2001-2002 Form 7 Mock Suggested Answers

8/2/2019 2001-2002 Form 7 Mock Suggested Answers

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Suggested Answers for 2001-2002 F7 Mock Examination

Paper I

1 a i HCCCCCN

ii High temperature; in the absence of O2

iii In the C-12 scale, the parent ion peak occurs at m/e = 12 60 = 720.

iv Even number / not less than 28

v

(a)

(b) In the cage structure, C atoms in six-membered rings are sp2hybridized C atoms located at the point where

three five-membered rings meet are sp3

hybridized. H atom will bond to the fourth electron.

vi Curved surface cannot be constructed from 6-membered rings containing only sp2

hybridized carbon

vii Bonding is exclusively sp3

/ SiSi bond (226 kJ mol-1

) is much weaker than CC bond (347 kJ mol-1

) / Si atom

has a large size cannot form stable bond.

viii Presence of delocalized electrons (linked systems of sp2carbons) so that it can conduct electricity along the

nanotube.

ix Sensible answer: Future computers (based on nanotubes rather than Si chips) because the electrical conducting

property of nanotubes can easily be modified, i.e. they can be made semi-conductors/super-conductors.

2 a Chromium(III) ion

b i Na(g) Na+

(g) + e-

ii Na+

(g) Na2+

(g) + e-

c i The size of the atom is the small ; The 2 electrons experience the strongest effect nuclear charge. The shielding

effect experienced by the 2 electrons are weak and is only secondary in nature. It has 2 protons comparing with

1 proton only in hydrogen. Hence the outermost electrons experience the strongest attraction.with the nucleus.

ii The outermost most electron of potassium is further from the nucleus and experience a weaker attraction from

the nucleus. Furthermore, the presence of the 2 more complete inner shells offers much stronger primary

shielding effect to the outermost electrons so that the attraction between the outermost electron and the nucleus

of potassium is much stornger.

3 a i Cl2(g) + 2NaOH(aq) NaCl(aq) + NaClO(aq) + H2O(l)

Cl2(g) + OH-(aq) Cl

-(aq) + ClO

-(aq) + H2O(l)

ii 3Cl2(g) + 6NaOH(aq) 5NaCl(aq) + NaClO3(aq) + 3H2O(l)

3Cl2(g) + 6OH-(aq) 5Cl

-(aq) + ClO3

-(aq) + 3H2O(l)

iii Ca3(PO4)2(s) + 3H2SO4(l) 2H3PO4(l) + 3CaSO4(s)

iv CaO(s) + H2O(l) Ca(OH)2(aq)

v 4H2O(l) + 2MnO4-(aq) + 3C2O4

2-(aq) 2MnO2(aq) + 6CO2(g) + 8OH

-(aq)

16H+

(aq) + 2MnO4-(aq) + 5C2O4

2-(aq) 10CO2(g) + 2Mn

2+(aq) + 8H2O(aq)


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b i Though the electronegativities of S, C and O are all different, SO2 is polar while CO2 is non-polar.

This is because SO2 has an angular structure while CO2 has a linear structure.

In SO2, the vector sum of the dipole moments is non-zero while the dipole moments of the 2 polar C=O bonds cancel out

each others and give a non-polar CO2 molecule.

SO O

OCO

ii The colour of cobalt(II) ions arises from d-d electron transition. In the d-d transition, certain wavelength in the

incident light is absorbed and the colours of the unabsorbed light is revealed. Cobalt(II) ion has the electronic

configuration 1s22s

22p

63s

23p

63d

7while Zinc(II) ion has the electronic configuration 1s

22s

22p

63s

23p

63d

10.

Cobalt(II) ion possesses partially filled d-orbital which allows electrons in the lower d orbitals to be promoted to

a higher d orbitals. However, Zinc(II) ions has a completely filled d-oribital where d-d transition is not possible.

4 a i

ii (I) As the temperature increases, the rates of all chemical reactions increase.

For an exothermic reaction as reaction I, an increase in temperature will cause a greater increase in the backward

rate that the forward rate as the backward reaction involves a higher activation energy.

Thus, the equilibrium position of reaction I will be shifted backward and the time require to reach equilibrium will be

shorted.

(II) As heat of reaction is only depending on the difference between the heat contents of the reactants and the products.

