief procedure of expt such as recording time taken for ppt formation at diff concentrations via dilution and comparing it with diff expt to determine order(2m)[I made naoh more concentrated but as my back part is right i probably lost 1 mark as only dilution is accepted to concentration lower than 2mol dm^-3].


      Graph shows a first order downward sloping curve with constant half life(reaction time) halves for a doubled increase in successive concentrations of naoh up to 2moldm^-3.[I labelled beyond 2 moldm3 so if lenient i lost 1 mark at most,if strict 2 marks lost].[I forgot whether this was part of qn 4 or 5]


      qn 5


      for the electrolysis part,delta g is -284.0kj mol-1(to 3 sf)=reaction thus spontaneous at r.t.p

      Define homogenous catalyst(2m),show how Fe2+ catalyses using appropriate equations in the reaction of S2O8^2- + 2I^- -> I2+ 2SO4^2- (2m) and state the property,which is able to form one or more stable ions with variable oxidation states(1m).


      Contrast between colour for the 3 mark question of iron compounds[solid hydrated fe3+,aqueous fe3+ and solid hydrated fe2+] is relatively ok


      Purpose of using c6h5cooh:It is very reactive/combustible.

      Water jacket purpose:Ensures temperature of surrounding water at same temperature as calorimeter thus no heat loss.

      Presence of black solid:Incomplete combustion of c6h5cooh forming carbon,thus suggest excess use of o2.


      to find heat capacity of calorimeter is -(q x average temp rise for sample 2 and 3)/amt of benzoic acid in moles=-3230kj mol^-1,where q is the heat capacity of calorimeter.One should end up with something like -(qx30)/0.05= -3230kj mol^-1 before eventually arriving at 5.38kj bla bla for q.


      the below 3 marks shld be fine.


      Napthalene(Moth Balls) qn:

      -590kj mol -1

      ai value higher thus molecule very stable due to p orbital overlap between singly bonded and doubly bonded carbon atoms results in extensive delocalisation of electrons(or words to the effect).


      sp2 hybridisation,draw out the hybridisation between two c atoms.c=c is 1 sigma 1 pi,c-c technically is 1 sigma only but cos of delcoalisation of electrons is somewhat 1 sigma 1 pi too for bonds.Thus only 1 diagram to show hybridisation is sufficient between sp2-sp2.


      Showing is c10h12 is fine.

      drawing of structures too and reason.

      Reagent and condition:H2(g),Nickel catalyst,200 degrees.



      Originally posted by MapPwner:

      Regarding the friend who found it easy,yeah when he stepped out of the hall he said only lose a few marks but after discussing most of the paper that snowballed to 25 marks.

      My class on average the B graders(only 1 person) lost about same marks as me(15-17,he thought the paper was easy as he felt he only lost 2 marks but guess not),the rest probably might have done worse,another i asked lost 22 marks supposedly(he got E for prelims).Rest lost 25-30 or so.But i estimate the rest to lose on average 25-30 or so in all seriousness,as one couldnt even decipher the skeletal formula given in one of the question to deduce proper reagents and conditions.

      Of course i have friends in AJ losing same marks as me or even more,up to 30.My Hwa Chong friend who did much worse than me for prac aced this paper,lost about 5-6 marks

      I too,felt the paper was actually simple(except for qn 6 and 7,but qn 7 was like a repeat question from my secondary sch sec 2 chem paper so it was actually familiar),but panic probably killed a lot of people.

      If i were to give an honest evaluation i would rate Qn 1 to 2 as easy(but i screwed up question 2 quite badly),qn 3 to 5 as moderate,qn 6 and 7 as moderate-challenging difficulty.

      Question 6 especially,as the question was unusual in the sense we need to state the purpose of benzoic acid that makes it useful for this calculation of heat capacity of calorimeter in a water,without considering about the water.

      Majority of the ppl were confused(>50% in my school) and other JCs,they didnt use -(q x(avg temp rise)/amt of benzoic acid in moles for enthalpy change of combustion,but other ways,which were wrong.

      Organic chemistry component was moderately difficult to me,yet a fun question to do even though it was a pain doing so as it was really interesting for me(only lost 1 mark for organic component part).




