Stpm Chem p1_p2 Skema 2011
-
Upload
acyl-chloride-hariprem -
Category
Documents
-
view
235 -
download
0
Transcript of Stpm Chem p1_p2 Skema 2011
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 1/21
,.,..-. ~ ~ ~ ~ " " ~ " "(] ' i f o ! ' l ~ [f;(gfl?f'lRI'YMIII'!Jii!)IIYid""'Xilfl;lJj}!f}V f O R f ' ! / O ' F i f O Y i J l l i i ' ! M I Y I ~ . i i ij;i 962/2 ~ , : . ; a ~ " : : ~ = = ~ a = = ~ = ' = : / : : ~ ~ tt; \1-'t2JlljiFUJ!JL;tlYJIIl!/J L f ! L ; # r ' D J P I ! ; i ! ~ J J : J f , . ' : f ~ · ~ i f ; l l J L \ A f ; f $ ! 1 i l l i U / f r ~ _ i t - / . 2 / A l ~ ~ , ~ ; l ! J l f f t q l j ® J i l / U J f O R f ' ! ! O ' F i i ' Y i X I I I ' ! I i l l i l r : l h ~ · f J j ) f j ~ ffJ{}jj},!O'Fii'YiXI"ffEiiJflj R O i f i J ! i Y J · 0 Y g l { l ~ l f l O i ( l i i . 1 L ; J l i t ~ l ~ J 7 . l ' ! i l N ' l i i l l f " @ ! ' . . CHEMISTRY . " i f ' L ! ' M J . l ' ; J G l ~ . . . r!'{.UiifJJJ.Ij{$'/}Ui'
~ f l 1 { { 1 f £ 1 f l f ! l f f l l f } f f J ~ V u i f c y f ' ' i i i l 0 ! ! 1 l ' ; J G ~ f%U:Jlf1Mf;cy'!;uV /JYlPJi1BijlfmjfffilJ}fl)flNf,1}j .. ·~ - ~ t l t f l ~ f f f ! J J ? ~ f m ; f % f l f f i ! l } j l J O J ! a ~ f ! J ! I P . ~ ~ @ i i f j l i 1 } i t i J J i M J ! ' .w ~ ~ • v f f { l } ~ ~ · ? J ! J a v ~ ~ ~ m , ~ • 1 l t r m ~ r ~ t J J J : f W ~ J i f f f f J ! M i l l : ~ f l j j , ~ I ! J W ~ I m i ~ ~ 1 M l I J & ? ~ ~ - ~ f f & P ~ J ~ J i J M 1 ! 7 ! t f l y V & W l l : f f f U ~ @ i i ! J l f i 1 J : t d E V ~
PEPERIKSAAN PERCUBAAN SETARA STPM
NEGERI SEMBILAN DARUL KHUSUS
TAHUN 2011
MARKING SCHEME
PAPER 1
PAPER2
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 2/21
, . · · · ~ · · · . . . . _c·
1 B
2 B
0 D
4 D
5 c
6 c
7 A
8 D
9 A
~ A
PEPERlKSAAN PERCUBAAN BERSAMA NEGERI SEMBILAN
STPM 2011
'"'
11 A
12 c
13 c
14 A
15 A
16 c
17 c
18 c
19 D
20 D
ANSWER SCHEME
PAPER 1 (962/1)
21 B 31
22 D 32
23 D 33
24 B 34
25 D 35
26 D 36
27 D 37
28 c 38
29 D 39
30 A 40
A 41 A
-B 42 B
c 43 B
D 44 A
D 45 B
c 46 c
A 47 B
B 48 B
B 49 A
D so c,
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 3/21
-" NUMBERI
2
"
4
5
6
7
PEPERIKSAAN PERCUBAAN BERSAMA NEGERI SEMBILAN ..
ANS\YE.R
B
B
.
