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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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    13,909 posts since May '05
    • Originally posted by theophilus:

      Yo ultima, did u sit for the pract ?? 


      I can neither confirm nor deny sitting for the pract.


      Originally posted by theophilus:

      LOL wattt ?? haha its was freaking hard sia jknfsdjkfn. But just asking if priv candidates technically take a completely different paper 5 from paper 4, it cant make sense to bell curve together ehhhhh .... i know abit futile to discuss but i figured with experience you might have noted some inidcators to back my hypothesis ??? otherwise... ALL THE BEST TO ME 


      Hard meh? Well I wouldn't know that either way (whether I took it or not). Lol.

      School candidates take SPA and not Pract paper, so it's meaningless to ask whether school candidates take the 'same' Pract paper or not.

      For 2017 syllabus onwards though, when there's no more SPA, school candidates and private candidates will take the same Pract paper.


      Originally posted by theophilus:

      LOLOL

      But essentially they are 2 different papers altogether ah and im not even gonna mention the help availed. So only have 3 conclusion

      1. They just hantam everything into one bell curve and ignore the difference 

      2. paper 5 candidates and paper 4 candidates individual bell curve

      3. Or somehow they bell curve only the theory papers together and separate paper 5 and 4 for inidividual bell curves and add them back separately


      Yeah, but it doesn't really matter, does it? Regardless of which case is true (a mere technicality), the only thing that's in your power, is you do the best you can. The rest is out of your hands.

      Though as an ex-MOE teacher, I can tell you there isn't a separate bell-curve for private candidates. The entire Singapore cohort is graded together. Cambridge sends MOE-SEAB the raw score, and MOE-SEAB assigns the grades based on the bell-curve of the Singapore cohort.


      Originally posted by theophilus:

      haixxx i guess you're right, just that i hear from afew people here and there la. wanted some assurance. Thanks though !! haha good luck for As which I may or may not be sure if you are taking LOL


      image

      Edited by UltimaOnline 25 Oct `16, 10:40PM
    • Originally posted by hoay:

      What would be the hybridization in CO, Al2Cl6 and NH4 +? 


      CO : C is sp, O is sp.

      Al2Cl6 : Al is sp3, Cl is sp3.

      NH4+ : N is sp3, H is s.

      That the bond is dative (or not) is a separate matter and hence irrelevant to the orbital hybridization.

      Further note : due to resonance and other factors, in some cases the orbital hybridization may not involve exact integer values, but that's for Uni level ; for A level purposes, choose the most accurate hybridization with integer values, based on the major resonance contributor.


      Originally posted by hoay:

      Thank you for ur answer.

      In CO2 and SO2 is it sp2? any explanation how to conclude which hybridization is there?


      CO2 : C is sp, O is sp2.

      SO2 : S is sp2, O is sp2.

      By observing the electron geometry.

      For instance, the N atom in the amide group is sp2 hybridized, with a trigonal planar electron geometry.

      The lone pair resides in an unhybridized p orbital, which allows for delocalization of the lone pair by resonance (which explains why amides are less nucleophilic and Bronsted-Lowry basic compared to amines and even imines), which is thermodynamically advantageous as it confers resonance stabilization energy by strengthening bonds in the resonance hybrid, resulting in a more exothermic enthalpy of formation, and hence a more negative Gibbs free energy of formation.

      image

      Edited by UltimaOnline 30 Oct `16, 12:39AM
    • Originally posted by hoay:

       

      1. Flask X contains 5 dm3 of helium at 12 kPa pressure and flask Y contains 10 dm3 of neon at 6 kPa

        pressure.
        If the flasks are connected at constant temperature, what is the final pressure?


        Ans. V1 = 5 dm3   ; P1 = 12kPa

                 V2 = 10dm3  ; P2 = 6kPa


        Do u have to use P1V1 = P2V2 or PV=nRT ?

         


      Use PV=nRT twice, to find the moles of gas (in terms of T) in each flask before connecting together.

      After connecting together, add up to find the total moles of gas. Plug the total moles of gas, together with the combined volume, into PV=nRT for the 3rd time, to solve for the final pressure.


      Originally posted by hoay:

      I got the answer 8 dm3 and that is correct.

      Many thank to you UltimaOnline.


