2016 H2 Chemistry JC1 & 2 students post your questions here
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Originally posted by ACA-Milkshake:
2012 TYS Paper 2
Qns 5b(i): The answer that my school gave was that "All molecules present in the same chirality, that us, only one enantiomer is present." But can I just say that the lactic acid contains chiral carbons that's why it causes the rotation of th eplane polarised light?
Your school teacher's answer and your own answer are both incomplete. You also need to define what a chiral carbon is, and/or what an optically active molecule entails, as part of your answer. -
MJC 2015 Prelims Paper 2
1e) For the safety hazard, can I write chemical reagents like ethanal are corrosive when it is in contact with the skin and eyes?
3a) Where did the ans get 2/4 from? I am also quite weak at approaching this type of qns where do you start?
3d) Why doesnt the ans include bond length? How do you know when the ans requires you the write bond length or bond energy?
Will I be penalised if I write PCl5 at rtp instead of anhydrous PCl5 at rtp?
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Originally posted by Fxwhy:
MJC 2015 Prelims Paper 2
1e) For the safety hazard, can I write chemical reagents like ethanal are corrosive when it is in contact with the skin and eyes?
3a) Where did the ans get 2/4 from? I am also quite weak at approaching this type of qns where do you start?
3d) Why doesnt the ans include bond length? How do you know when the ans requires you the write bond length or bond energy?
Will I be penalised if I write PCl5 at rtp instead of anhydrous PCl5 at rtp?
Q1e) Ethanal is toxic and carcinogenic, not corrosive.
https://en.wikipedia.org/wiki/Acetaldehyde#ToxicityQ3a) Instead of blindly memorizing and applying a mathematical formula, use common sense. For a uninegatively charged species : when mass is 37g, deflection is merely -1 degrees. Hence when mass is 1g (ie. much lighter), deflection should concordantly be -37 degrees (ie. larger magnitude deflected). Since mass of He is 4g, hence deflection should be -9.25 degrees (if uninegatively charged). Since He nuclei is dipositively charged, hence deflection should be +18.5 degrees.
Q3d) Bond length here is only indirectly relevant (since bond length is inversely proportional to bond strength, ceteris paribus). It is bond strength which is directly relevant. Hence, there is no need to (ie. no marks awarded for) mentioning bond length.
Q4ai) It's ok if you don't specify "anhydrous", as long as you do *NOT* specify aqueous (aq) (which would be marked wrong).
Same for all reagents which need to be anhydrous, eg. LiAlH4 reducing agent, acyl halides for nucleophilic acyl substitution, Grignard nucleophilic & reducing reagents, AlCl3 or FeBr3 Lewis acid catalysts for halogenation via electrophilic aromatic substitution and for Friedel-Crafts alkylation & acylation, etc.
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(George Chong Inorganic Chemistry pg 4) when to apply I3- (aq) and I2 (aq) ?
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Originally posted by LavXuan:
(George Chong Inorganic Chemistry pg 4) when to apply I3- (aq) and I2 (aq) ?
Pg 4 is the contents page lah.When you carry out *oxidation* of the iodine, you can oxidize either I- or I3- into I2 (balance accordingly). When you carry out *reduction* of the iodine, you can reduce I2 into I- or I3- (balance accordingly). Depends on which formula the question uses. But both are correct stoichiometrically, so Cambridge will accept either (do you understand why?).
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yes
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I love chem
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I love chemistry!
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In grp ll, there is greater covalent character in mgcl2 than those below it but in group Vll, there is greater covalent character in silver astatide than in silver iodide though astatine is below iodine. Is it something about those cations in group ll having smaller radius and hence higher polarising power while for the anions under group Vll, larger anions are more polarisable...?
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Originally posted by Ng.keebin:
In grp ll, there is greater covalent character in mgcl2 than those below it but in group Vll, there is greater covalent character in silver astatide than in silver iodide though astatine is below iodine. Is it something about those cations in group ll having smaller radius and hence higher polarising power while for the anions under group Vll, larger anions are more polarisable...?
Correct.Because Group II or 2 ions are the cations with the polarizing power, hence the higher the cationic charge density and concordantly the greater the polarizing power up the Group, the higher the covalent character for the ionic compound.
Conversely Group VII or 17 ions are the anions with the polarizable electron charge clouds, hence the lower the anionic charge density and concordantly the greater the polarizability down the Group, the higher the covalent character for the ionic compound.
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Originally posted by UltimaOnline:
Correct.Because Group II or 2 ions are the cations with the polarizing power, hence the higher the cationic charge density and concordantly the greater the polarizing power up the Group, the higher the covalent character for the ionic compound.
Conversely Group VII or 17 ions are the anions with the polarizable electron charge clouds, hence the lower the anionic charge density and concordantly the greater the polarizability down the Group, the higher the covalent character for the ionic compound.
AlI3 has weaker Al-hal bond(since more covalent) compared to Alcl3?
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Originally posted by Flying grenade:
AlI3 has weaker Al-hal bond(since more covalent) compared to Alcl3?
