2016 H2 Chemistry JC1 & 2 students post your questions here
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but ionic charge of M is unknown also
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Originally posted by Flying grenade:
but ionic charge of M is unknown also
Which is why in your final answer, the overall ionic charge (ie. sum of charges on metal ion + ligands) will remain unknown. To score full marks, you just need to write the formula of the coordination complex with the correct stoichiometry of metal ion to ligand, regardless of the charge.Eg. [M(L)]n-, [M(L)2]n-, [M(L)3]n-, [M(L)4]n-, [M(L)5]n-, [M(L)6]n-, etc.
Ok, enough with this question. Check your answer with your school teacher or private tutor. Don't ruin the fun for others. Leave my questions (lots more on my BedokFunland JC website) unsolved, like that more mysterious and more fun. To all Singapore JC students (especially JC1 students, coz JC2 students only left 3 months to this year's A levels liao), if your school teacher or private tutor can't solve my BedokFunland JC questions, you can consider joining my BedokFunland JC tuition.
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Originally posted by UltimaOnline:
Which is why in your final answer, the overall ionic charge (ie. sum of charges on metal ion + ligands) will remain unknown. To score full marks, you just need to write the formula of the coordination complex with the correct stoichiometry of metal ion to ligand, regardless of the charge.Eg. [M(L)]n-, [M(L)2]n-, [M(L)3]n-, [M(L)4]n-, [M(L)5]n-, [M(L)6]n-, etc.
Ok, enough with this question. Check your answer with your school teacher or private tutor. Don't ruin the fun for others. Leave my questions (lots more on my BedokFunland JC website) unsolved, like that more mysterious and more fun. To all Singapore JC students (especially JC1 students, coz JC2 students only left 3 months to this year's A levels liao), if your school teacher or private tutor can't solve my BedokFunland JC questions, you can consider joining my BedokFunland JC tuition.
AHHH i see now !! thanks for the guidance and tip
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I love chem
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I love chem
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CIE AS Level M/J 2016: 9701/22/M/J/16
There is no Marking Scheme for this Answer until Results Day on 11 August.
How do you do (i) and (iii)?
IE of … Y log 10 Difference by log Difference by I.E
fith 6542 3.82 +0.15 +2820
sixth 9362 3.97 +0.07 + 1656
seveth 11018 4.04 +0.49 + +22588
eighth 33606 4.53
I think my answer will be either Na/Mg/Cl
(iii) Big jump between sixth and seventh I.E.
So is it Sulfur? -
Originally posted by iSean:
CIE AS Level M/J 2016: 9701/22/M/J/16
There is no Marking Scheme for this Answer until Results Day on 11 August.
How do you do (i) and (iii)?
IE of … Y log 10 Difference by log Difference by I.E
fith 6542 3.82 +0.15 +2820
sixth 9362 3.97 +0.07 + 1656
seveth 11018 4.04 +0.49 + +22588
eighth 33606 4.53
I think my answer will be either Na/Mg/Cl
(iii) Big jump between sixth and seventh I.E.
So is it Sulfur?
(i) Group 17 or VII. The magnitude of increase from 7th to 8th ionization energy is significantly greater than preceding increases, because the 8th electron removed (ie. 8th ionization energy used) is from an inner electron shell closer to the nucleus, which is thus more tightly held electrostatically by the positively charged nucleus, as predicted by Coulomb's law, and also because of less shielding effect.
(iii) Correct, it is sulfur : 1s2 2s2 2p6 3s2 3p4
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Originally posted by UltimaOnline:
(i) Group 17 or VII. The magnitude of increase from 7th to 8th ionization energy is significantly greater than preceding increases, because the 8th electron removed (ie. 8th ionization energy used) is from an inner electron shell closer to the nucleus, which is thus more tightly held electrostatically by the positively charged nucleus, as predicted by Coulomb's law, and also because of less shielding effect.
(iii) Correct, it is sulfur : 1s2 2s2 2p6 3s2 3p4
So (i) Can't be Na because the relative similar shielding effect on 5th - 7th IE?
