German problems 2015

National German
Competition
Volume 21
47. International
Chemistry Olympiad
Azerbaijan 2015

Chemistry Olympiad 2015
4
National German Competition 2015, Volume 21
Translated and published by Wolfgang Hampe
Contact addresses:
IPN University of Kiel, z.H. PD Dr. Sabine Nick tel: +49-431-880-3116
Olshausenstraße 62 fax: +49-431-880-5468
24098 Kiel email: nick@ipn.uni-kiel.de
IPN University of Kiel, z.H. Monika Barfknecht tel: +49-431-880-3168
Olshausenstraße 62 fax: +49-431-880-5468
24098 Kiel email: barfknecht@ipn.uni-kiel.de
Wolfgang Hampe tel: +49-431-79433
Habichtweg 11
24222 Schwentinental email: Hampe@t-online.de
Association to promote the IChO
(Association of former participants and friends of the IChO)
Internet address: www.fcho.de
This booklet including the problems of the 47th
IchO and the latest statistics is available as
of September 2015 from http://www.icho.de ("Aufgaben")

Chemistry Olympiad 2015
3
Contents
Part 1: The problems of the four rounds
First round (problems solved at home) .................................... 6
Second round (problems solved at home) .................................... 10
Third round, test 1 (time 5 hours).......................................................... 18
Third round, test 2 (time 5 hours).......................................................... 24
Fourth round, theoretical test (time 5 hours).......................................................... 31
Fourth round, practical test (time 5 hours) ......................................................... 42
Part 2: The solutions to the problems of the four rounds
First round ................................................................................. 47
Second round ................................................................................. 53
Third round, test 1 ................................................................................. 60
Third round, test 2 ................................................................................. 66
Fourth round, theoretical test ................................................................................. 72
Part 3: The problems of the IChO
Theoretical problems ................................................................................. 81
Practical problems ................................................................................. 95
Solutions ................................................................................. 110
Part 4: Appendix
Tables on the history of the IChO ............................................................................ 120

Problems
5
Part 1
The problem set of the four rounds

Problems Round 1
6
First Round (homework)
Problem 1- 1 Quite Old!
Natural carbon consists of three isotopes.
a) Which are these isotopes? How do they differ in their atomic composition?
One of these isotopes is radioactive with a half-life of t½ = 5730 years.
b) How is this isotope generated in nature? Give a formation equation!
The radioactive carbon isotope is an  emitter.
c) Write down the decomposition equation.
d) What is the law for the radioactive decay?
e) What is the meaning of "half-life" of an isotope? Derive a general formula for the half-life, start-
ing with the law for the radioactive decay.
f) After which time does radioactive material stop to disintegrate?
g) What is the influence of temperature, pressure and similar conditions on the rate of radioactive
decay?
Radiocarbon dating is an important method to determine the age of carbon containing material.
h) Explain this method! Give the reason why examined materials have to be former living objects or
their derivatives.
The paper of a treasury map found in 2013 has a decay rate of 14.48 decays per g of carbon per mi-
nute. In natural carbon this decay rate is 15 decays per g of carbon per minute.
i) Calculate the age of the map? Should the finder go on a treasury hunt?
j) Account for the fact that the age of bones of dinosaurs cannot be determined by the method of
carbon dating.
Problem 1-2 Fats and Oils
Fats and oils are esters of carboxylic acids with glycerol. They can be solid, semisolid and liquid. Fats
which are liquid at room temperature are called oils.
Depending on their origin fats have a characteristic composition of fatty acids.
a) Explain what "characteristic composition of fatty acids" means and show such a characteristic
for an example chosen by you.
There is a simple code to characterize fatty acids by giving the proportion of the total number of
carbon atoms (m) and the number of double bonds (n) (as it is done in sport results). The code for
oleic acid is 18:1.
Fig. 1: Structure of oleic acid
C C C C C C C C C C C C C C C C C C
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H H H
H
H
H H H H H H
OHHHHH
OH

Problems Round 1
7
b) Write down the empirical formulae and plot the structures of the fatty acids below. Give their
code.
i) Capric acid ii) Linoleic acid iii) Linolenic acid iv) Erucic acid
Natural fats which contain (poly) unsaturated fatty acids are especially valuable for dietary intake.
c) Explain by considering intermolecular interaction why saturated fats are solid while unsaturated
fats are rather liquid.
More than 90 % of fats produced worldwide serve as foodstuff for men or animal feed, a small part
is raw material for producing other chemicals. There are special regions in a fat molecule where a
chemical attack is possible.
Fig. 2: Structural formula of ricinus oil
d) Mark the regions where a chemical attack is possible and give a name to it.
A possible reaction of a fat is the saponification with a solution of sodium hydroxide to give glycerol
and soap.
e) Give the reaction equation of the saponification of a fat (using structural formulae). Show the
detailed mechanism of the saponification of one of the three fatty acid residues. Explain why this
equilibrium reaction drifts to the products.
The saponification of copra oil needs several hours while the saponification of 3-nitrobenzoic-acid
methyl ester is finished after only ten minutes.
f) Account for this fact.
Soaps belong to the tensides.
g) What characterizes a tenside? Explain the cleaning effect of tensides.
h) To which kind of tensides do soaps belong? Which other kinds of tensides do exist? Give an ex-
ample to each kind and draw the structural formula.
Problem 1-3 Oxygen ...
Oxygen is the most frequent element on earth having a mass ratio of 49.4 %.
a) Give five natural deposits of oxygen.
The production of oxygen from air is technically performed by a special procedure in Germany
known as Linde operation.
b) How can oxygen be produced in a laboratory? Write down the names of three methods and the
equations of the reactions.
O
O
CH2
CHO
O
CH2O
O
OH
OH
OH

Problems Round 1
8
Besides oxygen nitrogen (consisting of N2 molecules) is the major component of air
c) Are there differences between O2 und N2 molecules concerning their magnetic properties? Ac-
count for your answer using MO diagrams.
Oxygen molecules (O2) are denoted as triplet oxygen. Moreover there is a more reactive species, the
so called singlet oxygen. These names arise from the total spin multiplicity M which can be calculat-
ed with the formula M = 2·S + 1 (S = total spin).
d) Account for the names of these two "kinds" of oxygen using this formula!
In text books you often find the Lewis structure for dioxygen (O2)
shown on the right hand side. Lewis structure of O2
e) Does this Lewis structure show the correct distribution of electrons in the O2 molecule? Account
for your answer.
Problem 1-4 ... and Oxides
Oxygen forms binary compounds with nearly all other elements.
a) From which elements no isolatable oxygen compound is known until now?
b) Write down the equations of the reactions of oxygen with
i) White phosphor ii) Yellow sulfur iii) Lithium iv) Calcium
Which acid/base properties do the aqueous solutions of the products show?
In its covalent compounds oxygen has in most cases the coordination number 1, 2 or 3.
c) Give one example for each of these three coordination numbers plotting its Lewis structure.
Besides some exceptions the oxidation number –II is assigned to oxygen.
d) Give four compounds in which oxygen has another oxidation number than –II. Do not name
more than one compound for each different oxidation number.
In quantitative analysis oxides can be used to determine the amount of other elements for example
iron, cobalt, nickel, tin and aluminum.
To determine the iron(III) content in a solution it is precipitated with ammonia, filtered through ash-
free filters, washed with water and at the end with ammonium nitrate solution.
The filter with the precipitate is given into a porcelain crucible
and heated with a Bunsen burner, at first slowly and then up to
maximal 700 °C. You have to pay attention that there are no
reduction processes and that besides Fe2O3 no Fe3O4 is formed
which would falsify the result of the analysis.
A sample of iron(III) chloride is weighed into a 250.0 mL measuring flask which then is filled with
water up to the calibration mark. Three samples of 50.0 mL each are taken and treated as described
above.
Weight of the ash: 0.2483 g / 0.2493 g / 0.2488 g.
e) i) Write down the equation for the formation of iron(III) oxide from iron(III) hydroxide.
O O

Problems Round 1
9
ii) What is the reason for washing with ammonium nitrate? Why should ammonium chloride
not be used in the last step? Account for your answer.
iii) Why is ash-free filter paper used?
iv) With which simple physical measurement could the unwanted formation of Fe3O4 be detect-
ed?
v) Does the formation of Fe3O4 lead to a higher or to a lower calculated content of iron? Ac-
count for your answer.
vi) Calculate the mass of the iron(III) chloride sample which was given into the measuring flask.
Problem 1-5 ... another Oxide - An Elaborate Determination
290.0 mg of a metal oxide (MeO2), concentrated hydrochloric acid as well as a solution of potassium
iodide are placed in an apparatus as shown below.
Fig. 3: Apparatus for a quantitative determination of a metal oxide
The hydrochloric acid in the dropping funnel is pressed into the reaction vessel A using carbon diox-
ide. This mixture is heated to maintain light boiling for 30 minutes.
During this time the apparatus is continuously purged with small amounts of carbon dioxide to re-
move the volatile content of vessel A.
When the reaction in vessel A has finished the content of the washing flask C is quantitatively trans-
ferred into the Erlenmeyer flask B.
The content of the Erlenmeyer flask B is titrated with a standard solution of sodium thiosulfate (c =
0.1 mol /L) until the mixture is nearly discoloured. Then some drops of solution of starch are added
and the titration is continued until the solution is totally discoloured.
24.25 mL of the thiosulfate solution are consumed for the titration.
a) Write down reaction equations for all reactions in the apparatus as well as for the reactions
during the following titration.
b) Why are drops of starch solution added?
c) Account for the reason of such an elaborate procedure. What is the influence of air and water
(moisture) on the accuracy of the determination?
d) Determine the metal in the oxide.

Problems Round 2
10
Second Round (homework)
Problem 2-1 Afraid of Water!
The nitrate of M, dissolved in a small amount of water at 50 °C, is added to a 7.5-fold excess of a
warm (also at 50 °C) aqueous solution of X. A solid Y forms which converts to compound Z · n H2O
when cooled down. The mass ratio of the metal M in Z amounts to 33.04 %.
The following information is given:
 The metal M dissolves is nitric acid of the concentration of c = 8 mol/L but not in sulfuric acid
or hydrochloric acid of the same concentration.
 The metal M dissolves in a hot solution of sodium hydroxide.
 Finely dispersed M ignites in air.
 If a solution of sodium hydroxide is added to an aqueous solution of M a precipitate forms
which dissolves in an excess of sodium hydroxide solution.
 M in an aqueous solution does not form a complex compound with ammonia.
 When dissolving X in water the solution cools down.
 Y dissolves slightly in water.
 As a solid Y has a layer structure.
 The conductivity of solid Y rises when heated.
 Y shows thermochromism.
 Z · n H2O can be dehydrated in a drying oven at 150 °C.
 Z and Z · n H2O decompose in water.
 When Z · n H2O is heated a gaseous compound G is formed besides water vapor and a metal-
lic mirror of M is formed
 X and G react in an aqueous solution with each other to yield compound H.
a) Determine M, X, Y and Z. Account for your decision.
b) Write down the reaction equation for the two step formation of Z using the compounds men-
tioned above.
c) Determine H. Write down the equation of its formation from X and G! Draw the Lewis structure
of H. Which structure do you expect according to the VSEPR model. Account for your answer.
The water containing compound Z · n H2O can be dehydrated in a drying oven. The image shows the
thermography (TG) plot.
Masspercentage
Temperature/°C

Problems Round 2
11
d) Determine the number of water molecules n (n integer) in one formula unit of Z · n H2O using
the observed decrease of mass.
Beside Z another compound was found, which consists of the same elements with the same oxida-
tion numbers as in Z. The mass ratio of the metal M, however, has only the value of 26.13 %.
e) Give the empirical formula of this compound.
f) Give at least three compounds which lead to a decrease of temperature when dissolved in wa-
ter. Give an explanation of this phenomenon!
g) Explain the formation of the metallic mirror when Z · n H2O is heated. Write down the reaction
equation and apply oxidation numbers.
h) Give the equation for the reaction of an aqueous solution of the nitrate of M with an excess of
an ammonia solution.
Problem 2-2 Organic Chemistry – Short and Crisp
5-Methylfuran-3-on (2) is needed as a reagent for the synthesis of a natural compound. It can be
produced in four steps starting with acetylacetone (1):
O O
A
i) LDA, THF,
-78 °C,
30 min
ii) TMSCl,
0 °C, 1 h
B
C11H24O2Si2
NBS, DCM,
Rückfluss,
2 h C
Et2O,
K2CO3,
RT, 12 h
Et3N, Hexan,
TMSCl, RT,
48 h O
O
1 2
Hints:
- In the reaction of 1 to A two isomers form.
- In the 1
H-NMR spectrum (A in CDCl3 at 298 K) of the mixture of isomers of A six singlets are
found. The intensities are written below the signals, the chemical shiftabove the signals.
Et3, hexane, NBS, DMC
reflux