D
D
.
c
c
A
-
STPM2011
ANSWER SCHEME
PAPER! (962/1)
EXPLANAT10N ·
Number of proton= Number of electron= 17
.. · · ~ ·
Number of neutron =Nucleon number-Number ofproton
= 35- 17
= 18
The smaller the m/e, the larger the angle of deflection
A 36 y + m/e = 36
B 36 y 2+ m/e = 3612 = 18
c 38 y + m/e = 38
D 38 y 2+.m/e= 38/2 = 19
28 X : 18 [ Ar ]3cfs- 4 s·
Y2+ ion has 28 electron since it is isoelectronic with X
Y atom should have 30 electrons
Y: 1s [Ar]3d10
4s2
Y2+: 1s [ Ar] 3d 10
Visible light region involved transition of electron from higher
energy level to n 2 ( energy level 2)
A= 400 nm involved n "' --+ n 2
A= 700 nm involved n 3 -- + n 2
Line J involved energy level in between n "' and n3
Physical properties for transition elements :High melting point, high
boiling point, high density & good conductor
Reason : Because of energy gap between 3d orbitals and 4s orbital is
very small, and hence the electrons from 4s and 3d orbitals are
availableto
become de1ocalized electron which contribute to strongmetallic bond and very good electrical conductivity.
The type of bond found in Caesium, Rubidium and Potassium is the
metallic bond. The melting point lowers while descending
Group 1
NH4+ has a tetrahedral shape with 4 bonding pairs whereas l of the
bonding pair is a dative bond.
SF 4 has a see-saw shape with 4 bonding pairs and l lone pair.
XeF4 has a square planar shape with 4 bonding pairs and 2 lone pair.
!Cl4' has a square planar shape with 4 bonding pairs and 2 lone pair.
2
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 4/21
8 D Rate of reaction(R) =k[P]
lsing (A) as the reference, the rates are
A is R, B is y, R, C is Y. R and D is 3/2 R
9 A l- t is consumed during the reaction (step 2 iuid3 j.}:!ence, !tcannot 1·· .
be a catalyst.. . . ' '- · · < ~ - - h · . ; .. . .. . ~ 4 . : ; , ,. -- . ' ' ,7]' ' ' : :;;c;u.;io\d'
•
10 A D The presence of catalyst increase the rate of forward reaction and
reverse reaction to the same amount, reduce the forward activation
energy and reverse activation energy to the same extent, and hence it
does not affect the equilibrium position and the composition of the
eq uilibri urn mixture.
c Decreasing pressure favour the direction which can produce
mere gas particles, and thus position of equilibrium shift to the right.
Hence, amount of PCh increases.
B When T increases, Kp increases. Increasing temperature favour
endothermic process so that heat is absorded to reduce the
disturbance of increasing temperature.
A Kp = Kc (RT)
11 A K= F;-·f:·;;:
' '7.83 atm = PF
2/ 0.200
PF = 1.25 atm
1.25 atm ofF atom come from dissociation of0.625 atm ofF2
molecule.
Hence, originally there is (0.625 + 0.200) atm of F2 molecule.
% ofF2 dissociate= (0.625/0.825) x l 00%
= 75.8%
12 c A Lewis
B Lewis
c Bronsted-Lowry and Lewis
D Arrhenius
3
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 5/21
13 c
14 A
15 A
I6 c
17 c
pH = - log[H+]
I 0.35 = - log[H+][H+] = I ~ 1 o 3 s [OR] = ~ 1 4 / IO103s
= 2.239 x 10·4 mol dm-3
·· . ...., Mg(OH)2 (s) -7 Mg3+(aq) + 20H-(aq)
Ksp = [Mg3+][0H·f
= (2.239 X l 0·4 /2)( 2.239 X i= 5.61 x 10"
12mol
3dm-9
NH4+ + oH· -7 NH 3 + H20Iniiial: 0.75 mol 0.50 mol
Final : 0.25 0 0.50 mol
[NH/] = 0.25/0.500
= 0.500 M
[NH3] = 0.50/0.500
= 1.00 M
pOH = pKb - lg [Base ]/[Salt]
= -JgJ.8 X 10"5
- lgJ.00/0.500
= 4.4437
pH = 14 - 4.443
= 9.56!':,,. = .'4u-<-'·,v
~ : i ,\f.:;.P,:.
mw (18.0)(99.2)=
I00 (169 .0)(1 .8)
mw= 587 g
X: R1 - 4.50/5.40-0.83 (Isoleucine)
Y: R1 = 1.80/5.40 = 0.33 (Taurine)
Z: Rt = 3.00/5.40 = 0.56 (Alanine)
B 2Cr'+ + 3Ba -7 2Cr + 3Ba-+
e.m.f= -0.74- ( -2.90) = +2.16V
A Co2+ + Ba -7 Co + Ba2+
e.m.f= -0.28- ( - 2.90) = +2.62V
C 2Co3+ + Ba -7 BaH + 2Co
2+
e.m.f= +1.82- ( - 2.90) = +4.72V
D 2Co3+ + Pb2+ + 2H20 -7 2Co2+ + PbO, + 4H+
e.m.f= +1.82- ( + I .47) = +0.35V
4
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 6/21
18 c
19 D
20 D
Gas at anode is oxygen gas.