So shifting from 2s to 2p orbital will not give such value of change in I.E?
Because CS Toh Step by Step Adv. Level Chemistry said so: (pg44)
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Originally posted by iSean:
So (i) Can't be Na because the relative similar shielding effect on 5th - 7th IE?
So shifting from 2s to 2p orbital will not give such value of change in I.E?
Because CS Toh Step by Step Adv. Level Chemistry said so: (pg44)
Shifting from 2s to 2p orbital will not give such a large increase in I.E, as the difference between an outer electron shell and an inner electron shell, is much larger than between an p orbital and an s orbital of the same electron shell.The difference between p orbital and s orbital is testable specifically under the anomalous pattern of Be to B, and Mg to Al.
Unless the question tells you specifically that the electrons are removed from the same electron shell, you needn't worry about that possibility, and focus on difference in electron shells, not difference in orbitals within the same electron shell.
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"CurryBoy" posted :
I have taken my A Levels in the year of 2015 and obtained a result of DDD/C, GP: B. I take PCME (H1 econs) May I know of the difference between school SPA and the practical Paper 5? I saw in a post (http://sgforums.com/forums/2297/topics/486763) that you said not to worry about the difference between SPA and P5, but may I know of more details as to why?
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Originally posted by UltimaOnline:
"CurryBoy" posted :
I have taken my A Levels in the year of 2015 and obtained a result of DDD/C, GP: B. I take PCME (H1 econs) May I know of the difference between school SPA and the practical Paper 5? I saw in a post (http://sgforums.com/forums/2297/topics/486763) that you said not to worry about the difference between SPA and P5, but may I know of more details as to why?
Why? Because in the first place, both SPA and P5, being a practical component, is difficult (in some ways impossible) to prepare for. So you should focus on preparing yourself for your theory components (ie. P1 P2 P3), because that's you can productively prepare for, and that's what can help pull up your grade (even if you don't do well for SPA or P5). As long as you do very well for your P1, P2 and P3, you can get a good overall grade, even possibly an A grade.But if you're still worried for P5, go download the UK Cambridge A level papers (go search online), and study their Practical Paper : depending on different syllabus in different years, their Practical Paper could be P5 or some other paper no.
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On the same topic about ionisation energy, i have this question that i need some clarifications.
Element Z is the successive element from potassium across the period. Compare and contrast the first ionisation energy of potassium and element Z (calcium). Support your answers with reasons.
At first, my answer was:
K - 1s2 2s2 2p6 3s2 3p6 4s1; Z (Calcium) - 1s2 2s2 2p6 3s2 3p6 4s2
Z < K as valence electron to be removed from Z is paired while that from K is unpaired, thus less energy is required due to inter-electronic repulsion.
However, after checking the Data Booklet, I realised that the first ionisation energy of calcium is higher than potassium, so I reconstructed my answer. My question is why, unlike sulfur and phosphorus, there is no inter-electronic repulsion that causes the first ionisation energy of calcium to be lower?
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Originally posted by Nikkilyx:
On the same topic about ionisation energy, i have this question that i need some clarifications.
Element Z is the successive element from potassium across the period. Compare and contrast the first ionisation energy of potassium and element Z (calcium). Support your answers with reasons.
At first, my answer was:
K - 1s2 2s2 2p6 3s2 3p6 4s1; Z (Calcium) - 1s2 2s2 2p6 3s2 3p6 4s2
Z < K as valence electron to be removed from Z is paired while that from K is unpaired, thus less energy is required due to inter-electronic repulsion.
However, after checking the Data Booklet, I realised that the first ionisation energy of calcium is higher than potassium, so I reconstructed my answer. My question is why, unlike sulfur and phosphorus, there is no inter-electronic repulsion that causes the first ionisation energy of calcium to be lower?
Chemistry is a microcosm of Real Life. When there are opposing factors, in some cases the 1st factor wins, in some cases the 2nd factor wins.Considering K vs Ca, the factor of increased nuclear charge outweighs the factor of inter-electron repulsion, consequently ionization enthalpy for Ca is more endothermic than for K. Considering P vs S, the factor of inter-electron repulsion outweighs the factor of increased nuclear charge, consequently the ionization enthalpy for S is less endothermic than for P.
