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!
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!
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-.
Originally posted by UltimaOnline:
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-.
Hello, thank you for your reply!
How do u determine which is a stronger activator or deactivator then? Am i allowed to just assume OH and NO2 are the strongest activator and deactivator respectively? And if I’ve got a Br2 aq do I need to find the common “intersection points” of OH and NO2 and sub the Br on it, or do i just tri sub it regardless of the position?
Originally posted by delurach:Hello, thank you for your reply!
How do u determine which is a stronger activator or deactivator then? Am i allowed to just assume OH and NO2 are the strongest activator and deactivator respectively? And if I’ve got a Br2 aq do I need to find the common “intersection points� of OH and NO2 and sub the Br on it, or do i just tri sub it regardless of the position?
Just base it on the OH group, as it outweighs the NO2 group.
Whether mono, di or tri bromination occurs, of course depends on the availability of the ortho and para positions.
Ok, I won't be available to answer any more questions for the day liao. Any more questions, pls go ahead and discuss amongst yourselves, students. Have fun with P3! :D
Paper 3 H2 Chemistry was considered more difficult than previous years for the theory questions,not because they were novel or what,but those content and commonly known questions were easily regurgitated but mark allocation was rather stingy,novel questions were killer for many,especially deduction surprisingly(it was tricky but simpler to deduce than previous years),and organic chem component was ridiculously easy,essentially free marks.
If i were to decide which were the hardest/killer questions in P3,it will definitely be Q2,4.Q3 can be killer if one fails to elucidate the compound and Q1 too if explanation of theoretical terms is sloppy.
Originally posted by MapPwner:Paper 3 H2 Chemistry was considered more difficult than previous years for the theory questions,not because they were novel or what,but those content and commonly known questions were easily regurgitated but mark allocation was rather stingy,novel questions were killer for many,especially deduction surprisingly(it was tricky but simpler to deduce than previous years),and organic chem component was ridiculously easy,essentially free marks.
If i were to decide which were the hardest/killer questions in P3,it will definitely be Q2,4.Q3 can be killer if one fails to elucidate the compound and Q1 too if explanation of theoretical terms is sloppy.
2017 H2 Chemistry Paper 3 Discussions on Reddit :
https://www.reddit.com/r/SGExams/comments/7blfjz/a_levels_h2_chemistry_paper_3_2017/
Of course, over at the Reddit thread, a lot of people have already posted their answers to almost the entire P3, so go check over there 1st. Still not sure, can ask here.
Overall quite an okay paper, definitely the hardest in recent years but i think i can manage around 64/80 in the worst scenario.
Btw, ultimaonline, I have PMed u the paper, mind giving some input on the Nitrogen question the part on Subliming A + Aq solution of A? thanks.
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.
Originally posted by do12453:Overall quite an okay paper, definitely the hardest in recent years but i think i can manage around 64/80 in the worst scenario.
Btw, ultimaonline, I have PMed u the paper, mind giving some input on the Nitrogen question the part on Subliming A + Aq solution of A? thanks.
Good to hear you're on track for your A grade! Well done! :)
Ok exactly which question, you want me to look at? Qn no, part no? But I'll need a while to reply you on that, a bit busy with other real-life stuff now. Thanks for ur patience.
haha alright, if you could just take a look at Q2c
Originally posted by do12453:haha alright, if you could just take a look at 2017 P3 Q2c
https://en.wikipedia.org/wiki/Ammonium_azide
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.
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 :))
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 :))
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 :))
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
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.
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.
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.
Difficulty of 2017 H2 Chemistry Paper 3 : http://www.strawpoll.me/14368146/r
Ask your school teachers, ask your private tutors, post here on SgForums to ask us for help (even if I'm away-from-forum sometimes, there are several helpful A grade students here who will also help you), and those with a Reddit account can and should ask for help there as well (there are also a couple of helpful private tutors there, 'screwloosely' and 'consca', who will help you out with your questions).
Originally posted by UltimaOnline:
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.
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.
Edit:Prac probably pulled down the overall so far.If i assumed P2 57-58/75,P3 65/80,P4 30/55,its probably 62.5/85 marks so far or so.
I also have some friends who didn't manage to finish the paper.
I feel that 70% for A this year may probably happen.
adding a buffer in for careless mistakes, it's probable that i'll obtain 50/75 for p2, 55/80 for p3, 30/55 for p4. if i score close to full or full marks in p1, it's just a borderline absolute A. what are my chances of an A? would the bell curve shift to 67-69%?
school's A rate for chem was almost 70% last year, and most of my friends who were comfortably in that range for prelims, aka 50th percentile and above found p3 yesterday much tougher than tys, mainly as it was a mad rush for time. we're expecting 55-60/80 for p3. even my friend who scored in the top 5% of the cohort (so way way within range) couldn't complete the paper. honestly for the other 2 papers we're expecting similar marks, prac was bad for everyone and our consensus for p2 is that it 'seems easy, but actually isn't', so our final grade might not even be an absolute A.
Originally posted by tokipaqur:
school's A rate for chem was almost 70% last year, and most of my friends who were comfortably in that range for prelims, aka 50th percentile and above found p3 yesterday much tougher than tys, mainly as it was a mad rush for time. we're expecting 55-60/80 for p3. even my friend who scored in the top 5% of the cohort (so way way within range) couldn't complete the paper. honestly for the other 2 papers we're expecting similar marks, prac was bad for everyone and our consensus for p2 is that it 'seems easy, but actually isn't', so our final grade might not even be an absolute A.