Equation: 40R -7 02 + 2Hz0 + 4e
112 0
. No ofmo1e of oxygen g a s = ~ ~ - = 5 X 10·o mol22400
From the above equation, 1 mole of oxygen gas produced by 4 F, .
hence"· · ········"·. ' 4F F'
5 X 10"3 mole of oxygen gas required (5 X 10"3
) =-1 50
Definition: The standard enthalpy change of formation of a
compound is the enthalpy change when l mol of the compound is
formed from its elements under standard condition.
A 2H(g) + \0 Oz(g) -7 HzO(g)Hydrogen gas must exist as molecule, H2
B CO(g) + \0 02(g) -7 C02(g)
Carbon monoxide, CO is not an element.
C 2N(g) + 3H2(g) -7 2NH3(g)
Nitrogen gas must exist as molecule, N2
D 2C(s) + 3Hz(g) -7 C2H6(g)
In the reaction between group 2 elements and water, the Group 2
elements act as reducing agents and reduce water to hydrogen .
Group 2 elements are reducing agents via the following process :
M -> M2+ + 2e
As the atomic size increases down the group , the ionization energy
decreases and the ability to lose electrons (oxidizing power)
increases. Hence, reactivity towards water increases.
Be Reacts with steam at very high
temperature
Mg Slow with hot water, but fast with steam
Ca Slow with cold water, but rapidly steam
Sr and Ba React vigorously with cold water
Going down Group 2, the thermal stability of the carbonate
increases, the solubility of the carbonate increases, the solubility of
sulphates decreases and the oxides become more basic in nature.
5
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 7/21
21 B Aluminium chloride can be prepared by action ofalu!Ylinium with
dry hydrogen chloride gas or dry chlorine gas.
2Al(s) + 6HCI (g)_, A]zCI6 (s) + 3 H2 (g)
2AI(s) + Cl 2 (g)_, A]zCI6(s).
. '··· .. . . . .. ~ · - - ··'· The apparatus used in the preparation have to be dry to prevent the :-:
hydrolysis of Aluminium chloride.
In the Friedel Craft's reaction , aluminium chloride acts as a lone-
pair electron acceptor (Lewis Acid) :
A]zCl6 + 2CI 2 _, 2AICI4 . + 2 Cl +
At room conditions, aluminium chloride exists as a dimmer so that
the aluminium atoms achieve octet configuration.
In the vapour state, aluminium chloride exists as discrete AICI 3
molecules.
A]zCI6 (s) _, 2AIC1 3 (g)
!!.
22 D Carbon, being a Period 2 element, does not have empty d orbitals in
its valence shelL
?O D PbCl4 (I) _, PbC]z (s) + Ch (g)
!!.
24 B Nitrogen gas can only be prepared by using the method of fractional
distillation of liquedfied air.
25 D Cl2 > Brz > I2
weaker oxidizing agent
c1· > Br · > I -
stronger reducing agent
26 D The coordination number of cobalt in the complex is 6
27 D A Lewis base : lone pair electron donor. Water acts as lone pair
electron donor, hence it is a Lewis base.
B Six ligands surround the central atom, hence its shape is
octahedral
C Bronsted Lowry acid : proton donor . Due to high charge density
and hence high polarizing powerof Ti3+ ion, 0-H is weaker and [
Ti (H20) 6] 3+ undergoes hydrolysis.
D 22 Ti : ls 2 2s22p6 3s2 3p63d24s2
6
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 8/21
Compound X shows both cis-trans m1d optical isomerism28 cFor the cis-isomer there are 2 optical isomers. For the trans
isomer,there are also 2 optical isomers. Hence the total number of
stereo isomers is 2+2 4. . .:·
29 D A is unlikely because the molec.ule is big with 8C atoms
.. ·. ,;:;: .. · · ~ · .J:l.js_correct for aliphatic alcohols and acids. Phenolic -OHgroups
react slowly with PC1 5
C is wrong because the group present is not methyl carbonyl but a
methyl amide CH3CONH-
Dis correct.. Ethanoil chloride reacts with the phenylamine to give
the amide acetaminophene
30 A I is an aliphatic alcohol and has the lowest acidity.
II, Ill, IV are all phenols which are more acidic than alcohols.