In Chemistry, including for A levels, you can only show your understanding of both (or all) opposing factors, then carry out experiments to determine which factors outweigh which. To discuss this deeper at higher levels (beyond A levels), yes it's possible to show with more involved calculations why this factor outweighs that factor at this point in the periodic table, but these would not be relevant or useful for A level purposes, and would only confuse the A level student and is therefore counterproductive.
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Thanks!
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Good Evening UltimaOnline, I have taken my A Levels in the year of 2015 and got sub par results May I know of the difference between school SPA and the practical Paper 5? I saw in a post (http://sgforums.com/forums/2297/topics/486763) that you said not to worry about the difference between SPA and P5, but may I know of more details as to why?
Thank you for taking your time to read this :) -
Originally posted by UltimaOnline:
"CurryBoy" posted :
I have taken my A Levels in the year of 2015 and obtained a result of DDD/C, GP: B. I take PCME (H1 econs) May I know of the difference between school SPA and the practical Paper 5? I saw in a post (http://sgforums.com/forums/2297/topics/486763) that you said not to worry about the difference between SPA and P5, but may I know of more details as to why?
Why? Because in the first place, both SPA and P5, being a practical component, is difficult (in some ways impossible) to prepare for. So you should focus on preparing yourself for your theory components (ie. P1 P2 P3), because that's you can productively prepare for, and that's what can help pull up your grade (even if you don't do well for SPA or P5). As long as you do very well for your P1, P2 and P3, you can get a good overall grade, even possibly an A grade.But if you're still worried for P5, go download the UK Cambridge A level papers (go search online), and study their Practical Paper : depending on different syllabus in different years, their Practical Paper could be P5 or some other paper no.
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Hi, qn1: while hcl reacts readily with alkenes under normal conditions, hcn does not. Explain using chemical bonding? i am not very sure but i think it has something got to do with extent of orbital overlap. C-hdond have greater extent of orbital overlap, stronger bond, require more energy to break, in order for the alkene to be protonated?? Qn2: explain why cr(nh3)6 3+ is coloured. as i explain, should i state in cr3+, the 5 d orbitals have the same energy level.... or should i state in cr(nh3)6 3+, the 5 d orbitals..... which phrasing is correct? qn 3: why doea cr3+ on reaction with na2co3 gives co2? is it because cr(h2o)63+ have high charge density, high polarising power, undergoes cation hydrolysis by polarising oh bond in h2o, causing h atom to ionise to form h+. Since solution is acidic it undergoes acid base neutralisation with na2co3 to form co2 Are my answers correct? Thanks!
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Originally posted by ArJoe:
Hi, qn1: while hcl reacts readily with alkenes under normal conditions, hcn does not. Explain using chemical bonding?
i am not very sure but i think it has something got to do with extent of orbital overlap. C-hdond have greater extent of orbital overlap, stronger bond, require more energy to break, in order for the alkene to be protonated??Qn2: explain why cr(nh3)6 3+ is coloured.
as i explain, should i state in cr3+, the 5 d orbitals have the same energy level….
or should i state in cr(nh3)6 3+, the 5 d orbitals….. which phrasing is correct?qn 3: why doea cr3+ on reaction with na2co3 gives co2?
is it because cr(h2o)63+ have high charge density, high polarising power, undergoes cation hydrolysis by polarising oh bond in h2o, causing h atom to ionise to form h+. Since solution is acidic it undergoes acid base neutralisation with na2co3 to form co2Are my answers correct? Thanks!
Q1. The C-H sigma bond is strong due to sp-s overlap, but more importantly, HCN is a weak acid because the conjugate base is significantly unstable due to the relatively low electronegativity of C atom that bears the negative formal charge (it is only the sp hybridization that just barely allows the C atom to bear the negative formal charge). The weakness of HCN as an acid results in high Ea for the alkene to act as a Bronsted-Lowry base to abstract the proton from HCN, hence the relative inertness.Q2. You can go straight into "In the presence of the 6 NH3 ligands, the d orbitals become non-degenerate... etc etc".