Any change in temperature will have no effect on the heat of reaction of reaction I.

b i 2Cl-(aq) Cl2(g) + 2e

-

ii Charge (Q) = Current (I) time (t) = 0.250 A (2.00 3600)s = 1800 C

no. of mole of electrons flowing through the circuit =1800C

96500Cmol-1 = 0.0187 mol

no. of mole of iron deposited =0.521 g

55.85 gmol-1 = 9.33 10

-3mol

Charge carried by the iron ion =

0.0187

9.33 10-3 = 2

Formula of the iron chloride is FeCl2

iii FeCl2(aq) Fe(s) + Cl2(g)


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iv 2Cl-(aq) Cl2(g) + 2e

-

no. of mole of Cl2 produced =no. of mole of electrons flowing through the circuit

2

= 0.0187 mol 2 = 9.35 10-3

mol

Volume of chlorine produced = 9.35 10-3

mol 24 dm3mol

-1= 0.22 dm

3or 220 cm

3

or

PV = nRT

V =nRT

P=

9.35 10-3

mol 8.314 JK-1

mol-1

298K

101 1000 Nm-2 = 0.000229 m

3

= 0.229 dm3

or 22.9 cm3

vno. of mole of Cl2(g) produced per hour =

3.00 g

35.5 2 gmol-1 = 0.0423 mol

no. of mole of Cl2(g) produced per second = 0.0423 mol 3600 = 0.0000118 mol

no. of mole of e-flowing through the circuit per sec = 0.0000118 2 = 0.0000236 mol

Current required = (0.000236 mol/s 96500 C/mol) = 2.27 A

5 a i carboxyl group

amide linkage / amide group / peptide linkage

iiC

O

OH NH2CH2 C

O

OH

iii Heat under reflux

Concentrated / Dilute hydrochloric acid / Alkali followed by acid

iv (I) Oxidation ; Heat, Alkaline KMnO4, HCl

II) No. of mole of methylbenzene in 1.0 g =1.0 g

(7

12.0 + 8

1.0) g

= 0.011 mol

no. of mole of hippuric acid produced

ass of hippuric acid produced

= 0.011 mol (9 12 .0 + 3 16.0 + 1 14.0 + 9 1.0) gmol-1

= 1.9 g

b i When infra red passes through a sample of organic compounds, certain wavelength would be absorbed by the

molecule and set the bond in vibration. As the energy is absorbed, absorption peaks are observed.

ii One peak at 3700 cm-1

corresponds to OH vibration

Another peak at 1000 cm

-1

corresponds to CO vibrationiii

R OH R ClLiAlH4 in ether

HCl(aq)

PCl5R H

6 a i Cu(s) + aq Cu+

(aq) + e-

ii Cu(s) + aq Cu2+

(aq) + 2e-

b i Hf[Cu+

(aq)] = [(+339) + (+745) + (-481)] kJmol-1

= +603 kJmol-1

ii Hf[Cu2+

(aq)] = [[(+339) + (+745) + (+1960) + (-2244)] kJmol-1

= +800 kJmol-1

c Hrxn = Hf(products) - Hf(reactants)

= [Hf(Cu

2+

(aq)) + H

f(Cu

(s))] - [(H

f(Cu

2+

(aq)) 2]

= {[+800 + 0] - [+603 2]} kJmol-1

= -406 kJmol-1

d Cu+

: 1s22s

22p

63s

23p

63d

10

Cu2+

: 1s22s

22p

63s

23p

63d

9


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e For the ions in vapour phase, Cu+

(aq) has a completely filled d orbital (1s22s

22p

63s

23p

63d

10) while Cu

2+(aq) doesn't

not has a complete filled d orbital (1s22s

22p

63s

23p

63d

9). Furthermore, the charge density of Cu

2+(g) is higher than

that of Cu+

(g). Therefore, Cu+

(g) is more stable that Cu2+

(g).

However, in the presence of water, Hrxn of 2Cu+

(aq) Cu2+

(aq) + Cu(s) is negative, -406 kmol-1

. Cu+

(aq) will

disproportionate to form Cu2+

(aq) and Cu(s) spontaneously. Therefore Cu2+

(aq) is more stable that Cu+

(aq) in aqueous

state.

7 a i Weigh the sample in the beaker by weighing by difference.

