2016 H2 Chemistry JC1 & 2 students post your questions here
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Originally posted by theophilus:
Yo is there anyway to get ready for private candidate practical paper 5 for chem?? Or is it smokable? And because private candidates do a different practical paper, are they marked similarly? :O Because major handicap as I heard in JCs they tend to get like full marks or sth for SPA
There isn't much you can do to prepare for Pract Paper. With a solid theoretical foundation (ie. as long as you know your Chem), just follow the Pract Paper instructions carefully, record down all observations carefully, etc.The only feasible way to prepare for Pract Paper, is by preparing for your Planning Qn, for obvious reasons (in fact, for the new syllabus, the Planning Qn is part of the Pract Paper, no longer Paper 2). For this purpose, buying and studying Chan Kim Seng and Jeanne Tan's H2 Chem Planning Book is the best (and kinda only) way for you to prepare for your Planning Qn + Pract Paper.
And no worries about losing out to school candidates. In practice (pun), all things considered (I won't go into details), they do not have any advantage (or disadvantage) over private candidates.
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Hi UltimaOnline, in George CHong's organic chem book pg 43, it gives an example of FRS in the methyl group in methylbenzene which makes me confused as I had always thought that FRS only occurs in alkanes.
If free radical substitution occurs can occur in alkene, how does it work?
Also for George Chong's Organic chem book pg 45, I dont get how the ratio is formed. Thanks
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Originally posted by MightyBiscuits:
Hi UltimaOnline, in George CHong's organic chem book pg 43, it gives an example of FRS in the methyl group in methylbenzene which makes me confused as I had always thought that FRS only occurs in alkanes.
If free radical substitution occurs can occur in alkene, how does it work?
Also for George Chong's Organic chem book pg 45, I dont get how the ratio is formed. Thanks
Alkyl side-chain of benzene is considered an 'alkane'.To determine product distribution of halogenation via free radical substitution, mathematically combine both factors :
No. of H atoms substitutable x Stability of alkyl radical intermediate (also known as Kinetic Reactivity of Pri vs Sec vs Tert H atoms)
Take the halogenation of butane for instance. In terms of "No. of H atoms substitutable", you'd expect 1-halobutane to be the major product (since 6 H atoms vs 4 H atoms). In terms of "Stability of alkyl radical intermediate", you'd expect 2-halobutane to be the major product (since the secondary alkyl radical intermediate to generate secondary alkyl halide 2-halobutane, is more stable than the primary alkyl radical intermediate to generate primary alkyl halide, 1-halobutane ; the more stable the intermediate, the lower the Ea, hence the faster the rate of reaction, resulting in the major product).
Overall, to determine which factor outweighs which to result in the major product, you'd have to combine both factors mathematically (don't memorize the relative stabilities or reactivities, the exam question will provide it, as this factor depends on the specific halogen, as well as specific temperature).
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2011 TYS question 6 part c v and d ii.
Which other reaction confirms W is aromatic? Explain your answer.
Why can't I use reaction 5 since reaction 5 proves that there is phenol functional group?
Explain clearly why you have placed each of the functional groups in their particular positions.
Why can't the chlorine atom be attached to the benzene ring? Why didn't they specify the adding of NaOH or KOH for reaction 1? Sorry if this question seemed too simple but I'm really confused over here. I thought NaOH/KOH is needed to substitute the chlorine no matter whether it is halogenoalkane or halogenoarene.
Edit. I think I get the answer. Reaction 3 specify that organic product produced has Mr of 138. Which means chlorine is removed. If reaction 3 is not there, can I assume that chlorine is attached to the ring instead of the alkyl group? The answer sheets say that reaction 1 shows that chlorine is attached to the alkyl group instead of the benzene ring, which doesn't make sense for me.
Do I need to remember how to produce phenol from benzene?
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Originally posted by senga:
2011 TYS question 6 part c v and d ii.
Which other reaction confirms W is aromatic? Explain your answer.
Why can't I use reaction 5 since reaction 5 proves that there is phenol functional group?
Explain clearly why you have placed each of the functional groups in their particular positions.