That, plus less effective overlap of hybridized orbitals to form the sigma bond due to more diffused valence orbitals of the halogen atom down the Group. -
so AlCl3 has lower BP, but AlI3 is more susceptible to thermal decomposition?
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Originally posted by Flying grenade:
so AlCl3 have lower BP than AlI3 (since less electrons, less extensive id-id forced of attn) ,but AlI3 more susceptible to sublime/undergo thermal decomposition due to weaker Al-hal bond?
Partly right, partly wrong. The chemistry (including physical properties) of the different aluminium halides are not so straightforward (hence not in H2 syllabus), and will be confusing for A level students.If you're still interested, go through with your private tutor or school teacher.
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nice! ! Thanks , Ultima.
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HCI 2014 paper 2 Q4a) why doesnt the reagent dilute HNO3 need conc H2SO4 for electrophilic sub? Why dilute HNO3 dont react with compound C?
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Originally posted by nicolemantou:
HCI 2014 paper 2
Q4a) why doesnt the reagent dilute HNO3 need conc H2SO4 for electrophilic sub? Why dilute HNO3 dont react with compound C?
If the reagent were conc HNO3 + conc H2SO4 (ie. strong nitrating agent), then all 4 molecules would be nitrated. The question deliberately specified dilute HNO3 (ie. weak nitrating agent), to see if you can tell which benzene rings are sufficiently activated to react with dilute HNO3 (ie. weak nitrating agent).Only D is sufficiently activated, because of the 2 electron-donating by resonance groups. If you're wondering why only mononitration occurs, it's because while OH and OR are both activating as they donate electrons by resonance, OH is a stronger activator than OR (why? I've already explained this somewhere among my 100+ questions on my BedokFunland JC website, go ask your school teacher or private tutor if you don't know), *and* NO2 group is strongly deactivating (ie. electron-withdrawing by *both* induction and resonance). Concordantly, while D is the only molecule sufficiently activated to undergo nitration with dilute HNO3 (ie. weak nitrating agent), but only mononitration occurs, ortho to the OH group.
C is not sufficiently activated (actually it's not activated at all), because by induction, the vinyl side-chain's sp2 C atom (ie. not being an sp3 C atom) does not donate electrons by induction. By resonance, the vinyl side-chain is in equal measure electron-donating and electron-withdrawing (vis-Ã -vis the benzene ring), and is hence neither. Concordantly, C is neither an activated or deactivated benzene ring, and requires conc HNO3 + conc H2SO4 (ie. strong nitrating agent) to undergo nitration.
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Hi UltimaOnline, I have 2 questions here:
Q1: In a question which requires “state what you would observe when A is converted to B” or “state the observations”, would a response of “white fumes are evolved” (instead of “white fumes of HCl (g) evolved”) suffice? I am of the opinion that given one cannot possibly observe the identity of the fumes evolved, a response of “white fumes are evolved” is sufficient and correct, and any further details as to the identity of the gas is superfluous (and wrong).
Q2: NYJC/2012/P1/Q18 [Ans: D]
When I mix A & B or A & C together, I will obtain AgCl (s), which will be the white ppt soluble in ammonia as shown in row 1. However, I do not understand the observations shown in row 2. Mixing A & C should produce iron (II) nitrate and silver chloride - so where is there no white ppt produced, as per row 1, and why is there a grey ppt (most likely it is iron) formed?
Thank you, UltimaOnline :)
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Originally posted by gohby:
Hi UltimaOnline, I have 2 questions here:
Q1: In a question which requires “state what you would observe when A is converted to B� or “state the observations�, would a response of “white fumes are evolved� (instead of “white fumes of HCl (g) evolved�) suffice? I am of the opinion that given one cannot possibly observe the identity of the fumes evolved, a response of “white fumes are evolved� is sufficient and correct, and any further details as to the identity of the gas is superfluous (and wrong).
Q2: NYJC/2012/P1/Q18 [Ans: D]
When I mix A & B or A & C together, I will obtain AgCl (s), which will be the white ppt soluble in ammonia as shown in row 1. However, I do not understand the observations shown in row 2. Mixing A & C should produce iron (II) nitrate and silver chloride - so where is there no white ppt produced, as per row 1, and why is there a grey ppt (most likely it is iron) formed?
Q1: In a question which requires “state what you would observe when A is converted to B� or “state the observations�, would a response of “white fumes are evolved� (instead of “white fumes of HCl (g) evolved�) suffice? I am of the opinion that given one cannot possibly observe the identity of the fumes evolved, a response of “white fumes are evolved� is sufficient and correct, and any further details as to the identity of the gas is superfluous (and wrong).
Q2: NYJC/2012/P1/Q18 [Ans: D]
When I mix A & B or A & C together, I will obtain AgCl (s), which will be the white ppt soluble in ammonia as shown in row 1. However, I do not understand the observations shown in row 2. Mixing A & C should produce iron (II) nitrate and silver chloride - so where is there no white ppt produced, as per row 1, and why is there a grey ppt (most likely it is iron) formed?
Thank you, UltimaOnline :)
Hi Gohby, no prob :)Q1. Yes, you're correct.