II is also a carboxylic acid- is the strongest acid of all.
IV is stronger acid than III.
IV has an electron withdrawing group -C l which reduces the -ve
charge density on phenoxide ion and stabilises it.
III is weaker because -CH3 group is electron donating, increases -ve
charge density on phenoxide ion and destabilise it.
31 A P contains Nand dissolves HCl r e a d i l y ~ > P is basic. P must be an
amine. Qwhen heated with NaOH produces alkaline gas (NH3) . Q
must be a primary amide( -CONH2).
A is correct.
32 B A is incorrect. Since the bonds in polymer chain is amide and ester,
both can be broken by hydrolysis with NaOH
B is correct because all along the polymer chain is observed -COO
(ester bond) and -CONH (amide bond). Hence the polymer is a
polyamide and a polyester.
C is incorrect. Only carbonyl compounds react with HCN. But the
chain is ester and amide bond and not a carbonyl bond
D is wrong. The first monomer is correct but the 2nd monomer
should only have 2C not 4C.
o oj j c The repeat unit comes from a monomer that must have 2 different
functional groups that can undergo condensation. Cis correct
because the acyl and hydroxyl group can condense to give the exact
repeat unit.
"L l, D R must have a carbonyl group joined to CH3 group. B and D has this.
But B is acidic because it is also a phenol whileD is neutral because
it is an alcohol.
D is conect .
7
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 9/21
35 D Step 4 is incorrect. The compound formed should have the formula
CH 3CONHCH2CH3 and not CH1NHCOCH2CH3
36 c To form a stable diazonium salt at 5'C, aniline or phenylamine must
be the reactant. Only Cis phenylamine. The rest are not.
..... 37 A CH3CH 2COOH is carboxylic acid while CH3CH(NH2)COOH is an
amino acid. An amino acid can exists as zwitter ions with strong
ionic bonds between them. Hence the carboxylic acid is more
volatile than the amino acid.
B CH3CH2CH2CH 3 is straight chain with a larger surface area.
CH 3CH(CH3) 2 has a branch- have a more spherical shape with
smaller surface area. Will have lower boiling point because slightly
weaker Vander waals forces and is more volatile
c first molecule has the polar C=O bond; stronger van der waals
forces.
D first molecule is cis (more polar) has higher boiling point than
trans.
38 B S is resistant to oxidation=> 3' alcohol or acid. Reacts with Nato
give Y, mol H2 gas=> S has only 1-OH group. Only C satisfies the
above properties.
39 B One CFC molecule can destroys 100,000 molecules OJ is because
there is a chain reaction where the free radicals Cl• and ClO• gets
regenerated.
B is the answer. Note 0• radicals are not generated in the chain
reaction
40 D Haloalkane undergoes nucleophilic substitution reaction with OR
41 A Only 1 is correct.
.The higher the place , the lower the atmospheric pressure due to the
low air density. The water boil when the water vapour pressure
equivalent to the atmospheric pressure, thus boiling point drop.
42 B The unit of the rate constant for a first order is time·'.
43 B 1 HN02 (aq) (Stronger oxidising agent than Sn"+)
2 Fe3+ (aq) (Stronger oxidising agent than Sn4
+)
0 Fe 2+ (aq) (Weaker oxidising agent than Sn 4
+)
8
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 10/21
44 A L\.E'- E' (reduction)+ E' (oxidation) e.m.f. must be positive.
1. Co 3+ (ag) + Cr2+ (ag) -7 Co 2+ (aq) + Cr
3+ (aq)
e.m.f. = (+ 1.81) + (+ 0.41) = + 2.22 (Feasible reaction)
2. Cr3+ (aq) + Ti2+ (aq) -7 Cr2+ (ag) + Ti1+ (aq)
e.m.f. = (-0.41) + (0.37) =- 0.04 (Not Feasible reaction) --
- . . ::·•
Ti3+ (aq) + Co 2+ (aq) -7 Ti 2+ (aq) + Co1+ (aq)~
e.m.f. = (-0.37) + (-1.81) =- 2.18 (Not Feasible reaction)
45 B 1 Electron affinity of bromine is -342 kJ mol''
Electron affinity of bromine= (280 -622) =- 342 kJ mor 1
2 The lattice energy of potassium bromide is - 672 kJ mor 1
Lattice energy= (-392- 280) = -672 kJ mor1
3 The enthalpy change for the reaction Br2(g) -7 2Br(g) is -224 kJ
Enthalpy of atomisation of bromine= 2(622- 51 0) =+224 kJ mor1
46 c Both the oxides of phosphorus are covalent .
