Thanks for making the effort and taking the trouble giving a
detailed look at the entire P2, MapPwner, appreciate it :)
Incompletely Compiled chemistry suggested answers so
far(Answers may not be fully correct so take it with a pinch of
salt):
Qn 1(Ionisation energies of element A(chlorine) and hydrogen
properties)
Increase in I.E as successive electrons are removed from element
A(1m),as it is increasingly difficult to remove a negatively
charged electron from a positively charged element A ion(1m).Large
increase from 7th to 8th i.e as the 8th electron is located in an
inner electron shell,which is closer to the nucleus and thus more
energy required to overcome the strong electrostatic forces of
attraction,resulting in large rise in I.E.(2m)
Element A is chlorine(1m).1s^22s^22^p63^s23p^5.(electronic
configuration-1m).
Hydrogen can be classified as a grp 1 element as it has 1
valence electron in its outermost shell(1m).
Cannot be classified:
Grp 1 elements-metal lattice structure vs simple covalent
molecule structure for H+describe(1m)
Contrasting properites=grp 1 higher melting/boiling point than
H/H2+accounting for the reason why(1m)
Qn 2(Acids and buffers of HCO2- and blood)
part i)ph=2.60+showing calculation(3m)
(ii)forgot the answer
(iii)ph will fall if no buffer system in a human while
exercising
Qn 3,4,5(C2H4O/cinammealdehyde and transition
elements)
qn3)
(i)showing molecular formula is c2h4o(relatively
straightforward)
(ii) structure is a triangular molecule with c-o-c bond and c-c
bond between both c for c2h4o,with remaining h bonded to the 2 c
ensuring octet rule fufilled.
Reaction with Br2(aq)=orange br2 decolourises rapidly to
colourless
Na2co3:No vigorous effervescence,production of co2 observed.
Kc is small which thus show most of the product is the (LHS
one-I forgot the name) and thus position of equilibrium lies more
to the left(1m).Kc shows the extent of favourability of product
formed in the equilibrium and can be used to determine the amount
of product formed based on Kc value.''If Kc small=poe left while if
Kc large=poe right.''(Stating something like this is 1m I
think)
(iv) memorise balanced eqn of triiodomethane with ch3cho
cant really remember much for qn 3...
qn(4)
for cinammealdehyde,1 structure still has c=c and aldehyde just
that it is arranged at different positions(swapped)=positional
isomerism,while the other only has an aldehyde group and no alkene
group=structural isomerism(Or show cis-trans version for one of
them).
After that is Hcn(aq),trace amount of nacn(aq),in the cold,10 to
20 degree.
when the nitrile compound is reduced as shown in question by
h2,alkene and nitrile is reduced,but for lialh4 alkene is not
reduced.
behind the mechanism is nucleophilic addition(relatively
straightforward)
Generation of Hydride ion:BH4- ->(reversible arrow) H^-
+BH3.
Then the slow and fast step relatively standard.
Why lialh4 a stronger red. agent than nabh4=orbital overlap of
Al and H in AL-H bond or different polarising power of Li+ and Na+
of anion electron cloud to release hydride ions.
the naoh thingy just dilute the thing(1m) and state how to
deduce order+brief procedure of expt such as recording time taken
for ppt formation at diff concentrations via dilution and comparing
it with diff expt to determine order(2m)[I made naoh more
concentrated but as my back part is right i probably lost 1 mark as
only dilution is accepted to concentration lower than 2mol
dm^-3].
Graph shows a first order downward sloping curve with constant
half life(reaction time) halves for a doubled increase in
successive concentrations of naoh up to 2moldm^-3.[I labelled
beyond 2 moldm3 so if lenient i lost 1 mark at most,if strict 2
marks lost].[I forgot whether this was part of qn 4 or 5]
qn 5
for the electrolysis part,delta g is -284.0kj mol-1(to 3
sf)=reaction thus spontaneous at r.t.p
Define homogenous catalyst(2m),show how Fe2+ catalyses using
appropriate equations in the reaction of S2O8^2- + 2I^- -> I2+
2SO4^2- (2m) and state the property,which is able to form one or
more stable ions with variable oxidation states(1m).
Contrast between colour for the 3 mark question of iron
compounds[solid hydrated fe3+,aqueous fe3+ and solid hydrated fe2+]
is relatively ok
Purpose of using c6h5cooh:It is very reactive/combustible.
Water jacket purpose:Ensures temperature of surrounding water at
same temperature as calorimeter thus no heat loss.
Presence of black solid:Incomplete combustion of c6h5cooh
forming carbon,thus suggest excess use of o2.
to find heat capacity of calorimeter is -(q x average temp rise
for sample 2 and 3)/amt of benzoic acid in moles=-3230kj
mol^-1,where q is the heat capacity of calorimeter.One should end
up with something like -(qx30)/0.05= -3230kj mol^-1 before
eventually arriving at 5.38kj bla bla for q.
the below 3 marks shld be fine.