Problems Round 2
12
- In the mass spectrum of C the following values for the molecular peak m/c are found (intensities
in brackets):
177.96 (100.0%), 179.96 (97.3%), 178.97 (5.6%), 180.96 (5.3%)
a) Draw the structural formulae of A, B and C and show how you derived the structures from the
hints and the reaction equation.
b) Which kind of isomers are formed in the reaction 1  A? Draw the structural formulae!
c) Analyze the 1
H-NMR spectrum of A: Assign all hydrogen atoms of b) to the corresponding sig-
nals. Determine the ratio of the two isomers of A. Calculate the intensities of the different sig-
nals using the ratio of isomers.
d) Why do you find two peaks with approximately equal intensities at 177.96 and 179.96 in the
mass spectrum of C? Explain!
e) In the first step of the synthesis of 1  A triethylamine, Et3N, is used. What is it needed for?
Explain!
f) In the second step of the synthesis of A  B lithium diisopropylamide, LDA, is used. Is it possible
to use Et3N here analogue to the reaction of 1  A? Account for your answer.
g) Would a reaction of acetylacetone with N-bromosuccinimide, NBS, in the presence of a base lead
to C, too? Why is the way via A and B used in the synthesis? Give an explanation!
In an aldol reaction followed by an elimination of H2O product 2 is linked to another structural ele-
ment. The unit 3 is formed which you find in the natural compound, too.
In the natural compound solved in a deuterized solvent an H/D exchange is observed which can be
explained by keto-enol tautomerism.
h) Show a mechanism of the H/D exchange at the methyl group of the structural unit 3. How many
of H/D exchange reactions are possible?
Problem 2-3 Hückel Theory
The Hückel theory can be used to describe π electron systems qualitatively. The p orbitals which are
involved in the π bonds (per definition pz orbitals) have to be taken into consideration. The π molec-
ular orbitals (MO) are linear combinations of them. By doing so bonding and antibonding MOs are
formed. If they are accommodated similarly a non-bonding interaction occurs. A simple example is
ethene (Fig. 2).
The energy levels of ethene result from bonding and antibonding interaction and contain the empiri-
cal parameters  (energy of the electron in the isolated atom) and β (coupling of the atom orbitals in
the molecule, β > 0).
The total π energy of ethene results from the occupation of the energy levels with two electrons as
follows
E = Σi ni εi = 2 (α + β) + 0 (α – β) = 2 α + 2 β,
with ni = number of electrons in MO i and εi = associated energy of the MO.

Problems Round 2
13
antibonding
bonding
Fig 2: π Molecular orbitals of ethene following the Hückel theory
A Hückel MO scheme can also be established for cyclic and planar π systems. There is a simple
“trick” for working out the orbital energies: Frost-Musulin diagrams. Starting from the energy level
of the atom orbitals () a circle with the radius 2β is drawn. The molecular framework of an n-cyclic
system is then drawn as a polygon into the circle with one atom put at the bottom. The atomic posi-
tions then map on to the energy level diagram of the π system. Fig. 3 shows the Frost-Musulin dia-
gram of benzene.
Fig. 3: Frost-Musulin diagram of benzene
If you compare the π bonding energy of benzene with that of hexatriene you find stabilization by
delocalization of 1.0β. This is often interpreted as the so called aromatic stabilization energy: Ben-
zene gains additional bond energy and is called an "aromatic" compound. Anitaromatic compounds
do not show such an effect, they are unstable.
a) Design the Frost-Musulin diagram of a planar 4- and 7-cyclus. Give the charge of the molecule
for different electron configurations (4-membered ring: 2, 4 and 6 π electrons; 7-membered ring:
6 and 8 π electrons) and calculate the stabilization β compared to the open chain versions of the
molecules (ions). Are these compounds (ions) aromatic?
(The π energy levels of the aliphatic versions are:
Butadiene:  ± 1,618 · β;  ± 0,618 · β;
Heptatrienyl cation: ;  ± 1,848 · β;  ± 1,414 · β;  ± 0,765 · β).
b) Which are the conditions a compound has to fulfill in order to be called aromatic?

Problems Round 2
14
c) Indicate which of the following molecules and ions are aromatic and which are not. Account for
your decision by using the conditions of b)-
i) Pyrrole ii) Allyl anion iii) Azulene iv) 1H-Pyrrolium cation v) Pyridinium cation vi) Caffeine.
The shape of the wave function of a special energy level can be found by linear combination of the pz
orbitals. The result for the cyclopentadienyl anion (Cp–
) is shown in fig. 4:
Fig. 4: Frost-Musulin diagram of cyclopentadienyl anion with the MOs
These MO diagrams are of great importance in coordination chemistry. The coordination mainly
takes place by HOMO (highest occupied molecular orbital) - LUMO (lowest unoccupied molecular
orbital) interaction.
d) Write down the HOMO and LUMO of the cyclopentadienyl anion!
An iron(III) cation is coordinated by a cyclopentadienyl anion (Fig. 5).
Fig 5: Orientation of the η5
-cyclopentadienyl iron(III) cation
e) Determine the interactions between the valence orbitals of the iron cation (3d, 4s und 4p) and
the frontier orbitals of the cyclodienyl anion. Use the orientation in the coordination system as
shown in fig. 5. Record your results in a table as shown below (x = interaction expected, – = no
interaction expected). Account for your assignment by one example for HOMO and LUMO each
with the help of orbital plots which illustrate the interaction.
In ferrocene two Cp-
ligands are coordinated to one iron(II) cation. In this complex iron has 18 va-
lence electrons and thus a stable noble gas configuration. This "magic" number is found in lots of
coordination compounds.
Orbital Fe HOMO (Cp-
) LUMO (Cp-
)
3dx2-y2
3dz2
...

Problems Round 2
15
f) Which compound is more stable: [Rh(η5
-Cp)2] or [Ru(η5
-Cp)2]? Which redox property should the
less stable compound have according to the configuration of the valence electrons at the metal
center? Account for your answer!
g) The complex [(η5
-Cp)2Ru2(CO)4] may have three diasteromeric structures. Draw 3-D structures
and give the number of valence electrons at each metal center.
Problem 2-4 Disproportionation of Copper
0.168 g of copper(II) nitrate is dissolved in water to yield 100 mL solution. The pH of this solution is
4.4.
a) Why does an aqueous solution of copper(II) nitrate react acidic? Write down the reaction equa-
tion! Calculate the pKa value of the first step of protolysis.
b) Determine the pH from which copper hydroxide precipitates in a solution with c(Cu2+
) = 1.03·10–2
mol/L. (Ksp, 25°C = 1.6 · 10–19
)
There are two redox equilibria for Cu+
ions:
Cu+
+ e–
Cu Eo
1 = 0.52 V
Cu2+
+ e–
Cu+
Eo
2 = 0.16 V
c) Write down the reaction equation for the disproportionation of Cu+
ions and calculate the equi-
librium constant for this reaction at 22 °C.
d) Which oxidation state of copper ions should be the most stable according to its electron configu-
ration? Which one is the most stable in an aqueous solution? Account for your answer.
e) 10 mmol of copper(I) nitrate are dissolved in 1 L of water at 22 °C. Calculate the composition of
the copper containing species in mol/L. (Use K =1.72 · 106
instead of the result in c). )
Copper(I) oxide is suspended in a copper(II) solution of c(Cu2+
) = 0.01 mol/L at 22 °C.
(Ksp (CuOH) = 1.0 ·10–15
)
f) Calculate the pH from which copper(I) oxide is stable in an aqueous solution. Which influence
does the temperature have in the range from 0 °C to 100 °C? Draw a graph!

Round 3 Test 1
16
Problems Round 3
The top 60 of the participants of the 2nd
round are invited to the 3rd
round, a
one-week chemistry camp.
Test 1 Göttingen 2015: Problems 3-01 to 3-09
Test 2 Göttingen 2015: Problems 3-11 to 3-20
time 5 hours.
your name write it on every answer sheet.
relevant calculations write them down into the appropriate boxes.
otherwise you will get no points
atomic masses use only the periodic table given.
constants use only the values given in the table.
answers only in the appropriate boxes of the answer
sheets, nothing else will be marked.
draft paper use the back of the pages of the problem
booklet, but everything written there will not
be marked.
problem booklet you may keep it.
Good Luck

Problems Round 3 Test 1 + 2
17
Useful formulas and data
G0
= H0
- T·S0
G0
= - E·z·F G0
= - R·T·ln K
G = G0
+ R · T· ln Q ln (Kp1/Kp2) =
−H0
R
·(T1
-1
- T2
-1
)
p·V = n·R·T for ideal gases and osmotic pressure
Nernst equation : E = E0
+
R ·T
z ·F
·ln (cOx/cRed)
for metals E = E0
+
R ·T
z ·F
·ln (c(Mez+
/c0
)
for non-metals E = E0
+
R ·T
z ·F
·ln (c0
/c(NiMez-
)
for hydrogen E = E0
+
R ·T
z ·F
·ln
c(H+)/c0
(p(H2)/p0)1/2
with c0
= 1 mol/L, p0
= 1.000∙105
Pa
Rate laws 0. order c = co - k·t
1. order c = co·e k t 1
2. order c-1
= k2·t + co
-1
Arrhenius equation: k = A ∙ e-Ea/(R∙T)
A pre-exponential factor
Ea activation energy
Law of Lambert and Beer: A = ·c·d A absorbance
 molar absorption coefficient
d length of the cuvette
c concentration
Transmission T =
I
I0
Absorbance A = lg
I0
I
with I = intensity
Freezing point depression T = K ·
n
m(Solvent)
n amount of particles dissolved
K cryoscopic constant
Speed of light c = 3.000∙108
ms-1
Gas constant R = 8.314 JK-1
mol-1
Faraday constant F = 96485 Cmol-1
Avogadro constant NA = 6.022·1023
mol-1
po
= 1.000·105
Pa 1 atm = 1.013·105
Pa 1 bar = 1·105
Pa
1 Å = 10-10
m
A periodic table was provided

Round 3 Test 1
18
Third Round Test 1
Problem 3-01 Basic Knowledge
A Fill in the missing numbers:
a)  Na2S2O3 +  I2  NaI +  Na2S4O6
b)  Ba2+
+  MnO4
-
+  CN-
+  OH-
 BaMnO4 +  CNO-
+  H2O
c)  ClO3
-
+  H3O+
+  Br-
 Br2 +  Cl-
+  H2O
B Match the possible colors with the given solid compounds and solutions of ions. Some colors
may be matched with more than one substance; some may not come into consideration.
Ions dissolved in water (c = 1 mol/L) Sample of solid of Possible colors
Fe2+
in acidic solution
Al3+
in acidic solution
Cu2+
in acidic solution
Cu2+
in ammoniac solution
Cl-
in basic solution
Na+
Iron sulfide (FeS)
Copper sulfate (CuSO4)
Silver iodide (AgI)
Potassium sulfate (K2SO4)
Potassium permanganate (KMnO4)
Potassium chromate (K2CrO4)
Black, white, colorless,
yellow, yellowish brown,
light green,
blue, deep blue,
red, violet,
C Which of the following compounds are sparely soluble?
Silver bromide, potassium nitrite, lead sulfate, calcium chloride, sodium fluoride, iron(II) sulfide.
D Write down the formulae of the following compounds:
Barium nitrate, potassium oxalate, aluminum oxide, potassium manganate(VI), potassium alu-
minum sulfate, sodium carbonate decahydrate.
E Given are sample of the elements below. Which sample of these are totally soluble in an excess
of dil. hydrochloric acid (c = 2 mol/L)?
Potassium, lead, aluminum, copper, zinc, silicium.
F Given are sample of the elements below. Which samples of these are totally soluble in an excess
of dil. nitric acid (c = 2 mol/L)?
Potassium, lead, aluminum, copper, zinc, silicium.
Problem 3-02
Given the following galvanic cell (T = 298 K):
Cu(s) | Cu2+
(aq) c = 1.00 mol/L || Ag+
(aq) c = x mol/L) | Ag(s).
a) Write down the equation for the cell reaction.
The voltage U for different values of x has been measured:
x 0.1000 0.0500 0.0100 0.0050 0.0010
U in V 0.403 0.385 0.344 0.326 0.285