Q3. Yes, that's correct. But in addition to a textual explanation, you also need to write out the relevant balanced equations to obtain full marks.
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C-H sigma bond of alkene is formed by sp2-s overlap
compared to what/which bond? bond between HCl? bond between HCN (sp-s) overlap?
does hybridisation apply for HCl and other molecules, or only for Carbon/nitrogen bonded molecules?
what does breaking the C-H bond got to do with the electrophilic addition of HCl or HCN to alkene? shouldn't it be breaking the pi bond between C atoms in the alkene?
damn stress now
thanks Ultima for explaining
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Originally posted by Flying grenade:
C-H sigma bond of alkene is formed by sp2-s overlap
compared to what/which bond? bond between HCl? bond between HCN (sp-s) overlap?
does hybridisation apply for HCl and other molecules, or only for Carbon/nitrogen bonded molecules?
what does breaking the C-H bond got to do with the electrophilic addition of HCl or HCN to alkene? shouldn't it be breaking the pi bond between C atoms in the alkene?
damn stress now
thanks Ultima for explaining
Compared to most other C-H bonds which are 2sp2-1s or 2sp3-1s overlap. H-Cl bond is even weaker because it is 3sp3-1s overlap.Orbital hybridization applies for all molecules, eg. 3sp3d2 hybridization for octahedral centers (eg. PCl6- ion).
If you draw out the mechanism (for the reaction that does not readily occur due to high Ea required), you'll see that you're breaking the 2p-2p pi bond of the alkene (ie. the weak Bronsted-Lowry base), as well as the 2sp-1s sigma bond of the H-CN (ie. the weak Bronsted-Lowry acid).
I don't want to elaborate further on these questions, so don't ask further. If you still don't understand, go ask your school teacher or private tutor. Besides, edart sterces.
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What is the expected volume of 0.1 mol/litre aqueous NaOH needed to reach the titration end point for 50 millilitre of 0.01 mol/litre mono-protic acid, HA?
A: 5 millilitre
B: 50 millilitre
C: 500 millilitre
D: depends on the strength of acid
Is the answer for this question option A?
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Originally posted by ACA-Milkshake:
What is the expected volume of 0.1 mol/litre aqueous NaOH needed to reach the titration end point for 50 millilitre of 0.01 mol/litre mono-protic acid, HA?
A: 5 millilitre
B: 50 millilitre
C: 500 millilitre
D: depends on the strength of acid
Is the answer for this question option A?
Yes, obviously. The only thing Singapore students must be familiar with to do this question, is British to American units, ie. cm3 = mL, and dm3 = L. And of course (even if not required for this question, you should still know that) molarity = M = mol/dm3 = mol/L. -
nice
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CIE W14 Q5.
Just wondering for this reaction besides Na, how does vanillin react for Br?
Why it react on C-5 and not C-3?As vanillin is named clockwised and not anticlockwised.
Also you mind explaining the other reactions that I'd highlighted? Especially RCOCl and Fe3+ thanks :)
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Originally posted by iSean:
CIE W14 Q5.
Just wondering for this reaction besides Na, how does vanillin react for Br?
Why it react on C-5 and not C-3?As vanillin is named clockwised and not anticlockwised.
Also you mind explaining the other reactions that I'd highlighted? Especially RCOCl and Fe3+ thanks :)
Halogenation via electrophilic aromatic substitution.As long as you correctly substitute the incoming electrophile onto an available ortho or para position relative to the phenolic group, Cambridge will accept.
Diazotization and diazonium coupling is not in the Singapore H2 syllabus. If you're interested, go read up on it yourself.
Acyl halides are electrophiles attacked by the most nucleophilic group, ie. amine on L-Dopa, and phenol in Vanillin.
Fe3+ (in neutral solution, can't be too acidic or too alkaline, interested students can ponder over why, and check your answer with your school teacher or private tutor) coordinates & complexes with phenolic ligands to generate a violet coloured coordination complex.