Dissolve the sample completely in distilled water with stirring.

Add BaCl2(aq) solution in 5 cm3 portion to the solution prepared until no more ppt. is formed.

Weigh the filter paper by weighing by difference

Filter the suspension using filter paper and funnel

To the filtrate, add a little BaCl2(aq) to test if all sulphate have been precipitated.

If not add more BaCl2(aq) to precipitate all the sulphate and filter again.

Wash the residue with a little distilled water.

Oven dry the residue with the filter paper.

Weigh the dried residue with the filter paper.

ii Mass of beaker

Mass of unknown sample with beaker

Mass of filter paper

Mass of dried precipitate with filter paper

iii Mass of barium sulphate = mass of precipitate with filter paper - mass of filter paper

Mass of unknown sample = mass of unknown sample with beaker - mass of beaker

No. of mole of barium sulphate precipitated =mass of barium sulphate

molar mass of barium sulphate

Mass of sulphate = no. of mole of barium sulphate molar mass of sulphate ion

Mass percent of sulphate in the unknown =mass of sulphate

mass of the unknown sample used 100%

iv No. As barium sulphate is insoluble in water while magnesium sulphate is soluble in water, therefore, MgCl2(aq)

cannot be used to precipitate the sulphate ions.

8 a imolar mass =

mass

no. of mole

Error would be introduced if the mass of HA is determined wrongly or the no. of mole of HA present isdetermined wrongly.

Too low.

As the NaOH(aq) is diluted by the distilled water present in the burette, a larger volume will be recorded in the

titration. A larger amount of NaOH will be pretended to be used. Thus a larger amount of HCl will be

pretended to be present.

ii Not affected.

The presence of extra water in the conical flask will not have any effect on the measurement of the mass of the

HA used and the amount of NaOH(aq) used.


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iii Too high

As a titration between a strong alkali and a weak acid, the equivalence point of the titration will be higher than

pH 7. If an indicator that changes color at pH 5 is used, the colour of the solution will change well before the

equivalence point and the amount of NaOH(aq) will be too small. Thus, the no. of mole of HA determined will be

smaller than the actual amount.

iv Too small.

When the air bubbles passes through the tip, the reading of the burette will drop suddenly and the volume of

NaOH(aq) recorded will be larger than the actual amount. Thus, the amount of HA present will be overestimated

and the value of molar mass of HA determined will be too small.

b Brown ring test

Dissolve the solid sample in a small amount of distilled water in a test tube.

To the dissolve sample, add a little freshly prepared FeSO4(aq) solution.

Run a little concentrated sulphuric acid to the test tube containing the sample carefully without disturbing the

aqueous layer.

A brown ring will be formed at the interface between the aqueous layer and the sulphuric acid layer if nitrate is

present.

(Sodium fusion test if for the presence of N, not specifically for NO3-)

Paper II

1 a i Temperature has no effect on rate of radioactive decay but the rate of a chemical reaction will increase exponentially with

temperature.

An increase in temperature speeds up the collisions among molecules and makes more molecule possessing an energy higher

than activation energy.

Radioactive decay is a change originated from the nucleus and occurs spontaneously. No collision between atoms are

involved. Therefore, its rate is independent of temperature.

ii Let IH0

be the original intensity of radiation from 3H present

Let IAr0 be the original intensity of radiation from39Ar present

Let IH

be the current intensity of radiation from 3H present

Let IAr

be the current intensity of radiation from 39Ar present

k be the decay constant of3H ; k' be the decay constant of39Ar

IH0

IH

= ektIAr

0

IAr

= ek'tIH

0

IAr0

=40

1= 40

IH

IAr

=1

1= 1

IH0

IAr0

I

H

IAr

=IH

0

IAr0

I

Ar

IH

=IH

0

IH

IAr

0

IAr

40 1 = ekt e

k't 40 = e(kt-k't) ln 40 = (k-k')t

t =ln 40

k - k'=

ln 40

5.8 10-2 - 2.1 10-3= 66 years

Assumption : All 3H and 39Ar are trapped in the meterorites.

b

SH H

105

XeF F

F F

90

PCl

Cl

Cl

107

B

Cl Cl

Cl

1204


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c CH3CH2CH2CH3, CH3COCH3, CH3CH2CH2OH, CH3COONa

CH3COONa has the highest boiling as it is an ionic compound while the others are molecular compounds. Ionic bond among

the ions is much stronger than van der Waals' forces among the molecules.