Due to the bigger size of, and more electrons in P40 10 compared to
P40 6 , the van der Waals forces between the P40 10 molecules are
stronger.
47 B l NH4 + ion : The lone pair electron on theN atom is used to form
dative bond with H +
2 CH 1NH 3 +ion : The lone pair electron on theN atom is used to
fonn dative bond with H + ·
" 2NH1+2e 2NH2. + H2 : The lone pair electron on theN
atom is not involved in the reaction.
48 B Step 1 is correct- oxidation by hot KMn04/H+ of side chain to
COOH group. Step 2 is coiTect- esterification reaction between-
COOH and CH1CH20H. Step 3 is reduction but the reducing agent
used is wrong. LiAIH4cannot be used. The reducing agent to convert
-N02 to -NH2 is metallic Sn with cone HCl
49 A 1 is correct. Y has the structure CH1CH(OH)- which gives yellow
ppt with aqueous alkaline iodine. X does not have any structure that
gives +ve result with this reagent So 1 can distinguish between X
andY.
2. Both X andY are phenols, so Br2(aq) will give +ve results with X
and Y and so cannot distinguish between them
3. Acidified KMn04 cannot distinguish because both X andY has
carbon side chains that can be oxidised.
50 c 1 is wrong. To give N2 gas with HN02 must have -NH2 group.
2 is correct because compound is a 3° amine which is basic.3 is correct because compound contains a carbonyl -C=O group.
9
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 11/21
(a)(ii)
(b) (i)
(b)(ii)
(c)
MARKING SCHEME
STRUCTURE QUESTIONS
..
••/H - ~ c ~ c ~ \ ~ · · / c=o
..../ H 2
H -O..
All oxygen atom, 0 have 2 pairs ofdot.
Two C atoms have a double covalent bond.
••
All carbons that have only single bonds are sp3 hybridized.The three carbons that have double bonds are sp
2hybridized.
Ionic : RbCl
Non-polar covalent : Ss, F2
Polar covalent : PF3, SCb, SF2
PCI5 :trigonal bypiramidal, 90° and 120°
PC!/ ion : tetrahedral, 109.5°
PCI6. ion : octahedral, 90°
l
1
l
1
1
1
2
3
1
3
Total 10
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 12/21
Q Answer . . . -· Mark I : Mark
2
(a)(i) Ksp = [Mg"+][OHr.
1 1
(a )(ii) Let [Mg(OH)2]=[Mg"+]=x .. - .
I• . . ' x,r:= 2:5cx· - 1 0 - ~ - = > · · ··· ·x = 8.55 x 10-5 mol dm-
31 1
(a)(iii) Less soluble 1 2
because of the common ion effect 1
(b) pOH = 14.0-9.0 = 5.0 1
[salt]pOH = pKb + log -[- )Base
-s [salt]5.0=-]oaL75x10 +log--
b 0.201
[salt]= 035 mol dm-3
Mass ofNH4CI in 250 cm3 = 0.35 x 53.5 x250/l 000 = 4.68g ·
1 ")
(c)(i)P, X,P,
0JO ~ 5 6 39 2 kPo}100
1
30P8 = X8 P8° -x 4 = 10.2 kPa
100
Total pressure= 39.2 kPa + 10.2 kPa = 49.4 kPa 1 2
(c)(ii) Dalton's Law
%XA vapour=392
xl00 =79.4%1
49.4
% Xs = (100 -79.4) = 20.6%1
The composition of the first vapour is 79.4% of A and 20.6% of B.
2
Total Max 10
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 13/21
Q Answer Mark I : Mark.
3
(a)(i) Giant covalent/macromolecule . . . . - . , ,J . I
I .. . . . . . . . . :.c .
(a)(ii) Na20 and MgO are basic (both) I
Si02, P40 10 and S02 are acidic oxides (any two) I 2
A)z03 is an amphoteric 1
(a)(iii) I••
/ s , .•• 0 •. 0••
or
••~ s ~ ..•• 0 •• () ..
