report micho selection 2013

7/29/2019 Report MIChO Selection 2013

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Preliminary Report on the Results of MIChO 2013

2

Inorganic Problem 1

For this problem, several students are too “shy” to show their working which could afford

marks in case the final answer is wrong. Some students lost a lot of marks for not showing the

essential steps.

(a) To balance the equation: 2

3 2 2 7 4CrI Cl O I lCr O C , we balance the three

equations separately:

3 2

2 2 72Cr 7 CrH 6O H1O 4 e [1]

2 4H O IO 8eHI 4 8 [1]

22e 2Cl Cl [1]

We combine the first two, with six equivalent of the second:

2

3 2 2 7 42CrI 31 Cr 6 62 54H O O IO H e

Finally, balance with 27 equivalent of the third:

2

3 2 2 2 7 42CrI 31 27 CH O Cl O IO Hr 6 l6 2 54C [1]

A common mistake for this problem is to have the third equation to be balanced as

2eCl Cl . Another common mistake includes the failure to balance the

combined first and second equation.

(b) To balance the equation: 4 2 2

4 2MnO UOU Mn , we balance the two equations

separately:

4 2

2 2O UOH 2eHU 2 4 [1]

2

4 28 5e MnMnO H O4H [1]

And combining them, with five equivalents of the first and two of the second, we have:

4 2 2

4 2 2MnO O U5 O HU 2 2H 5 2Mn 4 [1]

Then, add four equivalent of hydroxide:




3

4 2 2

4 2 2MnO O5U 2 4 5 2Mn O2UO HH [1]

Inorganic Problem 2

(i) More than one answer is possible for this question but no student have justified their

choice by calculating G or E for the disproportionation. Choose one will do.

Supposed we choose 2

2Ru(OH) . The half-equations are:

2

2 2Ru(OH) R H euO 2 1.5 VE

2 3

2 2H e H ORu(OH) 2 Ru 2 0.86 VE

And the balanced equation is:

2 3

2 2 22Ru(OH) RuO u H OR 2 0.64 VE

This is clearly not feasible.

If we choose 2Ru , the half-equations are:

2 2eRu Ru 0.8 VE

2 2

2 2Ru 2 Ru(H O 2eOH) 2H 0.24 VE

And the balanced equation is:

2 2

2 22Ru 2 Ru Ru(OH)H O H2 0.56 VE [3*]

This is a feasible reaction.

* 1 for choosing, 1 for balanced equation, 1 for emf calculated.

Other wrong choices are:

4 2 4 23RuO 4 RuO 2RuO 2H H O 1.6 2(0.99) 0.38 VE

(ii) Choose one that does not show H+ in the balanced half-equation. Only one will do.

For example, 2Ru because 2 2eRu Ru has no H+ in the expression. [2*]




4

* 1 for choice, 1 for equation.

Other possible choices are4 4

eRuO RuO .

A common mistake is that students failed to give the equation, thus losing some

essential marks.

(iii) 2(1.6) 1.5 0.86

1.39 V4

E

[2*]

* 1 for formula, 1 for answer.

Some students tend to forget to display the unit.

Inorganic Problem 3

(a) Coordination number of 2Be is 4. [1]

Coordination number of 2S is 4. [1]

Wrong answers could be attributed to: (1) the student does not comprehend the

meaning of coordination number; (2) the student could not visualize a unit cell as a “3D-

translational basic unit”, in which the coordination number for sulfide is commonly

given as 2.

(b) There are 4 2Be ions and1 1

82 8

6 4 2S ions. [2]

(c) There is a problem with the question. It should have been the maximum cation radii. In

case one has forgotten, this maximum ratio can be determined from a tetrahedron.




5

2 2 2

a a c a c

22

a a c

2

c

a

c

a

2 2 cos

12 1

3

21 1

3

3 1 0.2252

2r r r r r

r r r

r

r

rr

Therefore

2Be 0.225 1.24 0.279 Ar [2*]

* 1 for the factor / ratio, 1 for answer.