      Ahh, so your top 10% friend didn't do as well as he initially thought, but prolly still well enough to get A grade. In contrast, your HCJC friend really only lost about 5 marks for P2, enough to make up for his P4 errors. So MapPwner, yourself, you're well on track for A grade (so far overall score > 80%, yes?).

      Agree with you that the calorimetry question is a good challenging question for Singapore JC H2 students. I can actually think of 5 different reasons why benzoic acid is used as a standard for calorimetry (just as I can think of 3 different reasons why LiAlH4 is a stronger reducing agent than NaBH4), but Cambridge as expected, is already very lenient : any 1 reason (for both the benzoic acid and LiAlH4 questions) will suffice to score the 1 mark allocated. But the majority of Singapore JC H2 students are unlikely to have prior exposure to such questions, which is why these are, all things considered, good and fair distinction questions (not too unreasonable as to require 3 reasons for each question, but challenging enough to differentiate A graders from the rest).

      Similarly for the Naphthalene aromaticity question (from your description, the entire question is about Naphthalene being aromatic with resonance delocalization of pi electrons, while in the basic H2 syllabus only the simplest aromatic species, benzene, is taught), another excellent question that (from your description), didn't disappoint me, in the sense it is interesting and A grade differentiating.

      But as you (and your sarcastic school teachers) pointed out, Cambridge has already been very fair, since these are only a sprinkling of challenging questions amidst mostly easy or standard questions. So (unless P3 and P1 are hellish... hell yeah! I should set the A level papers Lol!) I doubt the bell-curve will deviate too much from past years.

      Keep up your excellent A grade performance, MapPwner! :)

      Originally posted by MapPwner:

      Overall score no lah how over 80% lol,my prac was rather bad.Prob for prac alone is 60%,this paper is 77-80%(pretty sure my estimates are very accurate,in line with my estimates for past chem papers for my exams this year,even if they deviate the score is higher than the estimated score).

      My prelim raw % scores for individual components were p1:80%,p2:72%,p3:76%,p4:56%,so more or less similar to a levels.

      As with regards about my HC friend,she probably will lose way more than 6 marks,i estimate eventually same marks lost as me as she only immediately excluded those she know is definitely wrong.


      Nice nice. So that means you're on par with, or better than, both your JC's top 10% friend, and your HCJC friend.



      Good good, as long as you don't cockup your P3 and P1 (ie. about same performance as your Prelims), your overall score as a safe estimate will be between 75% to 80%, so that's a guaranteed A grade (as I said before, when students are shocked they didn't get A grade, it's not that the A grade boundary > 75%, it's that they overestimated their own scores, but this shouldn't be an issue for you MapPwner).

      So P2 was mostly easy to average, with only a couple of tough questions. But since you're the top 20% of the national cohort, MapPwner, perhaps the average Singapore H2 student found this paper mostly average in difficulty, with quite a few tough questions. But that's still totally within expectations of a Cambridge paper, so the bell-curve shouldn't have too many surprises this year.

      Edited by UltimaOnline 10 Nov `17, 3:55AM
    • Originally posted by delurach:

      Hello, i’ve Got a query here! From Dunman high p3

      (b) A compound X, CH2CCH3(CH2)3CaHbO, is a constitutional isomer of Geraniol. It exhibits stereoisomerism and has a particular functional group that can be tested positively by Fehling’s solution, an alkaline solution of complexed Cu2+(aq), in a redox reaction.  In Fehling’s solution, the tartrate ion forms a complex with Cu2+ as shown below.    (i) Suggest why the copper ion needs to be complexed for the test to work.  [1]  

      how come only when it is complex then it wouldn’t form cu(oh)2? 

      Because when it's already coordinated with its non-H2O ligands, it's not available to form hydroxide ppt. Only metal ions that have H2O ligands can have their H2O ligands deprotonated into OH-, and thereby (subsequently generating) the hydroxide ppt.

      Edited by UltimaOnline 09 Nov `17, 9:19PM
    • Originally posted by delurach:

      Hello, I’ve got 2 questions..

      1) how do you determine which substituent “overpowers” the other? Say you have a phenol and an NO2 if you were to talk about electrophilic sub of No2, will you expect harsher or milder conditions? 

