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Theoretical Problems

44th International Chemistry Olympiad July 26, 2012 United States of America


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Instructions
Write your name and code on each page. This examination has 8 problems and Periodic Table on 49 pages. You have 5 hours to work on the exam problems. Begin only when the START command is given. Use only the pen and the calculator provided. All results must be written in the appropriate boxes. Anything written elsewhere will not be graded. Use the back side of the exam sheets if you need scratch paper. Write relevant calculations in the appropriate boxes when necessary. Full marks will be given for correct answers only when your work is shown. When you have finished the examination, put your papers into the envelope provided. Do not seal the envelope. You must stop working when the STOP command is given. Do not leave your seat until permitted by the supervisors. The official English version of this examination is available on request only for clarification.

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Physical Constants, Formulas and Equations
Avogadro's constant, NA = 6.0221 1023 mol­1 Boltzmann constant, kB = 1.3807 10 Speed of light, c = 2.9979 108 ms­ Planck's constant, h = 6.6261 10
­34 1 ­23

JK­

1 ­1

Universal gas constant, R = 8.3145 JK­1mol­1 = 0.08205 atmLK­1mol

Js
­31

Mass of electron, me = 9.10938215 10 Standard pressure, P = 1 bar = 105 Pa Atmospheric pressure, P
a tm

kg

= 1.01325 105 Pa = 760 mmHg = 760 Torr

Zero of the Celsius scale, 273.15 K 1 nanometer (nm) = 10­9 m 1 picometer (pm) = 10
­12

m
2

Equation of a circle, x2 + y2 = r r
2

r r3/3 r
2

Bragg's Law of Diffraction: sin = n/2d

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1 1 1.00794

Code:
18 2 4.00260

1
3

H
0.28

He
2 Atomic number 1 1.00794 Atomic weight Atomic symbol Covalent radius, å 5 10.811 13 6 14 15 16 17
1.40

4 6.941 9.01218

7 8 9 10 12.011 14.0067 15.9994 18.9984 20.1797

2

Li

Be

H
0.28

B
0.89

C
0.77

N
0.70

O
0.66

F
0.64

Ne
1.50

11 12 22.9898 24.3050

13 14 15 16 17 18 26.9815 28.0855 30.9738 32.066 35.4527 39.948

3

Na

Mg

Al

Si
1.17

P
1.10

S
1.04

Cl
0.99

Ar
1.80

3 4 5 6 7 8 9 10 11 12 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 39.0983 40.078 44.9559 47.867 50.9415 51.9961 54.9381 55.845 58.9332 58.6934 63.546 65.39 69.723 72.61 74.9216 78.96 79.904 83.80

4

K

Ca

Sc

Ti
1.46

V
1.33

Cr
1.25

Mn
1.37

Fe
1.24

Co
1.25

Ni
1.24

Cu
1.28

Zn
1.33

Ga
1.35

Ge
1.22

As
1.20

Se
1.18

Br
1.14

Kr
1.90

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 85.4678 87.62 88.9059 91.224 92.9064 95.94 (97.905) 101.07 102.906 106.42 107.868 112.41 114.818 118.710 121.760 127.60 126.904 131.29

5

Rb

Sr

Y

Zr
1.60

Nb
1.43

Mo
1.37

Tc
1.36

Ru
1.34

Rh
1.34

Pd
1.37

Ag
1.44

Cd
1.49

In
1.67

Sn
1.40

Sb
1.45

Te
1.37

I
1.33

Xe
2.10

55 56 57-71 132.905 137.327

72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 178.49 180.948 183.84 186.207 190.23 192.217 195.08 196.967 200.59 204.383 207.2 208.980 (208.98) (209.99) (222.02)

6

Cs

Ba La-Lu

Hf
1.59

Ta
1.43

W
1.37

Re
1.37

Os
1.35

Ir
1.36

Pt
1.38

Au
1.44

Hg
1.50

Tl
1.70

Pb
1.76

Bi
1.55

Po
1.67

At

Rn
2.20

87 88 89-103 (223.02) (226.03)

104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 (261.11) (262.11) (263.12) (262.12) (265) (266) (271) (272) (285) (284) (289) (288) (292) (294) (294)

7

Fr

Ra Ac-Lr
2.25

Rf

Db

Sg

Bh

Hs

Mt

Ds

Rg

Cn

Uut

Fl

Uup

Lv

Uus

UUo

57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 138.906 140.115 140.908 144.24 (144.91) 150.36 151.965 157.25 158.925 162.50 164.930 167.26 168.934 173.04 174.04

La
1.87

Ce
1.83

Pr
1.82

Nd
1.81

Pm
1.83

Sm
1.80

Eu
2.04

Gd
1.79

Tb
1.76

Dy
1.75

Ho
1.74

Er
1.73

Tm
1.72

Yb
1.94

Lu
1.72

89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 (227.03) 232.038 231.036 238.029 (237.05) (244.06 (243.06) (247.07) (247.07) (251.08) (252.08) (257.10) (258.10) (259.1) (260.1)

Ac
1.88

Th
1.80

Pa
1.56

U
1.38

Np
1.55

) Pu
1.59

Am
1.73

Cm
1.74

Bk
1.72

Cf
1.99

Es
2.03

Fm

Md

No

Lr

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PROBLEM 1
a­i 4 a­ii 2 a-iii 2 b 2

7.5% of the total
C 10 Problem 1 20 7.5%

a. Boron Hydrides and Other Boron Compounds Boron hydride chemistry was first developed by Alfred Stock (1876-1946). More than 20 neutral molecular boron hydrides with the general formula B xHy have been characterized. The simplest boron hydride is B2H6, diborane.

i. Using the data below derive the molecular formulae for two other members o f this series of boron hydrides, A and B (A and B). S ub s t a n c e A B S ta te (2 5 °C , 1 b a r) Liquid Solid Mass Percent Boron 83.1 88.5 Molar mass (g/mol) 65.1 122.2

A = _____ B5H11_______

B = ______ B10H14______

2 points each = 4 points

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ii. William Lipscomb received the Nobel Prize in Chemistry in 1976 for "studies on the structures of boron hydride s illuminating the problems of chemical bonding." Lipscomb recognized that, in all boron hydrides, each B atom has a normal 2 -electron bond to at least one H atom (B­H). However, additional bonds of several types occur, and he developed a scheme for describing the structure of a borane by giving it a styx number w h ere: s = number of B­H­B bridges in the molecule t = the number of 3-center BBB bonds in the molecule

y = the number of two -center B­B bonds in the molecule x = the number of BH2 groups in the molecule The styx number for B2H6 is 2002. Propose a structure for tetraborane, B4H10, with a styx number of 4012.