Inorganic Problem 4

(a) Hexaamminechromium(III) bromide: [Cr(NH3)6]Br3 [1]

(b) Tris(oxalate)cobaltate(III) ion: [Co(C2O4)3]3− [1]

Inorganic Problem 5

Three ions are formed: one2

3 4u( H )C N

and two Cl . [1]

Inorganic Problem 6

190 sin1

3

cr

ar




6

Any two of these: [6*]

2 6 2O)e BF (H r

52 2Fe(H B .rO) OHBr

Hexaaquairon(II) bromide Pentaaquabromoiron(II) bromide hydrate

2 2 24Fe(H BrO) 2 OH.

2 2 24Fe(H BrO) 2 OH.

cis-Tetraaquadibromoiron(II) dihydrate trans-Tetraaquadibromoiron(II) dihydrate

* 1 for each correct structure, 1 each for correct formula, 1 for each correct name.

Inorganic Problem 7

The electronic configuration of Co2+ at high-spin configuration is:

showing three unpaired electrons. [1]

The hybridized orbital is sp3d2. [1]

Magnetic moment is 3(3 2) 15 B.M. [1]

Fe

OH2

OH2

H2O

H2O

H2O

H2O

2+

2Br −Fe

Br

OH2

H2O

H2O

H2O

H2O

+

Br − H2O∙

Fe

Br

Br

H2O

H2O

H2O

H2O

2H2O∙ Fe

OH2

Br

H2O

H2O

Br

H2O

2H2O∙

t2g

eg




7

Inorganic Problem 8

The electronic configuration of Cr3+ is:

[1]

The crystal field stabilization energy (CFSE) is therefore

o

o

CFSE 3 0.4

1.2

[2]

Organic Problem 1

(a) [1]

(b) The resonance structures are as shown: [1]

Organic Problem 2

(a) True [1]

(b) False [1]

t2g

eg




8

(c) True [1]

Organic Problem 3

(a) [1]

(b) [1]

(c) [1]

(d) [1]

(e) [1]

Organic Problem 4

(a) The structures of A and B are: [3*]




9

A B

* 1 for each correct structure, 1 extra for correctly assigning between A and B.

(b) The mechanism of the reaction is shown below:

[1]

Two products are obtained as the nucleophilic attack may proceed from two different

faces (from the top, the re-face, and from the bottom of the page the si-face) of the planar

carbonyl group.

[1]

Student’s common mistakes are: (1) not realizing the attacking species is the cyanide ion,

some wrote HCN as the attacking species; and (2) when describing the formation of

enantiomers, students did not provide schemes to assist their explanation.

H+

O H

R

O H

R

or




10

(c) The structures of C and D are: [2]

C D

Unfortunately, some students do not know that lactones are cyclic esters and gave the

open-chain carboxylic acid as an answer. Some gave the structure of lactones without

indicating the configuration of the hydroxyl groups and here no marks will be given as

the structure of the lactone is already given in the problem. Mere copying does not

justify for any marks.

Physical Problem 1

(a) Energy,

34 9

24

6.626 10 2.4 10

1.59 0 J1

E h

[2]

9

2 1

10

2.4 10

2.998 10

8.0 10 cm

c

[2]

Some students do not realize (rather an error from the question) that the unit “joule”

cannot be capitalized as “Joule” which refer to the English physicist James Prescott Joule.

No penalty is awarded though.

(b) Microwave (300 MHz to 300 GHz). [1]

(c) A molecule will change its state of rotational motion upon absorbing a microwave

photon. [1]




11

(d) Nothing can be deduced. The range in spectrum shown is at a factor of 1000 times

greater than that of the wave emitted by the router. [1]

Except for only one student, the others did not observe this fact.

(e) Volume of the tank:

5 3

40 40

1.6

10

0

0

cm1

V

[1]

Mass of water in the tank:

5

5

1.6 101. g116 0

m [1]

Amount of water molecules:

5

3

18

1.6 10

8 mo.9 l10

n

[1]

Number of water molecules:

3

27

238.9 10 6.023 10

5.4 10

N

[1]

Number of molecules to be excited:

27

26

5.4 10

5.4

10%

10

N

[1]

Energy of a photon at 179.8 cm (read any value from spectrum): [1]

34 10

21

10 16.626 2.998 79.8

1.6

0

10 J

E

[1]

Total energy required to excite the water molecules:




12

26 21

5

2

5.4 1.6

8.5 J

8.5 k

10 10

10

10 J

E

[1]

Many students cannot give the correct energy because not realizing part (e).