Round 3 Test 1
19
b) Plot U as a function of lg x!
c) Write down the cell reaction for x = 0.0200 mol/L and determine the voltage.
d) Calculate the equilibrium constant K for the cell reaction.
3.00 g of potassium iodide are dissolved in water, the solution is filled up to 50.0 cm3
. 50.0 cm3
of
silver nitrate solution (c = 0.200 mol/L) is added. If this solution replaces the silver nitrate solution in
the galvanic cell the copper electrode becomes the cathode and a voltage of 0.420 V is measured.
e) Calculate the solubility product of silver iodide.
Cu2+
+ 2 e–
Cu E° = 0.34 V
Ag+
+ e–
Ag E° = 0.80 V
Problem 3-03 Interhalogen Compounds
Compounds between different halogens are called interhalogen compounds. Besides the diatomic
compounds XY there are compounds with more than two atoms. Their formulae are generally XYn,
where n = 1, 3, 5 or 7, and X is the less electronegative of the two halogens. The tendency to form
interhalogen compounds with more than two atoms rises with increasing atom mass of X and de-
creasing mass of Y.
a) Give an example for an interhalogen compound with more than two atoms which should be
existent and an example which should rather not be existent.
Diatomic interhalogens can be formed from the elements. All combinations are known.
There are the following natural isotopes of the halogens in existence:19
F, 35
Cl, 37
Cl, 79
Br, 81
Br, 127
I.
b) Write down the empirical formulae of the interhalogens XY and give the number of molecular
peaks in the mass spectrum of each compound XY.
c) Which shape should the molecules XY3, XY5 and XY7 have following the VSEPR model?
Sketch 3-D figures.
If exposed to water interhalogens disproportionate as halogens do, too. But also without water a
disproportionation of many of these compounds may take place.
d) Write down the equation of the reaction of XY with water (X is less electronegative than Y).
e) Write down the equation of a possible disproportion reaction of XY in the absence of water (X is
less electronegative than Y).
Problem 3-04 Hess's Law and Reaction enthalpies
A thermally very well insulated calorimeter was filled with water of 22.55 °C. When adding 1.565 g of
zinc sulfate the temperature went up to 23.52 °C after zinc sulfate had totally dissolved.
In a second experiment the same calorimeter was filled with water of 22.15 °C. After adding and
dissolving 13.16 g of zinc sulfate heptahydrate the temperature went down to 21.84 °C.
In both cases the heat capacity of the system was 0.900 kJ/K.

Round 3 Test 1
20
a) Calculate the enthalpy HR of the reaction ZnSO4 + 7 H2O  ZnSO4 · 7 H2O
b) Calculate the enthalpy of formation of nitrous acid (HNO2) in aqueous solution at constant pres-
sure and temperature from the given reaction enthalpies.
(1) NH4NO2(s)  N2(g) + 2 H2O(l) H1 = -307.4 kJ/mol
(2) 2 H2(g) + O2(g)  2 H2O(l) H2 = -571.7kJ/mol
(3) N2(g) + 3 H2(g) + aq  2 NH3(aq) H3 = -161.7 kJ/mol
(4) NH3(aq) + HNO2(aq)  NH4NO2(aq) H4 = -38.1 kJ/mol
(5) NH4NO2(s) + aq  NH4NO2(aq) H5 = +25.1 kJ/mol
Problem 3-05 Lithium
Lithium is the lightest metal and the least dense solid element. The small density is traced back to
the fact that it has as well as the other alkali metals a body centered cubic structure.
face centered cubic (fcc) body centered
cubic (bcc)
a) Calculate the packing fraction (in percent) of an fcc and of a bcc cell.
b) What is the percentage difference of the density between these two structures if the atoms are
assumed to be of the same kind?
Spodumene, LiAlSi2O6, is an important source of lithium. It is chemically opened up with CaCO3 and
converted into the oxides of aluminum and lithium. When treated with water lithium hydroxide crys-
tallizes as a monohydrate.
c) Write down the equation of the reaction of spodumene with CaCO3.
Using hydrochloric acid lithium hydroxide can be converted into lithium chloride which is taken to
obtain pure lithium by fused-salt electrolysis. In another process lithium can be obtained by electrol-
ysis of a solution of lithium chloride in pyridine or acetone. This solubility can be used to separate it
from sodium chloride and potassium chloride.
d) Give the reason why lithium chloride is soluble in solvents as pyridine, acetone and alcohols in
contrast to the other alkali chlorides.
In the qualitative analysis lithium is very difficult to detect because it forms nearly no poorly soluble
compound. Quantitatively it can be determined gravimetrically as sulfate or aluminate (formal: x
Li2O · y Al2O3). In literature (H. Grothe, W. Savelsberg (1937). Über die analytische Bestimmung des
Lithiums. Z. analyt. Chem. 110, 81 – 94) the following instruction is given (translated from German):

Round 3 Test 1
21
"III. Specification of a procedure to determine lithium.
A. Precipitating agent: 50 g of potassium aluminum sulfate are dissolved in 900 mL of warm water.
The solution is cooled down and a conc. solution of 20 g of sodium hydroxide is added while
stirring and cooling, until the formed precipitate is totally dissolved. After a long time (over-
night) the solution is filtered and the pH is brought to 12.6. Then the solution is filled up to 1 L.
B. Procedure of detection: The solution of lithium is brought to pH = 3. Then for each 10 mg of
lithium 40 mL of the precipitating agent is added. The solution is brought to pH = 12.6 again by
using some drops of sodium hydroxide solution (c = 1 mol/L). After a short time the solution can
be filtered. The precipitate is decanted with a lot of cold water, brought on a filter paper and
washed with cold water until the washing water does not redden phenolphthalein. The filter
paper is burned off and the residue (x Li2O · y Al2O3) weighed."
e) What could cause the turbidity when potassium aluminum sulfate ((KAl(SO4)2 · 12 H2O) is dis-
solved? Why can it be dissolved again by using sodium hydroxide solution? Give the equations
for the relevant reactions.
f) Why does the pH in this determination have be kept at exactly 12.6?
Exactly 0.1980 g of the residue of the ashing is brought into solution. The content of aluminum is
detected by complexation. 50.00 mL of a solution of Na2EDTA (c = 0.10 mol/L) is added to the solu-
tion and the Na2EDTA which has not reacted is titrated with zinc sulfate solution (c = 0.10 mol/L)
with xylenolorange (Ind) as indicator. Consumption: 15.25 mL.
g) Write down the equation for the complexation reaction of aluminum and Na2EDTA . Use H2Y2–
for the EDTA component.
h) Why are complexometric determinations often executed in buffer solutions (NH3/NH4Cl or
AcOOH/NaOOAc)?
i) Arrange the complex compounds of this determination (EDTA – Zn2+
, EDTA – Al3+
, Ind – Zn2
,+
Ind
– Al3+
) in the direction of decreasing stability.
j) In the original instruction the stoichiometric factors are replaced by x and y. Determine the em-
pirical formula of lithium aluminate in the residue of the ashing.
Problem 3-06
The dependence of the equilibrium constant on temperature of the reaction
PCl5 PCl3 + Cl2
can be expressed by the equation log Kp = -4374/(T/K) + 1.75·log(T/K) + 3.78.
a) Calculate Kp at 200 °C.
The reaction proceeds under isothermal-isobaric conditions at a temperature of 200 °C and a pres-
sure of 150 kPa in a vessel with variable volume until equilibrium is reached.
b) Calculate p(PCl5) and p(PCl3 in equilibrium. Use in this case Kp= 0.200.
c) Calculate the rate of conversion of PCl5 (in %).

Round 3 Test 1
22
Phosphorus pentachloride is a molecular compound only if heated above 160 °C. Below this temper-
ature it forms an ionic solid.
d) Which are the ions in the solid? Give the Lewis structure of these ions and draw their 3-D image.
Compounds as PCl5 and PF5 have a trigonal bipyramidal shape following the VSEPR model. In a trigo-
nal bipyramid you can distinguish axial and equatorial positions. Nevertheless only one single fluo-
rine signal in the 19
F NMR spectrum of PF5 is detected.
e) Account for this fact.
Problem 3-07 Isomerism
Draw the structures of all isomers with the empirical formula C3H6O and write down their names.
(Ignore stereo isomerism)
Problem 3-08 Aromatic Compounds
Given the following reaction scheme:
B
KMnO4
C
i / ii
iv / v
A
i / iii
H2 / Pd/C
Isomeri-
zation
F
vii
iv / v CH2O
D
1. NaNO2 /
vi
2. KI
- H2O
G E

Round 3 Test 1
23
a) Complete the structural formulae A – G as well as the empirical formulae of the reagents i - vii.
The isomerization is known by a special name, which one?
b) Which range of temperature has to be taken to produce G? Which product is formed if the solu-
tion (without KI) is heated to boiling?
c) What is the resonance effect of the substituent F in case of a second substitution? Show with the
help of resonance structures in which position the substituent directs. How does your result
match with the product which forms in the reaction of F with iv/v? Explain why the product
shown is formed.
Problem 3-09 Michael Reactions
α,β-unsaturated carbonyl compounds are typical Michael systems showing an interesting reactivity.
On one hand they decolor an aqueous solution of bromine as alkenes do and, like ketones, reactions
with nucleophiles occur. On the other hand the conjugated system of a keto group and an alkene
double bond shows some unique reactions.
a) Show the mechanism of the reaction of but-3-en-2-one with bromine. Give the (general) name of
the intermediate cation!
b) Write down the equation for the reaction of hydrogen cyanide (HCN) with the keto group of the
butenone.
The Grignard reaction of cyclohexenone and butyl magnesium bromide gives 3-butylcyclohexane-1-
one in addition to the classical product 1-butyl cyclohex-2-en-1-ol:
c) Using resonance structures account for the fact that there can be an addition not only at the
carbonyl group but also at the double bond. Is it an electrophilic or a nucleophilic addition?
Besides Grignard compounds many other organometallic reagents are used to form a C-C linkage e.g.
BuCeCl2 and Bu2CuLi both of which are produced in situ. The concept of hard and soft acids and ba-
ses (HSAB concept of Pearson) allows a good valuation whether the addition proceeds directly (at
the carbonyl carbon atom) or rather conjugated (at the β-carbon atom). The gist of this theory is that
soft acids react faster and form stronger bonds with soft bases, whereas hard acids react faster and
form stronger bonds with hard bases, all other factors being equal. Cer compounds are known to be
very hard Lewis acids while copper(I) compounds are rather weak.
d) Explain which C atom in cyclohex-2-en-1-one is harder, which one is weaker. At which C atom
would you expect a reaction with BuCeCl2 and Bu2CuLi, respectively?

Problems Round 3 Test 2
24
Third Round Test 2
Problem 3-11 Multiple Choice
With one or more correct answers even if the question is written in singular.
a) A member of which group of compounds is generated by the oxidation of a secondary alcohol?
A Aldehyde B Carboxylic acid C tert. Alcohol D Peroxide E Ketone
b) Which compound contains an element with the same oxidation number as chromium in
K2Cr2O7?
A Cl2O2 B Fe(CN)6
3-
C VO2
+
D K2MnO4 E H2S2O8
c) Which of the following mixtures is a buffer solution?
A CH3COOH (50 mL; 0.1 mol/L) + NaOH (50 mL; 0.1 mol/L)
B CH3COOH (50 mL; 0.1 mol/L) + NaOH (50 mL; 0.05 mol/L)
C CH3COOH (50 mL; 0.05 mol/L) + NaOH (50 mL; 0.1 mol/L)
D CH3COOH (50 mL; 0.05 mol/L) + NaOH (50 mL; 0.05 mol/L)
d) Which ion has unpaired electrons at its disposal?
A Cu+
B As4+
C Zn2+
D Ag+
E Mn5+
e) Which compound does puff up cakes?
A CaCO3 B (NH4)2CO3 C Ca(COO)2 D NaHCO3 E Yeast
f) The image1
shows the phase diagram of water. Which of the following statements is correct?
g) Which of the following formulae represent more than one compound?
A CH4O B C2H2Cl2 C Pt(NH3)2Cl2 D CuSO4·5H2O E C2H6O
h) Which of the following molecules and ions are planar?
A C2H4 B PH3 C COCl2 D PtCl4
2-
E CH4
1
Image from "chemikerboard.de"
With rising pressure
A the boiling temperature decreases and the
melting temperature increases slightly.
B the boiling temperature increases slightly and
the melting temperature decreases.
C the melting and boiling temperature increase.
D the melting and boiling temperature de-
crease.
E the melting and boiling temperature do not
change.
Temperature in °C
gaseous
fluid
solid
PressureinkPa

