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

last SG · 2010-07-26T23:40:35+00:00

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UltimaOnline SG

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.

UltimaOnline SG

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.

UltimaOnline SG

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.

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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.

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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 "
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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.

UltimaOnline SG

Sorry. Got tired during Question 19(c) will try to fix it another day. 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.

UltimaOnline SG

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.

UltimaOnline SG

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???

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

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

UltimaOnline SG

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).

UltimaOnline SG

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/

UltimaOnline SG

What is the composition of a buffer solution of pH 7.4 made by mixing 0.1 M NaH2PO4 and 0.1 M Na2HPO4. [Ka for NaH2PO4 = 6.22x10-8 M).

How to approach this question because evrything is given which can be pleaced in the Handerson-Hassalbach equation. But what we will get....?? Help please...

From the Henderson–Hasselbalch equation, plug in the pH and the pKa value, to solve for the mole ratio of conjugate base (HPO4 2-) to acid (H2PO4 -).

Since the molarities for both are the same, the ratio of moles = ratio of volumes, and that's the answer ie. 'composition' of buffer solution.

UltimaOnline SG

Hi UltimaOnline,

Here are some DHS MCQs on which I would like seek your help:

DHS14/P1/12 [Ans: D]

Remarks: I do not understand the titration curve. My understanding is as follows - the first unlabelled equivalence point between A and B signifies the end of neutralisation between maleic acid and sodium hydroxide, and at the point, only -O2CHCHCO2- is present. However, why is the second equivalence point is at C, (when an additional 10cm³ of maleic acid is added from the first equivalence point), instead of an addition 30cm³ of maleic acid (given that there is 30cm³ of -O2CHCHCO2- at first equivalence pt)? Can I confirm that at D, the only compound present is HO2CHCHCO2-? So where is “mixture” to resist the pH change? If there were an option further down the graph from D, I would have picked that, but D doesn’t seem correct to me.

DHS14/P1/20

Remarks: I understand that elimination is out because it wouldn’t yield a product which rotates plane-polarised light. I also understand that both SN1 (since an optically pure enantiomer of 2-bromobutane was used since an optically pure enantiomer of 2-bromobutane was used) and SN2 mechanisms will yield an optically active product. Am I supposed to draw any inference from the reduction of the angle of rotation or is that just a red herring?

DHS14/P1/25

Remarks: I can understand that C is not the answer because it cannot be deduced from the information above. However, how do I know whether C-16O bond is stronger than C-18O bond?

DHS14/P1/28 [Ans: C]

Remarks: I’m rather confused here. Do the arrows in the choices signify a successive addition of the compounds without any discarding of the aqueous layer? Why do I add water in the first and last steps and how do I go about ascertaining the sequence of the addition?

Once again, thank you very much for your help! :)

Hi Gohby,

DHS14/P1/12. Since you're adding a weak acid to a strong acid, hence maximum buffer capacity is double equivalence point instead of half equivalence point. Being a diprotic acid, there are of course, 2 maximum buffer capacities. The 1st being equal molarities of HA- and A2- (which occurs at [10+20]/2=15 cm3), and the 2nd being equal molarities of H2A and HA- (which occurs at 40 cm3). Clearly, the more effective buffer occurs at equal molarities of H2A and HA- (ie. weak acid and amphiprotic species) rather than equal molarities of HA- and A2- (ie. amphiprotic and strong base), as illustrated by the larger horizontal region of the titration curve. Besides, 15cm3 isn't even an option in this MCQ.

DHS14/P1/20. If it were SN1 only, and the reaction was complete (as implied), then the product would be optically inactive. If it were SN2 only, and the reaction was complete (as implied), then the product would rotate plane polarized light by 13.5 degrees in either direction. Hence a mixture of both SN1 and SN2 must have occurred.

DHS14/P1/25. Option D is correct because the reactant is cyclic, hence there is only 1 hydrolysis product, not 2. Addressing your specific query : heavier isotopes form stronger bonds. This is Uni level Chemistry, and for A level purposes, Cambridge can still tempt the students with beyond-syllabus MCQ choices, to test if the student is sufficiently proficient to recognize the within-syllabus correct MCQ answer, and disregard the other options.

DHS14/P1/28. Problem is alcohol, being part non-polar and part protic polar, is miscible with both water and alkyl halide. Addding [water and excess conc HCl] serves to protonate the alcohol, which then migrates into the aqueous layer. Discard the aqueous layer by decantation. Add excess Na2CO3(aq) to remove excess HCl(aq) in the form of CO2(g) and NaCl(aq). Add S2O3 2-(aq) to reduce all the I2(s/aq/alc) to I-(aq), forming S4O6 2-(aq). Add water to remove all ionic impurities, then carry out decantation to leave behind the alkyl halide.