      No prob, Hoay :)

      Edited by UltimaOnline 30 Oct `16, 6:23PM
    • Originally posted by gohby:

      Hi UltimaOnline,

      I've 1 question to ask here:

      N14/P2/Q6b(ii)

      In this case, there are 2 chiral centres, which means there are 4 stereoisomers. At the same time, however, it is also possible to for the compound to display geometric isomerism (E/Z) owing to the 3-D position of the Br & methyl substituents across the ring. Why aren't there 8 stereoisomers in all then?

      My thinking is that there can't be 8 stereoisomers because there can only be 4 permutations with which the substituents can be arranged around C1 and C2. However it seems odd that the number of stereoisomers is only 4 given that there are already 2 chiral centres - (and the geometric isomers seemingly don't contribute/coalesce with the optical isomers).


      Thank you! :)


      Hi Gohby,

      This stereochemistry involved in this question goes beyond the H2 syllabus, so notice that Cambridge doesn't specifically ask the candidate how many stereoisomers exist in total, or ask the candidate to draw all of them out.

      The reason for this, is that for cyclic molecules like this, some of the geometric isomers overlap with the optical isomers (eg. a trans isomer may also be an R isomer, with the exact same stereochemistry), and you cannot double count them.

      As an analogy, if you have an apple, a (red) pen, and a (round) pineapple, and you're asked to count how many objects you have, while it is true that apple is red, and the apple is round, that doesn't mean you have 4 objects (ie. you cannot double count apple as "red apple" and also as "round apple"), because you only have 3 objects in total (ie. there is overlap between the "red" and the "round" properties).

      In conclusion, due to overlap between optical and geometrical isomerism for this molecule (which H2 students needn't worry too much about, as it's beyond the syllabus), there are only 4 stereoisomers in total, not 8.

      No prob, Gohby :)

      Edited by UltimaOnline 30 Oct `16, 6:41PM
    • Originally posted by supercat:

      Hi, I have a question from JJC prelims

      http://m.imgur.com/nwWQpfl

      Why is sentence 3 true? Higher solubility means SrF2 is more likely to dissolve? Since Ksp of SrCO3 > SrF2, shouldn't statement 3 be false?? 


      Tis a common trick question.

      You have to use molar solubility to compare solubilities. You can only directly use solubility products as a shortcut to compare solubilities, provided the stoichiometry of the ions in the ionic compound are the same.

      Since SrCO3 (ie. 1 cation : 1 anion) has a different stoichiometry of ions, compared to SrF2 (ie. 1 cation : 2 anions), you cannot simply use solubility products as a shortcut to compare solubilities, but must instead separately calculate the molar solubilities of each ionic compound, to compare solubilities.

      Edited by UltimaOnline 30 Oct `16, 6:59PM
    • Originally posted by p0tat0:

      Hello, may i ask a question? It's from A level 2007 Paper 3 Question 3 d)i) Suggest the strucutres of the three carbonyl compounds, and the ratio in which they might be produced.

      I've gotten methanal and propanone but I don't know why ethanal is produced too. And how do we deduce the ratio?

      Thanks!


      In the original experiment, with calcium ethanoate (4 C atoms) reactant, the only products were propanone (3 C atoms) and carbonate (1 C atom). This gives you a clue that it's about balancing C atoms, ie. 1 C atom used to form the carbonate product, the remaining available C atoms used to form carbonyl compounds.

      In the new experiment, the reactants are 1:1 molar ratio of calcium methanoate (2 C atoms) and calcium ethanoate (4 C atoms). Since Cambridge stated that 3 different carbonyl compounds are produced, you must be able to deduce that the mechanism involves permutations from the available reactants (ie. 1 R group of the carbonyl may be from 1 reactant, while the other side R group of the carbonyl may be from the other reactant).

      Since both are equally available in a 1:1 ratio, in addition to methanal (from calcium methanoate alone) and propanone (from calcium ethanonate alone), it is also possible to obtain ethanal (from both calcium methanoate and calcium ethanonate, providing H and CH3 R groups respectively).

      As to the mathematical ratio, it's simple permutational probability. There's only 1 chance to obtain methanal (ie. using methanonate x methanoate) and also only 1 chance to obtain propanone (ie. using propanonate x propanoate), but there are 2 chances to obtain ethanal (ie. using methanonate x propanonate and propanonate x methanoate).