      2) I’m confused with the common ion effect.. and how you can compare just the solubilities of say Ca in Caco3 and Mg in Mgco3 to determine whether they precipitate or dissolve. Is there a way to understand this common ion effect? And if if were to be Na2co3 why wouldn’t it work this way? (similar to sajc p3 q5) 

      Much thanks! 

      Q1. Ortho-para directors outweigh meta-directors. When 2 ortho-para directors compete, the stronger activator wins. When 2 meta directors compete, the stronger deactivator wins. After which, take into consideration steric hindrance.

      In aqueous solvent, phenolic OH group is as strong an activator as NO2 is a deactivator, ie. consider the strength of the activator versus strength of the deactivator. Hence in your case, conditions remain unchanged, neither harsher nor milder.

      Q2. The common ion effect reduces the solubility of ionic compounds (that contain the common ion), as predicted by Le Chatelier's principle.

      All Na+ and K+ compounds are fully soluble, that's why these compounds are used to supply the common anion, and similarly, soluble nitrate(V) compounds are used to supply the common cation.

      If the stoichiometries of both ionic compounds involved are the same, eg. CaCO3 vs MgCO3, then comparing either the molar solubility or Ksp, will indicate which precipitates 1st, when increasing the molarity of the common anion, here CO3 2-.

      If the stoichiometries of both ionic compounds involved are different, as in the 2017 SAJC P3 Qn, then you'll have to manually calculate the molarity of the common ion (in the qn, CrO4 2-) required to initiate precipitation for both compounds separately (ie. using the Ksp value for each), to determine which precipitates 1st, when increasing the molarity of the common anion, here CrO4 2-.

      Edited by UltimaOnline 10 Nov `17, 4:33AM
    • Originally posted by delurach:

      Hello!! Do you mind explaining Q3 (c)? the question on oxidation state :) Why is it wrong if you assume the difference between the no. of moles used to oxidise G and H are 2e due to the Zn, and subsequently equating the difference to be for every +1 is O.S? thanks :))

      I'll give the correct approach, then you see if you can figure out whether your approach (which isn't very clear, without showing your actual working with values) is right or wrong. Did you get the final answer correct? if you did, you'll may even get full marks, because if you did get the final answer correct, and Cambridge doesn't see anything overtly wrong with your working, however weird it is, you may still get full marks.

      Your approach (since you just gave a generic description, it's difficult to explain what's wrong, if you gave full working with all values included, it's easier to point out which step you got wrong, and why that step is wrong, ie. your exact misconception at that step) seems to be wrong (I say seems, because your approach isn't very clear), because of a couple of reasons :

      Just because Zn is the reducing agent, you can't assume the OS of vanadium in G and H differs by 2, since excess Zn (ie. non-stoichiometry) is used to reduce the vanadium. You can only deduce such, only if the moles of Zn is the same as the moles of the vanadium species it reacts with, which is not the case here (you're using excess Zn, so you've no idea how much Zn was reacted).

      You idea of "subsequently equating the difference to be for every +1 in O.S" can be used only for comparing the relative amounts of MnO4- required for oxidizing the same amount of vanadium in G versus in H, ie. 24.6 - 16.4 = 8.2 cm3 of MnO4- magnitude difference, which implies 1.64 x 10-5 mol of MnO4- magnitude difference, which implies 8.2 x 10-5 mol of e- magnitude difference, which upon division by amount of vanadium species present 8.2 x 10-5 mol (see my working below), gives you 1.0 OS magnitude difference, which is correct, ie. the magnitude difference between V2+ is H and V3+ in G, is 1.0 OS magnitude difference.

      My BedokFunland JC approach for 2017 P3 Q3c

      Based on Data Booklet redox potentials, you can show (ie. by calculating standard cell potentials for each redox step) that under standard conditions, using Zn as the reducing agent, the vanadium species can be reduced to dipositive V2+, ie. OS of +2, and no further. This is the easy part, which almost all students should be able to get. (Though for 3 marks, Cambridge won't require you to waste time writing out all the individual redox steps and calculating all the standard cell potentials, just a brief statement explaining how you know the final OS of vanadium is +2 will suffice).

      The 1st portion of G reacts with 3.28 x 10-5 mol of MnO4-, ie. 1.64 x 10-4 mol of e- transferred.

