2012 H2 Chemistry Paper 1 (Answers)
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2012 H2 Chemistry Paper 1 (Answers)
Q1. C
Q2. C
Q3. D
Q4. B
Q5. C
Q6. A
Q7. D
Q8. C
Q9. A
Q10. D
Q11. D
Q12. C
Q13. B
Q14. A
Q15. A
Q16. D
Q17. D
Q18. C
Q19. C
Q20. C
Q21. D
Q22. B
Q23. B
Q24. D
Q25. C
Q26. A
Q27. C
Q28. B
Q29. B
Q30. A
Q31. B
Q32. D
Q33. A
Q34. B
Q35. B
Q36. B
Q37. D
Q38. A
Q39. B
Q40. B -
Hi sir can i please enquire on question 25. I always thought that in our syllabus we are supposed to know that its a basic hydrolysis reaction that involves water only and not oh- and the saponification mechanism? I struggled and chose to write 5 in the end as i thought the mechanism isnt taught in the syllabus:(
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Hiiii i'm sorry.
Do you have H1 chem answers??
I want it really badly!!!
Thanks..
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Dallas guo93, the line between what's testable and what's not testable, what's in the syllabus and what's not, is increasingly blurred every ongoing year for H2 Chem. So my advice is to just write what you feel is actually the most correct answer, and not worry whether it's in the syllabus or not.36/40 is certainly an A grade for Paper 1. As long as your SPA / Practical paper, Paper 2 and Paper 3 grades are also safely an A grade, then overall an A grade is well within your grasp, no worries.Mel12345, sorry I do not have the H1 answers.
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Hi thanks for uploading your answers!
For qn 15, why wouldnt changing the ligands result in a change in number of d orbital electrons?
For qn 31, since C loses one electron to the layer of delocalised electrons, why isnt the valency 3?
For qn 39, wouldnt option 2 affect both molecules as pi systems are present in both.
Thanks in advance!
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Originally posted by Chemquestions:
Hi thanks for uploading your answers!
For qn 15, why wouldnt changing the ligands result in a change in number of d orbital electrons?
For qn 31, since C loses one electron to the layer of delocalised electrons, why isnt the valency 3?
For qn 39, wouldnt option 2 affect both molecules as pi systems are present in both.
Thanks in advance!
For qn 31, since C loses one electron to the layer of delocalised electrons, why isnt the valency 3?
It depends on your interpretation or definition of "valency", and it's used rather loosely in Chemistry to mean either :
1) number of other atoms bonded with,
or
2) number of valence electrons (whether delocalized or not).
If the former, then statement 3 is correct. If the latter, then statement 3 is false.So it depends on Cambridge's idea of "valency", which is not actually specified in the H2 syllabus requirements.
Additionally, you've the argument that if one of the valence electrons is delocalized, it's not available for bonding. But Cambridge could argue that "delocalized" does not equate to "not available", afterall similar to benzene, it's possible to hydrogenate benzene into cyclohexane under extreme conditions.
This is a rather ambiguous question, and I disagree that Cambridge should use this ambiguous term. Regardless, most people (students and teachers) think the required answer is probably B (only statements 1 and 2 are correct). But then again, perhaps Cambridge takes one of the alternative viewpoints (as discussed previously), and the required answer might be A.
For qn 39, wouldnt option 2 affect both molecules as pi systems are present in both.
It's an issue of proximity. Just one atom away makes a lot of difference, when it comes to nucleophile & pi-bond repulsions, electron-withdrawing by induction effect, etc.
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For qns 15,
[Cu(H2O)6]2+ + 4Cl- ----> [CuCl4]2- + 6H2O
I thought the 6 H2O ligands would only be replaced by 4 Cl- lignads?
I remeber reading that Cl- ligands have a atomic mass of 35.5 which is much larger than that if water(Mr of 18). And due to the size of these larger Cl- ligands, only 4 of them are able to form dative bonds with the Cu2+ ion, giving rise to a Tetrahedral arrangement.
So there are only 4 chloride ligands that replace the water ligands.
Thus there should be a change in the no. of D electrons.
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And for question 31
I have also read that non metals like sulphur, phosphorus and carbon have a characteristic known as variable valencies.