••
I
(b )(i) 2Mg(N03) 2 (s) 2Mg0 (s) + 4N02(g) + 0 2(g) I I
(b)(ii) Brown fumes given of f I a white solid remains 1 I
(c) Decreases 1
The hydration energies of the ions decrease more than lattice I
energies decreases.I . . . M '+ C '+ S '+ B '+ I
0omc s1ze mcrease, < < < a-
Total 10
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 14/21
Q Answer Mark I:-Mark
4
(a) The amine group ( -NH- ) 1 -- -- -
CH3NH-®-CONHCH(CH20H)COOH+HCl -7 . - -_ :
-- 1
·. " -.· - : .[ C H 3 ~ I - - I 2 @ -CONHCH(CH20H)COOH] cr 2
(b)
The amide group (-CONH-); the hydroxyl group (-OH);} any
the carboxylic acid group ( COOH) two 1 l
(c) (i)CH 3NH-@-coo ·Na+ and H2N-CH-coo·Na+ 1+ 1
ICH 20H
(c)(ii)CH3NH-@-coNHCH-CH20H
I 1
CH20H
(c) (iii)
CH 3NH-@-CONHCH-COO'Na+ 1
ICH20.Na+
(c ) (iv) N H ~ -CONHC-COOH
1
IICH2 5
(d) (i)
CH 3NH -@ - c o c l + H2N-CH-COOH -7ICH 20H
CH 3NH-@--CONHCH(CH20H)COOH+ HCl 1
(d) (ii) N H - ~ C - O - C H - C H - C O O H II I I
0 NH2 2
Total 10
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 15/21
Question 5
Q5 Answer ---Mark :tMark
S(a) pV=nRT; n=rn!M, or Mr=mRT/pV 1
M, = 0.269 X 8.31 X (273+ 120) 1 .1.00 X 103
X 10.0 X J0-3 -- ~ - ~ - - - -- -- ···-··--- -- --
= 87.9 ( 3 sig figs) 1
. ·. ···· ·aas 'A behavesldeall y Becaus-e ·· 1
Pressure is low and temp is high 1 5
5 (b) M, of acid A is 90 1
identity of acid A is H2C204 or HOOC-COOH 1
mle suecies
17 OH+ 4,5-2
28 co+ 2,3- 1
45 COOH+ 0,1-0
56 O=C-C=O+ I C20 /
90 H2C20/ I HOOC-COOH +
4
5( c )(i) (CH3)3N and (CH3)2NH have simple molecular structures. 1
Between molecules (CH3)3N are van der Waals forces and 1
hydrogen bonding between molecules of (CH3)2NH
hydrogen bonds are stronger than van der Waals forces. 1
5( c)(ii) In polar solvent I water, HCOOH exists as sinzle HCOOH 1
molecules; RMM is thus 46
In a non-polar solvent I benzene, two HCOOH molecules 1bond with each other by hydrogen bond (to form dimer).
t Hydrogen bond
· 0:-----H-0 .1-e· · .·C-H
·0-H----- :0~ h y d r o g e n bondsDiagram 6
(hence RMM is 2x46 = 92)
Total 15
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 16/21
Question 6
Q6 Answer Mark z
Mark
6 (a) Electronegativity is the measure of the ability of an atom to l 1attract the electrons in a covalent bond to which it is .. .bonded.
(b) .,,,, ' Electronegativitjl increases " '
1
Across period:-
Atomic size decreases, l
the number of protons increases 1
thus the effective nuclear charge increases 1Max
the strength of attraction for electrons increases 1 4
(c) Across period 2:-
The number ofprotons increases and 1
screening effect almost constant 1 2
(d) The copper anode dissolves to form Cu"+ ions in solution 1
and
The silver impurities fall to bottom of container 1
The cathode has been deposited with a brown solid ofcopper. 1
The intensity of the blue colour of electrolyte solution 1
remains unchanged
Explanation:
At anode(+)
(i) Cu(s) -7> Cu2+(aq) + 2e E
0=-0J4V
(ii) Ag(s) -7> Ag\aq) + e E0=- 0.80V
>- 1(iii) 2H20(l) -7> 02(g) + 4H+(aq) + 4e E
0= -1 23 V
'Or 40H"(aq) -7> 2H20(l) + 02(g) + 4e E
0=-0.40V
The oxidation that is most favourable is that which has the 1least negative E0 value is copper anode dissolves to become
Cu2+ ions I Cu(s) -7> Cu2+(aq) + 2e
At the cathode (-22H+ ( aq) + 2e -7> H2(g) E0
= 0.00 V
}r 2H20(l) + 2e -7> H2(g) + 20H"(aq) E0
= -0.83V 1
(ii) Cu2+(aq) + 2e -7> Cu(s) E
0=+0.34V
Of the two above reduction reactions at the cathode, the
most favourable and the one having the most +ve or least-1 8
ve E
0
value is (ii)ICu
2
\aq) + 2e-7>
Cu(s)So copper is deposited on the cathode as a brown solid
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 17/21
.