      The 2nd portion has now been reduced to contain V2+ as solution H, and reacts with 4.92 x 10-5 mol of MnO4-, ie. 2.46 x 10-4 mol of e- transferred.

      Since the final oxidized vanadium species is stated as VO2+ (ie. OS of +5), and the initial vanadium species was V2+, for each mol of V2+, 3 mol of e- is lost.

      Since 2.46 x 10-4 mol of e- was lost, hence mol of V2+ oxidized = 8.2 x 10-5 mol

      This implies 8.2 x 10-5 mol of vanadium species was also present in the 1st portion (since both are of equal volumes).

      Hence, no. of mol of electrons lost per mol of the original vanadium species, in being oxidized to VO2 + (ie. OS of +5), must have been (1.64 x 10-4) / (8.2 x 10-5) = 2 mol of electrons.

      Since the final OS of vanadium in VO2+ is +5, the initial OS of vanadium must have been +3.

      Hence molarity of V3+ in G = mol / volume = (8.2 x 10-5) / (25 / 1000) = 3.28 x 10-3 M

      Edited by UltimaOnline 11 Nov `17, 4:30AM
    • Originally posted by do12453:

      haha alright, if you could just take a look at 2017 P3 Q2c


      In solution, ionic species that can conduct electricity is :



      Upon heating, Bronsted-Lowry acid-base proton transfer reaction occurs from the acidic NH4+ cation to the basic N3- anion, to generate the molecular species NH3(g) and N3H(g), the NH3(g) being Bronsted-Lowry basic and hence would turn moist red litmus paper blue.

      A BedokFunland JC enhanced perspective of this A Level Exam Qn : the molecular and ionic forms exist in equilibrium with each other ; in aqueous solution and at lower temperatures, the ionic form is favored by enthalpy ; while in gaseous state and at higher temperatures, the molecular form is favored by entropy.

    • Originally posted by MapPwner:

      If leniently marked,65-66/80,worst case 60-62/80.

      On average marks lost were about 25-30 out of 80 for paper 3 among other people.




      MapPwner and do12453's estimated scores (for P3, and prolly for all the other papers too) are excellent and approximately on par. As long as you keep it up and score close to full marks for MCQ (which I expect should be no problem for both of you), the 2 of you should be getting between overall 75% to 80%, a confirmed A grade (as I've said many times before, the A grade boundary almost never exceeds 75% ; if students expecting A grade fail to get A grade, it's because they had overestimated their scores ; don't believe that bullshit about A grade boundary for H2 Chem being 80+% or 85+%, that's just ridiculous).

      delurach, how is your overall % score so far? You shared on the Reddit thread that you didn't do very well for P3, but as long as you score close to full marks for P1, you should be able to get somewhere in between 70-75% overall score, yes? That would give you a good chance of scoring an A grade, and considering that this year's Papers are somewhat tougher than the last few years.

      Most people agree that this year's P3 is considered on the difficult side (relative to past few years TYS), mostly because it is rather time-consuming, so most students won't be able to finish P3.

      As such, I know most students are hoping that, (and this will indeed likely be the case, if P1 is of similar difficulty to the other 3 Papers so far), the bell-curve and A grade boundary will slide down a little more away from 75% and down towards 70% (eg. 72%). So if you manage an estimated 70-75% overall score, you've a very good chance of scoring an A grade this year.

      While hardly rigorous evidence, it is nonetheless indicative of the general sentiment from students across the entire cohort :

      Difficulty of 2017 H2 Chemistry Paper 3 : http://www.strawpoll.me/14368146/r

      Be constructive : the best thing you (ie. all J2 students reading this) can do now for the next 2 weeks, is to do *ALL* the 2017 Singapore JC Prelim Paper 1s, and if you've any query on any MCQ, be sure to post here on the forum to ask for help, and I (or MapPwner, do12453, etc) will give you our comments.

    • Originally posted by tokipaqur:

      Thanks for the reply! :) yeah I think I need a close to perfect, at the least >27 for mcq to scrape an A. No matter how easy/hard p1 is I personally feel the A grade range is not going to be high this year. Heard 2010 is tough but I scored much better for '10 p2 and p3 (>80%) compared to a levels this year, furthermore p4 was a nightmare for us.