25
Problem 3-12 Buffer Action and Acidity
A buffer solution with pH = 5.8 has to be made from a diluted acid (pKS = 6.5) and its sodium salt.
a) Calculate the ratio of amounts of acid and conjugated base.
b) Calculate the concentration of formiate ions at pH = 4.2.
There is 1 L of a buffer solution which contains 0.1 mol of NH3 and 0.1 mol of NH4Cl. The pH value of
this solution is 9.25.
c) Which volume of hydrochloric acid (c = 1 mol/L) and sodium hydroxide solution (c = 1 mol/L),
respectively, can be added with the result that the pH does not change more than 1?
A solution of 2.895 g of an unsubstituted weak carboxylic acid X in 500 g of water shows a freezing-
point depression of 0.147 K. If 0.957 g of sucrose (C12H22O11) is dissolved in 100 g of water a freezing-
point depression of 0.052 is detected.
d) Which saturated carboxylic acid was used?
e) Determine the acidity constant Ka of this acid and , the degree of protolysis.
Problem 3-13 Reactions
There are six aqueous solutions of the following compounds: NH4Cl, BaCl2, Na2S, Pb(NO3)2,
Na2SO4 and AgNO3.
a) Record in the table on the answer sheet whether a reaction takes place (e.g. "yellow prec." if a
yellow precipitate forms or "gas formation" ect.) or "n.r." if nor reaction takes place.
b) Write down the equations for all reactions (indicate the aggregate state and the hydration by
using (s), (l), (g), (aq))! Is it possible to identify the original solutions only by the results of the
reactions without using more utilities? Account for your decision!
Problem 3-14 Double-Contact Process
The double contact process is used to synthesize sulfuric acid on an industrial scale. During the deci-
sive step in this process sulfur dioxide is oxidized to sulfur trioxide:
2 SO2 + O2  2 SO3 (1).
Given are the following thermodynamic values* at 25 o
C and po
= 1.000·105
Pa (Standard pressure).
Hf
0
in kJ·mol-1
So
in J·mol-1
·K-1
Cp in J·mol-1
·K-1
SO2 (g)
O2 (g)
SO3 (g)
-297.00
0
-396.00
249
205
257
39,9
29,4
50,7
* values taken from Atkins, Physical Chemistry 3rd
Edition
Cp is the molar heat capacity at constant pressure. You can use it to determine the enthalpy of for-
mation and the entropy of formation of a compound at a temperature differing from the standard
temperature:  Hf
0
(T) =  Hf
0
+ Cp·(T – 298 K), S(T) = S0
+ Cp·ln(T/298 K).

26
a) Determine Kp at 600 O
C by using the Cp values.
To produce SO2 at first sulfur is burnt at 1400 °C to1500 °C under oxygen deficiency conditions. Af-
terwards it is oxidized completely to SO2 at 700 °C in an excess of air.
b) Why do they work at first at 1500 °C under oxygen deficiency conditions and why is the total
oxidation completed at 700 °C?
In doing so you get a mixture of 10% (V/V) of SO2, 11% (V/V) of O2 and 79% (V/V) of N2 which is prac-
tically free of SO3. This mixture is led through the contact reactor. At 600 °C and standard pressure
the equilibrium is established.
c) Calculate the volume percentage of the components of the gas mixture at equilibrium (in %).
Determine the degree of conversion of SO2 (in %).
(Use Kp = 65.00 here. At the end of the calculation there will arise an equation of third order
with only one real solution. The result should have two decimals.
Problem 3-15 A Well Near a Volcano
The water taken from a volcano well is tested for hydrogen sulfide. A sample of 10.0 mL is taken and
all hydrogen sulfide is removed with a stream of carbon dioxide. It is absorbed in bromine water and
then the excess of bromine is removed. The acidity of the sample is titrated with a solution of NaOH
(c = 0.100 mol/L) using methyl red as indicator. 19.95 mL are being consumed.
a) Write down balanced equations for all reactions of this method. Calculate the hydrogen sulfide
content of the water in g/L.
Another method to measure the hydrogen sulfide content works without a stream of carbon dioxide
using a different redox process. If a known amount of iodine is produced in a solution it will oxidize
hydrogen sulfide. The excess of iodine can be titrated with thiosulfate.
6.50 g of KIO3 are dissolved in water to give a solution of 1L. 10.0 mL of this solution are taken and
0.5 g of KI and 5 drops of starch solution are added. Then 10.00 mL of the volcano water are added,
too, and the mixture acidified with 10 mL of 10 % hydrochloric acid. After 5 minutes the mixture is
titrated with a solution of Na2S2O3 (c = 0.100 mol/L) until the blue colour disappears.
(Hint: Hydrogen sulfide is oxidized by bromine to form a sulfate and by iodine to form sulfur.)
b) Write down balanced equations for all reactions of this second method. Which volume of the
Na2S2O3 solution do you expect to be consumed in the titration?
c) The silver bracelet of the technician who worked with the samples has blackened. Explain by a
reaction equation.

27
Problem 3-16 Gaseous Compounds
A An airbag is a safety device in vehicles. It is an occupant restraint system consisting of a flexible
fabric envelope or cushion designed to inflate rapidly during an automobile collision. Older airbag
formulations contained sodium azide and other agents including potassium nitrate and silicium diox-
ide. An electronic controller detonates this mixture during an automobile crash forming sodium and
nitrogen. After the nitrogen inflated the cushion the temperature of the gas is appr. as low as 150 °C
due to the expansion.
a) Write down the reaction equation of the decomposition of sodium azide.
b) Draw the Lewis formula of the azide ion.
Since sodium metal is highly reactive the KNO3 and SiO2 react and remove it, in turn producing more
N2 gas following the (not arranged) equations (2) and (3):
Na + KNO3  K2O + Na2O + N2(g) (2)
K2O + Na2O + SiO2  K2SiO3 + Na2SiO3 (silicate glass) (3)
c) Arrange the equations (2) and (3).
d) Calculate the mass of sodium azide necessary to fill a cushion of 50.0 L (at 150 °C, 1300 hPa).
Until 1999 N2 and N3
–
were the only stable nitrogen containing particles which could be produced in
a larger scale. In 1999 the discovery of another such a non-cyclic particle was published: N5
+
.
The following structure was shown:
(K.O.Christie u.a. : N5
+
; ein neuartiges
homoleptisches Polystickstoff-Ion mit
hoher Energiedichte, Angew. Chem 111
(1999) 2112 – 2117)
e) Draw three mesomeric resonance structures of this particle which are consistent with this struc-
ture.
B 20 cm3
of a gas X are filled into a measuring tube. 80 cm3
of oxygen are added and the mixture
is ignited. When the pressure and the temperature of the beginning are restored you observe a de-
cline of volume of 10 cm3
. There is some oxygen left in the mixture after the reaction.
f) Which of the following gases could be X?
Hydrogen, ammonia, carbon monoxide, ethene, methane.
Distinguish ϑ > 100 °C and ϑ < 100 °C.
Problem 3-17 Decay of Arsine
At 500 °C arsine, AsH3, decomposes quickly and totally in a reaction of first order to give arsenic and
hydrogen.
a) Write down the reaction equation and the respective rate equation.
The kinetics of the decomposition was studied at a lower constant temperature in a closed tube.
108° 166°
1,12A
1,33 A

28
In the beginning of the experiment there was pure gaseous arsine in the tube with a pressure of p0 =
86.1 kPa.
After 120 minutes the pressure has increased to p120 = 112.6 kPa.
b) Determine the pressure at the end of the decomposition.
c) Determine the rate constant and the half-life of the reaction.
d) What amount of time is necessary to decompose 99 % of arsine?
(Use in this case k = 1.3·10-4
s-1
.)
Problem 3-18 Epoxides
Epoxides are very reactive heterocyclic compounds and therefore of great importance in the organic
synthesis.
a) Give a reason for the great reactivity of the epoxides.
They are divided into symmetric and asymmetric epoxides:
O O
A B
symmetrisch asymmetrisch
Compound B can be synthesized in the following sequence of reactions starting from dimethyl sul-
fide:
S
CH3I
C
Base
-HI D
H
O
-S(CH3)2
O
B
b) Complete the structural formulae of C and D (D is a sulfur ylide, a neutral dipolar molecule con-
taining a formally negatively charged atom attached to a heteroatom with a formal positive
charge).
c) Show the mechanism of the reaction of D to the epoxide B using structural formulae.
Epoxides can be opened by a lot of nucleophiles to form bifunctional alkanes. The reaction of oxira-
ne A with water, for example, catalyzed by an acid gives ethylene glycol. This reaction proceeds with
another mechanism than the reaction without an acid.
d) Show a possible mechanism of the ring-opening catalyzed by acid using structural formulae.
e) Complete the structural formulae of the products E – J of the following reactions:
symmetric asymmetric

29
If an epoxide is treated with Lewis acids (LSkat), such as BF3 or MgI2 isomerization to a carbonyl com-
pound takes place. Thereby a symmetric epoxide forms one isomer while asymmetric epoxides form
more of them.
f) Draw the structural formulae of K, L and M.
O LSkat. K
O LSkat. L M
Problem 3-19 Amino Acids
L-α-amino acids play an important role in all creatures. Histidine (on the right)
has a pKa2 value of 6.00. It is the only amino acid which contributes to the capac-
ity of the blood buffer (pH = 7.40). Glutamic acid (2-aminopentane diacid) and
aspartic acid (2-aminobutan diacid) are important neurotransmitter. Cysteine (2-
amino-3-mercapto propanoic acid, mercapto = thiol) plays an important role in
the structure of proteins because it can form the stabilizing disulfide bridges.
Two cysteine molecules can be linked together with a disulfide bridge to form
dicysteine (cystine).
a) Which species of histidine are present in the equilibrium at the pH values of 1.82 and 9.17, re-
spectively, which are the other pKa values of histidine? Only one of the two nitrogen atoms in the
ring can be protonated, which one? Rationalize your answer.
b) Draw L-glutamic acid in the Fischer projection.
c) Draw 3-D structural formulae of S-aspartic acid as well as of R-cysteine. ((Hint: in front of
the paper plane, behind the paper plane, in the paper plane)
In proteins and peptides the amino acids are linked with peptide bonds – the resulting compound is
in chemistry called a carboxylic acid amide. In some text books the reaction is shown in the following
way:
.
O NH3
E
O CH3OH / H+
F
O LiAlH4
G
O
H
H2S
O
I
HBr
O
J
1. RMgX
2. H2O/H+

30
This reaction proceeds actually in this way only under extreme conditions because the carboxylic
acid is not sufficiently reactive towards a nucleophilic attack (low carbonyl activity).
d) Which electronic effect is responsible for the low carbonyl activity of the carboxylic acids?
In order to raise the reactivity of the carboxylic acids they can be converted into halides or anhy-
drides.
e) Show the total mechanism of the formation of an amide between a reactive species R-COX and
an amine R‘NH2. Which geometrical structure shows the reaction center directly after the attack
of the amino group?
f) Write down the equation for the formation of dicysteine from cysteine (with structural formulae
without stereochemical concerns). Which type of reaction is it?
Problem 3-20 Organic Puzzle
a) Draw the structural formulae of the following compounds.
A Acetaldehyde B Acetone C Acetophenone
D Methanoic acid E Aniline F Benzaldehyde
G Nitrile of benzoic acid H 3-Nitrobenzoic acid I Phenol
b) Assign each compound (using the letters of the table above) to a cell of a 3x3 table given on the
answer sheet.
Consider thereby the criteria below.
1. Line: The aqueous solution shows an acidic reaction.
2. Line: The compound can react with itself in an aldol reaction.
3. Line: It in a monosubstituted aromatic compound.
1. Column:The compound contains an aldehyde group as a functional group.
2. Column:It is an aromatic compound, the substituent(s) of which directs newly incoming sub-
stituents into meta position.
3. Column:Compounds without such criteria.