Why can't the chlorine atom be attached to the benzene ring? Why didn't they specify the adding of NaOH or KOH for reaction 1? Sorry if this question seemed too simple but I'm really confused over here. I thought NaOH/KOH is needed to substitute the chlorine no matter whether it is halogenoalkane or halogenoarene.
Edit. I think I get the answer. Reaction 3 specify that organic product produced has Mr of 138. Which means chlorine is removed. If reaction 3 is not there, can I assume that chlorine is attached to the ring instead of the alkyl group? The answer sheets say that reaction 1 shows that chlorine is attached to the alkyl group instead of the benzene ring, which doesn't make sense for me.
Do I need to remember how to produce phenol from benzene?
Reaction 3 is a better choice, as only alkyl benzenes can be oxidized by KMnO4 to hydroxybenzoic acid. Reaction 5 is a weaker reason, as it could imply an acyl halide side chain (although the degree of unsaturation won't fit, but most H2 students won't know how to check for this anyway).Without OH-(aq), only alkyl halides can be hydrolyzed upon warming (adding OH- would of course speed up the reaction). If the Cl atom was attached to the benzene ring as you suggest, then due to the C-Cl bond strengthened by having partial double bond character in the resonance hybrid, concordantly and consequently not only would OH-(aq) be required (now you understand why Cambridge didn't use OH-(aq)? now you understand the answer sheet?), but a much higher temperature is required (the question specified 'warm'), and besides, the nucleophilic aromatic substitution hydrolysis (via the elimination-addition mechanism involving a benzyne intermediate) of aryl halides to phenol is beyond the H2 A level syllabus.
At University and professional levels, there are many different ways to generate phenol from benzene, but none of these are taught at A levels. In other words, within the H2 A level syllabus, there is no (within syllabus) method you can generate phenol from benzene. Of course, if you're aiming for a distinction A grade, it's always useful to expose yourself beyond the syllabus.
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Hi all, I know this is quite random (and not related to JC) but can anyone help us fill up the below survey form?
We are a group of students from MAD School trying to get survey answered for our marketing project.
Very short one, just 1-2 minutes can finish and capitaland vouchers and other gifts to be won!
Thanks a million!
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Hi UltimaOnline,
I would like to seek your help with regard to the following questions:
Q1: DHS/2015/P1/Q14
What’s wrong with Choice B, given that acid strength increases from HCl to HI? Interestingly when I researched on the pH of these compounds in a fixed concentrations their pH are the same. (Source: http://www.aqion.de/site/191)
Q2: DHS/2015/P1/Q34 [Ans B]
Why is the answer not A? I think choice 3 is correct - vanadium oxide gets reduced before being oxidised. (Source: http://www.chemguide.co.uk/inorganic/transition/vanadium.html) Or am I missing out on something here?
Q3: VJC/2015/P1/Q3 [Ans C]
Which one of the following is a possible configuration of a stable M3+ ion in the ground
state?
Remarks: I do not understand this question. Are they asking for the configuration of M or M3+? If it’s the latter, then why is there a need to add the words “in the ground state”?
Q4: VJC/2015/P1/Q14 [Ans C]
Remarks: I thought the answer should have been B instead. This is because at B, since the concentration of lead (II) containing species is at its lowest, [Pb2+] should be at its lowest at V. I didn’t think C was the correct answer, because at any point between 0cm3 to Vcm3, although the ionic product was initially greater than the solubility product, the ionic product being equal to the solubility product at any point through the formation of the ppt.Many thanks, UltimaOnline for your help! :)
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Originally posted by gohby:
Hi UltimaOnline,
I would like to seek your help with regard to the following questions:
Q1: DHS/2015/P1/Q14
What’s wrong with Choice B, given that acid strength increases from HCl to HI? Interestingly when I researched on the pH of these compounds in a fixed concentrations their pH are the same. (Source: http://www.aqion.de/site/191)
Q2: DHS/2015/P1/Q34 [Ans B]
Why is the answer not A? I think choice 3 is correct - vanadium oxide gets reduced before being oxidised. (Source: http://www.chemguide.co.uk/inorganic/transition/vanadium.html) Or am I missing out on something here?