Napthalene(Moth Balls) qn:
-590kj mol -1
ai value higher thus molecule very stable due to p orbital
overlap between singly bonded and doubly bonded carbon atoms
results in extensive delocalisation of electrons(or words to the
effect).
sp2 hybridisation,draw out the hybridisation between two c
atoms.c=c is 1 sigma 1 pi,c-c technically is 1 sigma only but cos
of delcoalisation of electrons is somewhat 1 sigma 1 pi too for
bonds.Thus only 1 diagram to show hybridisation is sufficient
between sp2-sp2.
Showing is c10h12 is fine.
drawing of structures too and reason.
Reagent and condition:H2(g),Nickel catalyst,200 degrees.
Regarding the friend who found it easy,yeah when he stepped out
of the hall he said only lose a few marks but after discussing most
of the paper that snowballed to 25 marks.
My class on average the B graders(only 1 person) lost about same
marks as me(15-17,he thought the paper was easy as he felt he only
lost 2 marks but guess not),the rest probably might have done
worse,another i asked lost 22 marks supposedly(he got E for
prelims).Rest lost 25-30 or so.But i estimate the rest to lose on
average 25-30 or so in all seriousness,as one couldnt even decipher
the skeletal formula given in one of the question to deduce proper
reagents and conditions.
Of course i have friends in AJ losing same marks as me or even
more,up to 30.My Hwa Chong friend who did much worse than me for
prac aced this paper,lost about 5-6 marks
I too,felt the paper was actually simple(except for qn 6 and
7,but qn 7 was like a repeat question from my secondary sch sec 2
chem paper so it was actually familiar),but panic probably killed a
lot of people.
If i were to give an honest evaluation i would rate Qn 1 to 2 as
easy(but i screwed up question 2 quite badly),qn 3 to 5 as
moderate,qn 6 and 7 as moderate-challenging difficulty.
Question 6 especially,as the question was unusual in the sense
we need to state the purpose of benzoic acid that makes it useful
for this calculation of heat capacity of calorimeter in a
water,without considering about the water.
Majority of the ppl were confused(>50% in my school) and
other JCs,they didnt use -(q x(avg temp rise)/amt of benzoic acid
in moles for enthalpy change of combustion,but other ways,which
were wrong.
Organic chemistry component was moderately difficult to me,yet a
fun question to do even though it was a pain doing so as it was
really interesting for me(only lost 1 mark for organic component
part).
Ahh, so your top 10% friend didn't do as well as he initially
thought, but prolly still well enough to get A grade. In contrast,
your HCJC friend really only lost about 5 marks for P2, enough to
make up for his P4 errors. So MapPwner, yourself, you're well on
track for A grade (so far overall score > 80%, yes?).
Agree with you that the calorimetry question is a good
challenging question for Singapore JC H2 students. I can actually
think of 5 different reasons why benzoic acid is used as a standard
for calorimetry (just as I can think of 3 different reasons why
LiAlH4 is a stronger reducing agent than NaBH4), but Cambridge as
expected, is already very lenient : any 1 reason (for both the
benzoic acid and LiAlH4 questions) will suffice to score the 1 mark
allocated. But the majority of Singapore JC H2 students are
unlikely to have prior exposure to such questions, which is why
these are, all things considered, good and fair distinction
questions (not too unreasonable as to require 3 reasons for each
question, but challenging enough to differentiate A graders from
the rest).
Similarly for the Naphthalene aromaticity question (from your
description, the entire question is about Naphthalene being
aromatic with resonance delocalization of pi electrons, while in
the basic H2 syllabus only the simplest aromatic species, benzene,
is taught), another excellent question that (from your
description), didn't disappoint me, in the sense it is interesting
and A grade differentiating.
But as you (and your sarcastic school teachers) pointed out,
Cambridge has already been very fair, since these are only a
sprinkling of challenging questions amidst mostly easy or standard
questions. So (unless P3 and P1 are hellish... hell yeah! I should
set the A level papers Lol!) I doubt the bell-curve will deviate
too much from past years.
Keep up your excellent A grade performance, MapPwner! :)
Overall score no lah how over 80% lol,my prac was rather
bad.Prob for prac alone is 60%,this paper is 77-80%(pretty sure my
estimates are very accurate,in line with my estimates for past chem
papers for my exams this year,even if they deviate the score is
higher than the estimated score).
My prelim raw % scores for individual components were
p1:80%,p2:72%,p3:76%,p4:56%,so more or less similar to a
levels.
As with regards about my HC friend,she probably will lose way
more than 6 marks,i estimate eventually same marks lost as me as
she only immediately excluded those she know is definitely
wrong.
Nice nice. So that means you're on par with, or better than,
both your JC's top 10% friend, and your HCJC friend.
:)
Good good, as long as you don't cockup your P3 and P1 (ie. about
same performance as your Prelims), your overall score as a safe
estimate will be between 75% to 80%, so that's a guaranteed A grade
(as I said before, when students are shocked they didn't get A
grade, it's not that the A grade boundary > 75%, it's that they
overestimated their own scores, but this shouldn't be an issue for
you MapPwner).
So P2 was mostly easy to average, with only a couple of tough
questions. But since you're the top 20% of the national cohort,
MapPwner, perhaps the average Singapore H2 student found this paper
mostly average in difficulty, with quite a few tough questions. But
that's still totally within expectations of a Cambridge paper, so
the bell-curve shouldn't have too many surprises this year.