H H B H H B H H B H B H H H H B H H H B B H H H B H H

H actual structure

unknown but acceptable structure

2 points for either of these structures

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iii. A boron-based compound is composed of boron, carbon, chlorine, and oxygen (B4CCl6O). Spectral measurements indicate the molecule has two types of B atoms, with tetrahedral and trigonal planar geometry, in a 1:3 r atio, respectively. These spectra are also consistent with a CO triple bond . Given that the molecular formula of the compound is B4CCl6O, suggest a structure for the molecule. S t ru ct u re :

O C Cl2B Cl2B B BCl2

2 points. Not required to show the stereochemistry

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b. Thermochemistry of Boron Compounds Estimate the B-B single bond dissociation enthalpy in B2Cl4(g) using the following information: B o nd Bond Dissociation Enthalpy (kJ/mol) B­Cl 443 Cl­Cl 242 Compound BCl3(g) B2Cl4(g) fH° (kJ/mol) ­403 ­489

A Born-Haber cycle gives a B-B bond dissociation enthalpy of 327 kJ/mol 2B + 6Cl

rHo = 6 x 443 kJ

rHo = -4 x 443 - 242 kJ - BB

2BCl3 rHo = +317 kJ

B2Cl4 + Cl2

2 points

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c. Chemistry of Diborane Give the structure for each numbered compound in the scheme below. Each numbered compound is a boron-containing compound.

NOTES: a. The boiling point of compound 5 is 55 °C. b. Excess reagents used in all reactions. c. The freezing point depression for 0.312 g of compound 2 in 25.0 g of benzene is 0.205 °C. The freezing point depression constant for benzene is 5.12 °C/molal

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Number 1

Molecular Structure of Compound
OCH3 B

B( O C H 3 ) 2
C6H5 B O O B B O

3

H3CO

OCH3

C6H5

C6H5

A dimer or tetramer [C6H5BO]x are also acceptable (x = 2 , 4 ) .
Cl B

3

BC l

3

Cl

Cl

4
H

H B H

H N H H

BN H

6

Formal charges not necessary

5
H H B N

H N B H B N H H

B3 N 3 H

6

Formal charges not necessary

10 points, 2 points each but only 1 point for formula only

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PROBLEM 2
a­i 4 a­ii 4 b -i 6 b-ii 1

7.8% of the total
c 5 Problem 2 20 7.8%

Platinum(II) Compounds, Isomers, and the Trans Effect. Platinum and other Group 10 metals form square planar complexes and the mechanisms of their reactions have been studied extensively. For example, it is known that substitution reactions of these complexes proceed with retention of stereochemistry.

It is also known that the rate of substitution of ligand X by Y depends on the nature of the ligand trans to X, that is, on ligand T. This is known as the trans effect. When T is one of the molecules or ions in the following list, the rate of substitution at the trans position decreases from left to right. CN­ > H­ > NO2­, I­ > Br­, Cl­ > pyridine, NH3, OH­, H2O The preparations of cis- and trans-Pt(NH3)2Cl2 depend on the trans affect. The preparation of the cis isomer, a cancer chemotherapy agent commo nly called cisplatin, involves the reaction of K 2PtCl4 with ammonia.

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i. Draw all possible stereoisomers for square planar platinum(II) compounds with the formula Pt(py)(NH3)BrCl (where py = pyridine, C 5H5N).

Cl Pt py

Br NH3

Br Pt py

Cl NH3

py Pt Cl

Br NH3

4 points. Penalty of -1 for excessive number of structures 3 D p ers p ect i v e s t ru ct u res n o t req u i red . N eed cl ear i n d i cat i o n o f rel at i v e l o cat i o n o f ligands.

ii. Write reaction schemes including intermediate(s), if any, to show the preparation in aqueous solution for each of the stereoisomers of [Pt(NH 3)(NO2)Cl2]-- using, as reagents, PtCl42-, NH3, and NO2-. The reactions are controlled kinetically by the trans effect .

cis-isomer:

2­
Cl Pt Cl Cl Cl

NH3

Cl Pt Cl

Cl NH3

NO2­

Cl Pt Cl

NO2 NH3

2 points

trans-isomer:

2­
Cl Pt Cl Cl Cl

2NO2
­

Cl Pt Cl

Cl NO2

NH3

H3N Pt Cl

Cl NO2

2 points

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b. Kinetic Studies of Substitution Reactions of Square Planar Complexes Substitutions of the ligand X by Y in square planar complexes ML3X + Y ML3Y + X can occur in either or both of two ways: · Direct substitution: The incoming ligand Y attaches to the central metal, forming a five-coordinate complex, which then rapidly eliminates a ligand, X, to give the product, ML3Y.

[ML3XY] ** ** = rate determining step, Rate constant = kY

ML3X

+Y

­X ML3Y

· Solvent-assisted substitution: A solvent molecule S attaches to the central me tal to give ML3XS, which eliminates the X to give ML3S. Y rapidly displaces S to give ML3Y.
+S ** ­X [ML3XS]
S

+Y [ML3S] ­S ML3Y

ML3X

** = rate determining step, Rate constant = k

The overall rate law for such substitutions is Rate = ks[ML3X] + kY[Y][ML3X] When [Y] >> [ML3X], then Rate = kobs[ML3X]. The values of ks and kY depend on the reactants and solvent involved. One example is the displacement of the Cl­ ligand in a square planar platinum(II) complex , ML2X2, by pyridine (C5H5N). (The ML3X scheme above applies to ML2X2.)