Q3: VJC/2015/P1/Q3 [Ans C]
Which one of the following is a possible configuration of a stable M3+ ion in the ground
state?
Remarks: I do not understand this question. Are they asking for the configuration of M or M3+? If it’s the latter, then why is there a need to add the words “in the ground state�?
Q4: VJC/2015/P1/Q14 [Ans C]
Remarks: I thought the answer should have been B instead. This is because at B, since the concentration of lead (II) containing species is at its lowest, [Pb2+] should be at its lowest at V. I didn’t think C was the correct answer, because at any point between 0cm3 to Vcm3, although the ionic product was initially greater than the solubility product, the ionic product being equal to the solubility product at any point through the formation of the ppt.Many thanks, UltimaOnline for your help! :)
Hi Gohby, no prob ;)Q1: DHS/2015/P1/Q14. While acid strength (indicated by pKa) increases from HCl to HI, but due to the leveling effect, in aqueous solutions they are all strong acids which dissociate completely and hence result in the same pH.
Q2: DHS/2015/P1/Q34. You're right. Perhaps (lame, but) the DHS teacher question-setter was nitpicking that "likely" should be replaced with "definitely", since the maximum OS of V is +5 (based on electron configuration).
Q3: VJC/2015/P1/Q3. The question is asking for the configuration of M3+. "Ground state" must always be specified, otherwise there would be many permutations and combinations possible for excited states.
Q4: VJC/2015/P1/Q11 (it's Q11, not Q14). "Ionic product" is always intended to be "initial before precipitation". This is a somewhat unfair question, as the mathematics involved goes beyond A levels, but conceptually still examinable as a sadistic distinction question.
The VJC teacher question-setter's thinking (and feeling rather proud of him/herself for setting such a tricky question, no doubt) is that in sufficiently large excess of halide ions, not only would all the precipitate dissolve, but even the molarity of aqueous Pb2+ ion, would eventually decrease to become even less than at point V, due to large excess of halide ions pulling the position of equilibrium over to the RHS to form the coordination complex ion, ie. [PbCl4]2- and [PbI4]2-, which are not considered as Pb2+(aq).
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Hi UltimaOnline,
I would like to seek your help on some of the HCI questions here:
HCI/2015/P1/28 [Ans A]
Remarks: I think the answer is C instead. PCl5 reacts with the alcohol and COOH (and not the phenol) and it reacts with sulphuric due to the close proximity between the COOH and OH.
HCI 2015/P1/25 [Ans C]
Remarks: I’m not entirely sure why the answer is C, but here are my deductions:
A is wrong because even though NO converts CO and unburnt hydrocarbons in the catalytic converter, the carbon dioxide formed is not harmless because it’s a greenhouse gas.
B is wrong because it should be N2O.
D is wrong because the reaction named is dissolution.
I referred to wikipedia to evaluate choice C [https://en.wikipedia.org/wiki/Acid_rain]:
In the gas phase sulfur dioxide is oxidized by reaction with the hydroxyl radical via an intermolecular reaction:
SO2 + OH· → HOSO2·
which is followed by:
HOSO2· + O2 → HO2· + SO3
In the presence of water, sulfur trioxide (SO3) is converted rapidly to sulfuric acid:
SO3 (g) + H2O (l) → H2SO4 (aq)
Nitrogen dioxide reacts with OH to form nitric acid:
NO2 + OH· → HNO3
However I’ve some disquiet about this choice:
Firstly, such detailed knowledge and pathway of acid rain formation seems to be beyond the reach of the syllabus.
Secondly, even if the student possesses knowledge on how exactly acid rain is formed, I wouldn’t say it catalyses the formation of acid rain from atmospheric sulphuric dioxide. This is because sulphuric acid is formed from sulphur dioxide, and the acid contributing to acid rain by nitrogen dioxide is nitric acid. Hence nitrogen dioxide doesn’t catalyse the formation of acid rain from sulphuric dioxide (unless it somehow forms sulphuric acid).
