Marking scheme

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Marking scheme

帖子 由 Giraffe 于 周日 七月 26, 2009 11:04 am

Trial STPM 2008 CHEMISTRY PAPER 2

No.

Answer
Marks

1(a)
Electron deflected the most
Because it is lightest

1m
1m

(b(i)
1s22s22p63s1

1m

(ii)
5

1m

(iii)
2s
2p


1m

(iv)
It is unstable

1m

(c )(i)

(ii)
Correct of drawing.
Correct of labeling.
Correct of electronic configuration(represent by arrow)

1m
1m
1m

(iii)
Fluorine has very high ionization energy/high electron affinity

1m

2(a)(i)
Al3++3e→Al

1m

(ii)
1000x103x3x96500

27

=1.07x1010C

1m




1m


(iii)
Needed less energy/lower temperature

1m

(b)(i)
Zn(s)+2Ag+(aq)→Zn2+(aq)+2Ag(s)

1m

(ii)
E.m.f=E reduction-E oxidation

=+0.80-(-0.76)


=1.56V


1m

(iii)
Ecell=Eѳcell-0.059/2
log[Zn2+] /[Ag+2




1m

(iv)
Ecell=1.56+ 0.059/2log [Ag+] 2/[Zn2+]

1.21=1.56+ 0.059/2log[Ag+]2/1

[Ag+]=1.17x10-6moldm-6


1m


1m

(v)
Voltage will decrease
Reason: Addition of hydrochloric acid decreases the conc.
of Ag+ ions.

*Ag+(aq)+Cl-→AgCl(s)

1m

3(a)(i)
2Mg(NO3)2→2MgO+4NO2+O2

1m

(ii)
1.
Charge density decreases going down the group
2.
Polarizing power decreases. Furthermore the lattice energy of the oxide formed becomes less exothermic.

1m
2m

(b)(i)
NH3+HCl→NH4Cl

1m

(ii)
Ammonia:sp3 hybridisation
Ammonium ion: sp3 hybridisation

1m
1m

(iii)
Correct of drawing.

1m

(iv)
Make fertilizers

1m

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回复: Marking scheme

帖子 由 Giraffe 于 周日 七月 26, 2009 11:05 am

Marking scheme Q4-5




4(a)(i)
Nitrogen atom in ethanamine has a lone pair of electrons that can be donated /used to form a coordinate band with a proton

1m

(ii)
The lone pair electron of the nitrogen atom is delocalized in the benzene ring. They are less readily donated.


1m
1m

(b)(i)
Substitution

1m

(ii)
(CH3CH2)2NH
(CH3CH2)2N
(CH3CH2)2N+Br


3correct:2m
2correct:1m
1correct:0m

(iii)
CH3CN+2H2
(CH3CH2)2NH2
Catalyst: Ni
Conditions:150°C

Or
CH3CN+4[H]
(CH3CH2)2NH2
Catalyst: LiAl4 in dry ether


Better because no other products will be formed.

1m

@
1m


1m


©
Add aqueous bromine/aqueous iron (III) chloride.
Phenylamine forms a white precipitate /purple solution whereas ethanamine
Give no visible change.

1m
1m

5(a)(i)
Hund’s
rule states that in a given set of orbitals of equilvalent
energy,electrons tend to occupy the orbitals singly first before
pairing up.

Pauli exclusion principle states that each orbital can be occupied by two electrons of opposite spin
only .

Aufbau principle
states that electrons must occupy available orbitals of lower energy
first before they fill orbitals of higher energy.


1m

1m

1m

(ii)
No. of electons in O2- ion=8+2=10
Step 1: Apply Pauli exclusion principle and Aufbau principle. Fill 1s orbital with 2 electrons.
Step2: Fill 1s orbital with 2 electrons.
Step 3: Apply Hund’s rule.

Fill 2px, 2py, 2pz orbitals with 1 electron respectively.

Step 4: Fill the remaining 3 electrons into 2px, 2py, 2pz orbitals respectively.


1m
1m
1m

(b)(i)
Equation for overall reaction
2O33O2
Consider step 1:
Rate of forward reaction:k1[O3]
Rate of reversed
reaction:k-1[O2][O]

At equilibrium, rate of forward reaction= rate of reversed
reaction

k1[O3]= k-1[O2][O]
[O]=
……….(1)
Consider step 2:
Rate of reaction:k2[O3][o]………(2)
Sub (1)into (2)
Rate of reaction=k2x
[O3]

=

[O3]2[O]-1

1m


1m
1m

1m





1m
1m

(ii)
Initial step
CFCl3
∙ CFCl2+∙Cl
Propagation step
∙Cl+O3→ClO∙+O2
ClO∙+O→ O2+∙Cl
(O3→ O2+∙O)



1m

1m
1m

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回复: Marking scheme

帖子 由 Giraffe 于 周日 七月 26, 2009 11:05 am

Marking scheme Q6-8




6(a)
C(s)+ O2→C O2(g)
Hf=-394kJmol-1

Heat released by burning 2.40g of carbon=
x394


=78.8kJ

Heat =heat capacity x increase in temperature
78.8kJ=10kJ°C-1x ѳ
Ѳ=7.88°C

1m
1m

1m


1m
1m


(b)(i)
Correct labeling of y-axis and x-axis
Correct labeling vapour and liquid
Draw the curve correctly.