N Pt N

Cl + Cl N

CH3OH

N Pt N

N + ClCl

Data for reaction at 25 °C in methanol where [pyridine] >> the concentration of the platinum complex are given in the table below.

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Concentration of pyridine (mol/L) 0.122 0.061 0.030

kobs (s­ 7.20 x 3.45 x 1.75 x

1

) 10 10 10

­4 ­4 ­4

i. Calculate the values of ks and kY. Give the proper unit for each constant. A grid is given if you wish to use it.

kY = 5.8 x 10­3 s­1M­1 kS = 0 s­1 (allow small range of values, ± 0.2 10­3) 6 points 1 point for each unit 1 point for each number 2 points for method

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kY = 5.8 x 10­3 s­1M­1 kS = 0 s­1 (allow small range of values, + or ­ 0.2 x 10­3) 6 points 1 point for each unit 1 point for each number 2 points for method

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ii. When [pyridine] = 0.10 mol/L, which of the following is true? (Tick the box next to t h e co rrect an s w er. ) Most pyridine product is formed by the solvent -assisted (ks) substitution pathway. X Most pyridine product is formed by the direct substitution ( kY) pathway Comparable amounts of product are formed by the two pathways. No conclusions may be drawn regarding the relative amounts of product produced by the two pathways. 1 point for (b) c. A chemotherapy agent In an effort to better target cisplatin to cancer cells, Professor Lippard's group at MIT attached a platinum(IV) complex to oligonucleotides bound to gold nanoparticles.

Gold nanoparticle Oligonucleotide

Pt(IV) complex attached

Experiments showed that a gold nanoparticle with a diameter of 13 nm. Attached to this nanoparticle are 90 oligonucleotide groups, with 98% of them being bound to a Pt(IV) complex. Suppose that the reaction vessel used for treating cells with the Pt(IV) nanoparticle reagent had a volume of 1.0 mL and that the solution was 1.0 x 10 ­6 M in Pt. Calculate the mass of gold and of platinum used in this experiment. (The density of gold = 19.3 g/cm3 and the volume of a sphere = (4/3)r 3 = 4.18879 r3.) 15

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Mass of platinum a) Amount of Pt used = (1.0 x 10­6 mol/1000 mL)(1.0 mL) = 1.0 x 10­9 mol Pt This is equivalent to 2.0 x 10­7 g Pt 1 point

Mass of gold b) (90 groups/nanoparticle)(0.98 Pt bound complexes) = 88 Pt complexes/nanoparticle or 88 Pt atoms per nanoparticle c) 1.0 x 10­9 mol Pt is equivalent to 6.0 x 1014 Pt atoms d) (6.0 x 1014 Pt atoms)(1 nanoparticle/88 Pt atoms) = 6.8 x 1012 nanoparticles e) Size of gold nanoparticles: Radius = 6.5 x 10­7 cm and volume of gold nanoparticle = 1.2 x 10 Mass of gold nanoparticle = 2.3 x 10
­17 ­18

cm

3

g
­19

Amount of gold in a nanoparticle = 1.2 x 10 f) Mass of gold:

mol

Atoms of gold in a nanoparticle = 7.1 x 104 atoms Total number of gold atoms = (6.8 x 1012 particles)(7.1 x 104 atoms/particle) = 4.8 x 1017 atoms of gold Equivalent to 1.5 x 10­4 g gold 4 points

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PROBLEM 3
a 4 b 12

7.5 % of the Total
c-i 6 c-ii 12 Problem 3 34 7.5%

Thiomolybdate ions are derived from molybdate ions, MoO42­, by replacing oxygen atoms with sulfur atoms. In nature, thiomolybdate ions are found in such places as the deep waters of the Black Sea, where biological sulfate reduction generates H2S. The molybdate to thiomolybdate transformation leads to rapid loss of dissolved Mo from seawater to underlying sediments, depleting the ocean in Mo, a trace element essential for life. The following equilibria control the relative concentrations of molybdate and thiomolybdate ions in dilute aqueous solution. MoS42­ + H2O(l) MoOS32­ + H2O(l) MoO2S22­ + H2O(l) MoO3S2­ + H2O(l) MoOS32­ + H2S(aq) MoO2S22­ + H2S(aq) MoO3S2­ + H2S(aq) MoO42­ + H2S(aq)
2­

K K K K

1 2 3 4

= = = =

1.3 1.0 1.6 6.5

â10 â10 â10 â10

­5 ­5 ­5 ­6

a. If at equilibrium a solution contains 1â10­7 M MoO would be the concentration of MoS42­?

4

and 1â10­6 M H2S(aq), what

Multiplying the mass action laws for the four given reactions produces:

[MoS42­] =

7â10­

12

Units: M
2­ 4

3 points for correct MoS

answer; 1 point correct units

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Solutions containing MoO2S22­, MoOS32­ and MoS42­ display absorption peaks in the visible wavelength range at 395 and 468 nm. The other ions, as well as H2S, absorb negligibly in the visible wavelength range. The molar absorptivities () at these two wavelengths are given in the following table: at 468 nm L mol­1 cm­1 MoS
2­ 4 2­ 3 2­

at 395 nm L mo­-1 cm­ 120 9030 3230

1

11870 0 0

MoOS

MoO2S2

b. A solution not at equilibrium contains a mixture of MoS42­, MoOS32­ and MoO2S22­ and no other Mo-containing species. The total concentration of all species containing Mo is 6.0â10­6 M. In a 10.0 cm absorption cell, the absorbance of the solution at 468 nm is 0.365 and at 395 nm is 0.213. Calculate the concentrations of all three Mo-containing anions in this mixture.
MoS
2­ 4

concentration is determined by absorbance at 468 nm: 0.365 = (11870)(10.0)(MoS42­). (MoS42­) = 3.08â10­6 M