No prob, Ghoby! :)

UltimaOnline SG

Hi, could I ask a question about ionic equilibria?

In JJC 2014 prelim mcq Q10,

Maximum separation occurs when IP = Ksp of Fe(OH)2 right? With the [Fe] being 0.05, I can calculate [OH-] but I keep getting pH of about 7. May I ask where did I go wrong??

A researcher accidentally mixed 50.0 cm3 of 0.100 mol dm–3 MgSO4 solution with 50.0 cm3 of 0.100 mol dm–3 FeSO4 solution in the laboratory. It is suggested that the metal ions can be separated through selective precipitation by adding solid sodium hydroxide. At 298 K, the solubility product of Mg(OH)2 is 1.8x10^–11 mol3 dm–9, and that of Fe(OH)2 is 4.1x10–15 mol3 dm–9. What would be the pH of the resultant solution when maximum separation has occurred?

Your error is assuming maximum separation occurs when Qsp = Ksp of Fe(OH)2, when it should be Mg(OH)2, ie. the more soluble ionic compound. You have to add as much OH- as possible to precipitate all the less soluble Fe(OH)2, without precipitating out any of the Mg(OH)2. Hence at that point, Qsp = Ksp of Mg(OH)2, not Fe(OH)2.

UltimaOnline SG

what is the product when butene reacts with IBr

Major product is 2-bromo-1-iodobutane, minor product is 1-bromo-2-iodobutane.

UltimaOnline SG

Hi, I have a question from MI 2014 prelim P1 Q24

Why are there only 2 optical isomers for R? Is it due to internal plane of symmetry? I counted a total of 4 chiral carbons in R. Did I do additional counting??

There are only 2 chiral C atoms. But due to an internal plane of symmetry, the R,S and S,R stereoisomers are the meso isomer and thus optically inactive, and hence the only 2 enantiomeric optical isomers are S,S and R,R.

UltimaOnline SG

Hello, I have another question regarding Ionisation Energy

For noble gases, as to why they have the highest Ionisation Energy among the elements, is it because they are at the end of the period(where they have increased nuclear charge) or is it because they have a stable electronic configuration hence making extraction of electrons difficult

Thanks :)

The former, of course. The latter is not a valid explanation, merely a desperate attempt at 'smoking'.

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Hello, I have a question regarding melting points of HI vs HCl

I know that melting point of HI is higher, is this due to the greater size of HI/number of electrons, less impact nuclear charge has on charge distribution hence greater polarizibility of electrons , leading to greater magnitude of partial charges on HI compared to HCl?

But the electronegativity difference of HCl is greater than HI, so what effect does this have on the strength of the Van der Waals forces?

The magnitude of electronegativity difference, ie. polarity, has the opposite effect (ie. for the same molecular size and no. of electrons, the more polar the molecule, the stronger the intermolecular van der Waals interactions) but is strongly outweighed by the greater molecular size and no. of electrons, and hence polarizability of electron charge clouds and hence the magnitudes of partial charges and dipoles, hence strength of the van der Waals interactions, for HI over HCl. Consequently and concordantly, all factors considered, HI has a higher melting and boiling point compared to HCl.

You mentioned 'the more polar the molecule, the stronger the intermolecular van der Waals interactions' Which van der Waals interactions are you talking about?( permanent dipole or instantaneous-induced dipole)

And do you mean that you can only consider electronegativity difference if both molecules have same molecular size/electrons?

Yes, of course.

Anytime a polar molecule is involved, all 3 types of van der Waals interactions are relevant : Keesom, Debye, and London Dispersion forces.

Sorry, i have another query regarding the abovementioned issue

Then in the case where electronegativity does play a role ( both molecules are of same size) then what actually happens?(as in does partial charges get stronger or something else comes into play)

Yes, of course. That's why (assuming similar molecular size) the more polar the molecule, the higher the melting & boiling points.

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Hello, would like to ask a question

As to why c=c is stronger than c-c bond, is it because of the additional pi bond, or because the sigma bond in c=c is shorter due to increased S character in the sp2 hybridization causing the bond to be closer to the nucleus?

Both, of course. In the A level exams, for any question, if you dunno which one Cambridge wants, always give both or all reasons.

UltimaOnline SG

kyungsoostan and Sir3,

A grade boundary for this year will most likely be approximately 75% (overall, combining all papers including SPA), depending on the entire cohort's performance.

However, note that your own (ie. all students') estimation of your marks may be inaccurate (eg. you don't actually know your SPA marks, you may have made careless mistakes you didn't realize, Cambridge's required answer may be different from your school's taught answer, etc).

Notwithstanding, assuming your estimation is accurate, then kyungsoostan's 77% is a safe A grade, and Sir3's 73.5% is a possible A grade. As long as you get > 70%, it's always *possible* to get an A grade (just that your chances increase dramatically at 75%), because the cohort's performance also has to be taken into account.