      Edit: just curious, are the % you mentioned your estimates? i do agree it's a reasonable range though, but just curious cos so far i haven't found anyone that knows how exactly the bell curve is applied. Thanks (:

      As 'longaf' said on Reddit : "no one can give an exact overall percentage" for the A grade boundary for this year, in fact not even Cambridge or SEAB, simply because the papers haven't been marked by Cambridge, and the raw scores haven't been submitted to SEAB, and no one right now knows how this year's cohort of students have performed (even after P1 is completed, let alone today is before P1), because that's the very idea of a bell-curve, ie. dependent on how every single student in this year's cohort has performed, which right now, neither Cambridge nor SEAB has any idea on this, because the papers aren't marked yet.

      And the astute will note that I'm certainly not giving an "exact overall percentage", I'm merely stating that, just as an atomic orbital is the region of space around a nucleus in which the probability is the highest (but never 100%) of finding an electron associated with that orbital, similarly, I'm merely stating from experience, that the A grade boundary has the highest probability of residing between 70% to 75% (hardly an exact percentage, you see), and of course it's entirely possible (just extremely unlikely), that this year's A grade boundary might go below 70% or above 75%, because nobody knows how easy / tough P1 is, and nobody knows how every single student in this year's 2017 cohort has performed.

      You're probably hoping for me to just directly reveal whether I have insider info from SEAB, or if these values are nothing more than my own personal estimates without evidence, but without revealing too much, let's just say it's 'somewhere in between', ie. I have very good reasons for giving this confident 70% to 75% estimate for the A grade boundary, unlike say, show-off school teachers who delight in sadistically bragging to their students, "I just saw this year's Paper and it's ridiculously easy! The A grade boundary for H2 Chem this year confirm 82.5% liao! Wahahaha!!!"

      Edited by UltimaOnline 12 Nov `17, 12:28AM
    • Originally posted by tokipaqur:

      Yep I get your point, was worried cos I'm just scraping an (absolute) A even with fullmarks in p1.

      Btw I have a few questions regarding mark allocation. For 1a, if I didn't explain in terms of atomic orbital/etc, but merely quoted values from the Data Booklet and described the trend, is it likely that I'll get just 1 mark or even 0?

      For the organic questions, would I be penalised for writing maintained at 55 degrees even though it was a deactivating group, and I didn't write limited Cl2(g) to ensure monochlorination? 


      Hate to be a bearer of bad news here, but you *might* get penalized for all 3 of the questions you asked on, but not for the reasons you think. See below.

      Q1. Similar to the discussion on how I had doubts over how most students chose to answer the acid-base nature of oxides across Period 3 (although on 2nd thoughts, I've recanted that one, Cambridge was probably just being generous with those 6 marks), it seems most students merely described the trend of thermal stabilities down Group 17, rather than the visual observation of their individual decompositions, as Cambridge probably wanted.

      As such, most students see 3 marks allocated and try to fill in more points relating to the trend (such as effectiveness of orbital overlap), but still missing out the individual thermal decomposition descriptions. Quoting Data Booklet bond dissociation enthalpy values for this question gets no marks, unfortunately for you (and many students).

      Based on my interpretation of what Cambridge probably wanted (as opposed to most students' interpretation), Cambridge could or should have fairly allocated 4 marks for this question (in contrast, Cambridge was inconsistently generous to allocate 6 marks for the acid-base nature of oxides across Period 3) ; that they stingily allocated only 3 marks here (and as such misleading many students) is likely thus : 1 mark for describing the trend down the Group (including a generic decomposition equation and stating that HCl is thermally stable and does not decompose), 1 mark for explaining the trend (just relationship between bond length and ease of bond dissociation will suffice, no need effectiveness of orbital overlap, no need quote Data Booklet values), and 1 mark for describing the different visual observations for HI and HBr upon heating.

      Organic Questions :

      "Limited" is actually indeed required to ensure monochlorination, and should have been included for a good answer. Cambridge has in some past papers let it slide, but in some other past papers penalized it. So you may or may not get away with this one.

      For nitration of a benzene (whether activated or deactivated), what Cambridge required was "heat under reflux" ideally, although they *may* accept "maintained at 55 degrees Celsius", so you *may* get away with this one.