4 points
6

From conservation of Mo, (MoOS32­) + (MoO2S22­) = Mo

Total

­ (MoS42­) = 6.0â10­6 ­ 3.08â10­6 = 2.9â10­ )

By rearrangement, ­ ­ (MoO2S22 ) = 2.9â10 6 (MoOS

2­ 3

From optical absorbance at 395 nm, 0.213 =(120)(10.0)(3.08â10­6)+ (9030)(10.0)(MoOS32­) + (3230)(10.0)(MoO2S22­) 0.213 = (120)(10.0)(3.08â10­6)+ (9030)(10.0)(MoOS32­) + (3230)(10.0)(2.9â10­6 (MoOS32­)) (MoOS32­) = 2.0â10­6 M 4 points (MoO2S22­) = 2.9â10­6 ­ (MoOS (MoO2S22­) = 0.9â10­6 M
2­ 3

) = 0.9â10­6 M MoO2S22 _____ MoOS32­ ______ MoS42­ _______
­

4 points

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c. A solution initially containing 2.0â10­7 M MoS42­ hydrolyzes in a closed system. The H2S product accumulates until equilibrium is reached. Calculate the final equilibrium concentrations of H2S(aq), and all five Mo-containing anions (that is, MoO42­, MoO3S2­, MoO2S22­, MoOS32­ and MoS42­). Ignore the possibility that H2S might ionize to HS­ under certain pH conditions. (One-third credit is given is given for writing the six independent equations that constrain the problem, and two-thirds credit is given for the correct concentrations.) i. Write the six independent equations that determine the system.
Mass balance for Mo: 2.0â10­7 = (MoS42­) + (MoOS Mass balance for S: 8.0â10 7 = 4(MoS42 ) + 3(MoOS Equilibrium constants: 1.3 1.0 1.6 6.5
­ ­
2­ 3 2­ 3

) + (MoO2S22­) + (MoO3S2­) + (MoO

2­ 4

)

2 points

) + 2(MoO2S22 ) + (MoO3S2 ) + (H2S)

­

­

2 points

â10­ â10­ â10­ â10­

5 5 5 6

= = = =

( ( ( (

MoOS32­)(H2S)/(MoS42­) MoO2S22­)(H2S)/(MoOS32­) MoO3S2­)(H2S)/(MoO2S22­) MoO42­)(H2S)/(MoO3S2­)

0.5 point each = 2 points Six equations in any format will be accepted provided they somehow introduce the four equilibrium constants and the two correct mass balance constraints.

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ii. Calculate the six concentrations making reasonable approximations, giving your answers to two significant figures. It is likely that multiple approaches will be found for solving these equations. Here is one approach: The maximum possible H2S concentration is 8.0x10­7 M, the amount formed if complete hydrolysis occurs. At this H2S concentration, MoO3S2­ is only about 12% of (MoO42­) and the remaining thio anions are much less abundant. Therefore, because the problem justifies a solution that is precise only to two significant figures, the mass balance equations can be truncated: 2.0â10­7 = (MoO3S2­) + (MoO42­) 8.0â10­7 = (MoO3S2­) + (H2S) (Mo mass balance) (S mass balance)

Subtracting the first from the second and rearranging gives: (MoO42­) = (H2S) - 6.0â10­7 Likewise, the S mass balance can be rearranged, (MoO3S2­) = 8.0â10­7 ­ (H2S) Employing the equilibrium constant for the reaction involving MoO [ [
4 2­

and MoO3S2 ] ]

­

Rearrangement and solution by the quadratic formula gives (H2S). Back substitution gives the remaining concentrations.

H2S __7.8â10­7 M __ MoO MoO2S22­ ___1.0â10­9 M

4

2­

___1.8â10­7 M
2­ 3

___
­11

MoO3S2­___2.1â10­8 M___
2­ 4

MoOS

___8.1â10

M___ MoS

___4.9â10­12 M___

2 points each answer; 12 points total

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PROBLEM 4
ab 12 14 c 10 d-i 4 d-ii 2 d-iii 2 d-iv 4 e-i 4

7.8% of the Total
e-ii 8 Problem 4 60 7.8%

In the 1980's a class of ceramic materials was discovered that exhibits superconductivity at the unusually high temperature of 90 K. One such material contains yttrium, barium, copper and oxygen and is called "YBCO". It has a nominal composition of YBa2Cu3O7, but its actual composition is variable according to the formula YBa2Cu3O7- (0 < < 0.5). a. One unit cell of the idealized crystal structure of YBCO is shown below. Identify which circles correspond to which elements in the structure.

= Cu

=O

= Ba

=Y

c b a

3 points each; 12 points total

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The true structure is actually orthorhombic (a b c), but it is approximately tetragonal, with a b (c/3). b. A sample of YBCO with = 0.25 was subjected to X-ray diffraction using Cu K radiation ( = 154.2 pm). The lowest-angle diffraction peak was observed at 2 = 7.450º. Assuming that a = b = (c/3), calculate the values of a and c. sin = n/2d d = (1)(154.2 pm)/2sin(3.725º) d = 1187 pm Lowest-angle => d = longest axis = c c = 1187 pm a = c/3 = 396 pm 8 points for calculating d; 6/8 points if student uses in radians and reports a positive value (0/8 points if negative distance); 6/8 points if uses 2 instead of . 6 points for correctly assigning a and c. 14 points total a = 396 pm c = 1187 pm c. Estimate the density of this sample of YBCO (with = 0.25) in g cm-3. If you do not have the values for a and c from part (b), then use a = 500. pm, c = 1500. pm.