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Guys I'm really trembling in fear... First i screw up paper2 (actually all my friends said it was hard) so i think i score around low B for that paper itself. Than for Paper 3, it was so easy! , but fml yeah i screw it up and my whole organic part was blank and i mixed up my R&C and haix. So my paper 3 was about low B too. yeah although i screw up, low B was quite possible. So based off p2 and p3 i got around 60-64%. Today's paper 2, i scored 27/40. First page all wrong idk what was i even thinking if not couldve 31 sia. but yeah its over. With those 3 papers i think i should be around 65 % AT MOST. Lets say spa affects my grades a little, will i drop to a D? I really do not want a H2 D... Haix. my uni course is AAB sia but i might end up with AAD if my chem really xiao suay.

Prepare for the worst, hope for the best. From your description, your most likely grade is a C, with a small chance of B or D.

Accordingly, be psychologically prepared to choose backup Uni courses with lower UAS requirements, and/or retake A levels as a private candidate (though if you have NS obligations, you've to decide exactly when to retake your A levels).

http://www.nus.edu.sg/oam/gradeprofile/sprogramme-igp.html

http://www3.ntu.edu.sg/oad2/website_files/IGP/NTU_IGP.pdf

This may sound harsh, but it isn't meant to demoralize you (of course you would rather hear soothing assurances like "no worries, you'll probably still get your B grade" but that's mere consolation not honest clarification), but rather to remind you, that whatever happens, make the most of it. Use every opportunity in life to grow, including growing in fortitude and resilience. Doing badly for A levels isn't the end, there are always 2nd chances in life, but only if you're willing to give them to yourself.

UltimaOnline SG

Flaws of the Singapore education system notwithstanding (and there are many), it's no mean feat to come in 1st place out of 57 countries worldwide (including USA, Hong Kong, Taiwan, South Korea, Japan, etc).

http://mothership.sg/2016/11/sporean-students-come-in-1st-out-of-57-countries-in-international-math-and-science-tests/

Questions like this one, seem like a joke, until you consider how sad it is that students from the countries at the bottom of the list have no idea how to answer it.

UltimaOnline SG

Mothership.sg translates for the general public how a trio of data scientists from GovTech solved the mystery of the "Rogue One MRT Train" (actually the Mothership article missed out on several key steps, read the original report written by the heroes themselves here). Daniel Sim, Clarence Ng and Lee Shangqian are 帅哥 Singapore heroes who are role models for a scientific career.

http://mothership.sg/2016/12/unsung-heroes-who-found-the-rogue-circle-line-train-at-3am-tell-the-real-story-behind-how-the-problem-was-solved/

UltimaOnline SG

A MCQ from CIE June 2016, which i cannot reproduce here, gives a mass spectrum showing only ONE isotope of Copper having mass of 63.5 showing 100% abundance.

The question was to prove which mass spectrum would give an Ar of 63.5.

The mass spectrum above would give 63.5 as the correct answer but this was not the answer. Another was the answer which different isotopes having different Whole number masses of isotope.

My question is all isotopes of an element have Whole number masses. We have observaed this so far. So we can eliminate any choice having an isotopic mass in decimal....is this correct?

Correct. The molar mass of any single isotope must be an integer (since there is no such thing as non-integer number of protons or neutrons), only averaged molar masses (ie. taking into consideration the relative abundance of various isotopes) can be non-integers.

UltimaOnline SG

Do Singaporeans realize that burning joss papers and joss sticks generates airborne carcinogens, in the same way as smoking cigarettes causes cancer? It's up to you to follow your own religious beliefs, but it's bad ethics and bad karma to cause harm to others.

Burning joss papers and joss sticks in HDB corridors is like forcing your neighbours to smoke carcinogenic cigarettes, and large scale burning during 7th month festivals is like forcing all other Singaporeans of other religions to smoke carcinogenic cigarettes.

And here's the irony : no Chinese deity or Buddha ever instructed their followers to burn joss papers and joss sticks.

It may be acceptable cultural practice back then, when there was no medical science to speak of, but today when medical science has clearly shown this activity's carcinogenic properties, you'd think humans would be intelligent enough and ethical enough to evolve their cultural practices to minimize harm to others and to themselves.

2016 case : Sg man with China wife vs Singaporean aunties (joss paper smokers)http://theindependent.sg/video-of-prc-scolding-locals-for-burning-joss-papers-goes-viral/

2015 case : Elderly Singaporean couple (joss sticks smokers) vs Young Singaporean couple with baby (helpless innocent victim)
http://news.asiaone.com/news/singapore/caught-camera-neighbours-lash-out-each-other-over-joss-sticks

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