      The irony is that while many (but certainly not all) Singapore students managed to correctly recognize that the aryl ketone substituent was deactivating, they erroneously thought that Cambridge was being sneaky-cunning and had required "more than 55°C" for the answer, when ironically, retired A level examiner-marker Jim Clark warns that :

      "Benzene is treated with a mixture of concentrated nitric acid and concentrated sulphuric (sic) acid at a temperature not exceeding 50°C. At higher temperatures there is a greater chance of getting more than one nitro group substituted onto the ring. You will get a certain amount of 1,3-dinitrobenzene formed even at 50°C."

      That's for benzene, so for deactivated benzene, 55°C is actually (and ironically for you) reasonable. But then again, that's Jim Clark's opinion (different A level syllabuses in the UK used slightly different temperatures) ; in the pre-2017 H2 syllabus, Cambridge actually wanted "50°C < T < 60°C" as the ideal student answer for nitration of unsubstituted benzene.

      But for the 2017 P3 Qn, Cambridge actually only required (and wanted) "heat under reflux" (more correct than just "heat", since this is a synthesis question, but fortunately for many students, Cambridge will probably accept just "heat", but seriously you students should have known better), since this new syllabus no longer requires (or even wants, ie. Cambridge will ignore) specific temperature values submitted by the student.

    • Originally posted by gabs1234:

      Why do you say that you recant your opinion on the 6m oxide qn? 

      Cambridge's fault lah.

      If they expected what most of you students wrote, then it doesn't seem worth 6 marks to me.

      If they expected what I would write if I took the A level paper this year (ie. beyond what JCs teach, ie. why ionic oxides are basic, why covalent oxides are acidic, why ionic oxides with covalent character are amphoteric), then that's worth more than 6 marks (because those points would be in addition to, not in exchange of, what you guys have written).

      So neither here nor there, just take it (ie. the more likely case is) that the standard answers most of you wrote can get the 6 marks lor.

      But Cambridge really nonsense this year, some questions so time-consuming only so few marks allocated, then some questions only got short answers yet so many marks allocated. Sheesh. *clap*clap* good job, Cambridge.

      But on an objective note, like MapPwner said, it's not that the paper's questions are super difficult, it's mostly that the entire paper (all the questions collectively) is time consuming, which results in a paper on the difficult side.

      So overall this year, the consensus amongst the general student population is that this year's Papers (assume P1 is of similar difficulty to all the other Papers) are significantly tougher than previous years, so yeah I expect the A grade boundary to slide down closer towards 70%.

      Then of course, like over at Reddit, you have students, or more accurately, friends of students, and school teachers, who brag that this is a ridiculously easy paper, it's a walk in the park, etc.

      Hey, the questions are easy for me as well, but you don't see me bragging (oh, I see what I did there) that this is an easy paper, because I think this is a time-consuming hence difficult paper, relative to previous years. So I agree with MapPwner that the A grade boundary should slide down towards 70%, but (let's be realistic), SEAB will resist lowering the A grade boundary to anything less than 70% (or the Universities will complain to MOE of grade inflation making it harder to select students for elite Uni courses).

      Thing is, this so-called grade inflation over the years is not because the papers are getting easier, it's rather that more Singaporeans are studying more desperately to do better for the sake of their futures, but that's harsh cruel reality of physical incarnation for you : like what elite PAP ministers say : Singapore can't have all doctors and lawyers and graduates, the economy needs lower non-graduate qualifications for lower pay jobs as well.

      C'est la vie.

    • Originally posted by MapPwner:

      Dang so for the p3 q3 if i wrote 'temperature more than 55 degrees' is 0?


      Cambridge probably will accept lah, no worries.

      At least for this year (1st year of new syllabus), they'll probably accept. If too many students all write the same answer which is not the Mark Scheme answer, especially if Cambridge recognizes it's a new syllabus teething problem caused by an entire country of teachers still teaching from the old syllabus, the Cambridge Markers will submit this issue to the Cambridge Chief Marker for him/her to make a decision on the matter, and the Cambridge Chief Marker will usually make a fair decision, so as not to unfairly penalize an entire country of students.

    • Suntory transparent (ie. chemistry terminology : 'clear and colorless', not 'transparent' ffs) milk tea will take S’pore by storm :



    • S2F10 (g) < -------- > SF4 (g) + SF6 (g)

      An unknown amount of S2F10 was placed in a 2.0 dm3 flask and heated to 100 oC. The equilibrium [S2F10] was found to be 0.5 M. More S2F10 was then added and the new equilibrium [S2F10] was 2.5 M.