HCI 2015/P1/18 [Ans C]
I can accept the answer because that forms an integral part of the definition of transition element. However, what’s wrong with B as the answer?
HCI 2015/P1/19 [Ans A]
Is there a manner for students to make an intelligent guess/extrapolate from their knowledge in the syllabus to obtain the answer without understanding and appreciating their individual mechanisms? Don’t think the Grignard reagent in step 1 is part of the syllabus so I’m thinking there should be a more efficient way of solving this question.
Many thanks Ultima :)
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Originally posted by gohby:
Hi UltimaOnline,
I would like to seek your help on some of the HCI questions here:
HCI/2015/P1/28 [Ans A]
Remarks: I think the answer is C instead. PCl5 reacts with the alcohol and COOH (and not the phenol) and it reacts with sulphuric due to the close proximity between the COOH and OH.
HCI 2015/P1/25 [Ans C]
Remarks: I’m not entirely sure why the answer is C, but here are my deductions:
A is wrong because even though NO converts CO and unburnt hydrocarbons in the catalytic converter, the carbon dioxide formed is not harmless because it’s a greenhouse gas.
B is wrong because it should be N2O.
D is wrong because the reaction named is dissolution.
I referred to wikipedia to evaluate choice C [https://en.wikipedia.org/wiki/Acid_rain]:
In the gas phase sulfur dioxide is oxidized by reaction with the hydroxyl radical via an intermolecular reaction:
SO2 + OH· → HOSO2·
which is followed by:
HOSO2· + O2 → HO2· + SO3
In the presence of water, sulfur trioxide (SO3) is converted rapidly to sulfuric acid:
SO3 (g) + H2O (l) → H2SO4 (aq)
Nitrogen dioxide reacts with OH to form nitric acid:
NO2 + OH· → HNO3
However I’ve some disquiet about this choice:
Firstly, such detailed knowledge and pathway of acid rain formation seems to be beyond the reach of the syllabus.
Secondly, even if the student possesses knowledge on how exactly acid rain is formed, I wouldn’t say it catalyses the formation of acid rain from atmospheric sulphuric dioxide. This is because sulphuric acid is formed from sulphur dioxide, and the acid contributing to acid rain by nitrogen dioxide is nitric acid. Hence nitrogen dioxide doesn’t catalyse the formation of acid rain from sulphuric dioxide (unless it somehow forms sulphuric acid).
HCI 2015/P1/18 [Ans C]
I can accept the answer because that forms an integral part of the definition of transition element. However, what’s wrong with B as the answer?
HCI 2015/P1/19 [Ans A]
Is there a manner for students to make an intelligent guess/extrapolate from their knowledge in the syllabus to obtain the answer without understanding and appreciating their individual mechanisms? Don’t think the Grignard reagent in step 1 is part of the syllabus so I’m thinking there should be a more efficient way of solving this question.
Many thanks Ultima :)
No prob Gohby :)Q28. You're right, both are possible products, depending on whether conc H2SO4 is in excess or limiting.
Q25. You're right (although D should be hydrolysis). For A level purposes, just use the simplified equations that NO2 oxidizes SO2 to SO3, which undergoes hydrolysis to generate H2SO4 (acid rain). The NO reacts with O2(g) to regenerate NO2, hence a catalyst.
Q18. The other options are *properties* of transition metals, not definition.
Q19. Even if the student isn't intimately familiar with the mechanism, but by inspection, should still be able to deduce whether addition, substitution, elimination, oxidation, reduction or hydrolysis has occurred. Also, the other options are inadmissible due to errors in steps 1, 2 or 3.
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Originally posted by UltimaOnline:
No prob Gohby :)Q28. You're right, both are possible products, depending on whether conc H2SO4 is in excess or limiting.
Q25. You're right (although D should be hydrolysis). For A level purposes, just use the simplified equations that NO2 oxidizes SO2 to SO3, which undergoes hydrolysis to generate H2SO4 (acid rain). The NO reacts with O2(g) to regenerate NO2, hence a catalyst.
Q18. The other options are *properties* of transition metals, not definition.