      Since both reactants are equally available in a 1:1 molar ratio, it's like tossing a coin twice, and the probability are as follows :

      Heads x Heads = 0.5 x 0.5 = 0.25
      Heads x Tails = 0.5 x 0.5 = 0.25
      Tails x Heads = 0.5 x 0.5 = 0.25
      Tails x Tails = 0.5 x 0.5 = 0.25

      In other words, the ratio of Heads x Heads : Heads x Tails or Tails x Heads : Tails x Tails, is 0.25 : 0.25 + 0.25 : 0.25, ie. 1:2:1

      Which is why methanal : ethanal : propanone is produced in the ratio of 1:2:1.

    • Originally posted by theophilus:

      Sgforums should make an app, i keep forgetting to check back. Thanks for reply, just on the side, but Pb has large ionic radius and oxidation number max +4 which then causes the magnitude of charge density to be small right? also it doesnt have any d orbitals vacant :O so how you extrapolate that it can form complexes ? 

      and so the answer for the TYS question is cus Pb forms a new complex and the conc of Pb2+ falls , then by lcp equi shifts to form more ions ?


      For Pb, the 5d orbitals may be filled, but the 6d orbitals are vacant and energetically accessible.

      The charge density of many metal ions lower in the periodic table are higher than you might think, vis-à-vis their high proton number, due to the d-block contraction effect and other factors (Uni level Chemistry).

      In addition, quite independent from charge density, different ligands have different affinities with different metal ions, ie. hard vs soft Lewis acids and bases (Uni level Chemistry).

      Correct, the shifting of equilibrium position (including any ligand substitution) is as predicted by Le Chatelier's principle (you're not allowed to use acronyms for A level exams).

    • Originally posted by p0tat0:

      Thank you so much!! :) Your explanation is really detailed, i finally undestand it after so long TuT 

      May i ask another question which is from A level 2008 paper 3? Question 2 e)iii) use these data to calculate the H:C ratio in alkene D, and hence suggest its molecular formula. 

      I've gotten the empirical formula which is C2H5 but i don't know how to get the molecular formula through calculation. 

      And another question from the same paper question 4b. Write equations, including state symbols for these two reactions and explain the observation. 

      The answer key used Al2Cl6 for its equation but can we use AlCl3 instead? 


      No prob :)

      The molecular formula is almost always 2 x empirical formula. Check using degree of unsaturation (you can google it out yourself and learn it if your school teacher or private tutor didn't teach you).

      The molar mass is specified in the question, hence your answer must be in concordance. In other questions, if molar mass or some other qualifying data isn't specified, then either AlCl3 or Al2Cl6 may be accepted (although you should take into consideration that different temperatures and pressures favor the monomer vs the dimer, you can google this out yourself for more info).


      Originally posted by p0tat0:

      ohh... so for alkane the degree of unsaturation is zero but C:H ratio (2:5) will give a magntiude of 1/2 using the formula for degree of unsaturation. Is that why i need to multiple by 2 to get C4H10 which will give me magntiude of zero?

      Thanks a lot! :)  


      Yes correct, degree of unsaturation should be zero for a fully saturated molecule (eg. alkane). Use degree of unsaturation to check your molecular formula.

      Edited by UltimaOnline 04 Nov `16, 1:02AM
    • Originally posted by supercat:

      Hi, I have a question from VJC Prelim 2016 P3 Q2bii,

       

      Why do we have to take 75 / 2 ?

      Is it because there's an equal chance of the molecule getting attacked, so a racemic mixture is formed??

       


      Yes, correct. This is concordant with the kinetics data that point to SN1 instead of SN2.

    • Originally posted by Ephemeral:

      Hi Ultima, I have some Qs from past yr papers and prelim papers :)

      2011 P2 Q4(c)(iii) What is 'inertness'? The suggested answers I have says that C does not have low-lying vacant d-orbitals to accept loe pair of electron from water. But I answered that energy released from weak VDW forces of attraction formed between ccl4 and h20 is insufficient to compensate energy required to break strong H-bonds between h20 molecules hence ccl4 is insoluble in h20 and hence inert.

      2011 P2 Q4(e) Since the electrode cell potential for oxidation of (VO)2+ to (VO2)+ and (VO3)- is the same, can we say it is oxidised to (VO3)- instead of (VO2)+? Also, when do we use (VO3)-? 

      2012 P2 Q2(c)(iii) My answer was that from reaction 1, position of equilibrium shifts left at higher temperature as high temperature favours backwards endothermic reaction hence SO2 is more stable at high temperature when heated hence SO2 is formed. The suggested answer says that formation of SO3 is a slow process with higher Ea and is hence not formed. Is my answer wrong?