Problems Round 4 (theoretical)
31
Fourth Round (theoretical problems)
(A periodic table and the same list of formulae and data as in the third round were provided)
The top 15 of the 3rd
round are the participants of the 4th
round, a one-week practical training.
Problem 4-01 Traces of Water
The compound KPbI3 can be used to detect minimal amounts of water qualitatively. To detect water
quantitatively other methods are used.
Method A: Heating / Annealing
The substance is heated until its mass is constant.
a) Record preconditions of the substance when using this method.
Method B: Gaseous (reaction) water
The enclosed water is bound chemically or physically. The increase of mass of the absorbing agent
such as sulfuric acid and anhydrous calcium chloride is determined. This method is also used to dry
gasses.
b) Give the reason why hydrogen sulfide, hydrogen iodide and ammonia should not be dried by
sulfuric acid. Write down respective reaction equations.
c) Give the reason why ammonia may not be dried by calcium chloride.
Method C: Calcium carbide (CaC2)
The water containing sample is brought to reaction with calcium carbide and the reaction product is
led into an ammoniac copper(I) chloride solution. The red precipitate is filtered off, dried to mass
constancy and weighed.
d) Draw the Lewis structure of the carbide anion. Give the formulae of three isoelectronic species.
e) How does copper(I) chloride exist in an ammoniac solution?
f) Write down the equation of all reactions of this method.
Method D: Iodometric (Karl Fischer method)
In 1935 Karl Fischer published a method to determine water based on the reaction between iodine,
sulfur dioxide and water previously published by Bunsen.
g) Give the equation of the "Bunsen" reaction!
In the Fischer method the water containing sample is brought to reaction with methanol, pyridine,
sulfur dioxide and iodine following formally the equation
H2O + SO2 + 3 C5H5N + I2 + CH3OH C5H5NHCH3OSO3 + 2 C5H5NHI .
The endpoint of the titration is reached when a permanent brown colour occurs.
h) What is responsible for the brown colour? Why can the iodine-starch reaction not to be used?
i) Which function has the pyridine?

32
To determine the water content the sample is added to a solution of iodine and sulfur dioxide in
water free methanol which contains pyridine and then titrated with a solution of iodine in alcohol.
As the change of colour at the endpoint (bright yellow  brown) is difficult to detect visually the
change is nowadays detected coulometrically. During the titration iodine is produced in an electro-
chemical cell until no more iodide is formed. The water content can be calculated by the used
amount of charge.
A B The titer of the Karl-Fischer solution is given in water equiva-
lents in mg/mL and amounts to t = 4.8 mg/mL. Samples of 10 g
of two different food oils are investigated. The results are given
in the table on the left.
Sample 1 1.65 mL 1.45 mL
j) Calculate the mass percentage of water in the oils.
In another oil C a water content of 0.09 % is found in a coulometric titration.
k) Which amount of charge (in coulomb) is consumed in an investigation of 10 g of oil C?
Problem 4-02 Electrochemistry
A Latimer diagrams are plots of reduction potentials of half reactions including each of the differ-
ent oxidation states of one element. Normally the species with the highest oxidation state is placed
on the left, going to the right the oxidation state decreases. The different states are connected with
arrows on which the reduction potential of the half reaction is written. These may refer to standard
conditions (25 °C, pH = 0, c = 1 mol·L−1
) or to any other condition (e.g. pH = 14).
Example: [AuCl4]–
 [AuCl2]–
 Au (Standard conditions)
a) Calculate the standard potential x = E°([AuCl4]–
/ Au).
Gold does not react with nitric acid but does with aqua regia, a 3:1 mixture of conc. hydrochloric acid
and conc. nitric acid, which was developed by alchemists to "dissolve" gold.
In this reaction with aqua regia the complex [AuCl4]–
is formed.
b) Calculate the complex formation constant of [AuCl4]–
, Kco =
c([AuCl4]−)/c0
(c(Au3+)/c0) ·(c(Cl−)/c0)4
using the
result of a) and E°(Au3+
/Au) = 1.50 V.
In an acidic solution (pH = 0) the following standard potentials are existent:
ClO4
–
/ ClO3
–
ClO3
–
/ ClO2
–
ClO2
–
/ HClO HClO/ Cl2 Cl2/ Cl–
E° in V 1.20 1.18 1.65 1.63 1.36
c) Draw the Latimer diagram.
Check whether ClO3
–
disproportionates under these conditions to form ClO4
–
and Cl–
. Write down
the equation for a disproportionation reaction if the case may be.
0.926 V 1.154 V

33
B
d) In which span of pH-values can hydrogen peroxide oxidize Cer3+
ions?
E°(Ce4+
/Ce3+
) = 1.61 V E°(H2O2, H+
/H2O) = 1.76 V
On the other hand hydrogen peroxide can be oxidized by strong oxidation reagents such as potassi-
um permanganate in acidic solution.
e) Write down the equation of this reaction.
Problem 4-03 Metals and Electrons
When atomic orbitals are filled up with electrons the lowest available orbital is fed first and then the
following in the direction of rising energies. Thereby Hund's rule as well as the Pauli principle has to
be considered. There are exceptions from these rules.
a) Give two examples for elements with electron configurations (in the ground state) other than
the expected regular one. Write down the expected and the observed configuration. (Use the
abbreviation for full electron shells such as [He] for 1s2
or [Ne] for s2
2s2
2p6
etc.)
The reason for this deviation is the stability of certain electron configurations.
b) Which electron configurations are energetically especially favoured?
These favoured electron configurations play a decisive role in the electron configuration of metal
cations.
c) Write down the electron configurations of the following metal cations.
(Use the abbreviation for full electron shells as mentioned above.)
i) Fe3+
ii) Mn3+
iii) Pd4+
iv) Cr3+
v) Fe2+
vi) Pb2+
vii) Au3+
viii) Co2+
ix) Cu+
x) Ti2+
Discrete metal ions as given in part c) exist only formally. In reality e.g. in solids or complex com-
pounds they always have a coordination sphere and are surrounded by other particles which form a
regular coordination polyhedron. You often find the coordination numbers four and six.
d) Draw 3-D figures of the polyhedrons for the coordination numbers four and six. (Don't consider a
hexagon and a three-sided prism)
e) Draw the theoretically possible stereoisomers of the complex compounds MX2Y2, MX4Y2 and
MX3Y3 and give their names (M: central particle; X and Y: monodentade ligands).
All degenerate systems try to reduce orbital degeneracy. The degenerated d-levels of an isolated
metal cation split as soon as they come under the influence of a ligand field into levels of higher and
lower energy.
f) Describe the splitting off of the d-electron energy levels in an octahedral ligand field. What are
the effects on the different d-orbitals? Draw an energy scheme which shows this issue as accu-
rately as possible.
g) With which d-electron configurations high- and low-spin configurations occur at all?
h) Give the number of unpaired electrons which exist in a high-spin and in a low-spin state for the
cations of part c) in an octahedral ligand field.

34
Problem 4-04 Calculations around Kinetic and Energetic
In a gas flow substance A exists with a partial pressure of p(A) = 8.9·10-4
bar. A is in equivalence with
A2 (2 A A2) which is gaseous, too, with Kp = 2.1·103
.
a) Determine the ratio p(A)/p(A2).
Given the reaction A + B  C + D with the rate constant keff.
The following reaction sequence is assumed to be the reaction mechanism
A + B AB  C + D .
The reaction rates v1 and v-1 are approximately of the same size. Additional is k1/k-1 = 15 (mol/L)-1
and k2 = 25 s-1
.
b) Calculate keff using the assumption of steady state equilibrium.
c) Which is the necessary precondition to use the steady state approximation?
The hydrolysis of urea follows the equation
(NH2)2CO(aq) + H2O(l)  2 NH3(aq) + CO2(aq) .
d) Calculate the equilibrium constant K for this reaction at 298 K.
e) Calculate G if the following concentrations exist at 298 K
c((NH2)2CO(aq)) = 0,85 mol/L c(CO2(aq)) = 0,097 mol/L c(NH3(aq))= 0,02 mol/L
In the equilibrium D + E F the reaction as well as the back reaction are elementary reac-
tions. c(D), c(E) and c(F) are time dependent concentrations in the process of the spontaneous reac-
tion of D and E.
f) Determine the dependency of G on v3 and v-3 which are the rates of the reaction and the back
reaction.
At a certain time in the course of the reaction D + E F from part e) the observed reaction
rate is defined as vob = v3- v-3 .
At a certain time let be
vob/ v3 = 0.5 c(D) = 0.4 mol/L c(E) = 0.9 mol/L c(F) = 1.8 mol/L.
g) Calculate the equilibrium constant K for this reaction at 298 K.
Problem 4-05 Three Compounds
Given are the compounds AB3, CA and CB2. The mass percentage of A in AB3 is 23.81 %, that of B in
CB2 73.14 %.
a) Calculate the mass percentage of C in CA.
(NH2)2CO(aq) H2O(l) NH3(aq) CO2(aq)
H° in kJ/mol -317.7 -286.0 -80.9 -413.1
S° in Jmol
-1
K
-1
176 68 110 121
keff
k1
k-1
k2
k3
k-3
k-3
k3

35
b) Determine the elements A, B and C. Write down the formulae and names of the given com-
pounds.
Problem 4-06 Complexes and More
A
Insoluble Prussian blue is a coordination compound. The crystal structure and the empirical formula
can be derived from one octant (on the left) of the cubic unit cell (drawn with iron ions only).
Indicate in all of the questions a) to d) the oxidation states of the iron ions as FeII
, FeIII
.
a) Determine the empirical formula of Prussian blue using the octant above. Show shortly how you
found your result.
Insoluble Prussian blue forms only at high concentrations of iron ions. At lower concentrations so-
called soluble Prussian blue K[FeIII
FeII
(CN)6] is formed.
It originates from the combination of an iron(II) solution with a solution of potassium hexacyanofer-
rate(III) as well as from an iron(III) solution with a solution of potassium hexacyanoferrate(II).
b) Write down the equation for these two reactions.
c) Taking these reactants find an equilibrium reaction which explains the formation of the identical
product.
Comparable with the reaction of part b) there is a reaction between a solution of iron (II) with potas-
sium hexacyanoferrate(II) which leads to a white precipitate.
d) Write down an equation of the reaction which leads to the white precipitate. Explain why this
compound is not colored in contrary to soluble Prussian blue.
The orientation of the cyano groups in the compound of part a) (see fig. above) can be derived with
the help of the HSAB concept. It says that the combinations of weak with weak and of hard with
hard acids and bases lead to more stable adducts than mixed combinations. Iron(III) is classified as
hard, iron(II) as weak. The hardness of bases decreases within the row
F > O > >N, Cl > Br, H > S, C > I, Se > P, Te > As > Sb.
(on the edges)
(H2O in the middle of the octant)
(on the edges)

36
e) Fill the cyano groups in their expected orientation following the HSAB concept into the empty
circles on the answer sheet (C for carbon, N for nitrogen). Determine the coordination spheres of
iron(II) und iron(III) in the compound.
B
A very diluted solution of copper(II) chloride in water has a light blue color. By adding hydrochloric
acid the solution turns green and with rising concentration of hydrochloric acid intensively green-
brown.
f) Which species of copper(II) chloride exist predominantly in the diluted aqueous solution and in
the solutions in half-concentrated and concentrated hydrochloric acid?
In analytical chemistry copper is detected in the form of a deep blue colored ammine complex. But it
can be recognized with a borax bead as turquoise metaborate (Cu(BO2)2), too. In literature you find
different formulae for borax with the empirical formula B4H20Na2O17:
As decahydrate Na2B4O7 · 10 H2O, octahydrate Na2[B4O5(OH)4] · 8 H2O
and as mixed oxide decahydrate Na2O · 2 B2O3 · 10 H2O.
g) Show with the help of a Lewis structure why the octahydrate is the most reasonable. Attach
charges to the atoms if necessary.
In the separation scheme of cations copper(II) ions are precipitated with hydrogen sulfide as black
copper sulfide. However, this compound is not copper(II) sulfide but a mixed compound of copper(I)
and copper(II) ions which contains another sulfur containing anion besides sulfide anions (S2–
).
h) Which other anion does the compound contain besides the sulfide anion? Give a more exact
formula than CuS.
In the Deacon process to produce chlorine by oxidizing hydrogen chloride using oxygen from the air
copper(II) chloride is used as catalyzer.
i) Write down the equation for the total reaction of the Deacon process.
j) Give reaction equations which show the catalytic effect as well as the regeneration of copper(II)
chloride.
Problem 4-07 Many Questions Concerning Thermodynamics
2 mol of oxygen at 273 K have a volume of 11.2 dm3
. The gas can be regarded as perfect with the
heat capacity of CV = 21.1 Jmol-1
K-1
which is supposed to be independent of temperature.
a) Calculate the pressure of the gas.
b) Give the different meanings of Cp and CV. Explain why these values have to be different.
Calculate Cp.
The sample of gas mentioned above is heated reversibly to 373 K at constant volume.
c) How much work is done to the system?

