stpm chem p1_p2 skema 2011

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PEPERIKSAAN PERCUBAAN SETARA STPM

NEGERI SEMBILAN DARUL KHUSUS

TAHUN 2011

MARKING SCHEME

PAPER 1

PAPER2



, . · · · ~ · · · . . . . _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,



-" 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 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



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



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



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



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



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



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



(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



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



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



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



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



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



.

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



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



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



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



[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

stpm chem p1_p2 skema 2011

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