      2012 P3 Q5(a) can the synthesis pathway be:

      1. HCN in trace amount of NaCn, cold at 10-20 degree celsius. 2. KCN in ethanol, heat under reflux. 3. HCl(aq), heat under reflux

      2014 P2 Q5(d) Will there be formation of white ppt or is phenylamine originally white?

      2015 P2 Q2(c) Is it enough to mention H-bonding and permanent dipole between methanol and only weak idid between methane or do I have to compare idid between both methanol and methane as well? 

      NYJC 2016 Prelim P2

      Q2(a) In step 2, Al(OH)3 is produced but in (b)(i), it was mentioned that excess NH3 is used. Shouldn't soluble complex Al(OH)4- be formed when excess NH3 is used?

      Q2(b)(iii) Why is CO2 produced? 

      NYJC 2016 Prelim P3 

      Q2(a) Should the bond angle be 109.5 degree of slightly less than that due to electronegative O atoms around S? For such Qs, sometimes answer keys take into consideration the electronegativity of atoms in the molecule but sometimes they don't. Do we always consider during the exam?

      Thanks :)


      Nope, your answer irrelevant and not acceptable. Inertness refers to chemically inactive. In this context, inert = resistant to hydrolysis (a chemical reaction). You're only talking about solubility, which doesn't involve a chemical reaction.

      At acidic pH, dioxovanadium(V) ion dominates. At alkaline pH, vanadate(V) ion dominates.

      Both are correct. When you're not sure which Cambridge wants, be exam-smart and write both. You'll only be penalized when one of your points directly contradict a required point, no penalty if the point is simply irrelevant to the mark scheme.

      Yes.

      It's not about the color, it's about solubility.

      The reason why you're confused, is becoz your school (ie. Singapore JCs), erroneously and artificially separate the different types of van der Waals.

      The 1st reason and more important reason is, as you've mentioned, hydrogen bonding exists between methanol molecules only (and not methane molecules).

      The 2nd reason is, the van der Waals forces between methanol molecules, are stronger than the van der Waals forces between methane molecules, for 2 reasons (yes, 2 sub-reasons within the 2nd reason) : the presence of a permanent dipole for the polar methanol molecule involving an electronegative O atom, as well as the greater number of electrons present in methanol from the additional O atom.

      But no worries, Cambridge is lenient when it comes to the (somewhat) erroneous way Singapore JCs teach this, so your answer should be fine.

      No, the pH from hydrolysis of NH3(aq) isn't sufficiently alkaline to generate the Al(OH)4 - coordination complex.

      It isn't. H2 is produced, not CO2.

      Again, you've been erroneously brainwashed by your school (ie. Singapore JCs), which teach wrongly for this subtopic.

      The correct reasoning for any observed deviation (eg. in H2S) from the basic geometries as predicted by period 2 VSEPR (eg. in H2O), is moreso because S is in period 3 (more diffused orbitals), hence reducing the degree (pun intended) of hybridization, compared to period 2 VSEPR geometries.

      Only when you're comparing across the same period (ie. period 2 vs period 2), then you look at other factors such as electronegativity. Note that electronegativity features primarily as the reason for angular deviation from VSEPR geometry, specifically when electronegative F atoms are present.

      In this case of H2SO4, the deviation is primarily due to greater electron repulsion between doubly bonded S=O groups and the greater magnitude of partial negative charge on these O atoms in the resonance hybrid, compared to moderate electron repulsion between the OH groups (due to steric strain), compared to least electron repulsion between the OH and S=O groups (favorable electrostatic interaction between partial positive H atom of OH groups and partial negative O atom of S=O groups).

      Be that as it may, for A level purposes, be BedokFunland JC exam-smart, and always write the basic H2 syllabus basic VSEPR geometry first, with qualification (ie. "assuming identical geometry as a period 2 central atom"), before writing the additional Uni level considerations stated above (if you suspect Cambridge is asking you for it, eg. "The angles actually deviate from the expected geometry, suggest reasons why", or if you've the time and you're bored and just feel like showing off your elite Uni level understanding, then go ahead and add these points in your A level paper answer script like I do, just for fun).