CH3CH2CH2CH3, CH3COCH3, CH3CH2CH2OH all have similar molecular sizes. The strength of the induced dipole-induced

dipole attraction should be similar. The difference in boiling point is due to the difference in polarity of the molecule.

CH3CH2CH2CH3 has the lowest boiling point as it is non-polar.

CH3COCH3 has a higher boiling point as it is polar.

CH3CH2CH2OH has an even higher boiling point as it can form hydrogen bond with the adjacent molecules through hydrogen

bond.

d i Standard electrode potential of an electrode is the e.m.f. of the cell measured under standard condition when standard

hydrogen electrode is used as the left hand cell and the unknown electrode is used as the right hand cell. The sign of the value

is given by the polarity of the unknown electrode.

ii (I) Ni(s) | Ni2+

(aq)Zn2+

(aq) | Zn(s)

(II) A salt bridge is constructed by using a piece of filter paper / cotton thread soaked in saturated KNO3(aq) or NH4Cl(aq)

solution.

(III) Ecell = Eright - Eleft = [(-0.76) - (-0.25)] V = -0.51 V

iii (I) According to the standard electrode potential, the reaction, Ni(s) + 2H+

(aq) Ni2+

(aq) + H2(g), has an overall electrode

potential (0 V) - (-0.25V) = +0.25V. As the overall electrode potential is positive, the reaction should be energetically

feasible.

(II) It doesn't not take place as the sign of the overall electrode potential only reflect the energetic feasibility of a reaction

while the reaction might be kinetically infeasible at the temperature at which the experiment is carried out.

2 a imole fraction =

no. of mole of CO

total no. of mole=

0.55 mol dm-3

(0.20 + 0.30 + 0.55 + 0.55) moldm-3= 0.34

ii Kc =[H2O(g)][CO(g)][H2(g)][CO2(g)]

= 0.55 moldm-3 0.55 moldm-3

0.20 moldm-3 0.30 moldm-3 = 5.0

iiiKp =

pH2O(g)

pCO(g)

pH2(g)

pCO2(g)

Assuming ideal gas behaviour, PV = nRT ; P =n

VRT = MRT

Kp =

pH2O(g)

pCO(g)

pH2(g)

pCO2(g)

=

MH2O(g)

RT MCO(g)

RT

MH2(g)

RT MCO2(g)

RT=

MH2O(g)

MCO(g)

MH2(g)

MCO2(g)

= Kc

iv Since there is no change in the number of mole of particles when the position of the equilibrium shifts, it is no need to

consider the change in pressure.

Concentration of difference species at the new equilibrium position[H2] = 0.20 moldm

-3 + (0.55 moldm-3 30%) = 0.365 moldm-3

[CO2] = 0.30 moldm-3 + (0.55 moldm-3 30%) = 0.465 moldm-3

[H2O] = 0.55 moldm-3 (1 - 30%) = 0.385 moldm-3

[CO] = 0.55 moldm-3 (1 - 30%) = 0.385 moldm-3

Kc =[H2O(g)][CO(g)]

[H2(g)][CO2(g)]=

0.385 moldm-3 0.385 moldm-3

0.365 moldm-3 0.465 moldm-3 = 0.87


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v Let x be the number of mole of CO(g) present at the equilibirum

Kc =[H2O(g)][CO(g)]

[H2(g)][CO2(g)]= 5.0 =

x

3.0moldm-3

x

3.0moldm-3

0.50 - x

3.0moldm-3

0.50 - x

3.0moldm-3

=x2

x2 - x + 0.25

4x2 - 5x + 1.25 = 0

x = 0.345 or 0.994 (rejected)

Equilibrium concentration of CO(g) is

0.345 mol

3.0 dm3 = 0.12 moldm

-3

b i C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

iino. of mole of propane in 10.0 gram of propane =

mass of propane

molar mass of propane

=10.0 g

(12.0 3 + 1.0 8) gmol-1=

10.0 g

44.0 gmol-1= 0.227 mol

no. of mole of oxygen required = 0.227 mol 5 = 1.14 mol

Volume of oxygen required =nRT

P=

1.14 mol 8.314 JK-1mol-1 (273 + 30) K

101 1000 Nm-2

= 0.0284 m3

Volume of air required = 0.0284 m3 21.0% = 0.135 m3 or 135 dm3

iii Hfoof propane 3C(s) + 4H2(g) C3H8(g)