Q19. Even if the student isn't intimately familiar with the mechanism, but by inspection, should still be able to deduce whether addition, substitution, elimination, oxidation, reduction or hydrolysis has occurred. Also, the other options are inadmissible due to errors in steps 1, 2 or 3.
Hello UltimaOnline,
For Q18 I can understand why the other choices (A and D) are not properties restricted to transition elements - for A, elements such as phosphorus and sulphur, amongst others exhibit more than 1 os in their compounds whereas for D, enzymes, for instance, are neither transition elements nor their compounds but are also used as catalysts. However, I cannot think of an example of a non-transition element which form many coloured compounds which is why I have difficulties eliminating B as a possible option.
For Q28 you mentioned that both are possible products - so that would include A as a possible option as well. Can I confirm that the product from the dehydration of A would be the acid anhydride? (albeit such knowledge would have been out of syllabus.)
Cheers! :)
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Originally posted by gohby:
Hello UltimaOnline,
For Q18 I can understand why the other choices (A and D) are not properties restricted to transition elements - for A, elements such as phosphorus and sulphur, amongst others exhibit more than 1 os in their compounds whereas for D, enzymes, for instance, are neither transition elements nor their compounds but are also used as catalysts. However, I cannot think of an example of a non-transition element which form many coloured compounds which is why I have difficulties eliminating B as a possible option.
For Q28 you mentioned that both are possible products - so that would include A as a possible option as well. Can I confirm that the product from the dehydration of A would be the acid anhydride? (albeit such knowledge would have been out of syllabus.)
Cheers! :)
Hi Gohby,Even if forming colored compounds are exclusively a property of transition metals, but property isn't the same as definition.
Yes, carboxylic acid anhydride.
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Thank you UltimaOnline.
I have some other questions here. I've penned down my thought process and doubts wrt some of the individual choices (pls pardon the lengthiness!). Thanks again.
DHS/2015/P1/11 [Ans B]
I’m not too sure what to make out of the choices but here are my workings:
A: Order of reaction wrt H2O2 can be done by juxtaposing the gradient of [I-] taken at 0s (initial rate of rxn)
B: the absorbance corresponds to the amount of I-(which doesn’t block out light) vs iodine (which blocks out light). But how do I know I can’t get the H2O2 order?
C: No idea on how to analyse - firstly, the 2 graphs look significantly different for a diffenence in 0.02M of hydrogen peroxide. Secondly I tried to establish the half life of iodine (0->0.5->0.75…) by making a scaled drawing but got nowhere.
D: I think the 2 graphs with differing [I-] is jus a red herring. I can note the rate of reaction at a particular [H2O2] at either of the graphs, double that [H2O2], and see to what extent the rate increases
DHS/2015/P1/16 [Ans D]
Regarding choice C, why does the addition of H+ in step 3 (to a mixture of [Cu(NH3)4]SO4 + Sn2+ + H2O) cause the formation of [Cu(H2O)6]2+?
VJC/2015/P1/8 [Ans C]
(i) I understand that the answer cannot be B because of the physiological environment (pH being between 7-8) in which the pacemaker is placed, but how exactly does that affect the Ecell, given that neither of the redox equations involve H+?
(ii) Is D incorrect because it is the LiI crystal which is being exhausted so the choice should have referred to an amount of LiI instead of Li(s) instead?
VJC/2015/P1/34 [Ans D]
Is choice 2 wrong because RCO2Na is oxidised to RCOO- and H2O is reduced to H radical? Although at first blush I might have accepted choice 2 because the R-R and H2 were formed thereafter from RCOO- and the H radical. What does choice 3 mean when it suggests that the redox/reduction potential is least/most positive?
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Hi Gohby,
DHS/2015/P1/11. For option B, the graph obtained is determined by stoichiometry, not by kinetics. Regardless of order of reaction for H2O2, you'll get the same graph anyway. For option C, you can't use the half-life method because H2O2 and H+ are not in large excess. Even if they were, you would only be able to determine the order of reaction of I- (which generates I2, the molarity of which is represented by the y-axis). But you can still compare gradient of tangent at initial to determine the order of reaction wrt H2O2 using option C. For option D, since the graphs are linear, H2O2 must be first order.