      You're welcome, Ephemeral (and all students), but in future pls post 1 qn at a time, and wait for my reply, before posting the next qn, instead of posting 10 qns at 1 shot (which may be more convenient for u, but more troublesome for me to ans so many qns at 1 shot).

      Edited by UltimaOnline 05 Nov `16, 2:29AM
    • Originally posted by Ien:

      Al2O3 has lower melting point than MgO due to partial covalent character. 

      Al2O3 does not react with water due to strong ionic lattice. 

      But doesn't induced covalent character means weaker ionic lattice?


      Having significant covalent character doesn't necessarily 'weaken' the lattice structure for all purposes in the same way (eg. melting vs hydrolysis).

      From the different melting points, you are expected to deduce the Ea required to melt MgO is higher than to melt Al2O3. This is due to the greater degree of covalency in Al2O3. Remember that compared to ionic lattice structures, covalent simple molecules have stronger intramolecular covalent bonds, but weaker intermolecular van der Waals forces. This explains why the greater the degree of covalency, the more unequal the distribution of bond strengths within the lattice structure, hence the lower the melting point.

      From the different solubilities, you are expected to deduce the Ea required to hydrolyze Al2O3 is higher than the Ea required to hydrolyze MgO. This is due to the lone pairs on the O atoms in Al2O3 being 'locked up' in the form of dative covalent bonds, and less available for hydrolysis.

    • Originally posted by Ien:

      Why isit that for di-substituted alkenes, they are energetically more stable than mono substituted alkene?

      ie cis-but-2-ene and trans-but-2-ene more energetically stable than but-1-ene.


      More substituted alkenes are more thermodynamically stable than less substituted alkenes, due to the greater no. of stronger C-C sigma bonds with greater % s orbital character, resulting in a more exothermic formation enthalpy.

    • Originally posted by Ien:

      For reaction of ROH and H-X gas, why is this reaction limited to only H-Cl and H-Br? What about H-I?

      and why the guidebook says "alkenes may also be formed due to competition from elimination reactions'? 

      please help me Ultima!! ): 


      The same method can be used to convert alcohols to alkyl iodides, except that non-oxidizing H3PO4 is used instead of oxidizing H2SO4 (which would oxidize HI to I2, thereby sabotaging the halogenation nucleophilic substitution mechanism).

      Upon protonation of the OH group (which is the Bronsted-Lowry acidic role of HX), E1 vs E2 vs SN1 vs SN2 all become possible (the 1st order reactions favoring tertiary alcohols, the 2nd order reactions favoring primary alcohols). Hence, "alkenes may also be formed due to competition from elimination reactions".

    • Originally posted by Ien:

      Hi Ultima, why is benzene ring a weak electron donating group? I thought it always 'withdraw' electrons since lone pair delocalise into benzene? 


      Depending on the group directly bonded to the benzene ring, the benzene ring may be any of the 4 :
      electron donating by induction
      electron donating by resonance
      electron withdrawing by induction
      electron withdrawing by resonance.

      While Singapore JCs do not teach their H2 students to be this explicit, but BedokFunland JC students and all A grade students are encouraged to specify the modality of electron donation or withdrawal.

    • Originally posted by Ien:

      When Cu2+ and NH3(aq) are mixed, will it form Cu(NH3)4(H20)2 ligand or Cu(OH)2 or both? 

      and how will we know its precipation of Cu(OH)2 and not form complex of Cu(OH)6?

      and Cu(OH)2 consist of Cu in =2 oxidation state, but I thought only Cu+ can form solid? 


      For A level exam questions on adding NH3(aq) to Cu2+(aq), requiring you to explain first the formation of the blue ppt, following by dissolving of the blue ppt to form a deep blue solution, this is exactly what Cambridge requires you to write (see bottom of the linked webpage) :

      http://www.chemguide.co.uk/inorganic/complexions/aquanh3.html

      OH- ligands are negatively charged, as such few coordination complexes will have 6 OH- ligands (due to destabilizing inter-anionic repulsions), with the notable exception of Cr(OH)6 3-.

      It's true that Cu+ compounds are mostly insoluble, unless complexed with negative ligands, eg. [CuCl2]-, but Cu2+ in the form of Cu(OH)2 is obviously a solid ppt, as are all d block hydroxides (only Group I / 1 hydroxides are soluble, and solubility of Group II / 2 hydroxides increase down the group), such as Fe(OH)2(s).