Vunit cell = a b c = 3a3 = 3(396 pm)3 = 1.863 10-22 cm3 munit cell = (1/NA)(88.91 + 2â137.33 + 3â63.55 + 6.75â16.00) munit cell = (662.22 g/mol)/(6.0221 1023 mol-1) = 1.100 10-21 g Density = (1.100 10-21 g)/(1.863 10-22 cm3) = 5.90 g cm-3 4 points for V 4 points for mu 2 points for 10 points total d. When YBCO is dissolved in 1.0 M aqueous HCl, bubbles of gas are observed (identified as O2 by gas chromatography). After boiling for 10 min to expel the dissolved 22

nit cell

Density = 5.90 g cm

-3

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gases, the solution reacts with excess KI solution, turning yellow-brown. This solution can be titrated with thiosulfate solution to a starch endpoint. If YBCO is added directly to a solution that 1.0 M in both KI and HCl under Ar, the solution turns yellow-brown but no gas evolution is observed. i. Write a balanced net ionic equation for the reaction when solid YBa2Cu3O7- dissolves in aqueous HCl with evolution of O2. YBa2Cu3O7- (s) + 13 H+ (aq) Y (aq) + 2 Ba (aq) + 3 Cu2+ (aq) + (0.25[1 ­ 2])O2 (g) + 6.5 H2O (l)
3+ 2+

2 points species, 2 points coefficients ii. Write a balanced net ionic equation for the reaction when the solution from (i) reacts with excess KI in acidic solution after the dissolved oxygen is expelled. 2 Cu2+ (aq) + 5 I­ (aq) 2 CuI (s) + I3­ (aq) ­or­ 2+ ­ 2 Cu (aq) + 4 I (aq) 2 CuI (s) + I2 (aq) 1 point species, 1 point coefficients. Iodo complexes of Cu(I) (e.g., CuI2­ ) will be given full marks as products iii. Write a balanced net ionic equation for the reaction when the solution from (ii) is titrated with thiosulfate (S2O32-). I3­ (aq) + 2 S2O32­ (aq) 3 I­ (aq) + S4O62­ (aq) ­or­ I2 (aq) + 2 S2O32­ (aq) 2 I­ (aq) + S4O62­ (aq) 1 point species, 1 point coefficients

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iv.

Write a balanced net ionic equation for the reaction when solid YBa2Cu3O7- dissolves in aqueous HCl containing excess KI in an Ar atmosphere. YBa2Cu3O7- (s) + (14 ­ 2) H+ (aq) + (9 ­ 3) I­ (aq) Y3+ (aq) + 2 Ba2+ (aq) + 3 CuI (s) + (7 ­ ) H2O (l) + (2 ­ ) I3­ (aq) ­or­ YBa2Cu3O7- (s) + (14 ­ 2) H+ (aq) + (7 ­ 2) I­ (aq) Y3+ (aq) + 2 Ba2+ (aq) + 3 CuI (s) + (7 ­ ) H2O (l) + (2 ­ ) I2 (aq)

2 points species, 2 points coefficients e. Two identical samples of YBCO with an unknown value of were prepared. The first sample was dissolved in 5 mL of 1.0 M aqueous HCl, evolving O2. After boiling to expel gases, cooling, and addition of 10 mL of 0.7 M KI solution under Ar, titration with thiosulfate to the starch endpoint required 1.542 10-4 mol thiosulfate. The second sample of YBCO was added directly to 7 mL of a solution that was 1.0 M in KI and 0.7 M in HCl under Ar; titration of this solution required 1.696 10-4 mol thiosulfate to reach the endpoint. i. Calculate the number of moles of Cu in each of these samples of YBCO. n
Cu

= nthiosulfate in the first titration nCu = 1.542 10­4 mol

4 points, errors in chemistry displayed in (d) will be carried forward without penalty nCu = 1.542 10­4 mol

ii. Calculate the value of for these samples of YBCO. Total Cu Cu(III) = (1.696 10­4 mol) So 90% of Cu For charge balance, 2(7 ­ ) = 4 points for partition of Cu(III)/Cu(II) 4 points for calculating Alternatively, using the balanced equations in (d): In the 1st titration, each mol YBCO = 1.5 mol I3­ = 3 mol S2O32­ In the 2d titration, each mol YBCO = (2­) mol I3­ = (4­2) mol S2O3 The 44 IChO ­ Theoretical Examination. The official English version
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= 1.542 10­4 mol ­ (1.542 10­4 mol) = 1.54 10­5 mol is Cu(II), 10% is Cu(III) 3 + 2â2 + 3â(0.90â2 + 0.10â3) = 13.30 = 0.35

2­

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Code: So (1.542 10­4 mol)/(1.696 10­4 mol) = 3/(4­2) = 1.5/(2­) 2­ = 1.650 = 0.35

4 points for translating (d) to a relation between titrations and 4 points for calculating = 0.35

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PROBLEM 5
a-i 2 a-ii 4 b 4 c 2 d 12 e 6 f 4

7.0 % of the Total
Problem 5 34 7.0%

Deoxyribonucleic Acid (DNA) is one of the fundamental molecules of life. This question will consider ways that DNA's molecular structure may be modified, both naturally and in ways devised by humankind. a. Consider the pyrimidine bases, cytosine (C) and thymine (T). The N-3 atom (indicated by *) of one of these bases is a common nucleophilic site in single strand DNA alkylation, while the other is not. i. Select (circle) which base, C or T, has the more nucleophilic N-3 atom.

(i) C 2 points ii. Draw two complementary resonance structures of the molecule you select to justify your answer. (ii) T

4 points (2 points each; 1 of these points for formal charges) Full marks for this part if student chooses T in part (i) but draws two valid resonance structures 3 points for valid resonance structures of the base not selected in part (i) 26

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b. One common modification of DNA in nature is methylation of the indicated (*) position of guanine (G) by S-adenosyl methionine (SAM). Draw the structures of both of the products of the reaction between guanine and SAM.

NH2 O O HO OH O S N N

N NH N
2

2 points; full marks if protonated; full marks for dimethylated

2 points; full marks if protonated

c. One of the earliest man-made DNA alkylating agents was mustard gas.