For example, suplhur has a variable valency of 4 and 6.
In the case of q 39, they were asking specifically for the valency of the carbon in the lattice. the definition of valency as quoted from wikipedia is :chemistry, valence, also known as valency or valence number, is the number of valence bonds[1] a given atom has formed, or can form, with one or more other atoms. For most elements the number of bonds can vary.The valence of an element depends on the number of valence electrons that may be involved in the forming of valence bonds.
In the case for graphite, only 3 electrons were involved in forming 3 valence bonds. The fourth was delocalised and not involved in bond formation.
So if we interpret this in context of the question, shouldn't the valency be 3?
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Agreed, you're right. This is probably the trickiest question in the entire Paper 1.
I've edited my answer to show that the answer to Q15 should be A, rather than C.
Regarding the valency question, I agree with you that there's definitely ambiguity over this debatable question (based on your definition of valency, C in graphite indeed has a valency of 3). Either A or B are both arguably correct. Some JC teachers will think A is the more correct answer, while other JC teachers will think B is the more correct answer.
Unfortunately, it's probable that Cambridge has already made up their minds on this issue, and will stick to whichever answer they decided upon in their mark scheme (even as some Cambridge examiners will probably personally disagree with the mark scheme).
I personally feel that such an ambiguous / debatable question shouldn't be asked by Cambridge at all. But if forced to choose between statement 3 as true or false, I personally agree with you that it should be true (though the majority of students and teachers think it should be false).
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Thanks so much for getting back to the questions so quickly. Your answers and explanations are greatly appreciated! 36/40 according to your updated solutions, managed to complete the paper under 40 minutes. I think the common consensus is that the mcq was relatively easier with majority with scores above 30.
Nonetheless, thanks for your effort!
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Hi! :) Can I just ask why question 16 the answer is D? Isn't magnesium hydroxide sparingly soluble in water so it is still (aq)? Shouldnt the answer be B? Calcium hydroxide is more soluble than magnesium hydroxide and is also like limewater which is (aq), isnt it?
Back to question 15, I don't really quite understand the explanation but isn't both Cu.6H20 and CuCl4, the Cu ion as a charge of 2+, so isnt the number of d-electrons around copper be the same? Even though the number of ligands reduce but Cl and H20 do not even have d orbitals so shouldnt we just look at Cu2+?
So question 31, the answer is confirm B? Because I put A... ...
Thanks! :)
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Originally posted by Tyuenkw:
Hi! :) Can I just ask why question 16 the answer is D? Isn't magnesium hydroxide sparingly soluble in water so it is still (aq)? Shouldnt the answer be B? Calcium hydroxide is more soluble than magnesium hydroxide and is also like limewater which is (aq), isnt it?
Back to question 15, I don't really quite understand the explanation but isn't both Cu.6H20 and CuCl4, the Cu ion as a charge of 2+, so isnt the number of d-electrons around copper be the same? Even though the number of ligands reduce but Cl and H20 do not even have d orbitals so shouldnt we just look at Cu2+?
So question 31, the answer is confirm B? Because I put A... ...
Thanks! :)
Q16. Sparingly soluble = relatively insoluble. Which means it's more correct to write Mg(OH)2(s) as the major product, and Mg(OH)2(aq) as the minor product. (And if you only write one product, it should be the major product, not minor). B can occur, depending on whether you add limiting amounts or excess amounts of water. If excess, you get limewater Ca(OH)2(aq). If limiting, you can slaked lime Ca(OH)2(s).Q15. Oxidation State is not the same as Formal Charge. Only when there are no ligands present, then the OS of a metal can be correctly determined from how many of its d orbitals are occupied (and whether singly or doubly). As part of a coordination complex, you cannot tell the OS from the number of electrons in its d orbitals, as some of these electrons would have been donated as dative bonds from ligands.
Since Oxidation State = Formal Charge + Electronegativity considerations, and ligands (being non-metals and electronegative) are always more electronegative compared to the metal ion (being metallic and electropositive), meaning that the dative bond electrons are polarized back to the ligands themselves and not 'counted' when determining OS, hence the OS of the metal ion is unchanged, but the number of electrons in the d orbitals of the metal ion does change, when accepting ligands, or when substituting a different number of donor atoms from different ligands (such as in this question).