      Finally, at A levels you only scratch the surface of transition metals coordination complex chemistry, which is more complex (weak pun I know) than can be handled at A levels. Why does Cu2+ form Cu(NH3)4 instead of Cu(NH3)6, and why is Cu(NH3)4 square planar instead of tetrahedral? These are due to Uni level concepts including the Jahn-Teller effect, which will not be asked at A levels.

      For A level purposes, you are expected to memorize a list of common transition metal coordination complexes (including their geometries and colours) that Cambridge can set questions on. Even if your school didn't give you such a list, you can look it up on the internet, or in Chan Kim Seng's and George Chong's H2 Chem books.

    • Originally posted by gohby:

      Hi UltimaOnline,


      I have the following questions to seek your clarification:

       

      1. N11/P3/5d(i)

       

      For stage I, should I include “followed by NaOH (aq)” in my answer? If I don’t include it, I wouldn’t have been able to form J because the NH2 substituent would have been protonated. However, if I were to include it, I still wouldn’t have been able to form J because the NaOH would have reacted with the phenol to form the phenoxide.

       

      2. N13/P3/5d

       

      Would a graph paper be given to candidates to so that an accurate initial rate of reaction could be obtained?

       

      3. N15/P2/4b(i)

       

      How do I determine the location of the partial charges at the transition state? I would think the δ+ would not reside on the N given its electronegativity. Besides, by referring to the SN2 of halogenoalkanes, there ar 2 δ- sites in the transition state - so how do I know if the partial charges are going to be both δ- or not?

       

      Many thanks, UltimaOnline!

       


      Hi Gohby,

      1. N11/P3/5d(i). Here, "limited NaOH(aq)" should be specified to get J (ie. just enough to deprotonate the aryl NH3+, but not enough to deprotonate the aryl OH), followed by "excess NaOH(aq)" to get K.

      2. N13/P3/5d. Yes, graph paper is provided.

      3. N15/P2/4b(i). It's not about electronegativity, but about formal charges. Take the average of the formal charges on that atom, in the reactant (ie. just before the transition state), and in the product (ie. immediately following the transition state).

      No prob, Gohby!

    • Originally posted by Ien:

      Thank you! 

      I have another question: why is the standard enthalpy change of hydration linked to ion-dipole interations? 


      Because that's the definition itself. For ionic compounds, enthalpy of hydration refers to the heat change when forming ion-permanent dipole interactions between gaseous ions and water solvent, ie. enthalpy change when state symbol changes from gaseous ions to aqueous ions.

    • Originally posted by Ien:

      Hi Ultima, 

      Is the concept of orbital overlap same as extent of distortion of electron cloud? 

      But why is it that i cannot seem to use electron cloud to answer the question "why the melting point decreases from carbon to germanium". The TYS answer is about orbital overlap. 


      "Electron cloud" is too vague, and is more relevant for intermolecular van der Waals interactions, which is relevant for simple molecules, not giant covalent structures like C, Si and Ge.

      Edit : you mentioned "distortion of electron cloud", that phrase is relevant specifically for trends of thermal stability (ie. temperature required to thermally decompose) for Group II compounds (eg. nitrates, carbonates, etc). Don't confuse this concept with trend of melting point and covalent bond strength down Group IV or 4, it's a totally unrelated matter. You need to revisit your CS Toh notes on Group II or 2 regarding "distortion of electron cloud".

      You *must* use the key phrase "effectiveness of orbital overlap" when answering such questions about bond strength.

      However, when explaining "why the melting point decreases from carbon to germanium", you first need to specify that C, Si and Ge are all giant covalent lattice structures, and melting each of them involves breaking of the multitude of covalent bonds in this giant covalent lattice structure.

      Then proceed to explain that going down Group IV or 14, atomic radii increases, hence bond length increases, hence electrostatic attraction between the positively charged nucleus and the bond pair of electrons, decreases. In addition, the valence hybridized orbitals used to form the sigma bonds between the atoms in the giant covalent lattice structures, become increasingly diffused going down the Group, hence effectiveness of head-on or end-on overlap to form the sigma bonds, decreases. Hence bond strength decreases, from C to Si to Ge. Hence melting point decreases.

      Edited by UltimaOnline 08 Nov `16, 1:20AM
    • Originally posted by iSean:

      Hi UltimaOnline, 

      Thanks for your prev. help in my CIE A-Level questions. 