Mustard gas acts by first undergoing an intramolecular reaction to form intermediate A which directly alkylates DNA, to give a nucleic acid product such as that shown in the equation above. Draw a structure for reactive intermediate A.

Cl

S Cl

A
2 points

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d. The nitrogen mustards react via an analogous pathway to the sulfur mustard of part c. The reactivity of the compound may be modified depending on the third substituent on the nitrogen atom. The reactivity of nitrogen mustards increases with increasing nucleophilicity of the central nitrogen atom. Select the most and least reactive from each of following groups of nitrogen mustards. i.
NO2 NO
2

Cl

N

Cl

Cl

N

Cl

Cl

N

Cl

I

II

III

MOST REACTIVE: II LEAST REACTIVE: I 4 points (2 points each) ii.
OCH3 NO2

Cl

N

Cl

Cl

N

Cl

Cl

N

Cl

I MOST REACTIVE: I LEAST REACTIVE: III 4 points (2 points each)

II

III

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Name: iii.
O Cl N CH3 Cl Cl Me N Cl

Code:

O Cl N

OCH3 Cl

I MOST REACTIVE: II LEAST REACTIVE: I 4 points (2 points each) e. Some potential duocarm product.

II

III

classes of natural products act as DNA alkylators, and in this way, they have the to serve as cancer therapies due to their antitumor activity. One such class is the ycins. Shown below are steps from an asymmetric total synthesis of the natural Draw the structures of isolable compounds J and K.

J

K

2 points; 1 point for other regioisomers

4 points (3 points for enantiomer, 3 points for epoxide opening and nosyl still present)

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f. Related small molecules were synthesized to study the way in which the duocarmycins work. One such example is the thioester shown below. Draw the structure of reactive intermediate Z.

Correct Answer

Full marks

4 points 1 point for elimination product

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PROBLEM 6
a 2 b 4

6.6 % of the Total
c 6 d 8 Problem 6 6.6% 20

Varenicline has been developed as an oral treatment for smoking addiction and can be synthesized by the route shown below. All compounds indicated by a letter ( A ­ H) are uncharged, isolable species.

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Name: a. Suggest a structure for compound A. A

Code:

Br

CN

2 points b. Suggest a structure for compound B consistent with the following 1H-NMR data: 7.75 (singlet, 1H), 7.74 (doublet, 1H, J = 7.9 Hz), 7.50 (doublet, 1H, J = 7.1 Hz), 7.22 (multiplet, 2 nonequivalent H), 4.97 (triplet, 2H, J = 7.8 Hz), 4.85 (triplet, 2H, J = 7.8 Hz) B
O O

CN

4 points (2 points for a feasible alternative if not NMR-consistent)
1

H NMR Chemical Shift Ranges*
Aromatics RHC=CHR PhO-CH R2C=CH
2

Alkyl-H RCCH ArCH R2C=CR-CH I-CH

RCH=O

F-CH Cl-CH

Br-CH RC(=O)-CH RCO2-CH O2N-CH RCONH RCOOH PhOH R2NH NC-CH R2N-CH ROH

12.0

11.0

10.0

9.0

8.0

7.0

6.0

5.0

4.0

3.0

2.0

1.0

0.0

d (ppm)

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Name: c. Suggest a structure for compounds C, D, and F. C
O NH

Code:

D

NH

2 points F
O2N N O2N CF3 O

2 points

2 points d. Suggest reagents X and Y to convert compound G into varenicline, and provide the isolable intermediate H along this route. X
O H O H

Y Aqueous NaOH or any other amide hydrolyzing reagents 2 points

2 points H
N N N CF3 O

2 points (Full credit given for whatever is the correct product of F and X) X and Y reversed receive full marks above, as long as G corresponds. 2 additional points for proper order of reagents.

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PROBLEM 7
a 9 b 15 c 8 d 6 e 8

7.5 % of the Total
f 6 Problem 6 52 7.5%

An artificial enzyme was designed to bind the two substrate molecules shown below (diene and dienophile) and catalyze a Diels-Alder reaction between them. a. There are eight potential products from a Diels-Alder reaction involving these two molecules in the reaction without any enzyme. i. Draw the structures of any two of the potential products that are regioisomers of each other, in the boxes that are given below. Use wedges ( ) and dashes ( ) to show the stereochemistry of each product in your drawings. Use R and R shown below to represent the substituents in the molecules that are not directly involved in the reaction.

H

N O

O O N

CO2diene

Me

Me

dienophile
O O N

R

H

N

R'
O

CO2

-

Me

Me

R'

R' R

R

1 point for any reasonable Diels-Alder product 2 points for regioisomeric relationship between compounds

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ii. Draw the structures of any two of the potential products that are enantiomers of each other, in the boxes that are given below. Use wedges ( ) and dashes ( ) to show the stereochemistry of each product in your drawings. Use R and R as in part (i).

R'

R'

R

R

1 point for any reasonable Diels-Alder product 2 point for enantiomeric relationship between compounds

iii. Draw the structures of any two of the potential products that are diastereomers of each other, in the boxes that are given below. Use wedges ( ) and dashes ( ) to show the stereochemistry of each product in your drawings. Use R and R as in part (i).

R' R

R' R

1 point for any reasonable Diels-Alder product 2 points for diastereomeric relationship between compounds

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b. The rate and regioselectivity of a Diels-Alder reaction depend on the degree of electronic complementarity between the two reactants. The structures of the diene and the dienophile from part a are given below. i. Circle the carbon atom in the diene that has increased electron density and therefore can act as an electron donor during the reaction. Draw one resonance structure of the diene in the box to support your answer. Indicate all non-zero formal charges on the atoms in the resonance structure that you draw.

H

N O

O

H

N O

O

CO2-

CO2-

5 points (2 points for circled carbon; 2 points for resonance structure; 1 point for charges)

ii. Circle the carbon atom in the dienophile that has decreased electron density and therefore can act as an electron acceptor during the reaction. Draw one resonance structure of the dienophile in the box to support your answer. Indicate all non-zero formal charges on the atoms in the resonance structure that you draw.