Hfoof propane = {3 Hfoof CO2(g) + 4 Hfoof H2O(l) }- Hcombustion [propane]

= {3 -393.5 kJ/mol + 4 -285.3 kJ/mol} - (-2220.1 kJ/mol) = -101.6 kJ/mol

ivno. of mole of propane in 30 grams =

mass of propane

molar mass of propane=

30.0 g

(12.0 3 + 1.0 8) gmol-1= 0.681 mol

Energy release when 30 grams of propane undergoes combustion

= 2220.1 kJ/mol 0.681 mol = 1510 kJ

The increase in temperature =q

mc=

1510 103 J

8.00 103 g 4.18 J/gK= 45.1 K

c i V is the triple point of the substance.

The 3 phases of the system, solid, liquid and vapour will only coexist at equilibrium this point.

ii It is the boiling curve of the substance. At each point, liquid is in equilibrium with the vapour.

iii At X, the substance is a solid.

When the temperature increases, the volume of the substance increase slowly under Y is reached.At Y, the solid sublimes and coexist with the vapour.

Upon further increase in temperature, all solid sublime

The vapour continue to expand and reach point Z.

iv The solid will sink.

Accordingly to the phase diagram, the substance has a freezing curve with a positive slope.

The substance will contract upon freezing (increase in density upon freezing).

An increase in pressure will cause a raise in melting point which implies that if a liquid with a temperature very close to

melting point is compressed, it will change to solid. Furthermore, an increase in pressure will cause a decrease in volume. This

implies that the solid is denser than the liquid for this substance.


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3 a Let the rate law be : Rate = k[ClO-(aq)]x[I-(aq)]

y

Rate1Rate2

=(0.0017)x(0.0017)y

(0.0034)x(0.0017)y= (

1

2)x =

1750

3500x = 1

Rate1Rate3

=(0.0017)x(0.0017)y

(0.0017)x(0.0034)y= (

1

2)y =

1750

3480y 1

From experiment 1, 1750 moldm-3s-1 = k [0.0017 moldm-3][0.0017 moldm-3] k = 6.1 108 dm3mol-1s-1

The rate law of the reaction is Rate = 6.1 108 dm3mol-1s-1[ClO-(aq)][I-(aq)]

bno. of mole of codeine in 5.0 mg =

5 10-3 g

(18 12.0 + 21 1.0 + 14.0 + 3 16.0) gmol-1=

5 10-3 g

299.0 gmol-1

= 1.67 10-5 mol

Concentration of the solution =no. of mole

volume of solution=

1.67 10-5 mol

10.0 1000 dm3 = 1.67 10

-4

B(aq) + H2O(l) d HB+

(aq) + OH-(aq)

initial concentration (M) 1.67 10-4M 0 0

concentration at equilibrium (1.67 10-4 - x) M x M x M

1.67 10-4 M

pKb = - log Kb

Kb = 10-pKb = 10-6.05 = 8.91 10-7

Kb =[HB+(aq)][OH

-(aq)]

[B(aq)]=

x x

1.67 10-3= 8.91 10

-7

x2 = 1.49 10-9

x = 3.86 10-5

pOH = -log (3.86 10-5) = 4.41

pH = 14 - pOH = 14 - 4.41 = 9.58c i Buffer solution is a solution with a pH resistant to change upon addition of a small amount of acid or base.

ii (I) NH3(aq) + H2O(l)d NH4+

(aq) + OH-(aq)

Kb = 1.70 10-5

no. of mole of NH4Cl in 50.0g =50.0g

(14.0 + 4 1.0 + 35.5)gmol-1=

50.0 g

53.5 gmol-1= 0.935 mol

Concentration with respect to NH4+ =

0.935 mol

1.00dm3= 0.935 M

Assuming the hydrolysis of NH3(aq) and NH4+

(aq) in the solution is very minimal

NH3(aq) + H2O(l)d NH4+

(aq) + OH-(aq)

Concentration 0.75 M 0.935

Kb = 1.70 10-5 =

[NH4+

(aq)][OH-(aq)]