37
d) Calculate the rise of the internal energy.
e) Calculate the heat which was added to the system.
f) What is the final pressure?
g) What is the increase in enthalpy H?
The gas sample (2 mol at 373 K in 11.2 dm3
) is allowed to expand maintaining the temperature
against a piston that supports a pressure of 2 atm.
h) Calculate the work done by the expansion.
i) What is the change in internal energy and in enthalpy of the gas?
j) Calculate the heat absorbed by the gas.
The dependence of the boiling point of methane on pressure is well described by the empirical equa-
tion: log(p/bar) = 3.99 –
443
Ts/K - 0,49
k) Determine the boiling point of methane at a pressure of 3 bar.
The difference in internal energy of liquid and gaseous methane at the boiling point of 112 K at at-
mospheric pressure is 7.25 kJ/mol.
An object is cooled by the evaporation of CH4(l).
l) What volume of CH4(g) at 1.000 atm must be formed by the liquid to remove 32.5 kJ of heat
from the object?
Problem 4-08 Synthesis of (–)-Muscone
(–)-Muscone is the primary contributor to the odor of musk. It is an oily liquid with a characteristic
smell. Natural muscone was originally obtained from musk, a glandular secretion of musk deer,
which has been used in perfumery and medicine for thousands of years. Since obtaining natural
musk requires killing the endangered animal, nearly all muscone used in perfumery today is synthet-
ic.
One synthesis of (−)-muscone begins with commercially available (+)-citronellal, a monoterpene
aldehyde, which is mainly found in citrus fruits. The scheme on the next page shows the way of syn-
thesis.
Hints:
- In this scheme compound A reacts exclusively to compound B and compound C exclusively to D
aas well as compound G exclusively to (–)-muscone.
- Compound V is a side product.
- Compound G exists as a mixture of two isomers.
- (–)-Muscone has the empirical formula C16H30O.

38
- TBDMSCl
Si
Cl
Protective reagent for hydroxy groups
- TBAF
N+
F-
splits up silyl ether
- THF O Solvent
- Grupps cat. I
Ru
PhP(Cy)3
P(Cy)3
Cl
Cl
Cy = Cyclohexyl-Rest
catalyzes the metathesis of olefins, e.g.:
R1
R2
R1
R2
R3
R4
R3
R4
Kat.
R1 R2
R3 R4
+
2
10-bromo-dec-1-ene
then r.t.then r.t.
r.t. = room temperature
Grupps cat. I
(-)-Muscone
Cat.

39
a) Determine and draw the structural formulae of the compounds A to G as well as the structural
formula of (-)-muscone.
During the reaction of (+)-citronellal to compound A in a so-called carbonyl-ene-cyclisation com-
pound V (isopulegol) forms as a side product. Under the terms given in the scheme above the Lewis
acid X is formed which catalyzes the carbonyl-ene-cyclisation (the stereoselectivity of this reaction
stays unaccounted):
H
(+)-Citronellal
OH
X
V, Isopulegol
O
b) Give the empirical formula of X and write down the reaction equation of the formation of X.
c) Map the mechanism of the carbonyl-ene-cyclisation taking the catalyzing effect of X into consid-
eration.
The formation of F succeeds with a (deep) red chromium(VI) species Y which is formed in situ. It
oxidizes E to compound F and is itself reduced to a green compound.
Y is formally the anhydrate of the chromium acid.
d) Write down the equations for the reduction and the oxidation as well as the complete equation
for the redox reaction between the compounds E and Y. Use the abbreviations R and R' for the
substituents outside the reactive center.
Step F to G is a so-called cyclisation metathesis. A gaseous compound Z forms besides compound G.
e) Give the name of compound Z?
Problem 4-09 Small but Powerful
Carbocyclic compounds are often found as basise of biologically active reagents and functional ma-
terials. Many ways are known to synthesize the especially stable five- and six-membered rings but
the preparation of larger and smaller rings is a synthetic challenge. The reason for the difficulties
with smaller rings is their low stability and thus their high reactivity.
The bonding situation in a cyclopropane ring is described by bent bonds. The overlapping orbitals
between two carbon atoms can't point directly to each other (as they do in alkanes); rather, they
overlap in an angle. The result is a banana-shaped bond shown on the next page.
a) Compare the bond angel and the bond length of cyclopropane and the more stable cyclohexane.
Draw a cyclopropane and a cyclohexane ring (in chair conformation with hydrogen atoms) in a
3-D plot. Account for the different stability of the C-C bonds in cyclopropane and cyclohexane.

40
Banana-shaped bond in cyclopropane
Because of the higher π-character of the C-C bonds in cyclopropane rings addition reactions may
proceed analogue to those of alkanes. The electrophilic hydrobromation is an example.
b) Write down the equation of the reaction of methyl cyclopropane with hydrogen bromide.
Give the mechanism of the hydrobromation and use it to explain which product is formed pref-
erentially.
What is the name of the rule of this regioselective formation?
c) Draw an energy profile of this reaction (ΔRH° < 0, energy depending on the reaction coordinate)
and attach reactants, products, transition states and intermediates (if existent).
Generally cyclopropane rings are generated via cycloaddition of carbenes and carbenoids to olefins.
Carbenoids are substances similar to carbenes and show a comparable reactivity.
d) Give a common form of the Lewis structure of a carbene (with substitutes = R). What is the hy-
bridization state of the carbon atom in the center of the carbene?
The best method for preparing cyclopropanes is by a process called the Simmons-Smith synthesis:
An addition of CH2-Zn-I to a C=C double bond. In this reaction the new C-C bonds form in cis position
of the original double bond.
e) Draw the structural formula and give the name of the product of the Simmons-Smith reaction of
(Z)-1,2-diphenylethene.
Another compound to synthesize cyclopropane rings starting from olefins is diazomethane (CH2N2),
which is also used for the preparation of the hydrocarbon 4 (C13H20). 4 has a tetrahedral shape. The
1
H- and 13
C-NMR spectra show two resp. three signals. The scheme of this reaction is shown below:
1 2
LiAlH4/Et2O
99%
Ph3P
.
Br2/DCM
-15° 20°C, 6h
95%
3
t
BuOK/DMSO
20°C, 6h
60%
4
CH2N2 (10 eq),
Pd(OAc)2
-20° 25°C
sechsfach wiederholt
92%
CO2Et
DCM = dichloromethane, DMSO = dimethyl sulfoxide
repeated six times

41
f) Give the structural formulae of the compounds 1 to 4.
g) Mark in a figure of 4 those C and H atoms which are responsible for the NMR signals.
Problem 4-10 Carbonyl Compounds as Reactants and Products
Carbonyl compounds are valuable resources for organic synthesis because they react with many
different compounds, often selectively. In the following scheme some reactions starting with a butyl
alcohol (1) are depicted:
iPr: Isopropyl
cHex: Cyclohexyl
DMP: m-CPBA:
a) Draw the structural formulae A to G.
b) Give a possible method for i) to form B directly from butyl alcohol.
Compound F is an epoxide.
c) Which reagents ii) are needed to synthesize the diol 2 and then to convert the diol 2 with iii) to
A? Give the name of the reaction of diol 2 to A.
(Help: Compound iii) contains a halogen atom in the highest oxidation state.)
d) Which product do you expect if compound C is brought to reaction with MeLi?
Draw the structural formula.
e) Why is it not possible to get compound D by bringing B directly into reaction with MeLi?
Finally compound D is converted to 5.
f) Propose the reaction conditions iv) for the reaction of D to 5.
g) Show the mechanism for this reaction and give the names of the class of substances of the in-
termediate and the product.

Problems Round 4 (practical)
42
Fourth Round (practical problems)
Problem 4-11 Gravimetric Determination of Zinc as Zn(NH4)PO4
Equipment:
2 x 400 mL beaker, 25 mL pipette with pipette control, 50 mL graduated cylinder, 100 mL narrow-
necked bottle, Bunsen burner with tripod and tile, glass rod, 2 glass filter crucibles, suction flask
with rubber ring, pressure tubing, vacuum attachment, desiccator with drying agent, precision bal-
ance, pen
Chemicals:
Test solution (100 mL volumetric flask)
Solution of ammonia, c(NH3) = 2 mol/L
Ammonium chloride, NH4Cl(s),
diluted hydrochloric acid, c(HCl) = 2 mol/L
Solution of diammonium hydrogen phosphate, w((NH4)2HPO4) = 10 %
Indicator solution of methyl red in Ethanol, w(C15H15N3O2) = 0,05 %
Demineralized water
Procedure:
The test solution has to be filled up with demineralized water to the calibration mark and mixed
well. 25 mL of this solution are pipetted into a 400 mL beaker. Approximately 150 mL of demineral-
ized water are added.
Then 25 mL of diluted hydrochloric acid, 2-3 full spatulas of ammonium chloride, 25 mL of a solution
of diammonium hydrogen phosphate and some drops of methyl-red indicator solution are added.
Heat to boiling and add dropwise diluted ammonia until a colour change to yellow-orange can be
observed. Then stir with the glass rod until the precipitate has formed crystals or is well sedimented.
While the solution cools down to room temperature the glass filter crucibles are marked with the
pen, weighed and the values are listed.
The precipitate is sucked through a glass filter crucible, washed with a small portion of cold demin-
eralized water and then dried at 130°C for one hour in a drying oven.
The crucible is cooled down to room temperature for about 20 to 30 minutes in an exsiccator and
then weighed again.
a) Write down the number of your sample on the answer sheet.
b) Write down the results of your weighing.

43
c) Calculate the mass concentration of zinc in mg/L in your test solution.
d) What kind of compound is zinc ammonium phosphate, a mixed crystal, a mixture of crystals, a
double salt or an alloy? Account for your decision.
e) What happens if you anneal the precipitate? Write down the reaction equation.
Problem 4-12 Standardization of a Na2EDTA Solution
This solution will be used in problem 4-13.
Equipment:
250mL volumetric flask with a solution of Na2EDTA (concentration unknown), stopper, 20 mL pipette
with pipette control, 2 Erlenmeyer flasks (wide mouth), spatula, 25 mL burette with funnel and
clamp, stand.
Chemicals:
Solution of Na2EDTA, (c(Na2EDTA) ≈ 0,1 mol/L when filled up)
Solution of ammonia, w(NH3) = 25 % (in the hood)
Indicator buffer pills
Standardized solution of zinc sulfate, c(ZnSO4) = 0.097 mol/L
Demineralized water
Procedure:
The flask with the solution of Na2EDTA of unknown concentration has to be filled up with demineral-
ized water up to 250 mL. Mix thoroughly. Transfer exactly 20 mL of this solution into an Erlenmeyer
flask and fill up to 100 mL. Add an indicator buffer pill and - after it is dissolved - 2 mL of the ammo-
nia solution (w(NH3) = 25 %). Titrate speedily with the standardized solution of zinc sulfate (c(ZnSO4)
= 0,097 mol/L). The end-point is given by the color change from green to red.
Problems:
a) Write down the label code of your volumetric flask on the answer sheet.
b) Note the consumption of the standardized solution of zinc sulfate.
c) Calculate the concentration of your Na2EDTA solution.

44
Problem 4-13 Complexometric Determination of Nickel
Equipment and Glassware:
Volumetric flask (100 mL) with stopper with test solution, volumetric pipette (20 mL), pipettes con-
trol, 1 burette (25 mL) with funnel, 1 measuring cylinder (50 mL), stand and clamp, spatula, 2 conical
(Erlenmeyer) beakers (300 mL, wide mouth)
Chemicals:
Test solution containing nickel a volumetric flask
Dil. solution of ammonia, c(NH3) = 2 mol/L
Trituration of murexide indicator
Solution of Na2EDTA·2 H2O, c(Na2EDTA)  0.1 mol/L (from problem 4-12)
Demineralized water
Procedure:
The flask with the test solution has to be filled up to 100 mL. The solution has to be mixed well.
20 mL of this solution are transferred with a pipette to a conical beaker (300 mL, wide mouth) and
15 mL of the solution of ammonia are added.
Drops of the indicator solution are added until intense yellow color occurs (approximately 8 drops).
If the solution in orange the pH value is not high enough and additional solution of ammonia has to
be added.
The mixture is filled up with demineralized water to about 100 mL and then titrated with the solu-
tion of Na2EDTA.
There will be a sharp change of colors from yellow to violet. This color has to persist.
Problems:
a) Write down the label code of your volumetric flask on the answer sheet.
b) Record the consumption of the standard solution of Na2EDTA.
c) Calculate the mass concentration  of nickel in your tested solution (in mg/L).
Problem 4-14 Qualitative Analysis
You find the following mixtures of salts in seven beakers:
BaCl2/NaCl - AgNO3/Cu(NO3)2 – FeCl3/CuCl2 – KSCN/KI – KIO3/K2SO4 – Na2CO3/NaOAc –
(NH4)2SO4/FeSO4.
Equipment:
25 test tubes, test tube holder, 3 Pasteur pipettes, pH paper, spatula
Chemicals:
Seven beakers with mixtures of salts (labelled from A to G)

45
Dil. solution of ammonia, c(NH3) = 2 mol/L
Dil. nitric acid, c(HNO3) = 2 mol/L
Demineralized water
Procedure:
Dissolve each mixture with about 30 mL of demineralized water. Combine at any time two of the
solutions in a test tube and note the observation. Repeat this procedure with all possible combina-
tions. You may use the dil. solutions of ammonia and nitric acid as aids as well as the pH paper.
Problems:
a) Write down the label code of your test mixture.
b) Report your observations on the answer sheet.
c) Assign the salt mixtures to the beakers.
The following reduction potentials are given (pH = 0):
E° in V E° in V
Cu2+
/ Cu+
0.16 Ag+
/ Ag 0.80
Cu2+
/ Cu 0.34 HIO3 / HIO 1.13
I2 / I–
0.54 Cl2 / Cl 1.34
Fe3+
/ Fe2+
0.77
Take possible acid/base reactions into consideration.
Additional safety precautions:
The mixtures contain toxic heavy metals, such as barium.
Do not pour the solutions into the sink.

