2010 H2 Chem P3 a killer paper
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I just heard my students say today's 2010 H2 Chem P3 was an absolute killer paper. Can someone post some of the killer questions here?
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- Draw dot-and-cross diagram for NO2 molecule. It contains a dative covalent bond.
- FO2 does not exist, but ClO2 does. By considering possible types of bonding in both compounds, suggest reasons for the difference.
- When CuO is stirred with NH3(l), it dissolves to give a coloured solution. Oxide ion is acting as a Bronsted base. Suggest the equation for the rxn and colour of soln.
Pissed off losing marks at the deductive question 3.
EDIT: Lost 22 marks minimum in total. Need to work for the other papers to get the 2.5% back
EDIT again: By the way for the CuO and NH3 question, it is in part (c). part (b) already has a question which tests on Cu2+ dissolving in NH3 (dropwise to excess) with equations + observations. So unless they are that kind to repeat the same answering content (which I highly doubt), the colour of soln should not be deep blue again.
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I'm hearing, "This year's 2010 Cambridge paper is tougher than ALL twenty JC's 2010 Prelim Papers!"
Unprecedented in history of Cambridge 'A' levels.
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Yes. It was tougher than RJC and HCl which I thought were very easy (paper 3 at least). But this A-level has set the standard way higher. Too high for me to reach :(
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Originally posted by Garrick_3658:
- Draw dot-and-cross diagram for NO2 molecule. It contains a dative covalent bond.
- FO2 does not exist, but ClO2 does. By considering possible types of bonding in both compounds, suggest reasons for the difference.
- When CuO is stirred with NH3(l), it dissolves to give a coloured solution. Oxide ion is acting as a Bronsted base. Suggest the equation for the rxn and colour of soln.
Pissed off losing marks at the deductive question 3.
EDIT: Lost 22 marks minimum in total. Need to work for the other papers to get the 2.5% back
EDIT again: By the way for the CuO and NH3 question, it is in part (c). part (b) already has a question which tests on Cu2+ dissolving in NH3 (dropwise to excess) with equations + observations. So unless they are that kind to repeat the same answering content (which I highly doubt), the colour of soln should not be deep blue again.
Yes, I agree that these 3 questions (and the other killer questions of the paper, if they be of similar difficulty) are indeed unfairly difficult for H2 'A' levels standard.-------------------------------------
Structure of NO2 :
NO2 has an unpaired electron (free radical) on N, which is doubly bonded to an O atom, and singly datively bonded to another O atom. The N atom has a +ve charge, the singly bonded O atom a -ve charge, and the doubly bonded O atom has no charge. N lacks a stable octet in NO2, and the molecule is a free radical.
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Why FO2 doesn't exist :
Reason #1
The problem with FO2, is that F (being in period 2), cannot expand its octet (since it doesn't have vacant, energetically accessible 3d orbitals to use). Hence, it cannot form a similar structure as ClO2, which has expanded its octet (4 bond pairs + 1.5 lone pairs on Cl = 11 electrons in terms of an expanded octet).
Reason #2
For FO2 to have single bonds between the F and two O atoms, F will gain a 2+ charge and each O atom with gain a -ve charge, and this separation of charge is destabilizing, since F is more electronegative than O.
Reason #3
In addition, in such a model, F atom would have an unpaired electron, making the molecule an unstable free radical.
Reason #4
And furthermore, the F atom wouldn't have a stable octet in such a model, which defeats the purpose of forming the structure in the 1st place.
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CuO reacting with NH3 :
The oxide anion (ie. the base) abstracts a proton from NH3 molecule (ie. the acid), generating first the OH- ion and NH2- ion. These two anions, together with four other NH3 molecules, act as ligands to donate dative bonds to the Cu2+ ion, generating an octahedral complex ion.
Another possibility is that the OH- abstracts a further proton from another NH3 molecule (since liquid ammonia used means NH3 is present in large excess), generating H2O and another NH2- ligand. In such an octahedra complex ion, two ligands are NH2-, one is the H2O generated, and 3 are NH3 molecules.
The resulting complex ion should still be deep blue in colour, or a slight variant (eg. cyan or violet) of the deep blue colour of the usual tetraaminecopper(II) complex ion, since the Cu2+ ion remains as Cu2+, and the ligands are similar.
CuO + 6NH3 --> Cu(NH2)(OH)(NH3)4or
CuO + 6NH3 --> Cu(NH2)2(H2O)(NH3)3
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Glutamic acid: HOOCCH2CH2CH(NH2)COOH (they gave skeletal but i cant draw it here)
There are 3 pKa values associated with glutamic acid: 2.1, 4.1, 9.5
Make use of these pKa values to suggest major species present in solutions of glutamic acid with the following pH values
pH1
pH3
pH7
pH11
Can't believe I did this question...