[NH3(aq)]=

0.935 [OH-(aq)]

0.75

[OH-(aq)] = 1.36 10-5 M

[H+(aq)] =Kw

[OH-(aq)]=

10-14

1.36 10-5= 7.35 10-10 M

pH = - log [H+(aq)] = -log (7.35 10-10) = 9.13


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(II) no. of mole of NaOH(aq) added =1.00 g

(23.0 + 16.0 + 1.0) gmol-1= 0.025 mol

NH3(aq) + H2O(l)d NH4+

(aq) + OH-(aq)

Amount present in1 dm3 0.75 + 0.025 mol 0.935 - 0.025 mol

0.725 mol 0.91 mol

pOH = pKb + logamount of NH3amount of NH4

+ = -log(1.70 10-5) + log

0.725mol

0.910 mol

= 4.77 + (-0.0987) = 4.67

pH = 14 - pOH = 14 - 4.67 = 9.33

d i

ii When the mixture is distilled, both water and nitric acid will vaporize. However, when the mixture vapourizes, the vapor will

be very rich in water, i.e. higher than 80% by mass. Eventually, as the distillation continue, the residual mixture will get richer

in nitric acid and the temperature of the mixture will increase until it reaches 122C.

iii (I) The partial vapour pressure of any volatile component of an ideal solution is equal to the vapour pressure of the pure

component multiplied by the mole fraction of that component in the solution.

(II) Nitric acid molecules form strong H-bond with water molecules. As a result, there is an overall strengthening in

intermolecular forces when nitric acid mixes with water and the resulting solution will become less volatile than

expected. Therefore, a positive deviation from Rauolt's law is observed.

e i MCO3(s)heat

MO(s) + CO2(g)

ii On moving down a group, the size of cations of group II metal gets larger and the polarizing power of the cation decreases

down the group.

As a resulting, the electron cloud of CO32- ions will get less distorted and the carbonate will get more thermally stable.

Furthermore, a big cation also do not favour the formation of the decomposition product group(II) metal oxide as the lattice

energy of the oxide gets smaller down the group.

iii The solubility of Group II metal carbonates decreases down the group.

CO3

2- ion is a big polyatomic ion. On moving down the group, the size difference between the cation and the carbonate ion

gets less. As a result, the magnitude of the lattice energy is rather constant on moving down the group and the magnitude of

the hydration energy decreases rapidly. Consequently, the heat of solution will get less exothermic down the group and the

carbonate will become less soluble.

4 a i (I) +4

(II) dichlorotetraammineplatinum(IV) bromide

ii 6

iii Octahedral

iv A yellow ppt will form.

v Aquation isomerism


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Page 10

vi (I)

(II) Geometrical isomerism

b i Transition metal ion : Vacant d orbital ; Higher charge density

Ligand : Availability of lone pair ; Carrying no or negative charge

ii The solution turns from green to blue.

CuCl42-

(aq) + 6H2O(l)d Cu(H2O)62+(aq) + 4Cl-(aq)

The blue solution turns paler with formation of pale blue precipitate

Cu(H2O)2+

6(aq) + 2OH-(aq)d Cu(OH)2(s) + 6H2O(l)

The pale blue precipitate redissolves to form a deep blue solution

Cu(OH)2(s) + 4NH3(aq)d Cu(NH4)42+

(aq) + 2OH-(aq)

iii Comparing with a monodentate ligand, when a bidentate ligand form a complex with a central atom, the total no. of molecule

formed will be more by 1. This gives an overall increase in disorderness of the system which is favorable in nature. Therefore,

bidentate ligand is ,in general, stronger.

c i This is because the energy of the ns electron and (n-1) d electron are similar. As a result, there is only a moderate increase in

successive ionization energy. As a result, there is no formation of any 1 oxidation state which is particularly energetically

favorable. Therefore, it may form more than 1 oxidation state.

ii Under the influence of ligands e.g. water , the 5 degenerate d-orbitals will become non-degenerated. Electrons in the lower

d-orbitals may be promoted to the upper d-orbitals by the absorption of visible light. As a result, certain visible light will be

absorbed by the compound and the colour of the light that passing through the compound will be revealed.

d Use Flame test

Use a platinum wire / nichrome wire.