1m
1m
1m

(ii)
Show the path of fractional distillation shown in the diagram.
The solutions boils at a temperature lower than azeotopic mixture.
The vapour will contain more than 68% HBr.
Distillate: pure HBr
Residue: azeotopic mixture.

1m
1m
1m
1m
1m

(iii)
Total pressure of the mixture is lower than that predicted based on Raoult’s law
Because intermolecular force in mixture is stronger than the pure liquids mixture (show negative deviation)

1m

1m

7(a)(i)
The d-orbitals are split into 2 groups of slightly different energies.
Electrons from lower energy orbitals absorb visible light.
Jump to a higher
energy orbitals in a d-transition ,and reflect blue light


1m
1m
1m

(ii)
Blue colour of the solution is due to the presence of [Cu(H2O)4]2+complex.
When aqueous
ammonia is added, substitution of H2O molecules in the complex by NH3
molecules take place forming [Cu(NH3)4]2+complex that is blue.

This show that NH3 is a stronger ligand than H2O

1m
1m
1m
1m



(b)(i)
Cobalt(II),
because standard electrode potential has a positive value.

1m
1m

(ii)
Fe3++e Fe2+
E=o.77V

1/2 O2+2H+ H2O
E=1.23V

4 Fe2++ O2+4 H+→4 Fe3++2 H2O
E=2.ooV

Valence electronic configuration of Fe2+ is 3d6and Fe3+ is 3d5
Fe3+ is more stable because its d-orbitals are filly filled.

1m
1m
1m
1m
1m

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回复: Marking scheme

帖子 由 Giraffe 于 周日 七月 26, 2009 11:05 am

Marking scheme Q8-9




8(a)
Lead-has metallic structure.

-consist of metallic bond between the lead atoms.


- The metallic bonds are strong and needs high energy to be overcome
thus high melting point

PbCl3-ionic lattice

-ions are held together by very strong ionic bond.

PbCl4-simple molecular structure

-molecules are held together by weak Van der waals forces.



1m
1m


1m
1m
1m
1m

(b)(i)
3 Cl2+6NaOH→5Na Cl+ Na ClO3
Disproportionation
reaction
take place.

Cl2 is oxidized to
ClO3- in sodium chlorate and reduce to Cl- in sodium chloride .Both
products are soluble in water forming colourless solution.



1m
1m
1m

(ii)
Cl2(g)+2KBr(aq)→ 2K Cl (aq)+Br2(l)
A displacement
reaction has taken place. This is also a redox reaction where chlorine
undergoes reduction (Cl2+2e→2 Cl-) and bromide ions have undergoes
oxidation(2Br → Br2 +2e)



1m
1m
1m

(iii)
CH2= CH2+ Cl2→ CH2 Cl CH2 Cl
An electrophilic addition reaction takes place/chlorination reaction take place.
Cl2 act as an electrophile.


1m
1m
1m

9(a)(i)
Rate=k[(CH3)3C-Br]
Mechanism:
CH3)3C-Br→ CH3)3C++Br-
(slow)

CH3)3C++OH-→ CH3)3C- OH
(fast){correct label slow@fast-1m}

The slow step is the rate determining step.
The reaction is a first order reaction.

1m

1m
1m
1m
1m

(ii)
Aqueous sodium hydroxide and ethanolic silver nitrate
1-
chlorobutane-No visible change
2-

2-chlorobutane-White precipitate after several minutes

3-

2-chloromethylpropane- White precipitate formed immediately.



1m
1m
1m
1m

(b)(i)
Saponification /alkaline hydrolysis
C17H35COO CH2
CH2OH



I


I

C17H35COOCH
+3KOH→ C17H35COOK+
CHOH



I



I


C17H35COOCH2
CH2OH

Importance of C17H35COOH :as soap

As milder and softer soap.



1m
1m





1m

(ii)
C17H35 CH2OH
CH2OH CHOHCH2OH

1m
1m

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回复: Marking scheme

帖子 由 Giraffe 于 周日 七月 26, 2009 11:05 am

Marking scheme Q10




10(a)(i)
2-amino-3-methylpentamide


1m

(ii)
It show optical isomerism
Because the presence of a chiral carbon.
Drawing of isomers.

1m
1m
1m

(iii)

CH3
O
CH3
O



I
I I
PCl5


I
I I


CH3CH2CHCHC – OH CH3CH2C-C-C – Cl

I


I


NH2
NH2


conc. ammonia


CH3
O


I
I I



CH3CH2CCHC – NH2



I


NH2


PCl51m
1m



1m

(b)(i)
The linking of monomers containing c=c bond to form one product.

1m

(ii)

Cl
H
Cl
H




I


I
I


I


n

C=C
→ -


-

I


I
I


I
n


Cl
H
Cl
H





1m
1m
1m

(iii)
Cl
aton is electon withdrawing group.

It stabilies the intermediate anion formed.

1m
1m

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