Depends on which school you were from + how many marks the cohort lost on average.
I'm from a mid-tier JC and most of the people i talked to lost 25-30 marks by superifical marking,some of them graded stringently but not all I suppose.
I enquired other friends from other JCs and even top JCs about the paper and all of them gave the same reply,it was difficult,couldn't finish,not easy etc.
So your current score I suppose has a fair chance of scoring A probably this year I think?I felt that this year was comparable to 2010.2010 the difficult components were paper 2 and maybe a bit of paper 3.But this year the harder than average papers were P4,and possibly P3 and also P2(even though it looks easy,but it really isn't unless your chem skills are really solid).Unless if many people got the same range of score as you,its probably a B as it sets a median score,I feel.So a near full mark for P1 is required.
Can I also just ask for Question 3(b) the 3 mark question on ligand,complex and about reaction between Cu2+ and NH3,is it fine to define ligand,complex in the context of cu2+ and nh3,and thereafter write an equation of cu(h2o)6 2+ +4nh3 ->cu(nh3)4(h2o)2 2+ +4h2o in presence of excess ammonia as stated,stating that nh3 a ligand displaces h2o ligands in cu2+ to form the product by forming a coordinate dative bond.
For careless mistakes wise,like the 4 marks question in Q1 asking about difference in melting point and accounting for it,i accidentally wrote between Si,cl atoms etc for the compound with simple covalent molecule structure despite correct identification of the different forces and how the id-id forces were strengthened(did write id-id in full word of course) for different molecules and their respective structure and bonding.So i assumed -1 mark out of 4 marks for 3/4 marks obtained.
The 3 mark question for deducing C6H5X and illustrating the electrophilic substitution mechanism has error carried forward i suppose?In that case i will probably only score 1/2 mark for outlining the mechanism.
For the 6 marks on describing and illustrating the acid-base nature of oxides,a brief mention on whether the oxide across the period is acidic/basic in nature due to presence of giant ion lattice/simple covalent structure and thus reacts with only acids/bases.Is that the general outline for the answer in this question,including the equations for the reactions?
I assume one also needs to know Al2o3 is amphoteric.
Originally posted by MapPwner:Can I also just ask for Question 3(b) the 3 mark question on ligand,complex and about reaction between Cu2+ and NH3,is it fine to define ligand,complex in the context of cu2+ and nh3,and thereafter write an equation of cu(h2o)6 2+ +4nh3 ->cu(nh3)4(h2o)2 2+ +4h2o in presence of excess ammonia as stated,stating that nh3 a ligand displaces h2o ligands in cu2+ to form the product by forming a coordinate dative bond.
For careless mistakes wise,like the 4 marks question in Q1 asking about difference in melting point and accounting for it,i accidentally wrote between Si,cl atoms etc for the compound with simple covalent molecule structure despite correct identification of the different forces and how the id-id forces were strengthened(did write id-id in full word of course) for different molecules and their respective structure and bonding.So i assumed -1 mark out of 4 marks for 3/4 marks obtained.
The 3 mark question for deducing C6H5X and illustrating the electrophilic substitution mechanism has error carried forward i suppose?In that case i will probably only score 1/2 mark for outlining the mechanism.
For the 6 marks on describing and illustrating the acid-base nature of oxides,a brief mention on whether the oxide across the period is acidic/basic in nature due to presence of giant ion lattice/simple covalent structure and thus reacts with only acids/bases.Is that the general outline for the answer in this question,including the equations for the reactions?
I assume one also needs to know Al2o3 is amphoteric.
Yes, should be just -1 mark.
Yes, you should get 2 out of 3 marks.
Yes.
Yes.
Hi, I'm from HCJC.
I can't say for sure the average marks lost for the cohort, but I did discuss the paper amongst my friends, who all got low to mid A for prelims, and I personally scraped an A, so we're around 70th percentile and above in sch (based on prelims results). With the 3 papers having concluded, we're expecting 55-60/80 for p3. p2 and p4 we are less certain of the exact marks but quite sure we're get somewhere in the score of 30/55 and 50-55/75 respectively. If coupled with a 25/30 for mcq, that's around 67-71% only which is lower than our prelims results.
Ultima, does this mean that there's a higher chance of us scoring a B than an A? Would really suck to get a lower grade in As than prelims. initially thought i screwed up bad but my schoolmates who scored As in prelim are expecting similar results...perhaps the bell curve would be 67/68%?
Originally posted by tokipaqur:Hi, I'm from HC.
I can't say for sure the average marks lost for the cohort, but I did discuss the paper amongst my friends, who all got low to mid A for prelims, and I personally scraped an A, so we're around 70th percentile and above in sch (based on prelims results). With the 3 papers having concluded, we're expecting 55-60/80 for p3. p2 and p4 we are less certain of the exact marks but quite sure we're get somewhere in the score of 30/55 and 50-55/75 respectively. If coupled with a 25/30 for mcq, that's around 67-71% only which is lower than our prelims results.
Ultima, does this mean that there's a higher chance of us scoring a B than an A? Would really suck to get a lower grade in As than prelims. initially thought i screwed up bad but my schoolmates who scored As in prelim are expecting similar results...perhaps the bell curve would be 67/68%?
Based on your estimates for the Papers completed, as long as you and your friends score near full marks for the upcoming P1, you guys will have a good chance of getting your A grade, no worries.