      What is the amount of S2F10 reacted (in terms of Kc), from the 1st equilibrium to the 2nd equilibrium?

      A : (0.5Kc)^0.5 − (2.5Kc)^0.5
      B : (2.5Kc)^0.5 − (0.5Kc)^0.5
      C : (2Kc)^0.5 − (10Kc)^0.5
      D : (10Kc)^0.5 − (2Kc)^0.5




      BedokFunland JC Solution :

      Based on the Kc formula, find molarity of either product (since molar ratio of products are 1:1, numerator becomes square of the unknown molarity of either product) at 1st equilibrium = (0.5Kc)^0.5, and at 2nd equilibrium = (2.5Kc)^0.5.

      Convert molarity to moles, you get (2Kc)^0.5 and (10Kc)^0.5 respectively.

      Hence amount of reactant used up = amount of (either) product generated from 1st to 2nd equilibrium = (10Kc)^0.5 - (2Kc)^0.5

      Edited by UltimaOnline 28 Nov `17, 12:13PM
    • Originally posted by delurach:

      Hello!! Please help me with this question. Thank you! It's from PJC P1 Q1

      1 Equal volumes of 1 mol dm−3 hydrogen sulfide and sulfur dioxide (each containing a

      different isotope of sulfur) are mixed to precipitate sulfur according to the equations

      shown below:

      H232S(aq) → 32S(s) + 2H+(aq) + 2e−

      34SO2(aq) + 4H+(aq) + 4e− → 34S(s) + 2H2O(l)

      What is the relative atomic mass of the sulfur precipitated?

      A 32.1

      B 32.7

      C 33.0

      D 33.3

      Based on the stoichiometry of the balanced redox, and considering the limiting reactant, the relative amount of products are 2 mol of 32g isotope for every 1 mol of 34g isotope, hence average molar mass of S(s) = (32+32+34) / 3 is your answer.

      Edited by UltimaOnline 29 Nov `17, 3:40AM
    • Ok, gone through all the trickiest / toughest MCQs of P1 liao. Thanks again for the uploads, MapPwner.



      My personal comments (this isn't reddit so can't be downvoted for speaking bluntly) : while this year's P1 MCQs may be considered at least as tricky / tough as previous years A levels (with possibly 2 MCQs being even trickier / tougher than previous years standards generally), but overall, this year's P1 is actually slightly easier than previous years, because of time allocation : 30 MCQs versus 40 MCQs for the same 1hr duration.

      This new syllabus arrangement (a result of repeated feedback complains to Cambridge) allows almost all students to be able to quite comfortably complete all 30 MCQs of P1 this year without needing to rush, while in all previous years (ie. old syllabus), with an additional 10 MCQs (ie. 40 MCQs) in the same 1hr duration, many / most students struggle to complete the P1, with many / most students rushing through the last 10 MCQs, and a significant % of students actually not having enough time to even attempt the last 5+ MCQs (which I'm sure is no longer a problem with the new syllabus).

      This factor of consideration, goes unnoticed for almost all new syllabus students, because when you guys practice the TYS old syllabus P1s, it's either without enforcing exam conditions of 1hr time duration, and/or because you guys simply skipped the out-of-syllabus MCQs, thus making it seem easier to complete the entire Paper 1s within 1hr.

      Nonetheless, notwithstanding, nice to see a general increase in quality and difficulty of questions across all 4 Papers starting from this year new syllabus. But as you already said it like it is (and kena downvoted for it), the overall score A grade boundary will definitely remain at (more likely slightly above) 70%. As the poll shows, majority of JC students scored 80+% for yesterday's P1 (in large part because now with the new syllabus you guys can actually complete the entire P1 without rushing).


      Edited by UltimaOnline 30 Nov `17, 9:20AM
    • A BedokFunland JC H2 Chemistry Qn

      How would the rate of a reaction that is -(1/2) order with respect to a particular reactant, change if you
      (i) increase the molarity of that reactant by 1/2?
      (ii) decrease the molarity of that reactant by 1/2?

      Edited by UltimaOnline 23 Dec `17, 6:44PM
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