Clean the wire by dipping it in conc. HCl and heating with a bunsen flame repeatedly.

Dip it into the sample of NaCl and heat the wire. The flame will be golden yellow.

Repeat the procedure by using KCl. The flame will be lilac / purple.

5 a i It is an aldose. as it is comprised of 2 glucose molecule which has a aldehyde acyclic form.

ii It is an reducing sugar as one of the glucose unit is interconvertible with the acyclic form / possesses hemiacetal group.

iii glycosidic linkage.It is formed by condensation between the 2 monosaccharides units with elimination of water.

iv Yes, it is soluble in water.

First of all it is a polar molecule.

Second, it possesses many hydroxyl group and is capable to form extensive hydrogen bonds with water.


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Page 11

b i Iodoform test / condensation with NaHSO4(aq)

Iodoform test

Iodine solution in KI(aq) with NaOH(aq) warm

yellow ppt. with CH3COCH2CH2CH3 and no pp. with CH3CH2COCH2CH3

CH3COCH2CH2CH3 + I2 + NaOH CHI3 + CH3CH2CH2COO-Na+

NaHSO4(aq)

NaHSO4(aq) white ppt. with CH3COCH2CH2CH3 and no pp. with CH3CH2COCH2CH3

ii Heat with acidified AgNO3(aq)

Formatioin of wit ppt with PhCH2Cl and no ppt. with CH3C6H4Cl.

PhCH2Cl + H2O PhCH2OH + HCl

Cl-(aq) + Ag+

(aq) AgCl(s)

iii HNO2(aq) at 0-5C / NaNO2(aq) with HNO2 at 05 C (Add phenol solution)

Compound I will not give bubbles immediately (and will give a orange red ppt. with phenol).

Compound II will give colour bubbles immediately (and will not give any ppt. with phenol).

PhCH2NH2 + HONO PhCH2N+N PhCH2+ + N2

cNH3CH3NH2 NH2 NO2 NH2

The basicity of a base is depending on the relative stability of the conjugate acid comparing with the base. The more stable the

conjugate acid, the more basic the base is.

Comparing the conjugate acids of the bases,

CH3NH2

His the most stable one as CH3 is an electron-donating group by inductive effect which helps to disperse the positive

charge on the acid. Thus CH3NH2 is the most basic.

ForNH3

H, H is neither electron-donating nor electron withdrawing. Therefore NH3 is less basic than CH3NH2.

ForNH2

H, Ph group is electron-withdrawing by inductive effect. This intensifies the charge density on N and makes

the ion less stable. Therefore, it is less acidic.

ConsideringNO2 NH2

H, on top of the negative inductive effect of the benzene ring, NO2 is also an

electron-withdrawing group by resonance effect. Therefore, NO2 NH2H

is the least stable conjugate acid and

NO2 NH2 would be the least basic base.

d

C C C

H

H

H

OH Br

H

H

H

CH

H

H

C C

H H

H

Br Br

C C C

H

H

H

Br

H

H

H

Cl


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Page 12

CC

Br

Br

+

bromoniumion

CC

Br

+

When bromine approaches the double bond of propene, the

bromine molecule is polarized by the electron rich C=C and

Br+ will be added onto the double bond to form a bromonium

ion intermediate.

The intermediate is vulnerable to the attack of different

nucleophile present to form different kind of product.

There are 3 kinds of nucleophile present in the solution H2O,

Br-, Cl-. They are added to the bromonium ion to form

different products respectively.

bromoniumion

CC

BrO

H

H

CC

Br

+

OH

H

CC

BrO

H

halohydrin

CC

Br

+ CC

BrCl

bromoniumionCl

-

CC

Br

+

Br-

CC

Br Br

bromoniumion

6 a i It will be a soft fatty solid. In the presence of double bond in one of the fatty acid, the molecule will become unsymmetrical

and cannot be packed effectively in the solid. Thus, it will be a softy solid instead of a hard one.

ii (I) In the presence of oxygen and moisture in air, lard may undergo oxidative rancidity or hydrolytic rancidity

HydroperoxideHydroperoxidefree radical

Aldehydes, ketonesand carboxylic acids

cleavage ofdouble bonds

The smell of rancidity is from the volatile fatty acids, aldehydes, ketones and carboxylic acids formed.