Question 7
. - - c Q c - 7 c c - - , - - - - - - - - - - A n - s w - e r - - - ~ = - ~ - - ~ · · = · · = · c c · c-c-c -= .x.. ~ : t v ~ l = a r ~ k = = - = · - ~ I - ~ - ·"--·------·----
7(a)
7(b)
7(c)(i)
7(c)(ii)
(i) An electrol)iic process to increase the thickness ofthe
aluminium oxide layer on aluminium.
(ii)
Method: electrolysis
Anode: aluininiilm and Cathode: graphite (Cu or Pb)
Electrolyte: dilute H2S04 (aq) (or dil HN03)
[diagram]
Al object - - + +- - - - graphite
-1-- dilute H,so,
diagram with label
At anode:
1
1
02 gas is liberated : 40H.(aq) -7 2H20(l) + 0 2(g) + 4e 1
Or 2H20 -7 02 + 4W +4e
The oxygen gas oxidizes the Al anode to give Ab03 :
4Al(s) + 302 (g) -7 2Ah0J(s)
The thermal stability decreases from CCl4 to PbC14
Size of atom increases from C to Pb
Length ofM-Cl bond becomes longer I weaker
CCl4, SiCl4 and GeCl4 are stable to heat
I will not decompose when heatedOr SnCl4and PbCl4 decompose on heating
I unstable to heat
PbCl4(l) -7 PbCb(s) + Ch(g)
or
Or SnCl4(l) -7 SnCb(s) + Cb(g)
CCl4
Carbon does not have any empty d orbital
SiCl4 [or other tetrachloride ofGroup 14]
SiCl4 + 2H20 -7 Si02 + 4HCl
[or equation using other tetrachloride of Group 14]
Total
1
1
1
1
1
1
1
1
mark
7
Max4
2
2
15
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 18/21
Question 8
8(a) Solid NaX when heated with cone. H2S0 4 produces T
gaseous hydrogen halide.
NaX(s)+
H2S04(aq)7
NaHS04(aq)+HX(g)
1
I (Na){: NaCl, NaBr or Nal is used)
Concentrated H2S04 is stronger oxidising agent than Br2 1
canoxidized HBr to Br2 ____
~ , _ , " ' ' ' " " " ' - • ~ " - " ' 2P!Br(g)'+HiS6:r-7 En( ) *:Sfh(g) + 2H20(/) 1 -- • { (}'" J ' ; - ,
Cone H2S04 as a weaker oxidizing agent than Ch so cannot 1
oxidise HCl to Ch Max4
8(b) (i) Electronic configuration for
Sc(protonno. 21) :1s2
2s2
2p6
3s2
3p6
3d1
4s2
3-2M
Sc3+ : ls
2 2s2 2p6 3s23p
61
Mn(proton no 25) : ls2
2s2
2p6 3s2
3p6
3d5 4s2
2-1MMn
2+ : ls 2 2s2 2p6 3s 23p
63d
51
I .
Zn(proton no 30): ls2
2s2
2p6
3s2
3p6
3d10
4s2
0 -OM
Zn2+ ; ls
2 2s2
2p6 3s23p
63d
101
8(b) Sc'+ and Zn"+ ions 2re colourless 1
(ii) Sc3+ has no d electrons and Zn2+, d orbitals fully filled 1
Both ions cannot have d--d transitions 1 5
8(c (i) Add aqueous ammonia to aqueous chromium(III) chloride 1
solution.
