2014 MIChO Report

8/12/2019 2014 MIChO Report

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Report and Analysis to the

Malaysian IChO Team Selection Quiz 2014

Third Phase

Prepared by:

Yau Ching Koon

Answers were accessed using points that were deemed suitable by theauthor and may not reflect the true marks given. Generally, marks will

not be awarded heavily to pure memory work but rather to the scienceand thinking skills displayed unless the nature of the answer required isso. Special thanks to Goh Jun Yan for his marking the papers and aidingin the arbitration process.

Version 2


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Suggested answer and analysis

Question 1

(a) The standard potential of a cell is the potential difference of that cell in standard conditions(pressure P = 1 bar, temperatureT = 298.15 K and activity, a = 1 for solutes in solution orfugacity, f =1 for gases in gaseous mixture) with respect to the standard hydrogen electrode(SHE). [2]

1 point is given for potential difference w.r.t. standard hydrogen electrode and 1 point formentioning that the cell is in standard condition.

Many students failed to answer this question correctly because they could not appreciate thatan electric potential cannot be measured but can be measured as a difference, hence the termpotential difference. Being a difference, another cell must be quoted naturally and in this case,the SHE.

All students do not know that the new standard pressure recommended by IUPAC is 1 bar!

(b) There are three modes of mass transfer: migration, diffusion and convection.

Migrationis the movement of a charged body under the influence of external electric field.

Diffusionis the net movement of a chemical species due to a gradient in chemical potential orconcentration.

Convection can be classified into two: anatural convectiondue to density gradient and aforcedconvectiondue to stirring introduced to the bulk liquid. [7]

1 point each for identifying each modes of mass transfer. 1 point each for explaining migrationand diffusion. 2 points for explaining convection. It is necessary to mention thecauseof themass transfer in the explanation to show that the students could distinguish them from oneanother.

Most students do not realize that the natural convection can be obtain as a result of densitygradient. During the operation of a electrochemical cell, the electrolyte will be heated becauseof its own resistance and due to inhomogeneous heating and cooling, different parts of the

electrolyte will attain different density. Density is dependent on temperature. Hence, thedensity gradient produces a natural convection.

(c) If the electron transfer process is the rate determining step, then other contributions to therate such as the rate of mass transfer and that of the preceding reactions are negligible. Therate of mass transfer can be kept at minimum by affecting an effective stirring while the rate of


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1 point for understanding the relation of rate determining electron transfer step to other effectsthat contributes to the overall rate. 1 point for giving examples on how the other rates can bekept at minimum. 1 point for producing Tafel or Butler-Volmer equation.

Most students fair poorly in this question showing that the students does not comprehendelectrode kinetics and in particular, the relationship between Butler-Volmer equation and Tafelequation.

(d) If the electron transfer is rapid, then mass transfer or preceding reactions could be the ratedetermining step. In either case, the concentration of the reactants near the electrode willdeplete and the product will increase. If we consider a reduction process (O becoming R),then [O] will decrease and [R] increase until the both are regulated by the rate mass transfer or

preceding reactions. The incorporation of mass transfer effect to Butler-Volmer equation gives

i

i0=

1 i

il,c

eF/RT

1 i

il,a

e(1)F/RT.

[2]

1 point for understanding the relation of electron transfer being the fast step to other effects

that contributes to the overall rate. 1 point for discussing the concentrations of oxidised andreduced species.

Most did not answer this question.

(e) Regretfully, answers to this part will not be produced because it is deemed as too much of pure memorywork to the extent that it may not worth any points by the authors standard.

Question 2

(a) dsp2: square planar, d2sp3: octahedral, dsp3: trigonal bipyramid or trigonal bipyramidal. [3]

1 point each for correct answer. Most students could answer this question. Some do not recognize dsp2 as square planar and

answered tetrahedral instead while some erroneously spelled trigonal bypyramid.

(b) The five d orbitals are shown below. According to the crystal field theory, six ligands ap-proached a metal ion from and along the +x, +y, +z,x,y, andz axes. All the five dorbitals will be destabilise because the metal is introduced into an environment with ligandsbut the orbitals dz2 and dx2 y2 would experience further repulsion because these orbitals are


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dz2 dx2y2 dyz

dxz dxy

Most students do not realize that the presence of ligands destabilised all the five d orbitals.Some did not draw the five d orbitals as required and some who had drawn did not label theaxes. Overall performance is good.