Answers
46
Part 2
The answers to the problems of the four rounds
The solutions are more detailed than expected from the pupils. That may facilitate their
comprehension in case of mistakes. Furthermore future participants should use this booklet
to become acquainted with the problems of the competition and their solutions.

Answers Round 1
47
Answers Round 1
Solution to problem 1-01
a) Natural isotopes: 𝐶6
12
, 𝐶6
13
und 𝐶6
14
. They differ in the number of neutrons in the nucleus.
b)
14
C is produced from nitrogen by cosmic radiation in the atmosphere: N + n0
1
→ C6
14
+ H1
1
7
14
c) C6
14
→ N7
14
+ e–
d) 𝑁 = 𝑁0 · 𝑒– 𝑡
N = number of nuclei at time t, N0 = number of nuclei at the begin (t = 0), λ = decay constant
e) The half-life is the time it takes for the number of nuclei to fall to half its initial value. It is constant for a
nuclide and characteristic for it.
The formula for half-life can be derived by setting N = N0/2:
N0
2
= N0 · e–t1/2  t½ =
ln 2

=
0,693

f) Never, N = 0 will never be reached.
g) Carbon has two stable, nonradioactive isotopes:
12
C and
13
C, and one radioactive isotope,
14
C. Radiocar-
bon dating is a radiometric dating technique that uses the decay
14
C to estimate the age of organic mate-
rials. The concentration of
14
C in the atmosphere might be expected to reduce over thousands of years.
However,
14
C is constantly being produced in the lower stratosphere and upper troposphere by cosmic
rays. Thus the proportion of radioactive to non-radioactive carbon in the atmosphere is constant and in
living organic material too. Once an organism dies the natural carbon exchange is not continued and the
ratio of
14
C in the material decreases. The comparison of the actual content of
14
C in some organic mate-
rial with its content in the atmospheric carbon leads to the age of the organic material. In doing this you
may compare the number of decays of 1 g of carbon per minute.
h) The decay rate is dependent on the number of atoms.
N = N0 · e–t
 t = ln
N0
N
·
1

and with  =
ln 2
5730
: t =
5730
ln2
years · ln
15
14,48
= 292 years
2013 – 292 = 1721
The treasury map originates from the year 1721, it could be authentic.
i) After 65 m years the ration of
14
C is so small that hardly a decay can be detected. Furthermore it cannot
be assumed that the ratio of carbon isotopes has been stable in such a long period.
Remark: Nevertheless the age of the bones of dinosaurs can be estimated by the decay of radioactive
isotopes of other elements with a longer half-life such as
238
U,
87
Rb und
40
K.
a) The "characteristic composition of fatty acids of a fat" shows the mean distribution of fatty acids in a fat.
For example, a fat could consist of the following three esters
3 x Oleic acid 2 x Oleic acid 1 x Stearic acid 1 x Oleic acid, 2 x Stearic acid
Then the characteristic composition of fatty acids is 66.7 % of oleic acid and 33.3 % of stearic acid.
b)
Capric acid
C10H20O2
10 : 0
Linoleic acid
C18H32O2
18 : 2
O
O
CH2
CHO
O
CH2O
O
O
O
CH2
CHO
O
CH2O
O
O
O
CH2
CHO
O
CH2O
O
COOH
COOH

Answers Round 1
48
Linolenic acid
C18H30O2
18 : 3
Erucic acid
C22H42O2
22 : 1
c) The residues of the fat molecules experience van der Waals forces. Because van der Waals forces operate
only at short distances they are strongest in molecules which chains can pack together closely. Saturated
residues can arrange linearly which leads to strong intermolecular forces. These fats are solid.
Residues with double bonds are mostly cis configurated. Thus a linear arrangement is not possible and
the interactions are lower. These fats are rather liquid.
d)
e)
The deprotonation of a carboxylic acid by hydroxide ions is almost irreversible, thus the reaction runs to-
wards the carboxylates.
f) The attack of the hydroxide is hindered by the long chain ester.
g) A tenside has two different ends, a hydophobic and a hydrophilic end. So they have the property to ac-
cumulate at the boundary surface of liquids and thus decrease the surface tension.
When tensides are dispersed in water the long hydrocarbon tails cluster together on the inside of a tan-
gled, hydrophobic ball, while the ionic heads on the surface of the cluster stick out into the water layer.
These mostly spherical clusters are called micelles. If there is a higher concentration cylindrical micelles
and block micelles may be formed.
Tenside molecule
COOH
COOH
R1
O
O
CH2
CHOR2
O
CH2OR3
O
–
OH
R1
O
O
CH2
CHOR2
O
CH2OR3
O –
OH
R1
O
O
CH2
CHOR2
O
CH2–
OR3
O
OH
–
OH
R1
O
O
CH2
CHOR2
O
CH2–
O
R3
O
O–
Na+
O
HH
R1
O
O
CH2
CHOR2
O
CH2
HO
R3
O
O–
Na+
+
+ NaOH
– Na+
+ NaOH
– H2O
+ H2O
– OH–
R1
, R2
, R3
= any alkyl- or alkenyl residue of a fatty acid

Answers Round 1
49
Spherical micelle Cylindrical micelle Block micelle
Grease and oil droplets are solubilized in water when they are coated by the nonpolar tails of the tenside
in the center of micelles. Once solubilized, the grease and oil can be rinsed away.
h) Soaps are anionic tensides. There are four kinds of tensides.
Anionic tensides: a long hydrophobic hydrocarbon tail is connected
with a hydrophilic negatively charged group.
–
Cationic tensides: a long hydrophobic hydrocarbon tail is connected
with a hydrophilic positively charged group.
+
Ampholytic tensides: a long hydrophobic hydrocarbon tail is connected
with an ampholytic group (positively and negatively charged).
+/–
Nonionic tensides: a hydrophobic residue is connected with an un-
charged group of polyglycol ethers.
Anionic tensides
Soaps
Alkylbenzene sulfonates (ABS, LAS)
Alkane sulfonates (AS)
-Olefin sulfonates (AOS)
Ester sulfonates (ES)
Fatty alcohol sulfate (FAS)
Fatty alcohol ether sulfate (FAES)
Cationic Tensides
Distearyl dimethylammonium chloride (DSDMAC)
Dodecyl dimethylbenzyl ammonium chloride
Esterquats (EQ)
Amphoteric Tensides
Betaines
R = C12 – C18
R CH2 COONa
CH SO3Na
R1
R2
CH SO3Na
R1
R2
R CH CH CH2 SO3Na
R CH COOMe
SO3Na
R CH2 O SO3Na
R O CH2 CH2 O SO3Na
n
N
CH3
CH3
H37C18
H37C18
Cl–+
N
H3C CH3
CH2H25C12
Cl–+
NH3C
CH2
CH2
CH2 O C R
CH2 CH2 OH
CH2 O C R
O
O
Cl–
+
R N+
CH2 COO–
CH3
CH3

Answers Round 1
50
Sulfobetaines
R = C12 – C18
Nonionic Tensides
Fatty alcohol polyglycol ether (FAE)
Alkylphenol polyglycol ether (APE)
Fatty alcohol polyethylenglycol polypropylenglycol ether
Fatty acid ethanolamide
a) Air, oxides, carbonates, silicates, sea water, water, biosphere or other compounds (or individual com-
pounds)
b)
Electrolysis of water: 2 H2O 2 H2 + O2
Thermic decomposition of peroxides: 2 BaO2 2 BaO + O2
Catalytic decomposition of hydrogen peroxide (Pt,
MnO2):
2 H2O2 2 H2O + O2
Catalytic decomposition of oxygen containing com-
pounds:
2 Ag2O 4 Ag + O2  > 160°C
2 Au2O3 4 Au + 3 O2  >
160°C
4 KClO3 3 KClO4 + KCl  > 400°C
KClO4 KCl + 2 O2  >
300°C
the decomposition of KClO3 proceeds already at
150 °C with MnO2 as catalyst:
KClO3 KCl + 1,5 O2
KClO3 + 3 MnO2 KCl + 3 MnO3
3 MnO3 3 MnO2 + 1,5 O2
c) Nitrogen is diamagnetic, oxygen paramagnetic. The reason is the existence of unpaired electrons in
the oxygen molecule.
R N+
CH2 CH2
CH3
CH3
SO3
–
R O (CH2 CH2O)n H
R O (CH2 CH2O)n H
R (CH2 CH2O)n (CH2 CO)m H
H
CH3
R C N
CH2
CH2
CH2
CH2
OH
OH
O
700 °C
500 °C

Answers Round 1
51
Oxygen Nitrogen
d)
Triplet oxygen Singlet oxygen
2 unpaired electrons with equal spin 2 unpaired electrons with different spin
Total spin S ½ + ½ = 1 –½ + ½ = 0
Multiplicity M 2 · 1 + 1 = 3 2 · 0 + 1 = 1
The names follow the multiplicity: M = 1 Singlet, M = 2 dublet, M = 3 triplet, M = 4 quartet etc.
Remark: The spin of the inner electrons does not have to be considered. They cancel each other because
of their opposite algebraic sign in double occupied orbitals.
e) Part c) and d) demonstrate that dioxygen is a diradical. The Lewis structure does not show any unpaired
electrons. Thus it does not show the correct distribution of electrons.
a) Helium, neon, argon
b) P4 + 5 O2 P4O10 acidic reaction 4 Li + O2 2 Li2O basic reaction
1
/8 S8 + O2 SO2 acidic reaction 2 Ca + O2 2 CaO basic reaction
c) Examples (R = Alkyl group)
KZ 1: KZ 2: KZ 3:
,
,
d)
e) i) 2 Fe(OH)3 Fe2O3 + 3 H2O
O C O
C O
N O N O
 
O
RR H
O
H
O
ClCl R
O
H
O
HH
H
+
T
Oxidation number Examples Oxidation number Examples
-I H2O2, Na2O2 +½ O2PtF6
-½ KO2 +I O2F2
-
1
/3 NaO3 +II OF2
0 HOF
Energy Energy

Answers Round 1
52
ii) Ammonium nitrate shows an acidic reaction. It should remove absorbed hydroxide ions and wash
the precipitate to be neutral. In the presence of chloride ions iron(III) chloride may evaporate and
thus the result is falsified.
iii) The filter shall be burned without any residue which could be weighed with Fe2O3 to give a wrong
result.
iv) Fe3O4 is magnetic and could be identified with a magnet.
v) M(Fe2O3) = 159.69 g/mol M(Fe3O4) = 231.54 g/mol
1 mol of iron(III) ions form 159.69g/2 = 79.855 g Fe2O3
or 231.54/3 g = 77.18 g Fe3O4, respectively
 If Fe3O4 is formed the mass after annealing is too low. Thus the formation of Fe3O4 leads to a
content of iron which is too low.
vi) Mean result of Fe2O3 after annealing:
(0.2483 g + 0.2493 g + 0.2488 g) : 3 = 0.2488 g .
Amount of iron in 50 mL solution:
n(Fe)50 =
0.2488 g ·2
M(Fe2O3)
=
0.2488 g ·2
159,69 g/mol
= 3.116 · 10
-3
mol
 Mass of iron(III) chloride in 250.0 mL:
m(FeCl3) = 5 · 3.116 · 10
-3
mol · M(FeCl3) = 0.01558 mol · 162.21 g · mol
–1
m(FeCl3) of the sample = 2.527 g.
a) MeO2 + 4 HCl MeCl2 + Cl2 + 2 H2O
Cl2 + 2 I
–
I2 + 2 Cl
–
I2 + 2 Na2S2O3 2 I
–
+ 4 Na
+
+ S4O6
2–
b) Starch solution is added to realize the change more easily. Iodine forms a clathrate with starch.
c) The oxygen of the air can oxidize iodide to iodine. Thus the consumption of thiosulfate would be too high
because more iodine has to be titrated.
In contact with steam chlorine might disproportionate into chloride and hypochlorite. Even if you assume
that the equilibrium of the formation of hypochlorite in the presence of hydrochloric acid lies on the side
of chlorine it cannot be excluded that some hypochlorite forms. As this reaction takes place in the vessel
with the metal oxide chlorine set free by the metal oxide would be lost to oxidize iodide to iodine. This
would lead to a smaller consumption of sodium thiosulfate.
d) c(thiosulfate) = 0.1 mol /L = 0.1 mmol/mL
24.25 mL of thiosulfate solution ≙ 0.1 mol/mL · 24.25 mL/2  1.21 mmol I2
 290 mg of MeO2 produce 1.21 mmol of I2
1 mmol of MeO2 ≙ 1 mmol of I2  1.21 mmol of MeO2 ≜ 290 mg
1 mol of MeO2 ≜ 290 g/1.21 = 239.7 g
M(Me) = (239.7 - 2 · 16.00) g/mol = 207.7 g/mol.  Me = Pb