[Cr(NH3
)6]3+ is more stable than [Cr(H,0)6]
3+so NH3 1
ligand displaces H20 ligand
[Cr(H20) 6]3\aq) + 6NH 3(aq) :;::::::::: [Cr(NH,)6]
3+ +6H20(l) l
8(c (ii) Cr'+ ions have empty 3d (4s or /and 4p) valence orbitals 1
Cr3+ ions have high charge density 1
Or small ionic size and high charge
8(c) [CrCh(NH3)4t Show geometrical isomerism
(iii)Cl C:l
N ~ ~ - J 7 : 1 H , ~ ~ J 1 \ n , 1
! Cr ' I Cr 1
H : ~ - r " / ~ , H , , ~ + ~ ' '0n, Cl 6
[both structures must be correct]
Total 15
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 19/21
Question 9
9(a)(i) Elimination reaction I.
(a)(ii) C!CH2CH2CH,CH2Cl -7 CH2=CH-CH=CH 2 + 2HCI 1
(a)(iii) Addi tion polymerization II
COOH c ~ . . :
I I' » " ~
- C H . C H ~ C H ~ C H · - C H - C - . . or , ___... ~ " ' " ~ " ; ~ ' . : " ' " : " ' ' ~ ; . "" · . · ·ecce··•· · · . , , . , , , , , . , , , : , J , ; ~ ~
I
COOH
I- C C H ~ C H - C H - C - C H -I I 4CK COOH
9(b )(i) Toluene undergoes electrophilic substitution reaction (to form I
ortho and para chlorotoluene)
F ~ l - ' 1 1 ""'
c1· - [FeCI,]' IElectrophile
< ~ c t - H
""';\ +) )-CH3
Cl · \....::::::_/ I
H ( ) ; · ~ ;;:::::-\"-.. ' ........, \' + ) ; - ~ C H , -:- FeCl-J· -?CI - I r ; ~ C H · +FeCI·
Cl/.- \ _ ~ _ / \ ~ / + H c l I 4
9(b )(ii) Name of mechanism: electrophilic addition I
Cl
0+ o· __/+\C H · - C ~ C H · + Cl-Cl 7 CH, -C -CH : +CI· 1
' 1 ~ _ / l ~ ICH: CH:
.c1) Cl
/+ \ I'·
CH 0-C-CH3 rCI' 7 CH;-C-CH: I 3
I vI
CH; Cl
9 (c)_ _ _ c . _ : ~ B is more acidic than A 1
R-OH + H20 - R-U + H,o+ 1
[R based on the example of A orB]
-N0 2 IS electron withdrawing, stabilized phenoxide ion, 1
equilibrium shifts more to the right
-CH3 is electron donating, destabilizing phenoxide 1011 thus 1 4equilibrium shifts more to the left.
Total 15
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 20/21
Question 10
1O(a)
(i) Orange precipitate with DNP- X is an aldehyde or ketone or a 1
carbonyl compound
No observable change with Tollen's reagent-X is a ketone
Yellow precipitate with !2/NaOH- X has the foil wing structure
CH 3 -CC'O
· .· • ........... ,, ......•...J ........ • • · ~ · " c · • · • · •.• ,_ •2· ~ .. ,.,,
1O(a)
(ii)
1O(b)(i)
(b )(ii)
(iii)
- ~ = 0 + H2N-NH-)Q>-NO,
CH 3 0 2N
CH,CH,CH, -C=N-NH-@-No ,
I ;CH 3 0 2N + H20
Amide
carboxyl
CHI3 + CH 3CH 2CH2COU
+r+H,O
(no need to be balanced)
C6H5COC1 and H2NCH2COOH
[condition] Heat with
[reagents] aqueous sodium hydroxide
[regants] Add dilute mineral (HCl, H2S04, HN03) acid
[equationJ
@-e-N- CH2COOH + 2NaOH 7 @-COONa
II I0 H + H2NCH2C00Na
@- c o oN a + HCI 7 @ - co oH + NaCl
1
r
1
1
1
1
1
1
1 + 1
1
1
1
1
.
5
2
8/4/2019 Stpm Chem p1_p2 Skema 2011
http://slidepdf.com/reader/full/stpm-chem-p1p2-skema-2011 21/21
[alternative]. - ···-- -- -
[condition]: beat 1[reagent]: dilute mineral (HCl, HN03, H2S04) acid 1 --
[equation]
@-e-N- CH,COO!-I +2HC1 -" @-COOH--1+1
_Max 8
II I0 H + +H3NCH2COOH
-:--- j-- "'"-1'··-:-;o:,-:-,-1 mark for equation
·----- ---·--- • ------ -- - · ~ a ··H----·-· --mark for H3NCH2C 0 -- -
Max 15