(c) The spectrochemical series is a list of ligands arranged in either ascending or descending orderof crystal field splitting energy produced by ligands in their complexes. This series is usuallyestablished with UV-Vis spectrophotometry. In the ascending order, the spectrochemical seriesfor the ligands given is

Cl, H2O,NH3, CN

Thus, the crystal field splitting energy for a complex, say[CoCl6]3 is the smallest while the

crystal field splitting energy in the complex[Co(CN)6]3is the largest. [3]


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a weak field ligand allows theb1genergy level to be occupied in[Ni(NH3)4]2+ where the elec-

tronic configuration eg4

a1g2

b2g1

b1g1

,

is assumed. On the other hand, cyanide ion is a strong field ligand in[Ni(CN)4]2and forces

theb1gelectron to pair up with the electron in b2gresulting in the electronic configuration:

eg4

a1g2

b2g2

.

Hence[Ni(NH3)4]2+ is paramagnetic while[Ni(CN)4]

2is diamagnetic. [4]

1 point for recognizing that the tetragonal distortion results in the splitting of the t2g and egorbitals toeg,a1g,b2gandb1gorbitals. 1 point each for electronic configuration of[Ni(NH3)4]

2+

and[Ni(NH3)4]2+ (may be given as energy level diagram). 1 point for drawing conclusion.

Many students either did not answer the question or gave irrelevant responses. Some, withoutPeriodic Table provided to them, used the wrong number of d electrons. Only one studentproduced the correct answer.

Question 3

(a) From Arrhenius equation, we have

ln k= ln A

EaR

1

T,


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From a graph of ln

k/mol1 L s1

against 1/ (T/K), the gradientmobtained is:

m=

Ea

R

K= 13605

Ea = 113kJmol1

[7]

1 point for effort to prepare the table, 1 point for conversion toTin kelvin, 1 point for effort toprepare graph, 2 point maximum for correct graph (penalty applied for any error), 1 point forcalculating gradient, 1 point for correct final answer including units.

Most students could do this question. The common mistake is not to realize that the unit iskJ mol1 and some forgetting the factor 103 in their calculation. Penalties were awarded fornot attempting to draw the line of best fit, graphs utilizing less than half of the graph paperprovided, and for not providing enough ticks that include the terminal points.

(b) From Arrhenius equation, we have

ln (0 079) = ln AEa 1


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The word change of rotational state is necessary because the molecule may be in a rotationalstate already.

No student could answer this question correctly. For those who has selected microwave inpart (b), many answered the molecule will rotate but that is insufficient because the moleculemay already be rotating initially. Upon receiving a photon in the microwave region, the initialrotational state is excited to another rotational energy level.

(d) Volume of the tank,V=100 40 40= 1.6 105 cm3.Mass of water in the tank,m = 1 1.6 105 =1.6 105 g.Amount of water molecules,n = 1.6 105/18= 8.9 103 mol.Number of water molecules,N=8.9 103 6.023 1023 =5.4 1027.Number of water molecules to be excited, N=10% 5.4 1027 =5.4 1026.Total energy required to excite the water molecules,

E=

5.4 1026

1.59 1027

= 0.85 kJ.

[6]

1 point each for each step. Many could not solve the entire problem, perhaps because they do not understand the basic

principles of quantum mechanics.

(e) The moment of inertia about the C2

axis,

I=2 1.0078

95.72 sin104.5

2

2,

=1.155 104 u pm2,1 917 1047 k 2


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(f) The first five energy levels are

F(0)= 0 (0 + 1)B= 0,

F(1)= 1 (1 + 1)B= 2B,

F(2)= 2 (2 + 1)B= 6B,F(3)= 3 (3 + 1)B= 12B,

F(4)= 4 (4 + 1)B= 20B,

where

B= h

82cI =

6.63 103482

3 1010

1.97 1047

=14.2 cm1.

[5]

1 point for calculating F, 1 point for drawing energy levels with increasing separation, 1 pointfor starting atJ= 0, 1 point for labeling correctly, 1 point for calculatingB.

Some students did not calculate the value ofB, some did not start with J = 0 and some didnot sketch the energy levels in increasing separation which is in clear violation to what one hascalculated forF.

(g) Infrared. [1]

Most students could answer this question.

(h) The maximum number of vibrational normal modes that can be observed is 3 (3) 6= 3. [2] 1 point for calculation with 3N 6, whereN= 3 and 1 point for final answer. Most students could answer this question.


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Most students could answer this question.