Q31. No, it's not confirmed. It's possible you may still get credit for this MCQ, if either :
a) Cambridge had decided right from the start that statement 3 should be correct (even if the majority of Singapore students and teachers think it should be wrong), or
b) Cambridge had initially decided that statement 3 should be wrong, but after realizing that many (albeit a minority) students and teachers think it should be correct, and after discussing amongst themselves, Cambridge may make a decision to accept both answers afterall.
Comments :
Q31 is the most ambiguous, debatable, and therefore flawed MCQ that Cambridge should never have included in the first place. In fact, this seems to be the most flawed question ever, even when compared to other ambiguous debatable Cambridge questions in past years. Now whichever answer Cambridge decides to accept (since it's highly unlikely for Cambridge to accept both answers), there will be many students who will be unfairly penalized, because "valency" can really be interpretated in two different ways even by professional chemists.
Q15 is the trickiest MCQ in Singapore-Cambridge A level history. I daresay over 90% of the entire Singapore cohort got tricked by this question. It's unclear if Cambridge knew exactly how tricky this question was when they set it, because listening to students feedback after the paper, I could tell that the majority of students choose the wrong answers, but for several totally different / distinct / separate reasons, making this MCQ extremely tricky on several different / distinct / separate counts / ways / levels.
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Ah slaked lime... forgot about that since never revise O'levels chem for a long time... :o
Okok thanks for the explanation! :) I get it already! Oh my goodness getting 31/40 is dangerous right? :( Or 32/40 (assuming q 31 is A)... :(
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Originally posted by Tyuenkw:
Ah slaked lime... forgot about that since never revise O'levels chem for a long time... :o
Okok thanks for the explanation! :) I get it already! Oh my goodness getting 31/40 is dangerous right? :( Or 32/40 (assuming q 31 is A)... :(
30/40 for Paper 1 should be approximately a borderline A grade (though this may be adjusted by the bell-curve). So it'll depend on how well you did for your SPA / Pract, Paper 2 and Paper 3. If you think (as many do) that you also got a borderline A grade for these other papers, then it's time to start praying to the God of the Bell-Curve.Huat ah!
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Originally posted by UltimaOnline:
Q15. Oxidation State is not the same as Formal Charge. Only when there are no ligands present, then the OS of a metal can be correctly determined from how many of its d orbitals are occupied (and whether singly or doubly). As part of a coordination complex, you cannot tell the OS from the number of electrons in its d orbitals, as some of these electrons would have been donated as dative bonds from ligands.
Since Oxidation State = Formal Charge + Electronegativity considerations, and ligands (being non-metals and electronegative) are always more electronegative compared to the metal ion (being metallic and electropositive), meaning that the dative bond electrons are polarized back to the ligands themselves and not 'counted' when determining OS, hence the OS of the metal ion is unchanged, but the number of electrons in the d orbitals of the metal ion does change, when accepting ligands, or when substituting a different number of donor atoms from different ligands (such as in this question).
For Question 15, d orbitals refer to the 4d orbitals am I right? During the formation of the 6 H2O ligands, the 4s,4p, and 4d orbitals are involved while during the formation of the 4Cl- ligangs only 4s and 4p orbitals are involved. Is this the reason why the number of electrons in the 4d orbitals change due to the 4d orbitals not being involved in the formation of Cl ligands?
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Originally posted by H4x0ru5:
For Question 15, d orbitals refer to the 4d orbitals am I right? During the formation of the 6 H2O ligands, the 4s,4p, and 4d orbitals are involved while during the formation of the 4Cl- ligangs only 4s and 4p orbitals are involved. Is this the reason why the number of electrons in the 4d orbitals change due to the 4d orbitals not being involved in the formation of Cl ligands?
Yes, you're correct.
Cu2+ is 4s0 3d9Accepting 6 dative bonds from 6 water ligands, we have
[Cu(H2O)6]2+ is 4s2 3d10 4p6 4d3
Substituting 6 H2O ligands with 4 Cl- ligands, we have
[Cu(Cl)4]2- is 4s2 3d10 4p5 4d0
Hence, the number of electrons in the 4d orbtals change from 3 to 0, and the electrons in the 4p orbitals change from 6 to 5.