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Originally posted by TenSaru:
Glutamic acid: HOOCCH2CH2CH(NH2)COOH (they gave skeletal but i cant draw it here)
There are 3 pKa values associated with glutamic acid: 2.1, 4.1, 9.5
Make use of these pKa values to suggest major species present in solutions of glutamic acid with the following pH values
pH1
pH3
pH7
pH11
Can't believe I did this question...
pH 1
HOOCCH2CH2CH(NH3+)COOH
pH 3
HOOCCH2CH2CH(NH3+)COO-
pH 7
-OOCCH2CH2CH(NH3+)COO-
pH 11
-OOCCH2CH2CH(NH2)COO-
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Originally posted by UltimaOnline:
pH 1
HOOCCH2CH2CH(NH3+)COOH
pH 3
HOOCCH2CH2CH(NH3+)COO-
pH 7
-OOCCH2CH2CH(NH3+)COO-
pH 11
-OOCCH2CH2CH(NH2)COO-
I got the 1st, 2nd and 4th correct... But the 3rd 1 I deprotonate the NH3+ instead of side chain COOH T_T
Anyway how does pKa relate to major species? And how to know which pKa belong to which?
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Originally posted by TenSaru:
I got the 1st, 2nd and 4th correct... But the 3rd 1 I deprotonate the NH3+ instead of side chain COOH T_T
Anyway how does pKa relate to major species? And how to know which pKa belong to which?
The idea is to, as you go from acidic environment to alkaline environment, deprotonate step-by-step one at a time (initially in very acidic conditions all groups are protonated; next deprotonate 1st the alpha carboxylic acid, then 2ndly the R group carboxylic acid, then 3rdly the ammonium group).The alpha carboxylic acid is stronger than the R group carboxylic acid, due to the electron-withdrawing by induction effect of the (+vely charged) amine group.
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Hm, but schools have been teaching the -NH2 group as strongly electron-releasing. Such as the case of phenylamine.
Looks like this part of the question requires critical application of context that it is -NH3+ and not NH2
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I find the CuO/ NH3 qn very interesting.
As said before, CuO + NH3 ---> CuOH + + NH2-
But I think OH- will react further with NH3 to produce H2O and NH2-
Furthermore, Cu, being d9 in most of its complexes is 6-coordinate (octahedral ideally)
So I think the final complex is Cu(NH3)5(H2O) 2+
Therefore the eqn is CuO + 8NH3 --> [Cu(NH3)5(H2O)]2+ + 2 NH2-
As for the colour, I don't know how to predict.
But since Cu(NH3)6 2+ is purple and Cu(NH3)4(H2O) 2+ is deep blue,
i dare say the colour is purplish blue...(haha)
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Originally posted by Chemfreak022:
I find the CuO/ NH3 qn very interesting.
As said before, CuO + NH3 ---> CuOH + + NH2-
But I think OH- will react further with NH3 to produce H2O and NH2-
Furthermore, Cu, being d9 in most of its complexes is 6-coordinate (octahedral ideally)
So I think the final complex is Cu(NH3)5(H2O) 2+
Therefore the eqn is CuO + 8NH3 --> [Cu(NH3)5(H2O)]2+ + 2 NH2-
As for the colour, I don't know how to predict.
But since Cu(NH3)6 2+ is purple and Cu(NH3)4(H2O) 2+ is deep blue,
i dare say the colour is purplish blue...(haha)
Agree that Cu2+ would probably be 6-coordinate and octahedral.OH- is a weaker base than NH2- (put in another way, H2O is a stronger acid than NH3) and under standard molarities, OH- wouldn't usually be able to abstract a proton from NH3. Furthermore, NH2- is a stronger ligand than NH3, and when generated, the NH2- is in close proximity to the Cu2+ ion and consequently has a greater propensity to complex with Cu2+. However with liquid ammonia (ie. extremely high molarity of NH3), your suggestion in regards to the above points may indeed be possible.
Your colour deduction is also reasonable.
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Hi, answers for today's paper 1 anyone?
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Since I don't have the question paper, I can't verify that these answers are entirely correct (there are probably a couple of questions in which the answers are debatable, even amongst teachers). But the MJC teachers probably checked the answers amongst themselves, so I expect these answers to be reliable.
Posted elsewhere on the internet :
as most of u shld know, mjc teachers have worked out their answers
BBBDB CAADB BDCBC BCCBC DBBDD ABCCD CBBAB BCAAC
these are not my answers, these are mjc's answers
but bear in mind, they are not cambridge's answers