(j) The vibrational mode of the OH bond is found typically at =3500cm1. We have =cand since the frequency, is inversely proportional to the square root of reduced mass,

under the assumption that the force constant for OH and OD bonds are similar, we have

OHOD

=OHOD

=

ODOH

=

1.77

0.94=1.37,

OD= 3500

1.37 =2555 cm1.

whereOD=2 16/18= 1.77 u andOH=1 16/17= 0.94 u. [8] 1 point for knowing a typical value offor OH bond, 1 point for knowing 1/, 1 point

for assuming kis constant, 1 point for knowing , 1 point each for calculating reduced massof OH and OD, 1 point for calculating OD, 1 point for correct answer.

Most students could not answer this question. Due to insufficient exposure to IR spectroscopy,they could not give a typical value ofand are not able to produce explicitly .

Question 5

(a) There are two chair conformers of piperidine, C5H11N:

There are four chair conformers of 1,4-disubstituted piperidine including their isomers.


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Many students did not answer this question, perhaps because they could not recall the struc-ture of piperidine.

(b) The aromaticy of a compound is determined by four criteria, all of which must be satisfied:

Compound Aromatic? Cyclic? Planar? Fully conjugated? Huckel rule?

(a) Y Y Y Y Y [2= 4(0) + 2](b) N Y Y N N [4= 4(?) + 2](c) Y Y Y Y Y [6= 4(1) + 2](d) N Y Y Y N [8= 4(?) + 2](e) Y Y Y Y Y [6= 4(1) + 2]

(f) Y Y Y Y Y [10= 4(2) + 2](g) Y Y Y Y Y [14= 4(3) + 2]

[7]

1 point for each correct answer along with justification. No student scored full marks for this question despite being simple, suggesting that the stu-

dents are not explicitly exposed to the criteria for determining aromacity.

Question 6

(a) (a) three, (b) two. [2]

1 point for each correct answer.

All students are able to do this question.

(b) (a) four, (b) three, (c) four, (d) four. [4]

1 point for each correct answer. Structures (c) with 19 wrong responses, followed by (d) with 14 wrong responses and finally

(a) with 11 wrong responses.

( ) ( ) H C

O

CH3

(b) B OCH [2]


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Results

Since 20 marks are allocated on each question, the total points for each questions is weighed equally to compute the final marks.

Name Institution Q1 Q2 Q3 Q4 Q5 Q6 TotalJames Loke Wen Liang Foon Yew High School 7.4 16 14 21 5 6 67.8Gan Kin Boon INTI 8.4 13 12 17 5 5 59.9Leong Sheng Yuan Foon Yew High School 7.7 17 9.7 22 0 6 59.3Tan Ze Kai Sunway University College 9 8 13 20 5 5 58.6Khu Wai Hoong Kolej Matrikulasi Pahang 8.7 10 14.7 17 3 5 58.3Ooi Kok Yong Kolej Matrikulasi Pulau Pinang 10 9 8.7 17 4 7 57.5

Ahmad Syahmi b. Adnan Kolej Mara Banting 7 16 11.7 11 0 6 54.2Marcus Lim Guozong Kolej Matrikulasi Pulau Pinang 6.7 9 13.7 10 0 8 53.2Tow Boon Chee Kolej Matrikulasi Pahang 7.7 9 14 12 3 5 52.9Denedy Wong Siong Yong SMK Batu Lintang 7.7 10 15 5 2 6 52.3Boon Zhen Hern SMJK Jit Sin 9 8 12 17 0 5 50.5Tan Yi Sheng SMK Temerloh 7 10 13.4 14 0 5 50.0Chen Yi Kit SMJK Jit Sin 7.7 5.5 11 16 0 6 47.1

Lim Chun Wei SMJK Heng Ee 8.7 6.7 10.7 15 2 4 46.6Ooi Lim Seong Liang SMJK Chung Ling 7.4 5.5 13 9.7 2 4 43.8Chang Ji Shen SMJK Chung Ling 7 8 10.7 6 0 4 39.0Nur Amalina binti Azmi Kolej Matrikulasi Perak 6.7 4 10 11 0 5 38.8Mohamad Syahmi bin Mohd. Hothzani Kolej Mara Banting 8.7 4 12 5 0 4 38.3Eow Wei Siang SMJK Heng Ee 4.7 7.7 8 11 0 5 37.6Mohamad Ashiqeen b. Anwar Ali Kolej Matrikulasi Pulau Pinang 4 7 8.7 10 1 4 35.4Tan Hooi Nee SMK Tsung Wah Kuala Kangsar 4 0 12 13 0 4 33.0Wan Jia Wei SMK Temerloh 8 10 4.7 4 0 3 32.3Tong Cher Ling INTI 6 9.5 0 4 0 6 30.5

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Documents

2014 MIChO Report