Q2. Ag+ from AgCl(s) is reduced to Ag(s) a grey ppt insoluble in NH3(aq), and Fe2+(aq) is oxidized to Fe3+(aq).
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Hi UltimaOnline - I've some further enquiries here:
[YJC 2012 Prelim]
I get that there are 5 dative bonds in this complex, but how does that point to the electronic configuration of the cobalt cation, given that in this case, I think it would have been the 1x4s, 3x4p and 1x4d orbitals which will hybridise to accept the lone pairs from the dative bonding?
NYJC/2012/P1/Q17
My thoughts: When I add iron (II) sulphate to bromine and chlorine, a redox reaction will occur and Fe3+ & chloride/bromide will be formed. When I then add sodium hydroxide to the mixture, iron (III) hydroxide will be formed. However, how can I distinguish between the solution containing chloride and that containing bromide? According to the E⦵ values, neither chloride nor bromide will be able to reduce iron (III) hydroxide to iron (II) hydroxide.NJC/2012/P1/Q15 [Ans: B]
Upon heating with barium carbonate, the barium oxide formed will react vigorously with water - however, what is the gas evolved during the "effervescence" stated in the question? Unless the question used the term "effervescence" to describe the vigorous reaction - which would be an inaccurate characterisation, no?
Thank you for your help. :)
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Originally posted by gohby:
Hi UltimaOnline - I've some further enquiries here:
[YJC 2012 Prelim]
I get that there are 5 dative bonds in this complex, but how does that point to the electronic configuration of the cobalt cation, given that in this case, I think it would have been the 1x4s, 3x4p and 1x4d orbitals which will hybridise to accept the lone pairs from the dative bonding?
NYJC/2012/P1/Q17
My thoughts: When I add iron (II) sulphate to bromine and chlorine, a redox reaction will occur and Fe3+ & chloride/bromide will be formed. When I then add sodium hydroxide to the mixture, iron (III) hydroxide will be formed. However, how can I distinguish between the solution containing chloride and that containing bromide? According to the E⦵ values, neither chloride nor bromide will be able to reduce iron (III) hydroxide to iron (II) hydroxide.NJC/2012/P1/Q15 [Ans: B]
Upon heating with barium carbonate, the barium oxide formed will react vigorously with water - however, what is the gas evolved during the "effervescence" stated in the question? Unless the question used the term "effervescence" to describe the vigorous reaction - which would be an inaccurate characterisation, no?
Thank you for your help. :)
NJC Qn : this question has come up time and time again, because it is, as you suspect, a faulty question. While there is a slight difference in the color of solution (because color isn't just dependent on cation, but on ligand substitution equilibria as well), it is not sufficient to reliably distinguish between iron(III) chloride vs bromide. And while the differing magnitudes of thermodynamic feasibility translates into differing extents of equilibrium positions (and thus differing amounts of ppt obtained), this isn't an appropriate pedagogical assessment context for the H2 syllabus. In summary, this is a faulty question, but which by elimination still finds that option to be the most feasible (for MCQs, choose the 'most correct' answer, even if it isn't 100% correct).YJC Qn : for such questions in the context of the A level H2 syllabus, instead of considering the transition metal orbitals in which the dative bond electrons reside in, treat such questions as simply asking you to deduce the OS of the metal cation, and concordantly, the electron-configuration. In other words, disregard the dative bonds residing in the transition metal orbitals.
NJC Qn : BaCO3 is significantly more stable than CaCO3, and at temperatures which decompose CaCO3, BaCO3 remains undecomposed and thus results in "effervescence" when mixed with HCl(aq).
No prob, Gohby :)
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from qifei: in the reaction between mg and hcl, why is the effective collisions between mg atoms and hydrogen ions but not mg and cl- ions?
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Originally posted by Ng.keebin:
from qifei: in the reaction between mg and hcl, why is the effective collisions between mg atoms and hydrogen ions but not mg and cl- ions?
Because the reaction is *not* an ionic precipitation reaction (since MgCl2 is soluble), but instead it is a redox reaction (H+ from acid reduced to H2 gas, Mg oxidized to Mg2+) between Mg and H+ (whether from HCl(aq) or H2SO4(aq) etc).Edited to add : and it also can't be a redox reaction between Mg and Cl-, because both Mg and Cl- can only be oxidized, but not reduced further. A redox reaction needs someone to be oxidized, and someone to be reduced. Mg and Cl- are both reducing agents, so no reaction btwn them.
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Originally posted by UltimaOnline:
Because the reaction is *not* an ionic precipitation reaction (since MgCl2 is soluble), but instead it is a redox reaction (H+ from acid reduced to H2 gas, Mg oxidized to Mg2+) between Mg and H+ (whether from HCl(aq) or H2SO4(aq) etc).Edited to add : and it also can't be a redox reaction between Mg and Cl-, because both Mg and Cl- can only be oxidized, but not reduced further. A redox reaction needs someone to be oxidized, and someone to be reduced. Mg and Cl- are both reducing agents, so no reaction btwn them.
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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