DHS/2015/P1/16. Adding H2SO4 protonates the NH3 ligands, consequently they can no longer function as ligands.
VJC/2015/P1/8. The conditions within the battery in the human body cannot possibly be at standard conditions (ie. body temperature isn't standard *this fact in itself will suffice for this MCQ*; furthermore molarities of Li+ and I- have no reason to be standard, even if they began as standard molarities, which is silly because that doesn't serve its medical function, they will no longer be standard as the reaction proceeds). For option D, a quick calculation reveals more than 9g of Li is required for the battery to last 5 years.
VJC/2015/P1/34. Option 2 is wrong because the OS of the alkyl group C atoms do not change, it's the OS of the carboxylic acid (acyl) C atoms which change. Option 3 is incorrect, because the redox potentials at the cathode and anode are determined by the voltage of the power supply, ie. this isn't a Galvanic-Voltaic cell, it's an electrolytic cell.
No prob, Gohby :)
Originally posted by gohby:Thank you UltimaOnline.
I have some other questions here. I've penned down my thought process and doubts wrt some of the individual choices (pls pardon the lengthiness!). Thanks again.
DHS/2015/P1/11 [Ans B]
I’m not too sure what to make out of the choices but here are my workings:
A: Order of reaction wrt H2O2 can be done by juxtaposing the gradient of [I-] taken at 0s (initial rate of rxn)
B: the absorbance corresponds to the amount of I-(which doesn’t block out light) vs iodine (which blocks out light). But how do I know I can’t get the H2O2 order?
C: No idea on how to analyse - firstly, the 2 graphs look significantly different for a diffenence in 0.02M of hydrogen peroxide. Secondly I tried to establish the half life of iodine (0->0.5->0.75…) by making a scaled drawing but got nowhere.
D: I think the 2 graphs with differing [I-] is jus a red herring. I can note the rate of reaction at a particular [H2O2] at either of the graphs, double that [H2O2], and see to what extent the rate increases
DHS/2015/P1/16 [Ans D]
Regarding choice C, why does the addition of H+ in step 3 (to a mixture of [Cu(NH3)4]SO4 + Sn2+ + H2O) cause the formation of [Cu(H2O)6]2+?
VJC/2015/P1/8 [Ans C]
(i) I understand that the answer cannot be B because of the physiological environment (pH being between 7-8) in which the pacemaker is placed, but how exactly does that affect the Ecell, given that neither of the redox equations involve H+?
(ii) Is D incorrect because it is the LiI crystal which is being exhausted so the choice should have referred to an amount of LiI instead of Li(s) instead?
VJC/2015/P1/34 [Ans D]
Is choice 2 wrong because RCO2Na is oxidised to RCOO- and H2O is reduced to H radical? Although at first blush I might have accepted choice 2 because the R-R and H2 were formed thereafter from RCOO- and the H radical. What does choice 3 mean when it suggests that the redox/reduction potential is least/most positive?
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Hi UltimaOnline,
I have some questions here which I'd like to seek your clarification -
Q1: Revisiting: HCI 2015/P1/18 [Ans C]
18 Which statement correctly defines a transition element?
A Transition elements exhibit more than one oxidation state in their compounds.
B Transition elements form many coloured compounds.
C Transition elements have partially filled d orbitals.
D Transition elements or their compounds are widely used as catalysts.
Although the definition talks about partially filled d orbitals, isn’t it an inaccurate definition of transition elements because it should have been the ions of the transition elements which have partially filled d orbitals (or else by this definition suggested in C, scandium would be a transition element and copper would not be regarded as one).
Q2: If I were to increase the concentration of the electrolyte, what will happen to the rate of electrolytic plating? On one hand there are sources to suggest that the rate of electrolysis will remain unchanged as the current remains unchanged, but this website (http://www.easychem.com.au/shipwrecks-and-salvage/3-electrolytic-cells/factors-that-affect-an-electrolysis-reaction) suggests that “Increased concentration of ions in electrolyte→ Increased current, and therefore increased rate of electrolysis”
Q3: In suggesting why phathalic acid is more acidic than terephthalic acid, the answer scheme suggests that the acid anion of phthalic acid is more stable than that of terephthalic acid as intramolecular H bonds can exist due to the close proximity of O with the negative chareg and H atom of the COOH group.