      Anyway would you mind helping me checking my suggested answers for my friends doing STPM (Malaysian 6th Form/A-Level Equivalent)

      These is their essay question

      I'll link my suggested answer in a PDF in another link.

       

      Suggested Solutions for

      Question 18.

      https://drive.google.com/file/d/0B3KZVrQaOQqbbWFIYVhmR2VkZEk/view?usp=sharing

      Question 19 - Still Compiling
      Question 20 - Still Compiling

      Regarding my suggested solution for 18(c) 

      Some of my friends argue that it should be only 1 major product, due to the lack of "s" in the question. 

      And I don't see how they can distribute 3 marks. 

      Mind giving your opinions. 

      Because some of them argue should use stability of alkyl free radical to explain. 

      Hope you can share some light.


      Sean, are you a student or a teacher? Your solutions machiam professionally prepared mark scheme! Even use software to draw out all structures somemore! Sean boleh!

      Q18(c) is a terribly phrased question. Just identifying the major product (for exactly which reaction isn't specified by the question either) is only worth 1 mark. Since 3 marks are given, the student is expected to give reasoning and justification.

      The major product for the halogenation free radical reaction is C6H5CHBrCH3, because it involves the significantly more stable benzyl radical (whose free radical character is delocalized by resonance over 4 atoms : the benzylic C atom, and the ortho, para, ortho C atoms of the benzene ring), hence lower Ea, hence faster rate of reaction, leading to the major product.

      As for alkaline hydrolysis of both products, SN2 predominates for the primary alkyl bromide (but since the reactant is achiral, so is the product), while SN1 predominates for C6H5CHBrCH3, due to both steric hindrance as well as the resonance stabilization energy of the benzyl carbocation. Since the halogenation via free radical mechanism generates a racemic mixture, so the final product for this pathway (involving mostly SN1 with a little bit of SN2, depending on molarity of OH-, solvent and temperature) will also be racemic.

      Edited by UltimaOnline 10 Nov `16, 8:18PM
    • Originally posted by iSean:

      I am a student haha,I'd finished all my CIE A Level papers yesterday, so pretty much helping out my sixth form friends for STPM Chemistry by trying to give out answers for today's paper. 

       

      Because the STPM paper is normally a hardest paper around compared to Australian WACE/SACE and UK CIE/EdExcel A-Levels, I was testing my limits on doing their papers, but failed badly to finish it under 30 minutes

      Should be equivalent to H2's Paper 3 Long Questions.

      -------------------------------------------

      Back onto topic, I kind of inspired by the way JC prepare their marking scheme so easily read. So this is why it looks so tidy. 

       

      The Diagrams are mostly photoshop. haha.

       

      --------------------------------------------

       

      Yes, I do agree with your statement, it is badly phrased, and they claimed that the papers was sent to Cambridge to be evaluated, and refined. 

       

      Anyways, as a tutor, what will you recomend students to write about the major product as the final answer tho? 

       

      The Answer based on P and Q (Free Radical Mechanism)

      Or R and S  (Alkaline Hydrolysis)

       

      Anyway, UltimateOnline, do you mind doing Question 19 for me? 

      Because this question is quite challenging to my opinion. 

      And I think there's more than one answer for X. 

       

      Oh yeah, feel free to adapt the question to your tuition group, and ask your students to try finishing it in 30 minutes for 2 questions. I would like to hear the feedback from Singaporean Students on our Paper, besides it being badly phrased all the time shade.png"
      ---------------------------

      Q19 https://drive.google.com/open?id=0B3KZVrQaOQqbckowWG1wempZQUE
      Q20 https://drive.google.com/open?id=0B3KZVrQaOQqbcGtObG9wS091SVE

      Question 19 should have many problems elucidating the structure. 

      Question 20 should be fine.

       

       


      Yo Sean, impressive, professionally done mark scheme for a student... until the last few pages for Q19. Lol!

      For Q18, I'd advise my BedokFunland JC exam-smart students to write out all possible answers with qualifications. In other words, since the question is ambiguous, write "If this, then this. If that, then that." to cover all bases. Which for this question, is unfortunately quite a handful, since there are several possible reactions involving several different reactants and products. No choice, it's the STPM question setter's fault.

      Q19 isn't tough at all, just that there will be several possible alternative answers (including yours), which might confuse the student into erroneously thinking there's only 1 correct answer and there's some trick to the question (but there isn't, just another lousily phrased STPM question).