(II) Add antioxidant e.g. BHT to lard / Keep it in dry place / Keep it in cool place / Keep it with drying agent. e.g silica gel.

b iCH3CH2CH2CHO CH3CH2CH2COOH

H2Cr2O7heat

PCl5

CH3CH2CH2CONH2CH3CH2CH2COClexcess NH3

ii

CH3CH CH2CH3CHCOOH

CH3

HCNCH3CH CH2

HC

N

dil H2SO4(aq)

reflux

iii conc HNO3conc H2SO4 55C

NO2 Sn/conc HCl

OHNH2i. HNO2(aq) 0-5C

ii. warm

iv

NO2 Sn/conc HCl NH2 HNO2(aq) 0-5C

N

N

ICuI

v

CH3CH2CH2OHP2O5

CH3 CH CH2conc. H2SO4(l)

CH3CHCH3

OH

H2O

heat

CH3CHCH3

O

S OO

OH


13/14

Page 13

c i OH is a very poor leaving group as OH- is a strong base. H2SO4 is added to protonated the OH group and covert it to

O+H2 so that the group can leave ready as H2O and substituted by Br- to form RBr.

ii Normally, 1 haloalkane will participate in SN2 reaction. However, the presence of bulky t-butyl group (CH3)3C

sterically hindered the -C and make it not vulnerable to the attack of the nucleophile.

iii In CH3CH=CHCl, the CCl bond possesses certain double bond character,

C C C Cl

HH

H

H

H C C C Cl

H H

H

H

H-

The hydrolysis of the chloride involves a higher activation energy and the rate of reaction would be slower.

CH2=CHCH2Cl, only possesses an ordinary CCl single bond.

7 a i Phosphorus(V) oxide

iiH C C C C

H

H

HH

O

H

H

H

HH

C C

H

CH3CH3

H

C C

H

CH3H

CH3

Q can exist as geometrical isomers as the double bond is not freely rotatable. This is because the bond in the C=C bond does

not have a cylindrical symmetry.

iii Step 2 NaCN

Step 3 H2SO4(aq) reflux

Step 4 CH3CH2OH + H2SO4(aq) reflux

iv

H C C C C

HHH

H

H

HH H Br

H C C C C

HHH

H

H

HH

H Br

H C C C C

HHH

H

H

HH H

Br-

C

O

R OHH+

fC

O

R OH

H

O H

R'

+fC

O

R OH

H

+H

O R'

r.d.s.

no chargeseparation involved

C

O

R OR'H2O H++ +f +C

O

R O

H

O

R'

H

H

f

In the acid catalyzed pathway, the carbonyl oxygen is first protonated. This activates the carbonyl group by increasing the

polarity of the C=O bond and makes the carbonyl carbon more positive and more vulnerable to the attack of the nucleophile,

alcohol. Furthermore, no reaction intermediate with charge separation is involved in the acid catalyzed mechanism and a

much lower activation energy would be involved.

v

H C C C C

HHH

H

H

HH H Br

S in chiral.

S possesses a chiral centre at 2nd carbon joining to 4 different groups CH3, Br, H, Et.

S does not possesses any plane of symmetry.


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P 14

b i SO2 + H2O H2SO3 Absorption

CaCO3 + H2SO3 CaSO3 + CO2 + H2O Neutralization

CaSO3 + O2 CaSO4 Oxidation

CaSO3 + H2O CaSO3H2O Crystallization

CaSO4 + 2H2O CaSO42H2O (gypsum) Crystallization

ii Mass of sulphur in the 6 million tonnes of coal = 6 106 103 kg 2% = 1.2 108 kg

S(s) + O2(g) SO2(g)

2SO2(g) + 2CaCO3(s) + O2(g) + 4H2O(l) 2CaSO42H2O(s) + 2CO2(g)

molar mass of sulphur dioxide = 32.1 gmol-1

molar mass of calcium carbonate = (40.0 + 12.0 + 16.0 3) gmol-1 = 100.0 gmol-1

The mole ratio of sulphur to calcium carbonate = 1: 1

Mass of limestone required = 1.2 108 kg 100.0 gmol-1

32.1 gmol-1= 3.74 108 kg = 3.74 105 tonnes

iii It can be used to make cement.

iv Use coal with lower sulphur content.

Documents

2001-2002 Form 7 Mock Suggested Answers