Answers Round 2
53
Answers Round 2
a) M: Pb, X: KI, Y: PbI2, Z: KPbI3
There are several ways to solve this problem. Only one of these will be described here.
Specification of M: The information about the solubility in acids indicates a passivation in the presence of
sulfate and chloride ions. These facts point to lead which forms a sparely soluble sulfate and a sparely
soluble chloride.
The information about the solubility in sodium hydroxide solution strongly limits the number of metals as
only a few metals can be dissolved in this solution (examples: beryllium, aluminum, tin, iron, lead).
The information of the pyrophoric property limits the selection, too. Metals which can ignite when finely
dispersed are magnesium, titanium, nickel, cobalt, iron and lead as well as rare earth metals of the inner
transition series.
When a sodium hydroxide solution is added, the aqueous solution of the cations forms an insoluble hy-
droxide which dissolves in an excess of it. Examples are aluminum, lead, zinc, beryllium, chromium, galli-
um, indium, copper and gold.
With ammonia there is no formation of an ammine complex: iron(III), aluminum, beryllium, titanium,
zinc, tin, lead.
Finding of X and M: The endothermic solvation hints to compounds which are asked for in part f). Com-
pound X must be able to react with the gaseous compound G. Possibilities are the formation of car-
bonates (from CO2), oxides (from O2), triiodates (from I2). An aspect in favor for the latter is the formation
of a metallic mirror when G is formed (reducing properties of iodine).
X or parts of X and the metal M must form a sparely soluble compound Y which has to have broader
properties: layer structure, properties of a semiconductor, thermochromism. Possibilities are sparely sol-
uble sulfates, chlorides, iodides.
Taking all statements into account only lead is remaining. Y must be lead(II) iodide. X may be sodium or
potassium iodide (see f)).
Part e) hints to a double salt. With the formation of the simplest double salt XPbI2 can the alkali metal be
determined by means of a calculation of the given mass ratio.
b) 2 KI + Pb(NO3)2 PbI2 + 2 KNO3
PbI2 + KI KPbI3
c) H = KI3 KI + I2 KI3
H is an ionic compound so the cation and the anion are shown separately. In the I3
–
anion the central
iodine atom is surrounded by 5 electron pairs resulting in a trigonal bipyramidal arrangement. There are
two bonded and three lone pairs. The lone pairs are equatorial arranged (according to VSEPR it is an
AX2E3 system). All these facts result in a linear structure of H.
d) The TG plot shows a difference in mass of 5 – 6 %. The molar mass of the water free compound is
M(KPbI3) = 627 g/mol. Then the mass loss is 627 g/mol · 5.5/94.5 = 36.5 g/mol  n = 2.
e) As the oxidation numbers and the elements are the same as in Z the wanted compound can only be
K2PbI4 with a mass ratio of lead of 207.2/793 · 100 % = 26.13 %
f) NH4Cl, KI, CaCl2 · 6 H2O, Na2SO4 · 10 H2O, NaI, KNO3, NaNO2.
I
I
I
–
K
+

Answers Round 2
54
The lattice energy is needed in order to dissolve a salt, the solvation energy is set free. As soon as the
needed energy (-lattice energy) is higher than the solvation energy the solution provides the difference
and cools down. Condition for cooling-down:
Lattice energy < Solvation energy or |Lattice energy| > |Solvation energy|
g) +I,+II,–I +I,–II +I,–II +I,–I 0 0
KPbI3 · n H2O n H2O + KI+ Pb + I2
h) Pb(NO3)2 + 2 NH3 + 2 H2O Pb(OH)2 + 2 NH4NO3 or as ionic equation:
Pb
2+
(aq) + 2 OH
–
Pb(OH)2(s)
Remark: Lead hydroxide is formed. There is no formation of an ammine complex in an aqueous solution.
a) In the reaction to A two isomers are formed. Acetylacetone is deprotonated by triethylamine to yield a
carbanion which is stabilized by mesomerism to an enolate. In doing so the two isomers are formed,
which are trapped by TMSCl.
Afterward the two isomers are deprotonated by LDA and again a carbanion or rather an enolate is
formed which is trapped again by TMSCl:
Both isomers of B react with NBS to yield C. The reaction takes place at the outer double bond.
T
free rota-
tability

Answers Round 2
55
b) E/Z-Isomers are formed:
c) Assignment and ratio of isomers:
δ = 5.24 ppm: 1 proton at position 2 (Isomer 1)
δ = 4.95 ppm: 1 proton at position 2' (Isomer 2)
Remark: The assignment to isomer 1 or isomer 2 is arbitrary and could be done the other way round but
it should be the same as in the following signal assignments. The signal at 5.24 refers to the Z-isomer, that
at 4.95 to the E-isomer but the students are not expected to know that.
From the ratio of isomer 1 to isomer 2 you find 2.11:1.
The signals of the hydrogen atoms of the methyl groups at position 3 and 3' are almost identical as well as
those of the methyl groups at position 1 und 1':
δ = 1.93 ppm: 6 protons at 3 and 3'
δ = 1.79 ppm: 6 protons at 1 and 1' .
The remaining signals belong to the hydrogen atoms of the trimethylsilyl groups.
δ = 0.00 ppm: 9 protons of the Si(CH3)'3 group
δ = -0.04 ppm: 9 protons of the Si(CH3)'3 group
Calculation of the intensities:
The protons at positions 3 and 3‘ as well as those at 1 and 1‘ are so similar that their signals are almost
identical and thus the signals appear as a singlet.
 The intensity for both singnals at δ = 1.93 ppm and  = 1.79 ppm is 3 · 2.11 + 3 · 1 = 9.33
For isomer 1 the intensity of 9 · 2.11 = 18.99 is expected
for isomer 2 the intensity of 9 · 1 = 9 is expected.
 δ = 0.00 ppm: 9 · 1 = 9 (Isomer 2) δ = 0.04 ppm: 9 · 2.11 = 18.99 (Isomer 1)
d) Compound C contains a bromine atom. Natural bromine consists of two isotopes,
79
Br und
81
Br (=2),
with nearly the same abundance. Thus in the mass spectrum two nearly identical peaks with =2 are
found at 177.96 and 179.96.
e) Et3N is used as a base. It deprotonates acetylacetone at the CH2 group between the two carbonyl groups,
so an enolate can be formed. This enolate reacts with TMS-Cl to yield TMS-enolether.
f) The base strength of Et3N does not suffice for a second deprotonation. Thus the stronger Base LDA is
used.
g) The reaction of acetylacetone with NBS in the presence of a base is not regiospecific and would not lead
to the desired product because the deprotonation of acetylacetone would take place at the CH2 group
between the two carbonyl groups.
The synthesis via A and B guarantees that only the desired regiospecific compound forms.

Answers Round 2
56
h)
The H/D exchange may happen up to three times so that a completely deuterized CD3 group may be ob-
tained.
a)
Cyclobutadiene
2 π electrons (charge: +2):
1P
4 π electrons (charge: 0):
1P
Remark: The degeneracy of 2 and 3 leads to a
diradical electron structure.
6 π electrons (charge: –2):
Tropylium ions
6 π electrons (charge: +1): 8 π electrons (charge: –1):
Remark: The degeneracy of 2 and 3 leads to a
diradical electron structure.

Answers Round 2
57
The calculation of the total energy is done as follows:
 The energy of the π system is compared with the aliphatic version. Therefore the respective energies
of the occupied orbitals have to be calculated with the given formulae and added.
 The energy levels of the Frost circle result from the following considerations:
 The height of the respective energy level with respect to the AO level (E = ).
 The difference from the origin to the energy level is 2β.
 The angle  is 360°/n; the angle between the energy level and the height amounts to m · , where
m = number of the energy level – 1 and n = number of carbon atoms in the n-cyclic system.
 To find the height the cosine function is used: cos(m · ) = H/2β  H = cos(m · ) · 2β
  E =  + H
The calculation for cyclobutadiene gives = 90°.
Number of electrons Bond energy Aliphatic Difference
2 E = 2 · (+2β) = 2.0+4.0β E = 2.0+3.2β ΔE = 0.8β
4 E = 2 · (+2β) + 2 ·  = 4.0+4.0β E = 4.0+4.5β ΔE = –0.5β
6 E = 2 · (+2β) + 4 ∙  = 6.0+4.0β E = 6.0+3.2β ΔE = 0.8β
 The cyclobutadienyl dication and the cyclobutadienyl-dianion are aromatic because of E > 0.
The calculation for tropylium gives  = 51.429°.
Number of electrons Bond energy Aliphatic Difference
6 E = 2 · (+2β) + 4 · (+cos () ∙ 2β) = 6.0+9.0β E = 6.0+8.1β ΔE = 0.9β
8 E = 2 · (+2β) + 4 · (+cos() · 2β) + 2 · (+cos(2) · 2β)
= 8.0+8.1β E = 8.0+8.1β ΔE = 0
 The tropylium cation is aromatic because of E > 0.
b) The compound has to be
1. cyclic, 2. planar,
3. totally conjugated 4. fulfill the Hückel rule: it must have 4n + 2 π- electrons.
c) The following compounds (ions) fulfill all conditions and are aromatic:
(i) Pyrrole (6 π electrons), (iii) azulene (10 π electrons), (v) pyridinium cation (6 π electrons), (vi) caffeine
(10 π electrons and 6 π electrons in the five membered ring).
Non-aromatic ore antiaromatic are:
(ii) Allyl anion (4 π electrons; violates rules 4 and 1), (iv) 1H-pyrrolium cation (N protonated pyrrole) (vio-
lates rule 3)
d) HOMO: 2 and 3; LUMO: 4 and 5
Remark: The orbitals are degenerate (having the same energy).

Answers Round 2
58
e) The decision is based on the number of nodal planes in the bond axis of the respective orbital. Orbitals of
same symmetry show interaction.
HOMO (Cp
–
), (2,3 ) LUMO (Cp
–
), (4,5)
3dx2-y2 – x
3dz2 – –
3dxy – x
3dxz x –
3dyz x –
4s – –
4px x –
4py x –
4pz – –
f) [Ru(η
5
-Cp)2] is more stable as it fulfills the 18 electron rule. [Rh(η
5
-Cp)2] has 19 electrons and should act
as a reducing reagent.
g) Possible structures:
18 Valence electrons each 18 Valence electrons each 18 Valence electrons each
Remark: The numbers of valence electrons at the metal centers are found as the sum of all electrons as-
signed to the metal. Each bond of the terminal CO groups counts for 2 electrons, the bridging CO groups
provide 1 electron for each metal center and the cyclopentadienyl ring 6 electrons. The 7 d-electrons of
the metal centers are added  Σ = 18.
a) [Cu(H2O)6]
2+
+ H2O [Cu(H2O)5(OH)]
+
+ H3O
+
(The tetraaqua complex is also accepted)
c1 c2 c3
M(Cu(NO3)2) = 187.56 g/mol  c(Cu(NO3)2) = 8.96 · 10
–3
mol/L.
Ka =
c2/(1 mol/L) · c3/(1 mol/L)
c1/(1 mol/L)
c2 = c3 =10
–4.40
mol/L c1 + c2 = 8.96 · 10
–3
mol/L
c1 = 8.96 · 10
–3
mol/L – 10
–4.40
mol/L = 8.92 · 10
–3
mol/L
Ka = 1.78· 10
–7
 pKa = 6.75
b) Ksp. 25°C = c(Cu
2+
)·c
2
(OH
–
)/(1 mol/L)3
 c(OH
–
) = √
Ksp .25°C · (1 mol/L)3
c(Cu2+)
x = interaction expected
- = no interaction expected
Orbital interaction between
Fe and Cp
–
is possible in each
column.

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