O N N

O

Me

Me

Me

Me

5 points (2 points for circled carbon; 2 points for resonance structure; 1 point for charges)

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iii. Based on your assignments in parts (i) and (ii), predict the regiochemistry of the uncatalyzed Diels-Alder reaction of the diene and dienophile by drawing the major product. You need not show the stereochemistry of the product in your drawing.

R' R

5 points. Stereochemistry not graded. Full marks as long as consistent with b(i) and b(ii)

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c. The figure below shows the Diels-Alder reactants as they are bound before they enter the transition state for product formation in the active site of the artificial enzyme. The gray area represents a cross-section through the enzyme. The dienophile is below the cross-section plane whereas the diene is above the cross-section plane, when the two molecules are bound in the active site that is shown. Draw the structure of the product of the enzyme-catalyzed reaction in the box given below. Show the stereochemistry of the product in your drawing and use R and R as you did for question a.

R' R

8 points; 4 points if wrong enantiomer; 2 points if wrong diastereomer; 0 points if wrong regioisomer

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d. Consider the following statements about enzymes (artificial or natural). For each statement, indicate whether that statement is True or False (draw a circle around "True" or "False"). i. Enzymes bind more tightly to the transition state than to the reactants or products of the reaction. True False ii. Enzymes alter the equilibrium constant of the reaction to favor the product. True False iii. Enzymatic catalysis always increases the entropy of activation of the reaction compared to the uncatalyzed reaction. True False 6 points; 2 points each

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e. Modified versions of the artificial enzymes with different catalytic activities were prepared (enzymes I, II, III, and IV, shown in the figure below). Two amino acid residues that differ among the different enzymes are shown. Assume that the enzyme functional groups shown are located in close proximity to the matching fragments of the reagents when they form the transition state in the enzyme active site. Of these four enzymes which one would cause the greatest increase in the rate of the Diels-Alder reaction compared to the uncatalyzed reaction? Enzyme # II 8 points

Enzyme I

Enzyme II

P he L eu
H N O O N O
NH
2

Tyr

G ln
O

H

N O O N O

H

O

CO O

CO O

Enzyme II I

En zyme I V

Tyr

P he

L eu
H N O O N O
O

G ln
H O NH
2

H

N O O N O

CO O

CO O

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f. The substrate specificity of the artificial enzymes V and VI (see below) was tested by using the dienophile reactants 1 - 6, shown below.
O N N CH3 CH3 O N CH3 CH3 O N CH3 CH3 O N O N O

H3C

CH3

CH3 4

OH

O 6

1

2

3

5

Dienophile #1 reacted most rapidly in the reaction catalyzed by artificial enzyme V (see below). However, artificial enzyme VI catalyzed the reaction most rapidly with a different dienophile. Of the six dienophiles shown above, which one would react most rapidly in the Diels-Alder reaction catalyzed by enzyme VI? Dienophile # 5 6 points

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PROBLEM 8
a 2 b -i 3 b-ii 4 b-iii 6 b-iv 4 b -v 2 c-i 5 c-ii 8

8.3% of the Total
c-iii 2 Problem 8 36 8.3%

Polycyclic aromatic hydrocarbons (PAHs) are atmospheric pollutants, components of organic light emitting diodes and components of the interstellar medium. This problem deals with so-called linear PAHs, i.e., those being just one benzene ring wide whereas the length is varied. Specific examples are benzene, anthracene and pentacene whose structures are shown below. Their physical and chemical properties depend on the extent to which the electron cloud is delocalized over the molecule.
benzene y x d da dp anthracene pentacene

a. The distance across the benzene ring is d = 240 pm. Use this information to estimate the distances along the horizontal (x) axis for anthracene and pentacene, da and dp, respectively. For anthracene, da = 3(240 pm) = 720 pm For pentacene, dp = 5(240 pm) = 1200 pm 1 point each 2 points total

b. Assume for simplicity that the electrons of benzene can be modeled as being confined to a square. Within this model, the conjugated electrons of PAHs may be considered as free particles in a two dimensional rectangular box in the x-y plane. For electrons in a two-dimensional box along the x- and y-axes, the quantized energy states of the electrons are given by

nx 2 n y 2 h 2 E 2 2 L Ly 8me x
42

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Code: In this equation, nx and ny are the quantum numbers for the energy state and are integers between 1 and , h is Planck's constant, me is the mass of the electron and Lx and Ly are the dimensions of the box. For this problem, treat the electrons of the PAHs as particles in a two dimensional box. In this case, the quantum numbers nx and ny are independent.

i. For this problem, assume that the benzene unit has x and y dimensions that are each of length d. Derive a general formula for the quantized energies of linear PAHs as a function of quantum numbers nx and ny, the length d, the number of fused rings w, and the fundamental constants h and me.

Ô n2 n2 E = à y + 2x à d2 w d Õ
3 points

2

Æ h2 Ô 2 n2 Æ h2 x   8m = à ny + w 2  8m d Üe ÜeÕ

2

ii. The energy level diagram below for pentacene shows qualitatively the energies and quantum numbers nx, ny, for all levels occupied by -electrons and the lowest unoccupied energy level, with the electrons of opposite spins represented as the arrows pointing up or down. The levels are labeled with quantum numbers (nx; ny). Pentacene: __ (3; 2) (9; 1) (2; 2) (1; 2) (8; 1) (7; 1) (6; 1) (5; 1) (4; 1) (3; 1) (2; 1) (1; 1)