Ok, I don’t dispute the veracity of the explanation, but how does it afford an explanation to the question, since phthalic acid itself can also form intramolecular H bonds, which therefore increases its stability as well. Hence, I don’t see why the ability of phthalic acid anion to form intramolecular H bonds a good reason why it’s more acidic than terephthalic acid, given that the ability of the phthalic acid to form H bonds could well work against the formation of phthalic acid anion through its increased stability.Thank you UltimaOnline! :)
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Originally posted by gohby:
Hi UltimaOnline,
I have some questions here which I'd like to seek your clarification -
Q1: Revisiting: HCI 2015/P1/18 [Ans C]
18 Which statement correctly defines a transition element?
A Transition elements exhibit more than one oxidation state in their compounds.
B Transition elements form many coloured compounds.
C Transition elements have partially filled d orbitals.
D Transition elements or their compounds are widely used as catalysts.
Although the definition talks about partially filled d orbitals, isn’t it an inaccurate definition of transition elements because it should have been the ions of the transition elements which have partially filled d orbitals (or else by this definition suggested in C, scandium would be a transition element and copper would not be regarded as one).
Q2: If I were to increase the concentration of the electrolyte, what will happen to the rate of electrolytic plating? On one hand there are sources to suggest that the rate of electrolysis will remain unchanged as the current remains unchanged, but this website (http://www.easychem.com.au/shipwrecks-and-salvage/3-electrolytic-cells/factors-that-affect-an-electrolysis-reaction) suggests that “Increased concentration of ions in electrolyte→ Increased current, and therefore increased rate of electrolysis�
Q3: In suggesting why phathalic acid is more acidic than terephthalic acid, the answer scheme suggests that the acid anion of phthalic acid is more stable than that of terephthalic acid as intramolecular H bonds can exist due to the close proximity of O with the negative chareg and H atom of the COOH group.
Ok, I don’t dispute the veracity of the explanation, but how does it afford an explanation to the question, since phthalic acid itself can also form intramolecular H bonds, which therefore increases its stability as well. Hence, I don’t see why the ability of phthalic acid anion to form intramolecular H bonds a good reason why it’s more acidic than terephthalic acid, given that the ability of the phthalic acid to form H bonds could well work against the formation of phthalic acid anion through its increased stability.Thank you UltimaOnline! :)
Hi Gohby,Q1. Yes, you're correct, it's the qn's fault, but you still gotta choose the best option, which remains as C.
Q2. No change in rate, the website author is erroneously lumping together galvanic-voltaic cells with electrolytic cells.
Q3. You're right that H bonding is also possible before 1st proton dissociation, which would have the effect of weakening / decreasing Ka1.
However, upon deprotonation, the COO- group (having a negative formal charge delocalized by resonance over both O atoms) is a far stronger hydrogen bond acceptor (compared to the neutral COOH group), thus more significantly stabilizing the amphiprotic conjugate base, with the overall effect of strengthening / increasing Ka1.
No prob, Gohby! :)
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CIE M/J/16/Q4(a) Regarding drawing the structure, A and B are my suggested structure. And the most right is Googled Structure.
Are my Drawn Structure Correct? Also why does NH2 groups are in different planes?
Does it mean the drug is Tetrahedral, so it is not superimposed?
Additional Question, why the molecule is neutral? Despite consisting of C2O4 2- ion?
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Originally posted by iSean:
CIE M/J/16/Q4(a) Regarding drawing the structure, A and B are my suggested structure. And the most right is Googled Structure.
Are my Drawn Structure Correct? Also why does NH2 groups are in different planes?
Does it mean the drug is Tetrahedral, so it is not superimposed?
Additional Question, why the molecule is neutral? Despite consisting of C2O4 2- ion?