(f) Moment of inertia about C2-axis of water:

2

4 2

4 27 24 2

47 2

1.2 u pm

1.2 1.661

104.

10 kg m

1.9 k

52 1.0078 95.72sin

2

10

10 10

1 g0 m

I

[4*+2**]

* 1 mark for calculating r , 1 mark for calculating I , 1 mark for factor 2, 1 for answer.

** This part of the mark may appear at the following parts too: 1 mark for converting

atomic mass unit to kg, 1 mark for converting pm to m.

(g) The first five energy levels are:

0 0 0 1 0

1 1 1 1 2

2 2 2 1 63 3 3 1 12

4 4 4 1 20

F B

F B B

F B BF B B

F B B

The energy level diagram is:

104.5

2

95.72 pm

r




13

[4*]

* 1 for correctly evaluating F , 1 for drawing a logical energy level (increasing separation),

1 for proper labeling, 1 for starting with 0 J .

(h) The rotational spectrum is as shown:

[2*]

* 1 mark for shape (several equally spaced peaks with attempt to label), 1 mark for label

of wavenumber. No marks will be awarded if only the spectrum is produced on mere

memory work.

0 0F J

1 2F J B

2 6F J B

3 12F J B

4 20F J B

1/ cm

0 1

1 2

2 3

3 4

2B 4B 6B 8B

1/ cm




14

(i) The corresponding transitions in the spectrum are shown: [1]

Physical Problem 2

5 0.200 9.81 1.55

8.04 J

U

[2]

Physical Problem 3

(a) Exothermic [1]

(b) Endothermic [1]

(c) Endothermic [1]

(d) Endothermic [1]

(e) Endothermic [1]

Physical Problem 4

(i) Heat:

3 29.4 285 260

2.21 kJ

Pq

[2]

0 0F J

1 2F J B

2 6F J B

3 12F J B

4 20F J B

0 1

1 2

2 3

3 4

1/ cmE




15

(ii) Enthalpy:

2.21 kJP

H q

[2]

(iii) Internal energy:

2205 3 8.314 285 260

1.58 kJ

H U

nR

P V

U H T

[3]

Physical Problem 5

135 C 7 atm

225 C 1.5 atm

7 atm

21.5 atm

7 atm

1.5 atm

d 1d d

d d 0

25 C 135 C,1.5 atm 135 C,1.5 atm 7 atm

d d

4082 29.4 ln

298

18.5

718.5 2 8.314 ln1.5

,d 0

d

PT T

dV n

S

R T PP P

T P U T

P

S S

nC U P V

P

T T

PnR

T

nRP

1

18.5 25.6

7.1 J K

[9*]

* 1 mark for calculating in steps, 1 mark for integrating temperature, 1 mark for calculating first

term, 1 mark for using first law, 1 mark for substituting dV , 1 mark for knowing d 0T implies

d 0U , 1 mark for integrating pressure, 1 mark for calculating second term, 1 mark for final

answer.

Physical Problem 6

For the reaction 2 22HI g Cl g 2HCl g I g , the free energy is




16

rxn f f 2 f f 2

1

HCl g I g 2 HI g Cl g

2 95.4 0 2 1.30 0

193.4 kJ m

2

ol

G G G G G

[3]

The reaction is spontaneous because rxn 0G . [1]

NameElectrochemistry

+ Solid state (21)

Complexes

(15)

Organic

(17)

Spectroscopy

(28)

Thermodynamics

(27)Marks

Chan Jer Yong 16 15 17 21 25 89.9

Kong Yan Xiang 15 14 13 15 13 69.3

Taufiq Hamidi 12 13 13 15 12 65.0

Liew Ziqing 17 10 8 23 14 63.1

Eng Kah Hoe 13 6 10 10 11 49.9

Chong Yue Linn 11 11 4 14 15 45.8

Tan Ke Leen 5 6 6 8 10 32.9Amirah 2 7 3 9 8 24.5

Jamalina 4 4 4 9 4 23.1

Mohd Ikhwan 5 6 1 6 5 18.8

The final marks are calculated by weighing each section to a maximum of (100/3)%.

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