2016 H2 Chemistry JC1 & 2 students post your questions here
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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.
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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!
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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!
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Thank you!
I have another question: why is the standard enthalpy change of hydration linked to ion-dipole interations?
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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. -
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.
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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.
IEn, this is the last question I'm answering. I'm logging off after this, and may not revisit the forum again until after the Paper 2."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.
Ok, logging off liao, will be back on the forum after Paper 2.
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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
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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.
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Originally posted by iSean:
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.
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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
"
---------------------------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.
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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
"
---------------------------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. -
Originally posted by UltimaOnline:
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.
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? -
Originally posted by iSean:
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.
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Originally posted by UltimaOnline:
You typed out the question wrongly ie. the question didn't specify OPTICAL isomer, since there isn't a chiral C atom in X.
Ah I see your point, apparently STPM made a huge mistake lol.
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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.
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Originally posted by UltimaOnline:
This is why Singapore pays Cambridge a lot of money to set the Singapore A level papers.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. -
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???Maybe Jho Low ran away with the fees meant to pay Cambridge.
If he's not careful, sekali he may kena blown up.
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Originally posted by UltimaOnline:
Yep. that was orginal print, I editted the question to C=O on photoshop.
I'm not into the whole politics thing, but politics around Malaysia. But currently everyone seems quite distracted on our Malaysia Original Silat Practices. lol
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what is the product when butene reacts with IBr
?
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Originally posted by baby6vat:
what is the product when butene reacts with IBr
?
Major product is 2-bromo-1-iodobutane, minor product is 1-bromo-2-iodobutane. -
Hello, I would like to ask something related to Ionisation Energy.
I understand that Xenon has a lower first Ionisation Energy as compared to the other noble gases. I find that there can be multiple reasons for which this happens( better shielding by inner shell electrons/ lesser attraction to nucleus due to larger atomic size)
So if a question were to ask for an explantion of why Xenon has a lower first Ionisation Energy as compared to other noble gases, should I put both of the reasons stated above or it can be a case of either/or
Thanks :)
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Originally posted by Jh2424:
Hello, I would like to ask something related to Ionisation Energy.
I understand that Xenon has a lower first Ionisation Energy as compared to the other noble gases. I find that there can be multiple reasons for which this happens( better shielding by inner shell electrons/ lesser attraction to nucleus due to larger atomic size)
So if a question were to ask for an explantion of why Xenon has a lower first Ionisation Energy as compared to other noble gases, should I put both of the reasons stated above or it can be a case of either/or
Thanks :)
Are you a J1 or J2 student in 2017?Always give all relevant factors for any question. Here, both are relevant and required.
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Originally posted by UltimaOnline:
Are you a J1 or J2 student in 2017?Always give all relevant factors for any quetion. Here, both are relevant and required.
J1
Hmm I see, thank you :)
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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 :)