The google image's stereochemistry is a result of the cyclohexane adopting a non-planar conformation to maximize stability (each C atom is sp3, not sp2 like in benzene).The coordination geometry about the Pt2+ ion (the dipositive charge or +2 oxidation state of Pt, explains why the coordination complex is neutral, asked in another of your questions) has to be square planar, as predicted by the Jahn-Teller effect (Uni level Chemistry, for A level purposes, just accept and memorize the geometries, either tetrahedral or square planar, of the commonly encountered 4-coordinate complexes).
And because each ligand is bidentate, hence only the cis-isomer is possible, not the trans isomer. Hence no cis-trans geometrical isomerism is possible.
You might be creative and force a trans isomer by placing 1 ligand above and 1 ligand below the square planar complex, but it would be too thermodynamically unstable due to severe deviations from the ideal bond angles about the donor atoms vis-Ã -vis orbital hybridization concordant to VSEPR theory.
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Originally posted by UltimaOnline:
The google image's stereochemistry is a result of the cyclohexane adopting a non-planar conformation to maximize stability (each C atom is sp3, not sp2 like in benzene).The coordination geometry about the Pt2+ ion (the dipositive charge or +2 oxidation state of Pt, explains why the coordination complex is neutral, asked in another of your questions) has to be square planar, as predicted by the Jahn-Teller effect (Uni level Chemistry, for A level purposes, just accept and memorize the geometries, either tetrahedral or square planar, of the commonly encountered 4-coordinate complexes).
And because each ligand is bidentate, hence only the cis-isomer is possible, not the trans isomer. Hence no cis-trans geometrical isomerism is possible.
You might be creative and force a trans isomer by placing 1 ligand above and 1 ligand below the square planar complex, but it would be too thermodynamically unstable due to severe deviations from the ideal bond angles about the donor atoms vis-à-vis orbital hybridization concordant to VSEPR theory.
Okay thanks for answering :)
But despite the NH2 group exists in on different planes, it joins coordinate bond as a square plannar? And not exists in Tetrahedral.... huh? -
Originally posted by iSean:
Okay thanks for answering :)
But despite the NH2 group exists in on different planes, it joins coordinate bond as a square plannar? And not exists in Tetrahedral.... huh?
Due to Jahn-Teller effect, the coordination-complex has to be square planar, which means the N atom has to be co-planar or at least peri-planar with the other 4 donor atoms. -
oh dang didn't know Jahn Teller effect causes Square planar. As in because the effect is more commonly known for
[CuCl4]2- Tetrahedral
[Cu(NH3)4]2+ Tetrahedral
what are the commonly encountered 4-coordinate complexes?
[CuCl4]2- Tetrahedral
[Cu(NH3)4]2+ Tetrahedral
is [Ni(CN)4]2- tetrahedral or square planar?? differing source different info
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[ZnCl4]2−
[CoCl4]2-
for Fe3+ ans Fe2+ cation respectively, [FeCl4]− and [FeCl4]2−
?
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Originally posted by Flying grenade:
oh dang didn't know Jahn Teller effect causes Square planar.
what are the commonly encountered 4-coordinate complexes?
[CuCl4]2- Tetrahedral
[Cu(NH3)4]2+ Tetrahedral
is [Ni(CN)4]2- tetrahedral or square planar?? differing source different info
Go google out for urself, don't waste my time.If you find different geometries from different sources on the internet, and both/all sources seem reliable, then it could be because such coordination complexes could adopt more than one geometry. If this is indeed the case (case by case basis), then Cambridge will accept more than 1 geometry.
Above all, read the exam question and be exam-smart. Cambridge will give clues if it's some obscure complex that you're not expected to be familiar within, within the H2 syllabus. Be exam-smart and psychologically profile what Cambridge wants.
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didn't know Jahn Teller effect causes Square planar. As in because the effect is more commonly known for
[CuCl4]2- Tetrahedral
[Cu(NH3)4]2+ Tetrahedral
using phone sucks, many problems, laggy and i type slower and affect my thought process as i should already have written some words for the previous posts above