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The energy level diagram for anthracene is shown below. Note that some energy levels may have the same energy. Draw the correct number of up and down arrows to represent the electrons in this diagram. Also, the blanks in parentheses within this diagram are the quantum numbers nx, ny, which you need to determine. Fill these blanks with the pertinent values of nx, ny for each filled and the lowest unfilled energy level(s). Anthracene: __ (__; __) __(_6_; _1_) __ (_3_; _2_) (_2_; _2_) (_1_; _2_) (_5_; _1_) (_4_; _1_) (_3_; _1_) (_2_; _1_) (_1_; _1_) 2 points for the correct placement of electrons and correct number of electrons 2 points for the correct assignment of quantum numbers 4 points total Note: Not penalty for labeling additional unoccupied energy levels iii. Use this model to create an energy level diagram for benzene and fill the pertinent energy levels with electrons. Include energy levels up to and including the lowest unoccupied energy level. Label each energy level in your diagrams with the corresponding nx, ny. Do not assume that the particle-in-a-square-box model used here gives the same energy levels as other models. Benzene: __ (_2_; _2_) (_2_; _1_) (_1_; _2_) (_1_; _1_) 2 2 2 6 point point point point s s s s for the correct energy diagram for the correct placement of electrons and correct number of electrons for the correct assignment of quantum numbers total

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iv. Often the reactivity of PAHs correlates inversely with the energy gap E between the highest energy level occupied by -electrons and the lowest unoccupied energy level. Calculate the energy gapE (in Joules) between the highest occupied and lowest unoccupied energy levels for benzene, anthracene and pentacene. Use your result from parts ii) and iii) for anthracene or benzene, respectively, or use (2, 2) for the highest occupied energy level and (3, 2) for the lowest unoccupied energy level for these two molecules (these may not be the true values).

E for benzene: E E2;2 E1;2 3
Alternate solution:

h2 3.14 101 8 J 2 8med

h2 E E3;2 E2;2 5 5.23 10 8m e d 2

18

J

E for anthracene: E E 6;1 E 2;2
Alternate solution:

5 h2 5.81 1019 J 2 9 8 me d
19

5 h2 E E3 E2;2 ;2 9 8m e d

2

5.81 10

J

3 h2 E for pentacene: E E3;2 E9;1 1.26 101 9 J 2 258med

1 point for each, 3 points total Rank benzene (B), anthracene (A), and pentacene (P) in order of increasing reactivity by placing the corresponding letters from left to right in the box below. B A P Least reactive -----------------------------------> Most reactive 1 point for correct ranking

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v. The electronic absorption spectra (molar absorptivity vs. wavelength) for benzene (B), anthracene (A), and pentacene (P) are shown below. Based on a qualitative understanding of the particle in the box model, indicate which molecule corresponds to which spectrum by writing the appropriate letter in the box to its right.

A

P

B

2 points: 0 correct = 0 points, 1 correct = 1 point , all correct = 2 points

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c. Graphene is a sheet of carbon atoms arranged in a two-dimensional honeycomb pattern. It can be considered as an extreme case of a polyaromatic hydrocarbon with essentially infinite length in the two dimensions. The Nobel Prize for Physics was awarded in 2010 to Andrei Geim and Konstantin Novoselov for groundbreaking experiments on graphene. Consider a sheet of graphene with planar dimensions of Lx=25 nm by Ly=25 nm. A section of this sheet is shown below.

i. The area of one hexagonal 6-carbon unit is ~52400 pm2. Calculate the number of electrons in a (25 nm 25 nm) sheet of graphene. For this problem you can ignore edge electrons (i.e., those outside the full hexagons in the picture). The number of hexagonal units in the graphene sheet:
Nu A r egar
aphene

ni t s

A r euan



25000 2 pm
52400 2 pm

12000n i t s u

it

2points Since each carbon atom in a graphene sheet is shared by three hexagonal units, each unit of the area 52400 pm2 contains 6/3=2 carbon atoms contributing 2 -electrons total. 2points Therefore, 12000 units contribute 12000 pairs of -electrons. Answer: 24,000 electrons. 1 point; total is 5 points ii. We can think about the electrons in graphene as being free electrons in a 2dimensional box. In systems containing large numbers of electrons, there is no single highest occupied energy level. Instead, there are many states of nearly the same energy above which the
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remaining are empty. These highest occupied states determine the so-called Fermi level. The Fermi level in graphene consists of multiple combinations of nx and ny quantum numbers. Determine the energy of the Fermi level for the 25 nm â 25 nm square of graphene relative to the lowest filled level. The lowest filled level has a non-zero energy; however, it is negligible, and can be assumed to be zero. To solve this problem it might be helpful to represent the (nx,ny) quantum states as points on a 2-D grid (as shown below) and consider how the energy levels are filled with pairs of electrons. For the number of electrons use your result from part i or use a value of 1000 (this may not be the true value).

Two electrons fill each state, so the Fermi level has 12000 filled levels. This corresponds to the number of (nx,ny) pairs that are occupied. Since Lx=Ly and the lowest energy level's energy is approximated as zero,

E E
1 point

hi ghest _ occupi ed

nx ny



2

2



h2 8me L2
n



This is rearranged to the equation of a circle.

y

R 2 nx 2 n



2 y





E8me L2 cons tan t h2
n
x

2 points

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2 The area of the populated grid is Areag ri d R . 4 The area of each quantum number pair is 1.

Code:

1 point Therefore, the number of points is given as
Np
ot n is



Ar e a g Ar e a p

ri d ai r



R 2 Ns 4

t at es

12000 .

1 point Rearranging and solving for energy yields the Fermi energy. R2 8meL2E Ns t at es 12000 4 4h2 1.48 1018 J 4h2 12000 E 2 8meL 3 points Total: 8 points Alternate solution: R 2 8m eL2E N s ta te s 1000 4 4h 2 4h 2 1000 E 1.23 10 19 J 2 8m eL

iii. The conductivity of graphene-like materials correlates inversely with the energy gap between the lowest unoccupied and highest occupied energy levels. Use your analysis and understanding of electrons in PAHs and graphene to predict whether the conductivity of a 25 nm â 25 nm square of graphene, at a given temperature, is less than, equal to or greater than the conductivity of a 1 m â 1 m square of graphene (which is the largest obtained to date). Circle the correct answer: less equal greater

The energy gaps decrease with the graphene sample size increase and the conductivity increases as the energy gap decreases. 2 points

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