      In addition, your E2 mechanism is a bit wrong, but the question didn't ask for the mechanism anyway.

      Q20. This one more jialat, got more errors.

      Since the question states "a bright coloured product is formed" implying that azo coupling has occurred, you're wrong to say in your mark scheme "Allow : Bonding between aromatic rings".

      In addition, "either 2 OR 3 Position on Z" is also wrong, because of both electronics (OH is a stronger activator and ortho-para director than the tert-butyl group, because OH donates electrons by resonance, while tert-butyl group donates electrons by induction) and sterics (tert-butyl group poses significant steric hindrance).

      Finally, and this last error is either your fault (you typed out the question wrongly) or the STPM question setter's fault. If the question really specified "optical isomer", then there isn't any simple chemical test, but rather a simple physical test : just use a polarimeter. But there isn't a chiral C atom in X, so either the question was wrong to say "optical isomer" or you typed out this part wrongly.

      Still, it's nice of you to put in so much effort to help out your sixth form friends for STPM Chemistry.

      Edited by UltimaOnline 10 Nov `16, 11:25PM
    • Originally posted by iSean:

      Sorry. Got tired during Question 19(c) will try to fix it another day. confused.png My digital stylus ran out of battery I need to do draw on my touch screen. 

      Thanks for the notation on where's the error at fault. :) 

      It seems I still have a long way to go. 

      Okay will look onto the  E2 mechanism, I kind of gave up during halfway Question 19. :S 

      But can you further elobarate on it tho for Q20? 
      I don't technically get it, because Resonance and Induction wasn't really taught by my lecturers.

      So The Bonding location is wrong in the figure? Where should it be?


      Simplified :

      Since azo coupling occurred, hence benzene ring to benzene ring bond not acceptable (you wrote in your mark scheme to accept this as alternative answer).

      During electrophilic aromatic substitution, the electrophile is attacked by benzene ring on ortho position relative to OH group, not alkyl group (ie. for directing effects, follow the stronger activator, ignore the weaker activator).

      You typed out the question wrongly ie. the question didn't specify OPTICAL isomer, since there isn't a chiral C atom in X.

      Edited by UltimaOnline 10 Nov `16, 11:22PM
    • Originally posted by iSean:

       

      Ah I see your point, apparently STPM made a huge mistake lol.


      This is why Singapore pays Cambridge a lot of money to set the Singapore A level papers.

      image

      Edited by UltimaOnline 10 Nov `16, 11:44PM
    • Originally posted by iSean:

      Before, I quit STPM, and joined CIE A-levels, a STPM Board representative came during an orientation week and told us, their papers was sent to Cambridge Assesement (Cambridge/UCLES) to be checked and qualified for suitability before aproval to mass print. 

      Apparently someone is either lying or not doing their job. haha.
      Quite fortunate my parents let me go for CIE haha.


      Woot! Is that a double bond H (ie. =H) in the STPM question paper???

      image

      Maybe Jho Low ran away with the fees meant to pay Cambridge.

      If he's not careful, sekali he may kena blown up.

      Edited by UltimaOnline 11 Nov `16, 12:01AM
    • Originally posted by hoay:

      If we calculate gibbs energy change from the cycle pattern according to hess's law and then by the relation G = H -TS, will it give the same value ?? 


      If you intend to apply Hess' Law to Gibbs free energy cycle, note that all steps in the cycle must all be in terms of Gibbs free energy (and not some steps inolving enthalpy, some entropy and some Gibbs free energy).

      Provided it's for the *same* reaction under the exact *same* conditions (ie. temperature, pressure, solvent, etc), then you should get the same value, whether you calculate the Gibbs free energy change for a particular reaction by applying Hess' Law (and given Gibbs free energy change values for several *different* but related equations in a cycle), or by applying delta G = delta H - T x delta S (and given the enthalpy change and entropy change values for the exact *same* reaction as what you're calculating Gibbs free energy change for).

      Edited by UltimaOnline 14 Nov `16, 12:41AM
    • Man completely dissolved (no remains found) in boiling acidic hot spring (machiam heated under reflux with concentrated acids until he totally dissolved).

      http://www.grindtv.com/random/man-boiled-to-death-in-yellowstone-hot-spring-attempting-to-hot-pot-video/

      Edited by UltimaOnline 13 Oct `17, 2:18AM
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